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R infinite space endowed with the infinite norm and fixed point theorem of Schauder | Define a sequence $(p^{(n)})_{n \in \mathbb{N}}$ in $C$ by $p^{(n)} = (\delta_{in})_{i \in \mathbb{N}}$. In other words,
\begin{equation*}
p^{(n)}_{i} = \left\{ \begin{array}{r l}
1, & \text{if} \, \, i = n, \\
0, & \text{otherwise.}
\end{array} \right.
\end{equation*}
$(p^{(n)})_{n \in \mathbb{N}}$ does not have any accumulation points in $C$ with the given norm topology. The idea is this: $\|p^{(n)} - p^{(k)}\|_{\infty} = 1$ for all $n,k \in \mathbb{N}$ so no subsequence is Cauchy.
Edit: Given $m \in \mathbb{N}$, if we define $C^{m}$ by
\begin{equation*}
C^{m}_{*} = \{x \in \mathbb{R}^{\infty} \, \mid \, \sum_{k = 1}^{\infty} |x_{k}| \leq 1, \, \, |x_{k}| \leq \frac{m}{k} \},
\end{equation*}
then $C^{m}_{*}$ is compact in $\mathbb{R}^{\infty}$ with the norm $\|\cdot\|_{\infty}$ given above. Note that the set $C^{m}$ defined by Ari.stat below is simply $C^{m} = C^{m}_{*} \cap [0,1]^{\infty}$. Since $[0,1]^{\infty}$ is a closed subset of $\mathbb{R}^{\infty}$ with respect to $\|\cdot\|_{\infty}$, $C^{m}$ is a closed subset of $C^{m}_{*}$ and, thus, is itself compact.
To see this, first, notice that if $x \in C^{m}_{*}$, then $\|x\|_{\infty} \leq \sum_{k = 1}^{\infty} |x_{k}| \leq 1$.
To see that $C^{m}_{*}$ is compact, I will prove that any sequence in $C^{m}$ has a subsequence that converges. Suppose that $(x^{(n)})_{n \in \mathbb{N}} \subseteq C^{m}_{*}$. Since $|x^{(n)}_{k}| \leq 1$ independently of $n$ and $k$, we can employ a diagonalization argument to find a subsequence $(n_{j})_{j \in \mathbb{N}} \subseteq \mathbb{N}$ and a sequence $x \in \mathbb{R}^{\infty}$ so that, for each $k \in \mathbb{N}$, $\lim_{j \to \infty} x_{k}^{(n_{j})} = x_{k}$. In fact, we can show that $\lim_{j \to \infty} \|x^{(n_{j})} - x\|_{\infty} = 0$ must hold.
Indeed, given $\epsilon > 0$, if we choose $K_{0} \in \mathbb{N}$ such that $\frac{m}{k} \leq \epsilon/2$ for each $k \geq K_{0}$, then the inequality $\max\{|x^{(n)}_{k}|, |x_{k}|\} \leq \frac{m}{k}$ implies
\begin{equation*}
\|x^{(n_{j})} - x\|_{\infty} \leq \max \{ |x^{(n_{j})}_{1} - x_{1}|, \dots, |x^{(n_{j})}_{K_{0}} - x_{K_{0}}|, \epsilon \}.
\end{equation*}
Therefore, by pointwise convergence, $\limsup_{j \to \infty} \|x^{(n_{j})} - x\|_{\infty} \leq \epsilon$. We conclude by sending $\epsilon \to 0^{+}$. |
How to evaluate $\lim _{x\to \infty }\left(x^3\ln\left(\frac{x}{x-1}\right)-x\sqrt{x^2+x}\right)$? | About what you did: In your attempt, where did the $\ln$ go?
An approach: Here is a solution using Taylor expansions: we will use the fact that, when $u\to 0$,
$$\begin{align}
-\ln(1-u) &= u + \frac{u^2}{2} + \frac{u^3}{3} + o(u^3) \\
\sqrt{1+u} &= 1 + \frac{u}{2} - \frac{u^2}{8} + o(u^2)
\end{align}$$
and the fact that $\frac{1}{x} \xrightarrow[x\to\infty]{} 0$ (this will be "our $u$").
(Intuitively, we go to order $3$ in the first and only order $2$ in the second since, for our application, we will multiply by ${x^3}=u^{-3}$ and $x^2=u^{-2}$, respectively: we want to do an expansion of both terms of the difference, up until we get a constant.)
First term:
$$x^3\ln \frac{x}{x-1} = -x^3 \ln \left(1-\frac{1}{x}\right)
= x^3 \left(\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{3x^3}+o\left(\frac{1}{x^3}\right)\right)
= x^2 + \frac{x}{2} + \frac{1}{3} + o(1)$$
Second term:
$$
x\sqrt{x^2+x} = x^2\sqrt{1+\frac{1}{x}} = x^2\left(1+\frac{1}{2x}-\frac{1}{8x^2}+o\left(\frac{1}{x^3}\right)\right)
= x^2 + \frac{x}{2} - \frac{1}{8} + o(1)$$
Now, the difference of the two:
$$x^3\ln \frac{x}{x-1} - x\sqrt{x^2+x} = \frac{1}{3} + \frac{1}{8} + o(1) = \frac{11}{24} + o(1) \xrightarrow[x\to\infty]{} \frac{11}{24}.
$$ |
Show that maximum modulus of complex function assumes on saddle point. | This is not a true answer, but it may help to understand what is "under the hood". Here is the graphical representation of function
$$Z=\phi(X,Y)=|f(z)| \ \ \text{where} \ \ z=X+iY.$$
(caution : $Z$ represents the 3rd component in this $\mathbb{R}^3$ representation ; different from $z$ !) with some of its level lines.
One sees
the two roots $\zeta_1:=-1$ (simple root) represented by a kind of cone, whereas root $\zeta_2:=1$ (double root) is represented by a kind of paraboloid.
the pole $\zeta_3=0$ which "induces" the central peak (truncated on the figure).
The $(X,Y)$ coordinates of the saddle points of the surface have been represented by little stars. |
Prove there are $b,c$ such that $f(x)=\frac{a}{2}x^2 + bx + c$ | I wouldn't approach this problem with the mean value theorem at all, and I'm not sure how one would use the mean value theorem to finish your argument (except in the sense that the fundamental theorem of calculus follows quickly [caution: self-promotion, mildly] from the mean value theorem and this problem is trivial with integration).
Theorem: If $f'(x) = 0$ for all $x$ in $[d,e]$, then $f(x) = c$ on $[d,e]$ for some constant, i.e. $f(x)$ is a constant function there.
Proof:
By the mean value theorem, $\displaystyle \frac{f(x) - f(x_0)}{x - x_0} = f'(\xi) = 0 \implies f(x) - f(x_0) = 0$, so that $f(x) = f(x_0). \;\diamondsuit$
Corollary: If $f'(x) - g'(x) = 0$ for all $x$ in $[d,e]$, then $f(x) = g(x) + c$ on $[d,e]$ for some constant $c$.
Proof:
From above, $f(x) - g(x)$ is constant, meaning that $f(x) - g(x) = c$ for some constant. $\diamondsuit$
Theorem: If $f''(x) - g''(x) = 0$ for all $x$ in $[d,e]$, then $f(x) = g(x) + bx + c$ on $[d,e]$ for some constants $b,c$.
Proof:
Using the above corollary on the functions $f'(x)$ and $g'(x)$, we know that $f'(x) = g'(x) + b$ for some constant $b$. Let $G(x) = g(x) + bx$. Then $f'(x) - G'(x) = f'(x) - (g'(x) + b) = 0$. By the corollary above, we see that $f(x) = G(x) + c = g(x) + bx + c$ for some constant $c$, which is what we wanted to show. $\diamondsuit$
Claim: If $f''(x) = a$ for all $x \in [d,e]$, then $f(x) = \frac{a}{2}x^2+ bx + c$ on $[d,e]$ for some constants $b,c$.
Notice that $g(x) = \frac{a}{2}x^2 + \beta x + \gamma$ satisfies $g''(x) = a$. Suppose $f$ is another function with $f''(x) = a$. Then as $f''(x) - g''(x) = 0$, we must have that $f''(x) = g''(x) + b'x + c'$ for some constants $b', c'$, which gives us exactly what we wanted to show. $\spadesuit$
In a certain sense, I used the MVT once to show the first Theorem. The corollary follows, implicitly using the MVT once. Then the second theorem uses that corollary twice, and therefore implicitly uses the MVT twice. So in a very weak sense, this does use the MVT twice. |
The proof of the kernel and cokernel formula of a certain map. | Unfortunately, the formula you have is not right. Consider the case where $a=b$ and $s=0$. Then $n=d$ and $k=1$, so the map you've defined is the identity map and hence has trivial kernel and cokernel, but your formula would give its kernel to be $\mathbb{Z}/p^{a+1}$ and its cokernel to be $\mathbb{Z}/p^a$.
If you don't mind, I'll use the notation $\phi_s$ for what you refer to by $k\Lambda$. So $\phi_s$ takes $\overline{1}\in\mathbb{Z}/p^a$ to $\overline{p^{s+b}/d}\in\mathbb{Z}/p^b$, where $d=\gcd(p^a,p^b)$. Note that the only positive divisors of $p^a$ and $p^b$ are powers of $p$, so in fact $d=p^{\min\{a,b\}}$, and we can write $\phi_s(\overline{1})=\overline{p^{s+b-\min\{a,b\}}}$. To see that $\phi_s$ is a well-defined map, you need to show that $p^a\phi_s(\overline{1})=\overline{0}\in\mathbb{Z}/p^b$. This amounts to showing that $p^ap^{s+b-{\min\{a,b\}}}$ is divisible by $p^b$ in $\mathbb{Z}$, ie that $a+s+b-\min\{a,b\}\geqslant b$. Since $s\geqslant 0$, this is clear.
Now, $\overline{n}\in\ker\phi_s$ if and only if $\overline{0}=\phi_s(\overline{n})=\overline{n{p^{s+b-\min\{a,b\}}}}\in\mathbb{Z}/p^b$, which is in turn true if and only if $np^{s+b-\min\{a,b\}}$ is divisible by $p^b$ in $\mathbb{Z}$. If $p^k$ is the highest power of $p$ dividing $n$, then this is true if and only if $k+s+b-\min\{a,b\}\geqslant b$, ie if and only if $k+s\geqslant\min\{a,b\}$. You are assuming that $\phi_s$ is not the zero map, so we know $s<\min\{a,b\}$. Thus a generator for $\ker\phi_s$ is given by $\overline{p^{\min\{a,b\}-s}}$. The order of $\overline{p^{\min\{a,b\}-s}}$ is $p^{a-(\min\{a,b\}-s)}=p^{a+s-\min\{a,b\}}$, and subgroups of cyclic groups are cyclic, so $\fbox{$\ker\phi_s\cong\mathbb{Z}/p^{a+s-\min\{a,b\}}$}$. In particular, in the case you brought up, where $b\leqslant a$ and $s=b-1$, we have $\ker\phi_s\cong\mathbb{Z}/p^{a-1}$.
