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Prove that two subspaces are not equal under a given condition | This is not true. The error is in this line:
Since $U_1$ and $W$ have no elements in common, $u_1 \not\in W$. Therefore $u_1 \in U_2$.
It is not true in general that if $x \in A \oplus B$ then $x \in A$ or $x \in B$. (Perhaps you are confusing the properties of the direct sum $U_2 \oplus W$, with the properties of the union $U_2 \cup W$, which in general is not a subspace.) An instructive example is taking $A$ to be the $x$-axis in $\Bbb R^2$ and $B$ the $y$-axis. Then, $\Bbb R = A \oplus B$ but, e.g., $(1, 1)$ is in neither $A$ nor $B$. |
What $f(t)$ satisfies the inverse Laplace transform $\mathcal{L}^{-1}\left\{\frac{p'(s)}{p(s)}\right\}=f(t),$ where the polynomial $p$ is given. | Let $n=\text{deg}(p)$ be the degree of the polynomial $p$. Then by partial fractions we may write $$\frac{p'(s)}{p(s)}=\sum_{k=1}^n\frac{A_k}{s-\sigma_k},$$ for some $A_k$, where $p(\sigma_k)=0$ for all $1\leq k\leq n$. But since $\frac{1}{s-a}=\mathcal{L}\left\{e^{-at}\right\}$, then
$$\sum_{k=1}^n\frac{A_k}{s-\sigma_k}=\sum_{k=1}^n A_k\mathcal{L}\{e^{-\sigma_k t}\}=\mathcal{L}\left\{\sum_{k=1}^n A_ke^{-\sigma_kt}\right\}.$$ Hence, $$\mathcal{L}^{-1}\left\{\frac{p'(s)}{p(s)}\right\}=\sum_{k=1}^n A_ke^{-\sigma_kt}.$$ It may be that the right hand side has closed form due to the exponential forms of the sine and cosine functions, depending on our choice of $p$.
Note - this generalises the original question since $p'/p\equiv q/p$, where $\text{deg}(q)<\text{deg}(p)$.
However, in the original case for $p'/p$, it appears that $A_k=1$ for all $k$, in which case there will indeed be closed fom in terms of sines, cosines, and their hyperbolic equivalents. |
Is there another proof of $V=U+U^\perp$ especially when $V$ is infinite-dimensional? | Here's the proof given in Conway's A Course in Functional Analysis, 2nd ed. (Theorems 2.5 and 2.6).
Step 1: If $\mathscr{H}$ is a Hilbert space, $K$ a closed, convex subset of $\mathscr{H}$, and $h \in \mathscr{H}$ then there is a unique point $k_0 \in K$ such that $$ \lVert h - k_0 \rVert = \operatorname{dist}(h,K) \equiv \inf \{ \lVert h - k \rVert : k \in K \}. $$
Proof. We start by noting that by translating $K$ to $K - h \equiv \{k - h : k \in K\}$ we may assume $h = 0$. Next, let $d = \operatorname{dist}(0,K)$. Pick some sequence of points $(k_n) \in K$ such that $\lVert k_n \rVert \to d$. We will use convexity to show that $(k_n)$ is a Cauchy sequence and then use the closedness of $K$ to conclude that $k_n \to$ some $k_0 \in K$.
We note that, by the Parallelogram Law, $$\left\lVert \frac{k_n - k_m}{2} \right\rVert^2 = \frac{\lVert k_n \rVert^2 + \lVert k_m \rVert^2}{2} - \left\lVert \frac{k_n + k_m}{2} \right\rVert^2. $$
Now, since $K$ is convex, $\frac12 (k_n + k_m) \in K$ (you can also take this as the definition of convexity if you are unfamiliar). Therefore, since $d = \operatorname{dist}(0,K)$, we have $\lVert \frac12 (k_n + k_m) \rVert^2 \ge d^2$.
Now let $N$ be such that $n \ge N$ implies $\lVert k_n \rVert^2 < d^2 + \frac14 \varepsilon^2$. We can do this since $\lVert k_n \rVert \to d$. It follows that
$$ \left\lVert \frac{k_n - k_m}{2} \right\rVert^2 < \frac{2d^2 + \frac{1}{4}\varepsilon^2}{2} - d^2 = \frac14\varepsilon^2. $$
Therefore, $\lVert k_n - k_m \rVert < \varepsilon$. I.e. the sequence $(k_n)$ is Cauchy.
Since $\mathscr{H}$ is complete, and $K$ is closed, there is a limit point $k_0 \in K$ such that $k_n \to k_0$. Since the norm is continuous, $$\lVert k_0 \rVert = \lim_{n \to \infty} \lVert k_n \rVert = d. $$
This shows existence.
For uniqueness, suppose $k_0'$ is a second point in $K$ such that $\lVert k_0' \rVert = d$. Then
$$ d \le \