Now, the cokernel of $\phi_s$ is the quotient of $\mathbb{Z}/p^b$ by the image of $\phi_s$. Quotients of cyclic groups are cyclic, and cyclic groups are uniquely determined by their cardinalities, so the cokernel of $\phi_s$ will be determined by its size. Now, by the first isomorphism theorem, the image of $\phi_s$ is isomorphic to the quotient $\frac{\mathbb{Z}/p^a}{\ker\phi_a}$. Since $|\mathbb{Z}/p^a|=p^a$, and $|\ker\phi_s|=p^{a+s-\min\{a,b\}}$ by the computation above, we have $$|\operatorname{im}\phi_s|=\frac{p^a}{p^{a+s-\min\{a,b\}}}=p^{\min\{a,b\}-s}.$$ (Again, you are assuming $s<\min\{a,b\}$, so this number makes sense.) Since $|\mathbb{Z}/p^b|=p^b$, we can hence compute $$|\operatorname{coker}(\phi_s)|=\frac{|\mathbb{Z}/p^b|}{|\operatorname{im}\phi_s|}=\frac{p^b}{p^{\min\{a,b\}-s}}=p^{b+s-\min\{a,b\}},$$ and so $\fbox{$\operatorname{coker}\phi_s\cong\mathbb{Z}/p^{b+s-\min\{a,b\}}$}$. In particular, in the case you brought up, where $b\leqslant a$ and $s=b-1$, we have $\operatorname{coker}\phi_s\cong\mathbb{Z}/p^{b-1}$. |
Special Elements: Spectrum | You already found a non-normal operator $N$ with spectrum $\{0\}$. Now simply note that
$1+N$ will be another non-normal operator with spectrum $\{1\}$. |
Showing $\mathrm{int}(A-B)=\mathrm{int}(A)- \bar{B}$ | int($A -B$) = int($A \cap B^c$) = (int $A$) $\cap$ (int $B^c$)
= (int $A$) $\cap(\bar B)^c$ = (int $A$) $-\bar B$. |
If $f(x^2)$ is $\mathscr{R} [0,a]$, then prove $f(x^2)$ and $xf(x^2)$ are $\mathscr{R}[-a,a]$. | If we know that $\int_{0}^{a}f(x^2) \> dx < \infty$, what can we say about $\int_{-a}^{0} f(x^2) \> dx$? In particular, if we write $g(x)=f(x^2),$ is there a relationship between $g(c)$ and $g(-c)$? If that hint doesn't make sense maybe try graphing some functions of $x^2$. For the second one perhaps you can make a substitution. |
How could I approach $\lim_{n\to\infty}\left(\frac{1+\cos\left(\frac{1}{2^{n}}\right)}{2}\right)^n$? | \begin{align*}
\left(\dfrac{1+\cos(1/2^{n})}{2}\right)^{n}\leq\left(\dfrac{1+1}{2}\right)^{n}=1.
\end{align*}
Now $\cos u\geq 1-u^{2}/2$ for small $u\geq 0$, then
\begin{align*}
\left(\dfrac{1+\cos(1/2^{n})}{2}\right)^{n}\geq\left(1-\dfrac{1}{2^{2(n+1)}}\right)^{n}\rightarrow 1.
\end{align*} |
Solve the dirichlet pde with the given conditions. | You are more or less on the right track, but you have committed some errors. The first one is that the eigenvalues for the equation for $X$ are $\lambda_n=(n\,\pi/3)^2$ and the corresponding eigenfunctions $X_n=\sin(n\,\pi\,x/3)$. The corresponding equation for $Y$ is $Y''-(n\,\pi/3)^2Y=0$, whose solution is $Y_n=A_ne^{n\pi y/3}+B_ne^{-n\pi y/3}$. The solution is then of the form
$$
u(x,y)=\sum_{n=1}^\infty\sin\Bigl(\frac{n\,\pi\,x}{3}\Bigr)\bigl(A_ne^{n\pi y/3}+B_ne^{-n\pi y/3}\bigr).
$$
Imposing $u(x,0)=0$ we get $B_n=-A_n$ and
$$
u(x,y)=\sum_{n=1}^\infty A_n\sin\Bigl(\frac{n\,\pi\,x}{3}\Bigr)\bigl(e^{n\pi y/3}-e^{-n\pi y/3}\bigr).
$$
Now you can find $A_n$ using the condition for $y=7$. |
Solving $a^5=a^3bc+b^2c$ in integers | Hint:
$cb^2+a^3cb-a^5=0 \Rightarrow a^6c^2+4a^5c$ is a square number,
so $a^2c^2+4ac$ should be square number. |
Is the product of countable metrizable spaces with box-topology still metrizable? | No, if $X$ is any non-discrete metric space, $X^\omega$ in the box topology will not be first countable, so in particular not metrisable.
Sketch of proof: let $p \in X$ be a non-isolated point and pick a local base $U_n$ which obeys $\overline{U_{n+1}} \subsetneq U_n$ for all $n$ and then prove (via a diagonal argument) that $(p,p,p,\ldots)$ does not have a countable local base in $X^\omega$. For more info on box products, see the chapter about them in the Handbook of Set-theoretic Topology. You can also see there that $X^\omega$ will rarely be compact, connected and many other nice properties. It's a rarely used topology for that reason, except to construct examples. |
Polynomial basis - why orthogonal? | from definition of linearly independence, all you have to show is follows..
(Linear independence(LI) in infinite dimensional space (in fact, general version) is defined by LI of its finite dimensional subspaces.)
$$
a_n t^n +...+a_0 = 0 \quad where\, t \in [a,b]\,\, \Rightarrow\,\, a_0 = a_1 = ...= a_n = 0
$$
if coefficient are not - all zeros, it is contradiction by Fundamental theorem. |
Partial fraction of $\frac{x^n}{(1-x)(1-2x)(1-3x)\ldots(1-nx)}$ | First you have to take out a multiple of the denominator to make the degree of the numerator smaller than that of the denominator:
$$\frac{x^n}{(1-x)(1-2x)(1-3x)\ldots(1-nx)}$$
$$= \frac{(-1)^n}{n!} + \frac{x^n-\frac{(-1)^n}{n!}(1-x)(1-2x)(1-3x)\ldots(1-nx)}{(1-x)(1-2x)(1-3x)\ldots(1-nx)}\;.$$
Now you can deal with the remaining fraction. You already have the denominator in factorized form, so you know the partial fraction expansion will have the form
$$\frac{a_1}{1-x}+\frac{a_2}{1-2x}+\frac{a_3}{1-3x}+\ldots+\frac{a_n}{1-nx}\;.$$
To get the coefficients $a_i$, you can use the fact that the residues of the poles must be equal to the corresponding residues in the fraction. |
Difficulty understanding statistics | Think about how your random variables look like if you want to model the answers of one person or 500 people.
Regarding c): Be $X_i$ iid then how is $$\sum^{500}_{i=1}X_i$$ approximately distributed? You might want to center and normalize it. |
Equivariant maps for modules over different rings | These maps are called $R$-module homomorphisms, where $M'$ is regarded as an $R$-module via restriction along the homomorphism $\phi$. They are of course non-trivial and very interesting objects, in general.
For instance, Frobenius reciprocity is the statement that the obvious map
$$\mathrm{Hom}_R(M, \mathrm{Res}(M')) \to \mathrm{Hom}_S(S \otimes_R M, M')$$ is an isomorphism. The domain of this map is the set of maps you are looking at. |
Proving $ \left\{\frac{y^2+z^2+2}{yz}\ \mid\ y,z\in\mathbb{N}\right\}\cap\mathbb{N}=\{4\} $ | This is precisely the type of question that Viète jumping solves.
$$\frac{y^2+z^2+2}{yz}=k$$
If $y=z$, $\frac{2y^2+2}{y^2}=2+\frac2{y^2}$ and $y$ can only be 1, in which case $k=4$. Now let $y>z$, fix $z$ and $k$ and let $y$ vary as a variable $q$; we may write
$$q^2-kz(q)+(z^2+2)=0$$
One root of this is $y$ and the other is
$$q'=kz-y=\frac{z^2+2}y$$
so $q'$ is a positive integer. That $y>z$ implies that $q'=\frac{z^2+2}y<z$ as long as $z>2$. Hence $q'$ may replace $y$ to obtain a smaller valid solution.
Because of Viète jumping, now only the $z=1$ and $z=2$ cases need be considered.
If $z=1$, $\frac{y^2+3}{y}=y+\frac3y=k$ and $y$ can only be 3, whence $k=4$.
If $z=2$, $\frac{y^2+6}{2y}=\frac y2+\frac3y=k$ and a contradiction is reached because the first term implies an even $y$ and the second an odd $y$.
Hence $k$ is always 4 when it is a natural number. |
Contours and Closed Path | It seems that a contour is defined as "made up of a finite number of smooth paths which have non-zero continuous derivatives". See p.91.
Your path is continuous (note that $\lim_{t \to 0} \gamma(t) = \lim_{t \to 1} \gamma(t) = 0$). However it is not differentiable at $t = 0, 1$ since
$$\dfrac{t + it\sin(\pi/t)}{t-0} = 1 +i\sin(\pi/t)$$
does not have a limit as $t \to 0$. |
Determining which values to use in place of x in functions | To find the constants in the rational fraction $$\frac{x^2+3x-4}{x^3-4x^2+4x}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2},$$
you may use any set of 3 values of $x$, provided that the denominator $x^3-4x^2+4x\ne 0$.
The "standard" method is to compare the coefficients of $$\frac{x^2+3x-4}{x^3-4x^2+4x}=\frac{A}{x}+\frac{B}{x-2}+\frac{C}{(x-2)^2},$$ after multiplying this rational fraction by the denominator $x^3-4x^2+4x=x(x-2)^2$ and solve the resulting linear system in $A,B,C$. Since
$$\begin{eqnarray*}
\frac{x^{2}+3x-4}{x\left( x-2\right) ^{2}} &=&\frac{A}{x}+\frac{B}{x-2}+%
\frac{C}{\left( x-2\right) ^{2}} \\
&=&\frac{A\left( x-2\right) ^{2}}{x\left( x-2\right) ^{2}}+\frac{Bx\left(
x-2\right) }{x\left( x-2\right) ^{2}}+\frac{Cx}{x\left( x-2\right) ^{2}} \\
&=&\frac{A\left( x-2\right) ^{2}+Bx\left( x-2\right) +Cx}{x\left( x-2\right)
^{2}} \\
&=&\frac{\left( A+B\right) x^{2}+\left( -4A-2B+C\right) x+4A}{x\left(
x-2\right) ^{2}},
\end{eqnarray*}$$
if we equate the coefficients of the plynomials
$$x^{2}+3x-4\equiv\left( A+B\right) x^{2}+\left( -4A-2B+C\right) x+4A,$$
we have the system
$$\begin{eqnarray*}
A+B &=&1 \\
-4A-2B+C &=&3 \\
4A &=&-4,
\end{eqnarray*}$$
whose solution is
$$\begin{eqnarray*}
B &=&2 \\
C &=&3 \\
A &=&-1.
\end{eqnarray*}$$
Alternatively you could use the method indicated in parts A and B, as an example.
A. We can multiply $f(x)$ by $x=x-0$ and $\left( x-2\right) ^{2}$ and let $%
x\rightarrow 0$ and $x\rightarrow 2$. Since $$\begin{eqnarray*}
f(x) &=&\frac{P(x)}{Q(x)}=\frac{x^{2}+3x-4}{x^{3}-4x^{2}+4x} \\
&=&\frac{x^{2}+3x-4}{x\left( x-2\right) ^{2}} \\
&=&\frac{A}{x}+\frac{B}{x-2}+\frac{C}{\left( x-2\right) ^{2}},\qquad (\ast )
\end{eqnarray*}$$
if we multiply $f(x)$ by $x$ and let $x\rightarrow 0$, we find $A$: $$A=\lim_{x\rightarrow 0}xf(x)=\lim_{x\rightarrow 0}\frac{x^{2}+3x-4}{\left(
x-2\right) ^{2}}=\frac{-4}{4}=-1.$$ And we find $C$, if we multiply $f(x)$ by $\left( x-2\right) ^{2}$ and let $x\rightarrow 2$: $$C=\lim_{x\rightarrow 2}\left( x-2\right) ^{2}f(x)=\lim_{x\rightarrow 2}\frac{%
x^{2}+3x-4}{x}=\frac{2^{2}+6-4}{2}=3.$$
B. Now observing that $$P(x)=x^{2}+3x-4=\left( x+4\right) \left( x-1\right)$$ we can find $B$ by making $x=1$ and evaluate $f(1)$ in both sides of $(\ast)$, with $A=-1,C=3$: $$f(1)=0=2-B.$$ So $B=2$. Or we could make $x=-4$ in $(\ast)$ $$f(-4)=0=\frac{1}{3}-\frac{1}{6}B.$$ We do obtain $B=2$.
Thus $$\frac{x^{2}+3x-4}{x\left( x-2\right) ^{2}}=-\frac{1}{x}+\frac{2}{x-2}+\frac{3%
}{\left( x-2\right) ^{2}}\qquad (\ast \ast )$$
Remark: If the denominator has complex roots, then an expansion as above is
not possible. For instance
$$\frac{x+2}{x^{3}-1}=\frac{x+2}{(x-1)(x^{2}+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{%
x^{2}+x+1}.$$
You should find $A=1,B=C=-1$:
$$\frac{x+2}{x^{3}-1}=\frac{1}{x-1}-\frac{x+1}{x^{2}+x+1}.$$ |
Expressing eigenvalues of a symmetric matrix in terms of its entries | This is the same as saying that $aI< A< bI$ where $A<B$ means that $B-A$
is positive definite. A symmetric matrix is positive definite iff each
matrix formed by intersection of the first $k$ rows and
first $k$ columns has positive determinant for all $k$ from $1$ to $n$.
So we can express your condition in terms of a finite set of strict inequalities. |
Prove that there exists an ordered basis $\gamma$ for which $[T^*]_\gamma$ has a column of $0$s. | Since $T^*$ is not an isomorphism, there is some $V\ni a\not=0$ such that $T^*(a)=0$. Now try to find a basis containing $a$. |
Rotation about z axis using quaternions | So I unravelled the mystery. The rotation of a vector defined as:
$$\mathbf{v'} = \mathbf{q} \oplus \mathbf{v} \oplus \mathbf{q^*} = \mathbf{q}^+ \mathbf{q^*}^\oplus\mathbf{v}$$
It is actually the active rotation of a vector $\boldsymbol{v}$. Hence, in my example I am rotating the vector to a new vector that I wrongly called $\boldsymbol{v_B}$, but is still expressed in the reference frame $\boldsymbol{A}$, therefore it should be called $\boldsymbol{v'_A}$.
The passive rotation of a vector is defined as:
$$\mathbf{v'} = \mathbf{q^*} \oplus \mathbf{v} \oplus \mathbf{q} = \mathbf{q^*}^+ \mathbf{q}^\oplus\mathbf{v}$$
That gives the correct result. |
find the volume of a solid limited by a paraboloid and a cylinder | The region projected onto the $x-y$ plane gives a circle of radius $1$, and $z$ is integrated from $0$ to $4-x^2-y^2$. Therefore, using cylindrical coordinates, the integral becomes
$$V = \int_0^{2\pi}\int_0^1 \int_0^{4-x^2-y^2} r\, dz\, dr\, d\theta$$ |
Is the mean of the truncated normal distribution monotone in $\mu$? | To "address all the concerns" we prove two results: first that the derivative
of the truncated mean w.r.t. $\mu$ has an upper bound of unity. Second that this derivative is equal to the ratio of the truncated variance over the untruncated variance (and hence is always positive).
$$\text{Result A:} \frac{\partial E[x|a\le x\le b]}{\partial \mu} \le 1$$
We will use results related to log-concave functions (i.e. functions whose logarithm is a concave function). Please see link and references therein.
Obviously, $\beta\,\gt\alpha$. Define then the non-negative bi-variate function $F(\beta,\alpha)\equiv {\Phi(\beta)-\Phi(\alpha)}$. It can be equivalently written as $$F(\beta,\alpha)\,=\,\int_{\alpha}^{\beta} \phi(\tau)d\tau=\,\int_{-\infty}^{\infty} \phi(\tau)I(\alpha\le\tau\le \beta)d\tau$$
where $I()$ is the indicator function. The indicator function is log-concave. The standard normal pdf $\phi$ is log-concave. The product of two log-concave functions is log-concave. If we integrate a log-concave function over one of its arguments, the resulting function is log-concave w.r.t. to the remaining variables. So $F(\beta,\alpha)$ is log-concave in $(\beta,\alpha)$ (this is a reproduction of a proof found in Pratt, J. W. (1981). Concavity of the log likelihood. Journal of the American Statistical Association, 76(373), 103-106.) Consider now the uni-variate function
$$\ln H(\mu)=\ln F(\beta(\mu),\alpha(\mu))$$
Note that both $\beta$ and $\alpha$ are linear functions of $\mu$. Then $\ln H(\mu)$ is a concave function in $\mu$ (see for example p. 86 eq. [3.15], Boyd & Vandenberghe (2004). Convex optimization, noting that the 2nd derivatives of $\beta$ and of $\alpha$ w.r.t. $\mu$ are zero).
Now
$$\frac{\partial \ln H(\mu)}{\partial \mu}\,=\,\frac{\phi(\beta)-\phi(\alpha)}{\Phi(\beta)-\Phi(\alpha)} \left(\frac{-1} {\sigma}\right)= \frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)} \left(\frac{1} {\sigma}\right)$$
Multiply throughout by $\sigma^2:$
$$\sigma^2\frac{\partial \ln H(\mu)}{\partial \mu}\,= \frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)} \sigma$$
Then we can write
$$E[x|a\le x\le b]=\mu+\sigma^2\frac{\partial \ln H(\mu)}{\partial \mu}\tag{1}$$
and therefore
$$\frac{\partial E[x|a\le x\le b]}{\partial \mu}=1+\sigma^2\frac{\partial^2 \ln H(\mu)}{\partial \mu^2}\tag{2}$$
Since $\ln H(\mu)$ is concave, its second derivative is non-positive. So we have established:
$$\frac{\partial E[x|a\le x\le b]}{\partial \mu}\le 1 \qquad \forall \;(\mu,\sigma,a,b, a \lt b) \tag{3}$$
Intuitively, the mean of the truncated distribution never changes as much as the underlying location parameter (don't expect the equality to hold).
$$\text{Result B:} \frac{\partial E[x|a\le x\le b]}{\partial \mu}=\frac{\operatorname{Var}_{tr}(x)}{\sigma^2}\gt 0$$
For compactness we will use the following shorthands: ${\Phi(\beta)-\Phi(\alpha)}\equiv Z$, which is a function of $\mu$, $E[x|a\le x\le b]\equiv E_{tr}(x)$, $\frac{\partial \ln H(\mu)}{\partial \mu}\equiv h'$. Expressing the truncated mean in integral form we have
$$E_{tr}(x)=\int_a^bx\frac{1}{Z\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)dx $$
$$\Rightarrow \frac{\partial E_{tr}(x)}{\partial \mu} = \int_a^bx\frac{1}{Z\sigma}\phi'\left(\frac{x-\mu}{\sigma}\right)\left(\frac{-1}{\sigma}\right)dx\;+\;\left[\phi(\beta)-\phi(\alpha)\right]\frac{1}{Z\sigma}\int_a^bx\frac{1}{Z\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)dx$$
Now
$$\phi'\left(\frac{x-\mu}{\sigma}\right)=(-1)\left(\frac{x-\mu}{\sigma}\right)\phi\left(\frac{x-\mu}{\sigma}\right)$$
Also, the last integral equals $E_{tr}(x)$, while $\left[\phi(\beta)-\phi(\alpha)\right]\frac{1}{Z\sigma}=-h'$. So we have
$$\frac{\partial E_{tr}(x)}{\partial \mu} = \left(\frac{1}{\sigma^2}\right)\int_a^bx\frac{1}{Z\sigma}\left(x-\mu\right)\phi\left(\frac{x-\mu}{\sigma}\right)dx\;-\;h'E_{tr}(x) \tag{4}$$
Denote the remaining integral $I$ and break it in two:
$$\left(\frac{1}{\sigma^2}\right)I=\left(\frac{1}{\sigma^2}\right)\int_a^bx^2\frac{1}{Z\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)dx\;-\;\left(\frac{\mu}{\sigma^2}\right)\int_a^bx\frac{1}{Z\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)dx$$
The first integral is the second raw moment of the truncated distribution, while the second is $E_{tr}(x)$. So
$$\left(\frac{1}{\sigma^2}\right)I=\left(\frac{1}{\sigma^2}\right)E_{tr}(x^2)\;-\;\left(\frac{\mu}{\sigma^2}\right)E_{tr}(x)$$
Inserting into eq.(4) we have
$$\frac{\partial E_{tr}(x)}{\partial \mu} = \left(\frac{1}{\sigma^2}\right)E_{tr}(x^2)\;-\;\left(\frac{\mu}{\sigma^2}\right)E_{tr}(x)\;-\;h'E_{tr}(x)$$
$$\Rightarrow\frac{\partial E_{tr}(x)}{\partial \mu} = \left(\frac{1}{\sigma^2}\right)\left[E_{tr}(x^2)\;-\;\mu E_{tr}(x)\;-\;h'\sigma^2E_{tr}(x)\right]$$
$$\Rightarrow\frac{\partial E_{tr}(x)}{\partial \mu} = \left(\frac{1}{\sigma^2}\right)\left[E_{tr}(x^2)\;-\;\left(\mu +h'\sigma^2\right)E_{tr}(x)\right]$$
From eq.$(1)$ we have $\mu +h'\sigma^2=E_{tr}(x)$. Substituting we obtain
$$\Rightarrow\frac{\partial E_{tr}(x)}{\partial \mu} = \left(\frac{1}{\sigma^2}\right)\left[E_{tr}(x^2)\;-\;\left(E_{tr}(x)\right)^2\right]=\frac{\operatorname{Var}_{tr}(x)}{\sigma^2}\gt 0$$
which is what we wanted to prove. |
How can I write the statement "the product of two negative integers is positive" using quantifiers? | $\forall x \forall y ((x<0) \land (y<0) \implies (x \times y > 0))$ |
boolean algebra simplification to remove extra term | $$\begin{align*}
AB+\lnot AC+BC&=AB+\lnot AC+(A+\lnot A)BC\\
&=AB+\lnot AC+ABC+\lnot ABC\\
&=(AB+ABC)+(\lnot AC+\lnot ACB)\\
&=AB+\lnot AC
\end{align*}$$ |
Proving $(5-\frac5k )(1+\frac{1}{(k+1)^2}) \le 5 - \frac{5}{k+1}$ | If you are still stumped about where the minus sign is coming from, here is a more detailed look at the manipulations and properties involved to get from Step 3 to Step 4 in Gigili's answer.
$$\eqalignno{
&\phantom{=}\thinspace 5\left(1 -{k^2 + k + 2\over k{(k + 1)}^2} \right)&(1)\cr
&=5\left(1 + \left(-{k^2 + k + 2\over k{(k + 1)}^2}\right) \right)&(2)\cr
&=5\left(1 + \left(-1\left({k^2 + k + 2\over k{(k + 1)}^2}\right)\right) \right)&(3)\cr
&=5\left(1 + \left(-1\left({k(k + 1) + 2\over k{(k + 1)}^2}\right)\right) \right)&(4)\cr
&=5\left(1 + \left(-1\left({k(k + 1)\over k{(k + 1)}^2} + {2\over k{(k + 1)}^2}\right)\right) \right)&(5)\cr
&=5\left(1 + \left(\biggl(-1\biggr)\left({k(k + 1)\over k{(k + 1)}^2}\right) + \biggl(-1\biggr)\left( {2\over k{(k + 1)}^2}\right)\right) \right)&(6)\cr
&=5\left(1 + \left(\left(-{k(k + 1)\over k{(k + 1)}^2}\right) + \left(- {2\over k{(k + 1)}^2}\right)\right)\right)&(7)\cr
&=5\left(1 + \left(-{k(k + 1)\over k{(k + 1)}^2}\right) + \left(- {2\over k{(k + 1)}^2}\right)\right)&(8)\cr
&=5\left(1 - {k(k + 1)\over k{(k + 1)}^2} - {2\over k{(k + 1)}^2} \right)&(9)\cr
}$$
(1) Given (Step 4 in the question / Step 3 in Gigili's answer).
(2) Definition of subtraction in terms of addition.
(3) Multiplication property of negative one.
(4) Associative property of addition; factoring.
(5) Addition of fractions with like denominators.
(6) Distributive property.
(7) Multiplication property of negative one.
(8) Associative property of addition.
(9) Definition of subtraction in terms of addition (Step 4 in Gigili's answer). |
My solution to optimizing a utility function does not agree with the textbook answer. | My error lies in the following step:
$C_1 = (V_1 + w(1-\tau)(1-F) - C_1)(1+\phi)$
$\implies\ C_1 = \displaystyle\frac {(V_1 + w(1-\tau)(1-F))(1+\phi)}{2+\phi}$
$\ $
In any case, a quicker method would be to realise that $$0 = U_F = \frac{\beta}F - \frac {w(1-\tau)}{C_1}$$ and then move on from there. |
If $\lim_{n \rightarrow \infty} (a_{n+1}-\frac{a_n}{2})=0$ then show $a_n$ converges to $0$. | Let $b_n = a_{n+1} - \frac{1}{2}a_n$. Then
$$ a_n = b_{n-1} + \frac{b_{n-2}}{2} + \cdots + \frac{b_1}{2^{n-2}} + \frac{a_1}{2^{n-1}}. $$
Since $(b_n)$ converges, there exists $M$ such that $|a_1| \leq M$ and $|b_n| \leq M$ for all $n$. Thus for any fixed $m$ and for any $n > m$, we have
$$ |a_n| \leq \Bigg| \underbrace{b_{n-1} + \cdots + \frac{b_{n-m}}{2^{m-1}}}_{\text{(1)}} \Bigg| + \underbrace{\frac{|b_{n-m-1}|}{2^m} + \cdots \frac{|a_1|}{2^{n-1}}}_{(2)}.$$
Note here that
$\text{(1)}$ consists of fixed number of terms, each tending to zero as $n\to\infty$.
$\text{(2)}$ is uniformly bounded by $\frac{M}{2^m} + \frac{M}{2^{m+1}} + \cdots = \frac{M}{2^{m-1}}$.
So, taking limsup as $n\to\infty$ yields
$$ \limsup_{n\to\infty} |a_n| \leq \frac{M}{2^{m-1}}. $$
Since the LHS is a fixed number and $m$ is arbitrary, letting $m\to\infty$ proves the claim. |
How do I show R is regular? | It's a metric space, so it's even normal. The open sets are
$$\{x\in\mathbb R\mid\inf_{y\in F_1} d(x,y)<\inf_{y\in F_2} d(x,y)\}\text{ and }\{x\in\mathbb R\mid\inf_{y\in F_1} d(x,y)>\inf_{y\in F_2} d(x,y)\}$$ |
Polynomial-form - proper naming | A polynomial $f(z)\in R[z]$ is defined over, say, a ring $R$, and is written as in $(2)$, where the coefficients $a_i$ belong to $R$. So this has the name "polynomial of degree $n$ over $R$". $(2)$ is a special decomposition into "linear factors", assuming that all $z_i\neq 0$. For $R=\mathbb{C}$ the general decomposition is given by
$$
P_n(z)=a_n(z-z_1)\cdots (z-z_n).
$$
The name for this is the fundamental theorem of algebra. $(3)$ has no special name.
Sometimes different names are used for polynomial rings over various rings, e.g., polynomials $f\in \mathbb{Z}[x]$ are called "integer polynomials", and the equation $f(x)=0$ then is called a "Diophantine equation". |
Evulating $\int_I f$ by using Darboux Sum convergence Criterion | The volume of $J$ should be $\frac{1}{n^2}$. So, you have the sum $$\frac{1}{n^5}\sum_{k,l=1}^nk^2l=\frac{1}{n^5}\sum_{k=1}^nk^2\sum_{l=1}^nl=\frac{1}{n^5}\frac{n(n+1)(2n+1)}{6}\frac{n(n+1)}{2},$$ which converges to $\frac{1}{6}$ as $n\to\infty$. |
Are there any common functions having order of magnitude between $x$ and $\log x$? | Sure: for instance, $\sqrt{x\log x}$. |
Problem Solving - Bouncing a billiards ball between two rays | I take it that $\angle R'AX = \beta$. Every time the ball hits a wall, we get a new line ($AX, AM, AN...$) which swings out by angle $\beta$. The ball will stop bouncing when the line swings out when the next line swings 'past parallel' to line $RX$, so that they don't intersect. As you can see in the diagram, this happens when the next line has rotated at least $180-\alpha$ degrees. Hence we seek the smallest $k$ such that $k \beta \geq 180 - \alpha$. This gives the formula. |
Continuity of functions in topology | Your argument (well, you don't prove that the inverse image is not open) shows that $\Phi$ is not globally continuous, but it does not study the pointwise continuity of $\Phi$, as you asked.
It's easy to see that each point has one minimal neighbourhood: the circle around the origin going through it, so:
A function $f(\mathbb{R}^2, \sigma) \to X$ is continuous at $(x,y)$ iff for every open set $O$ of $X$ with $f(x) \in O$ we have $f[B_r] \subseteq O$ where $r = \sqrt{x^2 + y^2}$.
As $B_0 = \{(0,0)\}$ this implies that all $f$ are continuous at $(0,0)$.
But as to $\Phi$, this criterion shows that $\Phi$ is only continuous there.
Also a function $f:X \to (\mathbb{R}^2, \sigma)$ is continuous iff $f^{-1}[B_r]$ is continuous for all $r \ge 0$.
Let $x$ be a point in $\Psi^{-1}[B_r]$. This means that $x^2 + a^2 = r^2$, which means that $r^2 \ge a^2$ and $x \in \{-\sqrt{r^2-a^2},\sqrt{r^2-a^2}\}$.
So $\Psi^{-1}[B_r]$ can be empty, but if it is not, then it is a finite subset of $\mathbb{R}$, so never open. In particular $\Psi^{-1}[B_{a^2}] = \{0\}$ for all $a$. So whatever $a$ is, $\Psi$ is not continuous. |
Multiple integral 3 dimension | Hint
The graph shows in a 2D way the graph of $z$ (vertical axis) against $r=\sqrt{x^2+y^2}$. The blue shaded area is $z\ge r^2$, the other area is the other inequality. So you are interested in the overlap. You have to rotate it about the vertical axis (through the centre) to get the solid, whose volume you have to find. |
What did I do wrong with this logarithmic equation? | The calculation (last step) step is wrong.
$$ x = \frac{\ln(4e) + 2}{2} = \frac{\ln 2^{2} + \ln e + 2}{2} = \frac{2\ln 2 + 3}{2} = \ln 2 + \frac{3}{2} \approx 2.193 $$
(Also, you should have used parentheses the third step down.) |
Prove that trace of a matrix is $0$. | The given equality is equivalent to $AC=CA$. According to the Jacobson lemma, $C$ is nilpotent and we are done.
cf. page 1 of https://jankobracic.files.wordpress.com/2011/02/on-the-jacobsons-lemma.pdf
EDIT. to @ George R. Assume that we replace the underlying field $\mathbb{C}$ with a commutative ring $R$. Then the previous reference shows that there is $k$ s.t. $n!A^k=0_n$; if $n!$ is not a zero divisor in $R$, then $A^k=0$ and $A$ is nilpotent. Unfortunately, we know little about the traces of $(A^j)_{j\leq k}$ and even $A^n$ is not necessarily $0$ ! For instance, consider $A=2I_2$ over $R=\mathbb{Z}/16\mathbb{Z}$; $A$ is nilpotent because $A^4=0$ but $tr(A)=4\not= 0,tr(A^2)=8\not=0$ and $A^3\not=0$.
Now, we replace $\mathbb{C}$ with a field $K$ s.t. $charac(K)>n$; then $n!$ is invertible and $A^k=0$; if $spectrum(A)=(\lambda_i)_i$ (the eigenvalues), then $spectrum(A^k)=(\lambda_i^k)_i=(0,\cdots,0)$. Thus, for every $i$, $\lambda_i=0$, and moreover, for every $i,p\geq 1$, $\lambda_i^p=0$. Finally, for every $p\geq 1$, $tr(A^p)=\sum_i\lambda_i^p=0$.
Conversely, we can prove that if $tr(A)=\cdots=tr(A^n)=0$, then $spectrum(A)=\{0,\cdots,0\}$ and $A$ is nilpotent. |
How should I proceed with this inequation? | HINT: after expanding we obtain $$2x^2+5x+2<0$$ this can be factorized into
$$(x+2)(2x+1)<0$$ |
How to define a lattice as an abelian group? | As Tobias Kildetoft says. A lattice is the same thing as a finitely generated free $\mathbb{Z}$-module, i.e. a finitely generated free abelian group.
This is equivalent to the definition you gave:
If I have a free $\mathbb{Z}$-module it is isomorphic to $\mathbb{Z}^d$ for some $d\in{\mathbb N}$. It is clear that $\mathbb{Z}^d\subset \mathbb{R}$ is $\mathcal{L}(B)$, where $B$ is the standard basis of $\mathbb{R}$.
Conversely, every $\mathcal{L}(B)$ is clearly finitely generated by definition. To see that it is free, observe that it is torsion-free (as a sub-$\mathbb{Z}$-module of $\mathbb{R}^d$ which is torsion-free). It is a general fact that torsion-free finitely generated $\mathbb Z$-modules are free. |
Cofusing partial order "implies", on logic and that on sets | First we need to review the definition of "implies".
Suppose $P,Q$ are two sentences. $P\implies Q$ is exactly the case that if $P$ is true then $Q$ is also true.
In your example, $x\ge 1$ implies $x\ge 0$ because $1\ge 0$ and $\ge$ is transitive, that is if $x\ge y$ and $y\ge z$ then $x\ge z$.
Suppose now that you have $\mathcal P=\{P_1,\ldots\}$ a set of sentences (not necessarily finite), and $Q$ another sentence. We say that $\{P_1,\ldots\}\implies Q$ exactly if all the sentences in $\mathcal P$ are true, then $Q$ is also true.
An example is:
$$\{"x\ge2", "x\text{ is odd}"\}\implies x\ge 3$$
Neither $x\ge 2$ nor $x\text{ is odd}$ imply $x\ge 3$ on their own, however combine the two facts and you have that $x\ge 3$.
This can be extended into replacing $Q$ by some $\mathcal Q=\{Q_1,\ldots\}$ another set of sentences, in which case we say that $\mathcal P\implies\mathcal Q$ exactly whenever all the sentences in $\mathcal P$ are true, then all the sentences in $\mathcal Q$ are true as well.
For example:
$$\{"x\ge 1", "x\text{ is even}", "x\text{ can be divided by } 6"\}\implies\{"x\text{ can be divided by } 3", "x\ge 6"\}$$
There is no strict requirement that a "big" sets have to imply the truth of some "smaller" or "bigger" set than itself. One sentence can imply many sentences, and it may be the case that several sentences are required to imply a single sentence.
A mistake in your post is that $\implies$ is not a partial order, but rather a quasi-order. That means reflexive and transitive. It is not antisymmetric.
For example: $$"x<1 \text{ and } x>-1"\implies "x=0"$$ as well: $$"x=0"\implies "x<1 \text{ and } x>-1"$$
However formally speaking, $x=0$ is not the same sentence as $"x<1 \text{ and } x>-1"$. |
Maximum of trivial function $(y-x^T\beta)^2$ | I would not expect a simple expression. You are asking for the two support hyperplanes for $B$ that are orthogonal to $x.$ If $B$ is not convex, it is possible that a support hyperplane intersects $B$ in more than one point, but that will not change $y - x^T \beta$ or $(y - x^T \beta)^2$
https://en.wikipedia.org/wiki/Supporting_hyperplane |
Using Cauchy Integral Formula To Solve Integrals | Yes, the solution would be $2\pi i$. Letting $z=e^{i\theta}$, we have $\text{d}z=ie^{i\theta}\text{d}\theta$, and your integral becomes
$$i\int_{0}^{2\pi}e^{2\cos\theta}\cos\left(2\sin\theta\right)\text{d}\theta-\int_{0}^{2\pi}e^{2\cos\theta}\sin\left(2\sin\theta\right)\text{d}\theta.$$
Notice that the right integral equals $0$ because it is odd with a period of $2\pi$. Thus,
$$\int_{\left\lvert z\right\rvert=1}\frac{e^{2z}}{z}\text{d}z=i\int_{0}^{2\pi}e^{2\cos\theta}\cos\left(2\sin\theta\right)\text{d}\theta=2\pi i,$$
giving
$$\int_{0}^{2\pi}e^{2\cos\theta}\cos\left(2\sin\theta\right)\text{d}\theta=2\pi.$$ |
Electric field off axis inside a charged ring. | So if we take a very small gaussian pillbox centred on the origin of height $2z$ and radius $r$ in the limit the field out of the top and bottom surfaces is: $$\frac {Qz\pi r^2}{4\pi\epsilon_0(a^2+z^2)^{3/2}}$$
therefore as the total charge enclosed is zero and we know the field through the sides of our pillbox is radial and of constant magnitude we can arrive at: $$ \frac {2Qz\pi r^2}{4\pi\epsilon_0(a^2+z^2)^{3/2}} - 4\pi rzE = 0$$ This rearranges to: $$E=\frac{Qr}{8\pi\epsilon_0(a^2+z^2)^{3/2}}$$ which is: $$\frac{Qr}{8\pi\epsilon_0a^3}$$ if we take the taylor series and say the $\frac{z^2}{a^2}$ and above terms are negligible in the limit $z<<a$. Seems to be quite wishy washy but it does get the right answer.
Thanks to wltrup for help! |
How to show this function is monotonic strictly increasing? | For sure $f$ is increasing in the sense that $f(x) \geq f(y)$ whenever $x > y$. Now assume $f(x)=f(y)$ for $x > y$. Clearly, then $f$ must be constant on $[y,x]$, contradicting the fact that the derivative vanishes at only one single point. |
Conditional probability of continuous throwing a dice | In a single roll of the dice, the only way a prime number does not appear is, if the die lands $2,4,6$. Restricting our attention to these three outcomes, the probability that the game does not end, conditional on a prime not appearing is $2/3$, as you are interested only in $2,4$. All trials are independent. So, if we extend this to $5$ trials, we get $(2/3)^5$. Further, in the sixth trial, you would like a $6$. So, that $(1/3)$.
Hence, it is $(2/3)^5 (1/3)$. |
for which conditions of postive integer $n, m >0$ :$\dfrac{\sigma{(n)}}{n}\leq\dfrac{\sigma{(n+m)}}{n+m}$ hold? | Since the divisors of $n+m$ have little to do with the divisors of $n$, I doubt that you'll find a nice necessary and sufficient condition. The function $f(n) = \sigma(n)/n$ is quite irregular: here is its graph for
$1 \le n \le 1000$.
One sufficient condition: if $m = k n$, then $\sigma((k+1)n) \ge (k+1) \sigma(n)$ so
$$ \dfrac{\sigma(n+m)}{n+m} = \dfrac{\sigma((k+1)n)}{(k+1)n} \ge \dfrac{\sigma(n)}{n}$$
A necessary condition is that $m+n$ is composite, since if $p$ is prime, $\sigma(p)/p = 1 + 1/p < \sigma(t)/t$ for all $t < p$. |
Evaluate $\int\sec^4(u) \operatorname d \!u$ | $$\int \sec^4(u) du = \int \sec^2(u)\cdot \sec^2(u) du = \int \sec^2(u)(1+\tan^2(u))du$$
Now let $x=\tan(u)$ so that $dx = \sec^2(u)du$ thus transforming the integral to:
$$
\int (1+x^2)dx
$$ |
Is it possible to find an exact solution for $x$ in $\frac{x^2+9}{6 x}=\frac{\sin (\pi -1) x}{\cos \pi x}$? | Here, I have a sort of "moral" argument for why there should be no poles. To begin, if there is a closed for for solutions to
$$\frac{(x^2+9)\cos(\pi x)}{x^2}=6\sin(\pi-1)$$
(which is equivalent to your problem) then we should expect that this neat formula will also give solutions to the equations
$$\frac{(x^2+9)\cos(\pi x)}{x^2}=c$$
for any constant $c$ since the constant $6\sin(\pi-1)$ is not nice at all and I cannot conceive of a formula that would solve this equation for only the value $c=6\sin(\pi-1)$. This means that we are looking for a "nice" function $f(c)$ such that
\begin{equation}
\frac{(f^2(c)+9)\cos(\pi f(c))}{f^2(c)}=c\tag{1}
\end{equation}
If $f(c)$ is a nice function, then surely a derivative of this function must exist (otherwise, I would not qualify $f(c)$ as a nice function). Thus,
$$\left(\frac{(f^2(c)+9)\cos(\pi f(c))}{f^2(c)}\right)^{'}=1$$
taking the derivative, we can then substitute in from (1) to write $\cos(\pi f(c))$
and $\sin(\pi f(c))$ in terms of a polynomial in $f(c),c$. Overall, this yields that
$$-18cf\left(c\right)^{-3}f'\left(c\right)=1+\frac{9}{f\left(c\right)^{2}}+\pi f'\left(c\right)\left(1+\frac{9}{f\left(c\right)^{2}}\right)^{2}\sqrt{1-\frac{c^{2}f\left(c\right)^{4}}{\left(f\left(c\right)^{2}+9\right)^{2}}}$$
Which, after expanding, is a polynomial in $f$ and $f'$. Apart from that, this differential equation has no special symmetry. It has been shown that general polynomials in $f$,$f'$, and $c$ have no closed form solution and so since we have no reason to expect that something special is going on here I would say that it is highly unlikely that your equation has a closed form. Not impossible, but definitely improbable. |
$g(f(n))\in o(g(n)/n)$ for any $f(n)\in o(n)$ | $f(n)\in o(n)$, hence $f(n)\leq n/h(n)$ for some function $h(n)\to +\infty$. Thus $h(n)\geq 2$ for any large $n$. Take $g(n)=e^n$: $g(f(n))\leq e^{n/2}\in o(e^n/n)$. |
Dual vector space properties ($\forall \phi\in E^*,\langle\phi,x\rangle=0\implies x=0$) | Let $dim(E)=n$ and $A=\{v_1,v_2...v_n\}$ a basis of $E$.
Now let $x \in E$,then $x=a_1v_n+a_2v_2+...a_nv_n$ for some $a_1...a_n \in F$
Take the set of linear functionals $B=\{f_1,f_2...f_n\}$ where $f_i(v_j)=0$ if $i \neq j$ and $f_i(v_j)=1$ if $i=j$.
We have that $$0=f_i(x)=a_1f_i(v_1)+a_2f_i(v_2)+...+a_if_i(v_i)+...+a_nf_i(v_n)=a_i,\forall i \in \{1,2....n\},$$
Thus $a_i=0,\forall i \in \{1,2....n\} \Rightarrow x= \bar{0}$ |
Where am I wrong when calculating $\sum_{k=2}^{\infty}5k \frac{1}{6^{k-1}}$ | Can you differentiate a constant? Yes, but the result is zero. Sorry, but your method is completely wrong.
You can leave out the constant factor $5$ for the moment and consider
$$
f(t)=\sum_{k=2}^\infty kt^{k-1}
$$
If we integrate from $0$ to $x$, we get, for $|x|<1$,
$$
\int_0^x f(t)\,dt=\sum_{k=2}^\infty x^k=\frac{1}{1-x}-1-x
$$
so if we differentiate
$$
f(x)=\frac{1}{(1-x)^2}-1
$$
With $x=1/6$ and deploying the factor $5$ we obtain
$$
\sum_{k=2}^\infty 5k\frac{1}{6^{k-1}}=\frac{5}{(1-1/6)^2}-5=5\cdot\frac{36}{25}-5=\frac{11}{5}
$$ |
Applying sine orthogonality | Your eigenfunctions are solutions of
$$
y''+\lambda y = 0,\;\;\; 0 \le x \le a \\
y(0) = 0,\;\; y(a)=0.
$$
That is,
$y_n(x)=\sin(n\pi x/a)$ for $n=1,2,3,\cdots$ are the solutions where $\lambda_n=n^2\pi^2/a^2$. These are automatically orthogonal because of the selfadjoint nature of this ODE. And they form an orthonormal basis of $L^2[0,a]$.
Likewise, $y_n(x)=\cos(n\pi x/a)$ for $n=0,1,2,3,\cdots$ are solutions of
$$
y''+\lambda y= 0,\;\;\; 0 \le x \le a \\
y'(0)=0,\;\; y'(a)=0
$$
So these functions also form an orthogonal basis of $L^2[0,a]$.
There are also orthogonal bases of $\sin$ functions with non-harmonic arguments that form an orthogonal basis. For example, consider the more general problem
$$
y''+\lambda y = 0,\;\;\; 0 \le x \le a \\
y(0)=0,\;\;\; Ay(a)+By'(a)=0.
$$
The functions $\sin(\alpha_n x)$ are solutions where
$$
A\sin(\alpha_n a)+B\alpha_n\cos(\alpha_n a)=0
$$
The solutions $\alpha_n$ are not evenly spaced, but the corresponding $\sin(\alpha_a x)$ are mutually orthogonal. |
Related to Hall's Theorem | Since $S\subset X$, $N(S)\subset N(X)$ and hence $\vert N(S)\vert\leq \vert N(X)\vert$. |
I don't understand completely partial derivatives | The $f$ in your question really stands for two different functions. Let’s separate them. Letting $\phi:(u,v)\mapsto(x,y)$ be the rotation, the second function that you’re dealing with is really $g = f\circ\phi$. The partial derivative ${\partial g\over\partial u}$ should then make perfect sense in terms of coordinates, and can be computed via the chain rule. That’s what the somewhat imprecise notation ${\partial f\over\partial u}$ means here.
If you’re familiar with directional derivatives, here’s another way of looking at it. Partial derivatives are just special directional derivatives taken in the directions of the coordinate axes. From that point of view, you can think of ${\partial f\over\partial u}$ as equivalent to $D_uf$, the directional derivative of $f$ in the $u$ direction. |
How to interpret the fact $P(\limsup A_n) \ge \limsup P(A_n)$ | $P(\limsup A_n)$ is the probability that infinitely many of the $A_n$ occur. If $\limsup(P(A_n)) = a$ that means (roughly) that there are infinitely many events that have probability $a$. So at any point in the sequence, there is at least probability $a$ of another event occurring. Thus the probability of infinitely many events is at least $a$.
This matches up to the line
$$
P\left(\bigcup_{k \geq n}A_k\right) \geq \sup_{k\geq n}P(A_k)
$$
in the proof, where you then take limit $n\rightarrow\infty$ to get the theorem. |
Help with short exact sequence | The splitting lemma tells you that $G$ is a direct sum of the other two groups iff the sequence splits (either on the left or on the right, since we are in the abelian case). In general there would be no reason for this to happen, however your case is special.
Pick an element $g\in f^{-1}(1)$ and define $\psi(1)=g$, extending to a homomorphism $\psi\colon\mathbb{Z}\to G$. It is easy to see that $f\circ \psi$ is the identity, so your sequence splits on the right, hence $G=\mathrm{ker}(f)\oplus \mathbb{Z}$.
The proof generalizes easily to the case where $\mathbb{Z}$ is replaced by any free abelian group (if we were dealing with modules over some ring, a projective module would've sufficed, but, as was pointed out in the comments, there is no distinction between free and projective $\mathbb{Z}$-modules). |
How to find y given an x of this implicit equation | If you have not done numerical methods so far (Regula Falsi, Newton Raphson &c.) then an approximate way is to paper plot the graph of (inverse) function and from it read off approximate $y$ for a given $x:$
$$ x= \sin^{-1}( \log |y| +y^2-1) $$
The online graphing calculator Desmos can do this for us (click on image for larger size):
Keep in mind that by convention $\sin^{-1}$ here only returns $x \in [-\pi/2,+\pi/2]$. The two branches correspond to using the absolute value of $y$, but since the initial point $(0,1)$ is on the upper branch, the continued solution remains on that branch. Visualize the complete solution by reflecting that upper branch backward and forward to yield a periodic "wave". |
Use strong induction to prove that n is congruent. | This game is sometimes known as Nim. At any rate, here's a hint:
Suppose it's Steve's turn with $3$ coins on the table. If he takes $1$ coin, then Bill can take $2$ coins to win. If he takes $2$ coins, Bill can take $1$ coin to win. In this manner, Bill can win no matter what Steve does, as long as he can bring the number of coins down to $3$.
So, if it's Bill's turn and there are $3+1 = 4$ or $3+2 = 5$ coins, then Bill can take $1$ or $2$ coins to put himself in a winning situation.
Now, suppose it's Steve's turn with $6$ coins on the table...
Here's what an inductive proof might look like:
Base case: If it's Bill's turn with $1$ or $2$ coins, then Bill can take the remaining coins in his next turn. So, Bill can win no matter what Steve does in these cases.
Inductive step: Suppose that Bill can win no matter what Steve does if there are $3k+1$ or $3k + 2$ coins left for all integers $k$ from $0$ to $n$.
Now, suppose there are $3(n+1) + 1$ or $3(n+1) + 2$ coins left. Then Bill can take either $1$ or $2$ coins so that there are $3(n+1)$ coins left.
When Steve takes his turn, he take $2$ or $1$ coins, leaving either $3n+1$ or $3n+2$ coins. By our inductive hypothesis, Bill can now win no matter what Steve does.
The conclusion follows. |
How to solve this combinatorial question (Phone number consists of 7 digits) | I'll just get you started.
To count the number of phone numbers with exactly $n$ $0$s, we first pick $n$ spots to be $0$s in ${7}\choose{n}$ ways and then fill the rest out with non-zeros in $(7-n)^{9}$ ways, for a total of $(7-n)^{9} {{7}\choose{n}}$ ways.
For B you will reason similarly, but replace ${7}\choose{n}$ with the total number of ways to pick $n$ zeros such that no two are adjacent.
For C, you know that there are $7^{10}$ possible phone numbers. How many ways can you partition $7^{10}$ objects into three parts? Such numbers are known as Stirling numbers of the second kind. |
The precise relationship between conic sections and parabolas, circles, etc. explained intuitively? | This question is answered in James Tanton's companion guide to quadratics. Although it is tempting to simply copy and paste his explanation here, that would be stealing! He clearly put a lot effort into making the explanation as accessible as possible and so I think he deserves the small compensation of not having his material stolen!
If you are quite new to this as you say you are, here is his open course on quadratics: http://gdaymath.com/courses/quadratics/
And the explanation you are looking for is in this book here:
http://www.lulu.com/shop/james-tanton/companion-guide-to-quadratics/ebook/product-20965967.html |
A particular euclidean function implies the domain is local? | Since $R$ is already given to be an Euclidean domain , so it is a PID , so a non-zero ideal is maximal iff it is prime iff it is generated by an irreducible element iff it is generated by a prime element . Now $p$ is already a given prime , so if we can show that any prime $b\in R$ is associated to $p$ , we are done . Consider $f(b)$ ; if $f(b)>0$ , then $p|b$ , since both $p,b$ are irreducibles , hence this implies $(p)=(b)$ ; now if $f(b)=0$ , then dividing $p$ by $b$ , by Euclidean algorithm , $\exists q,r \in R$ such that $p=bq+r$ , where $r=0$ or $f(r)<f(b)=0$ , but $f(x)\ge 0,\forall x \in R\setminus \{0\}$ , hence $f(r) <0$ is impossible , so $r=0$ , so $b|p$ , then again we get $(b)=(p)$ , thus there is a unique non-zero maximal ideal . |
Let $F = \mathbb{Q}(\sqrt{2}, \sqrt[3]{2})$. Show that $F$ is not normal over $\mathbb{Q}$ | You did it well. Since $\mathbb{Q}\bigl(\sqrt2,\sqrt[3]2\bigr)\subset\mathbb R$ and since $x^3-2$ has one but not all roots in $\mathbb{Q}\bigl(\sqrt2,\sqrt[3]2\bigr)$ (since the other roots are non-real), $\mathbb{Q}\bigl(\sqrt2,\sqrt[3]2\bigr)$ is not normal. |
How to show the congruence involving the order | Hint: Suppose $\textrm{ord}_m a = r$ and $\textrm{ord}_m b = s$; then $a\cdot b\equiv 1\mod m$ implies that $a^rb^r\equiv 1\mod m$, so that $b^r\equiv 1\mod m$. What does that tell you about $r$ as it relates to $s=\textrm{ord}_m b$? Now repeat the process, raising both sides to $s$. |
Real Analysis, related to Arzela-Ascoli theorem | Hints:
1) Show that $g_n$ converges uniformly to $g$.
2)
since $f_n$ are equicontinuous on a compact interval, they are uniformly equicontinuous. |
Generalized Schur-Weyl Duality | As Aaron has pointed out, there are generalizations of the Schur-Weyl duality to certain reductive groups. But as far as I know, there is no unified theory. This paper of Stephen Doty seems to give a good overview of what can be done (although it is most probably not exhausting). |
Solving an inequality with square root | The root square exists if $$x\in ]-\infty,0]\cup [7,+\infty[ $$
we also must have $$x-4\geq 0$$
thus $x $ must necessarly be in $D_1=[7,+\infty[ $.
in $D_1$, we can take the square, to get
$$x^2-8x+16>2x^2-14x $$
$$\iff x^2-6x-16<0$$
$$\iff (x-8)(x+2)<0$$
$$\iff x\in D_2=]-2,8 [$$
Finally
$$x\in D_1\cap D_2=[7,8[$$ |
Continuous Dependence of $x''+K\sin x=0$ | This is the pendulum equation and there isn't a simple solution. If you make the assumption that $x$ is small, you can approximate $\sin x$ with its Taylor expansion and getting a simple solution in terms of trig functions.
enter link description here |
Locally finite vs. Borel measures on $\sigma$-compact Polish spaces | I think that indeed the idea from the common thread works:
Let $e_n, n \in \Bbb N$ be the standard orthonormal base of Hilbert space $\ell^2$. Let $X = \{\lambda e_n \mid n \in \Bbb N, \lambda \in \Bbb R\}$. Then $X$ is $\sigma$-compact Polish but not locally compact (at $0$).
Let $L_k = \{\lambda e_k: \lambda \neq 0\}$ and $\delta_A$ be the Dirac measure with carrier $A$: $\delta_A(B) = 1$ iff $A \cap B \neq \emptyset$ and $0$ otherwise, we can define $\mu = \displaystyle_{k=1}^\infty \delta_{L_k}$, which is a Borel measure (not $\sigma$-finite, and not locally finite at $0$), but (I think) finite at compacta.
I haven't checked all the nitty gritty details. Others might feel inclined to add to this. |
How many $n$-disk legal configurations are there for the Tower of Hanoi? | Each disk can go on one of three needles, so there are $3^n$ ways to assign disks to needles. Once you have a set of disks assigned to a needle, the order is determined. Now we have to worry about what of these are "the same" configuration. If the needles are labeled, we are done. If consider the needles interchangeable, most configurations have $3!=6$ alternatives, but the three configurations with all the disks on one needle are special, so we have $\frac {3^n-3}6+1=\frac {3^{n-1}+1}2$ configurations. |
How to recover a shuffled matrix | That is NP-problem about graph isomorphism.
What kind of algorithm you consider "efficiently"? Some polynomial algo? You can sure use some heuristics, if you know something about $A$, but there is no polynomial algo for general case. |
pullback of canonical divisor | It's exactly not true that $div(f^*\omega) = f^*div(\omega)$. (And your claim that it follows from the definition of the pull-back of Cartier divisors is simply not true.)
Let's just consider the case of one point being blown up, and look locally around the point being blown up; let's say we have local coords. $x$ and $y$, and $x = y = 0$ is the point being blown up.
Suppose that $\omega = dx \wedge dy$ locally, so it has trivial divisor (at least
in a n.h. of $(0,0)$).
The blow-up map is $(x,y) \mapsto (x,xy),$ and so this differential pulls back to
$dx \wedge d(xy) = x dx \wedge dy,$ which now has a non-trivial divisor, namely
$x = 0$, which is exactly the exceptional divisor.
This computation shows that indeed
$$div(f^* \omega) = f^*div(\omega) +\text{ the exceptional divisor}.$$ |
If an abelian group has subgroups of orders $m$ and $n$, respectively, then it has a subgroup whose order is $\operatorname{lcm}(m,n)$. | $\DeclareMathOperator{\lcm}{lcm}$One possibility is to use the formula
$$
o(H K)
=
\frac{o(H) \cdot o(K)}{o(H \cap K)},
$$
then note that since $H \cap K \le H, K$, by Lagrange
$$
o(H \cap K) \mid o(H), o(K),
$$
so that
$$
o(H \cap K) \mid \gcd(o(H), o(K)),
$$
and finally recall that $\gcd(x, y) \cdot \lcm(x, y) = x y$.
With this you obtain that $G$ has a subgroup of order a multiple of the lcm, so you will have to know that the converse of Lagrange's theorem holds in a finite abelian subgroup. |
A question in a theorem of section 6.6 of Hoffman Kunze | They did. They first established that
$$\alpha = E_1\alpha + \cdots + E_n\alpha$$
is one desired decomposition with $E_i\alpha \in W_i$.
Then they took a different decomposition
$$\alpha = \alpha_1 + \cdots + \alpha_n$$
with $\alpha_i \in W_i$ and showed that actually it must be $\alpha_i = E_i\alpha$ for $1 \le i \le n$.
Therefore the decomposition $\alpha = E_1\alpha + \cdots + E_n\alpha$ is unique. For any other decomposition $$\alpha = \beta_1 + \cdots + \beta_n$$
with $\beta_i \in W_i$ we would also get $\beta_i = E_i\alpha$ and hence $\beta_i = \alpha_i$ for all $1 \le i \le n$. |
Definition of $Y^f$ in category theory: exponential pullback. | $Y^f$ is precomposition, internalized via the tensor-hom adjunction.
By the adjunction, to give an arrow $Y^B\to Y^A$, it suffices to give an arrow $Y^B\otimes A\to Y$. We have an arrow $\mathrm{eval}\colon Y^B\otimes B\to Y$. So we can get the desired arrow as $\mathrm{eval}\circ (\mathrm{id}_{Y^B}\otimes f)$. Intuitively, this says "apply $f$ to the input in $A$ before evaluating the map in $Y^B$".
Similarly, $g^A$ can be viewed as internalized postcomposition. By the adjunction, to give an arrow $X^A\to Y^A$, it suffices to give an arrow $X^A\otimes A\to Y$. We have an arrow $\mathrm{eval}\colon X^A\otimes A\to X$. So we can get the desired arrow as $g\circ \mathrm{eval}$. Intuitively, this says "apply $g$ to the output in $X$ after evaluating the map in $X^A$". |
Distance between color change in playing cards | This is about the average length of runs in binary words of length $2n$ having exactly $n$ zeros and $n$ ones. The total number of admissible words is given by $A_n={2n\choose n}$, and the numbers $a_k$ of admissible words with exactly $k$ runs are given by
$$\eqalign{a_{2r}&=2{n-1\choose r-1}^2\qquad\qquad\quad(1\leq r\leq n)\ ,\cr
a_{2r+1}&=2{n-1\choose r}{n-1\choose r-1}\qquad(1\leq r\leq n-1)\ .\cr}$$
(For the first line we can choose to begin with a $0$ or a $1$. Then we have to put $r-1$ separators in the $n-1$ spaces between $n$ stars; this separately for the zeros and the ones. The second line is explained similarly.)
The expected length $E(n)$ of the observed runs is then given by
$$E(n)={2n\over A_n}\left(\sum_{r=1}^n {a_{2r}\over 2r}+\sum_{r=1}^{n-1}{a_{2r+1}\over 2r+1}\right)\ .$$
The following figure shows a plot of the resulting values. As expected one has $\lim_{n\to\infty}E(n)=2$. In particular $E(26)=1.96151$. |
Variable in the exponent? | $$(-4x)\log_{10}10=\log_{10}62\\
\implies x=-\dfrac{\log_{10}62}{4}$$
Now, use a calculator. |
Does the property hold almost everywhere? | Suppose there is a set of positive measure in which you have not convergence. Then it exists a compact set $[ \epsilon, 1-\epsilon]$ such that its intersection with the set considered before has positive measure. But the function converges on $[ \epsilon, 1-\epsilon]$, and you have a contradiction. This should actually prove that the function converges everywhere, and it is based on the fact that you can view your open interval as a countable increasing union of compact sets |
What is this set notation called? | In general, if $A$ and $B$ are subsets of some set with an addition structure, you can write $A+B = \{a+b \mid a \in A, b \in B\}$. Formally what the professor wrote is $$\beta + 2\pi \Bbb Z = \{ \beta + 2\pi n \mid n \in \Bbb Z \},$$and what you're proposing is also correct: $$e\Bbb Z + \pi \Bbb Z = \{ae+b\pi \mid a,b \in \Bbb Z \}.$$The point is precisely that $e\Bbb Z + \pi \Bbb Z \neq (e+\pi)\Bbb Z$. I do not think that this notation has any specific name, though. |
why is $\sqrt{16}$ not equal to $-4$? | By convention, $\sqrt{a}$ denotes the positive* solution to the equation $x^2=a$. So, in your example, $\sqrt{16}$ is the positive solution to $x^2=16$. Then,
\begin{align}
x^2 &= 16 \\
x^2 - 16 &= 0 \\
(x-4)(x+4) &= 0 \\
\end{align}
So $x=4$ or $x=-4$. But since $x$ must be positive, we are left with $x=4$.
We adopt this convention so that $\sqrt{}$ is a function. A function is like a machine that takes in a number and spits out another. It is often represented using an arrow: the function $x \mapsto x+5$ takes in a number $x$, and adds $5$ to it. For example, if we inputted $4$ into the function $x \mapsto x+5$, then the output would be $9$. Note that the '$x$' is simply a piece of notation used to illustrate the rule that the machine is following.
There is one very important requirement that functions must satisfy: the machine can only produce one output. Since each positive number has two square roots, the function $x \mapsto \sqrt{x}$ doesn't really make any sense unless we require that $\sqrt{x}$ denotes the positive square root of $x$.
*To be precise, I should really say nonnegative, since the square root of a number can also be $0$. |
Cross products and determinants in $\mathbb{R}^3$ | They were invented independently.
The determinant was first made to analyze how many solutions a system of equations could have as well as for Cramer's Rule (see http://www-history.mcs.st-and.ac.uk/HistTopics/Matrices_and_determinants.html , for example).
The cross product arose from Hamilton's extension of the complex numbers, the "quaternions" (see http://www.math.mcgill.ca/labute/courses/133f03/VectorHistory.html , for example). |
If $\|f_n\| \leq c$ and $f_n \rightarrow f$ in weak-star topology, then $\|f\| \leq c$? | You do not need heavy machinery like Banach-Alaoglu.
Let $x \in E$ be arbitrary. From (1) you get
$$|f_n(x)| \le c \, \|x\|_E.$$
Due to weak-star convergence, you can pass to the limit and get
$$|f(x)| \le c \, \|x\|_E \qquad\forall x \in E.$$
By definition of the dual norm
$$\| f \|_{E'} = \sup\{ |f(x)| \;\mid\; \|x\|_E \le 1\} \le c.$$ |
Find the splitting field of $x^3-1$ over $\mathbb{Q}$. | Convince yourself, by using the quadratic formula to solve:
$x^2 + x + 1 = 0$
that the splitting field of $x^3 - 1$ is $\Bbb Q(\sqrt{-3})$.
Note: it is customary to denote this field as $\Bbb Q(\omega)$, where:
$\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2}$
Therefore, convince yourself that $\Bbb Q(\omega) = \Bbb Q(\sqrt{-3})$.
It may be helpful to recall the formula for the discriminant of a quadratic equation. |
Lebesgue integral with repect to counting measure | As $\mathcal{A} = \mathcal{P}(\Omega)$, we know that any function $f: \Omega \to \mathbb{R}$ is measurable. By the Sombrero lemma, there exists a sequence $(f_n)_{n \in \mathbb{N}}$ of non-negative simple functions such that $f_n \uparrow f$. Applying Beppo Lévy's theorem yields
$$\int f \, d\mu = \sup_{n \in \mathbb{N}} \int f_n \, d\mu. \tag{1}$$
As $f_n$ is a non-negative simple function, we can choose $c_j^n > 0$ and $\emptyset \neq A_j^n \in \mathcal{B}(\mathbb{R}) $ such that
$$f_n(x) = \sum_{j=1}^{m_n} c_j^n \cdot 1_{A_j^n}(x).$$
Now we consider two cases separately:
$A_j^n$ is a finite set for all $n \in \mathbb{N}$, $j=1,\ldots,m_n$. Then, as $\mu$ is the counting measure, we have $$\int f_n \, d\mu = \sum_{j=1}^{m_n} \sum_{x \in A_j^n} c_j^n.$$ As $f_n \leq f$, we know that $c_j^n \leq f(x)$ for any $x \in A_j^n$. Hence, if we set $F_n := \bigcup_{j=1}^{m_n} A_j^n$, $$\int f_n \, d\mu = \sum_{j=1}^{m_n} \sum_{x \in A_j^n} c_j^n \leq \sum_{x \in F_n} f(x) \leq \sup\left\{ \sum_{x \in F} f(x); F \subset \Omega \, \text{finite} \right\}.$$
There exists $n \in \mathbb{N}$ and $j\in \{1,\ldots,m_n\}$ such that $A_j^n$ is infinite. Then $$\int f \, d\mu \geq \int f_n \, d\mu \geq \int_{A_j^n} f_n \, d\mu = \infty.$$ On the other hand, we can choose finite sets $F_m$ such that $F_m \subseteq F_{m+1} \subseteq A_j^n$ and $\bigcup_m F_m$ is an infinite set, then $$\begin{align*} \sup \left\{ \sum_{x \in F} f(x); F \subset \Omega \, \text{finite} \right\} &\geq \lim_{m \to \infty} \sum_{x \in F_m} f(x) \\ &\geq \lim_{n \to \infty} \sum_{x \in F_m} \underbrace{f_n(x)}_{c_j^n} \\ &= c_j^n \lim_{m \to \infty} \sum_{x \in F_m} 1 = \infty. \end{align*}$$
Consequently, in both cases,
$$\int f \, d\mu \leq \sup \left\{ \sum_{x \in F} f(x); F \subset \Omega \, \text{finite} \right\}.$$ |
Concerning the limit of the numerator of a rational function | No.
What about $$\left( a_n \right)=\left( \frac{1}{n}\right)$$
where $a_n=\frac{1}{n}$,
$ b_n=1$ and $ c_n=n$
Edit-Edited to answer your additional question.
Again No.
What about $$\left( a_n \right)=\left( \frac{n}{n^2}\right)$$
where $a_n=\frac{n}{n^2}$,
$ b_n=n$ and $ c_n=n^2$ |
Is $k[x,y,z]/(x^2+y^2-z^2)$ a UFD? | The problem is more difficult than it first appears. Easy enough it is to say "Oh in the quotient ring we just have $z^2 = (x+iy)(x-iy)$ and so it is not a unique factorization domain". Though, one needs to prove that $x+iy, x-iy$ and $z$ are really irreducible elements in the quotient ring.
However you can reduce your problem to what you already know about $k[x,y,z]/(xy - z^2)$: I claim that in fact your ring is isomorphic to this! Indeed define variables
\begin{eqnarray*} u&:=& x+iy\\
v &:=& x-iy .\end{eqnarray*}
and note that $k[x,y,z] = k[u,v,z]$ by using that $x = \frac{u+v}{2} $ and $
y = \frac{u-v}{2i}$. Then $$k[x,y,z]/(x^2 + y^2 - z^2) \cong k[u,v,z]/(uv - z^2)$$
where the ring on the right hand side you already know is not a UFD, so you have reduced your problem to what you already know! |
Show that the series $\sum 3^n \sin(\frac{1}{4^n} x)$ converges absolutely and uniformly on $(a,\infty)$, where $a>0$. | $\sin x$ always less than $x$ for $x>0$
$\implies \sin(1/4^n x) < 1/4^n x$
$3^n \sin(1/4^nx) < (3/4)^n \cdot 1/x$
$\max{ (3/4)^n \cdot 1/x} = (3/4)^n \cdot 1/a$ which converges
Hence by Weierestrass theory the given series converges uniformly. |
Consider the equation $y''+\omega ^2y=A\cos (\omega x)$ where $A, \omega$ are positive constants. | $\frac{Ax\sin \omega x}{2\omega}$ can be arbitrarily large by choosing large value of $x$.
That’s enough for part b.
Try plotting graph on geogebra or other graphing tool. |
Manifold Orientability Definition | Let me speak of a vector bundle $E \to M$ over a smooth manifold $M$, and let $F$ be the typical fiber of the bundle $E$. In particular, for each point $p \in M$ we can find an open neighborhood $U \ni p$, such that $E|_U$ is diffeomorphic to $U \times F$. A choice $\varphi \colon E|_U \to U \times F$ is called a local trivialization of $E$.
The points $p,q \in M$ are "sufficiently near" if there is an open set $U \subseteq M$, such that $p,q \in U$ and there is a local trivialization $\varphi \colon E|_U \to U \times F$.
In the trivial vector bundle $V \times F$ all the fibers are canonically isomorphic via the map $\tau \colon (p,v) \mapsto (q,v)$. Using this we can compare the orientations of the fibers $E_p$ and $E_q$ for "sufficiently near" points. If the canonical isomorphism $\tau$ is orientation-preserving for these points, the chosen orientations in the fibers are called coherent.
One can find some insights how to make the full statements, for instance, in this discussion, and in textbooks, e.g. Loring Tu, "An Introduction to Manifolds", p. 240. |
Compute the preimage of dyadic interval via binary expansion map. | There are always numbers in any interval of the form $\ \left(\frac{k}{2^n},\frac{k+1}{2^n}\right)\ $ which don't have binary expansions of finite length (in fact, there are always an uncountable number of irrational numbers in such an interval). Your question about the binary expansion of points in a dyadic interval having only a finite number of entries therefore implicitly assumes a false proposition.
A number $\ \alpha\ $ will lie in the interval $\ \left[\frac{k}{2^n},\frac{k+1}{2^n}\right]\ $ if and only if $\ 2^n\alpha\ $ lies in the interval $\ [k,k+1]\ $—that is $\ 2^n\alpha=k+\beta\ $, where $\ \beta\ $ is a number in the interval $\ [0,1]\ $. If $\ \displaystyle k=\sum_{i=0}^r k_i2^i\ $ is the terminating binary expansion of $\ k\ $, then $\ k+\beta\ $ will have a binary expansion $\ \displaystyle k=\sum_{i=0}^r k_i2^i+\sum_{i=1}^\infty\frac{\beta_i}{2^i}\ $ for some sequence $\ \left(\beta_1,\beta_2,\dots, \beta_j,\dots\right)\in\Omega\ $, and $\ \alpha\ $ will have a binary expansion $\ \alpha=$$\displaystyle \sum_{i=0}^r \frac{k_i}{2^{n-i}}+\sum_{i=1}^\infty\frac{\beta_i}{2^{n+i}}\ $.
Thus, corresponding to your map $\ f\rightarrow \left[0,1\right]\ $, there is also a map $\ g:\Omega\rightarrow \left[\frac{k}{2^n},\frac{k+1}{2^n}\right]\ $ given by
$$
g(\omega)= \sum_{i=0}^r\frac{k_i}{2^{n-i}}+\sum_{i=1}^\infty\frac{x_i}{2^{n+i}}=\frac{k+f(\omega)}{2^n}
$$
for $\ \omega=\left(x_1,x_2,\dots,x_j,\dots\right)\ $.
Addendum:
If $\ k\ge 2^n\ $, then no element of $\ \left(\frac{k}{2^n},\frac{k+1}{2^n}\right)\ $ lies in the range of $\ f\ $, so $\ f^{-1}\left(\left(\frac{k}{2^n},\frac{k+1}{2^n}\right)\right)=\phi\ $ in that case.
If $\ 0\le k<2^n\ $ and $\ x\in \left(\frac{k}{2^n},\frac{k+1}{2^n}\right)\ $ is not a dyadic rational, then $\ x\ $ has a unique binary expansion $\ \displaystyle \sum_{i=0}^r\frac{k_i}{2^{n-i}}+\sum_{i=1}^\infty\frac{x_i}{2^{n+i}}\ $ where $\ \left(x_1,x_2,\dots,x_j,\dots\right)\in\Omega\setminus\{\mathbf{0},\mathbf{1}\}\ $, with $\ \mathbf{0}=$$(0,0,\dots,0,\dots)\ $ and $\ \mathbf{1}=(1,1,\dots,1,\dots)\ $. Therefore $\ x=f\left(\underbrace{0,\dots,0}_{n-r-1},k_r,k_{r-1},\dots,k_0,x_1,x_2,\dots\right)\ $ in this case.
On the other hand if $\ x=\frac{\ell}{2^m} \in \left(\frac{k}{2^n},\frac{k+1}{2^n}\right)\ $ is a dyadic rational with $\ \ell\ $ odd, then we must have $\ m>n\ $ and $\ 2^{m-n}k<$$\ell<$$2^{m-n}(k+1)\ $, and $\ x\ $ has exactly two binary expansions:
\begin{align}
x&=\sum_{i=0}^r\frac{k_i}{2^{n-i}}+\sum_{i=1}^{m-n-r}\frac{\ell_{m-n-r-i}}{2^{n+i}}&\text{ and}\\
x&=\sum_{i=0}^r\frac{k_i}{2^{n-i}}+\sum_{i=1}^{m-n-r-1}\frac{\ell_{m-n-r-i}}{2^{n+i}}+\sum_{i=m-n+1}^\infty \frac{1}{2^{n+i}}\ .
\end{align}
Thus, if $\ x=f(\omega)\ $, then we must have either
\begin{align}
\omega&=\\
&\big(\underbrace{0,\dots,0}_{n-r-1},k_r,k_{r-1},\dots,k_0,\ell_{m-n-1},\ell_{m-n-2},\dots,\ell_1,1,0,0,\dots\big)\\
\text{or}\\
\omega&=\\
&\big(\underbrace{0,\dots,0}_{n-r-1},k_r,k_{r-1},\dots,k_0,\ell_{m-n-1},\ell_{m-n-2},\dots,\ell_1,0,1,1,\dots\big)
\end{align}
Thus, $\ x=f\big(\underbrace{0,\dots,0}_{n-r-1},k_r,k_{r-1},\dots,k_0,x_1,x_2,\dots\big)\ $ for some $\ \left(x_1,x_2,\dots,x_j,\dots\right)\in\Omega\setminus\{\mathbf{0},\mathbf{1}\}\ $, in this case also.
Therefore, if $\ 0\le k<2^n\ $, then
\begin{align}
f^{-1}&\left(\left(\frac{k}{2^n},\frac{k+1}{2^n}\right)\right)=\\
&\bigg\{\big(\underbrace{0,\dots,0}_{n-r-1},k_r,k_{r-1},\dots,k_0,x_1,x_2,\dots\big)\,\bigg|\left(x_1,x_2,\dots\right) \in\Omega\setminus\{\mathbf{0},\mathbf{1}\}\bigg\}
\end{align} |
Contour integral calculation 2 | The Laplace transform is a self-adjoint operator, i.e.
$$ \int_{0}^{+\infty} f(x)\cdot(\mathcal{L} g)(x)\,dx = \int_{0}^{+\infty} g(x)\cdot(\mathcal{L} f)(x)\,dx$$
by Fubini's theorem. For any $a>0$ we have
$$ \mathcal{L}(\sin x)(s) = \frac{1}{s^2+1},\qquad \mathcal{L}^{-1}\left(\frac{1}{x+a}\right)=e^{-ax} $$
hence
$$ \int_{0}^{+\infty}\frac{\sin x}{a(x+a)}\,dx = \int_{0}^{+\infty}\frac{e^{-ax}}{a(x^2+1)}\,dx\stackrel{x\mapsto t/a}{=}\int_{0}^{+\infty}\frac{e^{-t}}{t^2+a^2}\,dt$$
QED. |
Proof of the Conditional Exchange | Here is a proof of going from the conditional to the disjunction in the system I like to use myself: |
What is conjugate in group theory? | As some comments mentioned, conjugation is only really useful in non-abelian groups. Here are a few other things that may be useful to know:
We say "conjugation by $u$" for the action of taking some element, $g$ say, to $u^{-1}gu.$ It is easy to see that this is an isomorphism (automorphism if you like).
The relation "$a$ is conjugate to $b$" is an equivalence relation. We call the classes conjugacy classes.
An intuition for conjugation is that $u^{-1}gu$ is looking at $g$ from the point of view of $u$. For example you may know how to solve some problem in some special case (e.g. The North Pole of a sphere or the point $\infty$ on the projective plane) and then you can use conjugation to solve the problem more generally (i.e. Conjugating by the element which moves your point of interest to the North Pole or $\infty$ in the vague examples I gave). |
Establishing continuity for a complex variable function | In fact, it's almost the same as the real case since
$$\frac{z^4-1}{z-i}=\frac{(z^3+iz^2-z-i)(z-i)}{z-i}=z^3+iz^2-z-i$$
for $z\ne i$. |
Terminology of homocyclic groups | In this case, $n$ is also the rank of $G$ (the minimum number of generators). So "the homocyclic group of $\mathbb{Z}/p\mathbb{Z}$ of rank $n$" would define the given group. |
Merge two linear functions | Let $f_1=ax+b$ and $f_2=cx+d$ with $a \neq c$, then what you call Plot2 is given by the graph of the function $f$ defined as follows: $f(x)=f_1(x)$ for $x \leq (d-b)/(a-c)$ and $f(x)=f_2(x)$ else. |
Convergence of Lax-Wendroff method | By construction, the Lax-Wendroff method
$$
U_i^{n+1} = U_i^{n} - \frac{\Delta t}{2\Delta x} A (U_{i+1}^{n}-U_{i-1}^{n}) + \frac{{\Delta t}^2}{2{\Delta x}^2} A^2 (U_{i+1}^{n}-2U_{i}^{n}+U_{i-1}^{n})
$$
is consistent with the linear system of conservation laws $\partial_t U + A \partial_x U = 0$ (cf. this post and related ones), where $U_i^{n} \simeq U(i\Delta x, n\Delta t)$. It remains to prove stability to have a convergent method. The Lax-Wendroff method is not a monotone scheme, therefore the technique consisting in proving directly $\|U^{n+1}\| \leq \|U^{n}\|$ will not work. However, for linear problems, the stability condition can be obtained using Fourier analysis (von Neumann approach). To do so, the matrix $A$ is diagonalized as $A = R\Lambda R^{-1}$, where
$$
\Lambda = \text{diag}(\lambda_1,\lambda_2,\lambda_3) =
\begin{pmatrix}3&0&0 \\ 0& \frac{-1 - \sqrt{13}}{2}&0 \\ 0&0&\frac{-1 + \sqrt{13}}{2}\end{pmatrix} .
$$
One can show that $V = R^{-1}U$ satisfies the diagonal linear system of conservation laws $\partial_t V + \Lambda \partial_x V = 0$. Therefore, the stability of the Lax-Wendroff method for $\partial_t U + A \partial_x U = 0$ is deduced from the stability of the Lax-Wendroff method for all scalar conservation laws $\partial_t V_p + \lambda_p \partial_x V_p = 0$. The von Neumann analysis gives the Courant-Friedrichs-Lewy (CFL) stability condition
$$
\left|\frac{\lambda_p \Delta t}{\Delta x}\right| \leq 1
$$
for all $p$, i.e. ${\Delta t} \leq \frac{1}{3}{\Delta x}$. |
Prove or disprove every positive fraction less than 1 can be expressed with $\frac{1}{a+b} + \frac{1}{a+c}$ | I will call
$$r := \frac{1}{a+b} + \frac{1}{a+c} \qquad \qquad\qquad\qquad\qquad (1)$$
I see two main cases:
First suppose $a+b>0$ and $a+c>0$. In this case, we also have $a+b>1$ and $a+c>1$ since $r≤1$.
If $a+b = a+c = 2$, then $r=1$.
Else, if $a+b ≥ 2$ and $a+c≥ 3$, then $r≤ \frac{5}{6}$.
Thus, we see that the rationals in $(\frac{5}{6},1)$ are not covered.
If it is not the case, we can suppose $a+b>0$ and $a+c<0$.
If $a+b > 1$, then $r≤\frac{1}{a+b}≤ \frac{1}{2}$, so that the rationals in $(\frac{5}{6},1)$ are not covered.
Else, if $a+b = 1$, then $r = 1 - \frac{1}{|a+c|}$. The only point of accumulation of such a set is $1$, so that for every $\varepsilon\in(0,\frac{1}{6})$, only a finite number of rationals are covered in $(\frac{5}{6},1-\varepsilon)$.
This proves (if I made not mistakes) that every every positive fraction less than $1$ cannot be expressed in the form $(1)$. There is perhaps a more straightforward proof?
$~$
If you are looking for a specific example of number, we see that the numbers of the form $1 - \frac{1}{|a+c|}$ are $\{0,\frac{1}{2}, \frac{2}{3},\frac{3}{4},\frac{4}{5},\frac{5}{6},\frac{6}{7}, ...\}$, all the remaining numbers are smaller than $\frac{6}{7}$. Therefore, all the numbers in $(\frac{5}{6},\frac{6}{7})$ can not be written on the form $(1)$. In particular, the following fraction
$$r = \frac{11}{13}$$
is a counterexample! |
normal distribution help | Note that $8,2$ is the standard desviation and not the variance. So it is
$$P(X>115)=P\left(Z=\dfrac{X-110}{8,2}>\dfrac{115-110}{8,2}\right).$$ |
velocity with height of time for an arbitrary projectile being launched from a cliff | Unless you have a picture or anything else to indicate horizontal motion, I'd take it that the motion occurs only up-and-down and that the (upward) launch from the cliff is just a way of getting an initial non-zero displacement. And presumably the cliff is perfectly vertical and the projectile motion takes place an infinitesimal distance away from the cliff so the projectile doesn't graze it on the way down.
In this scenario, the velocity is $h'(t)$. |
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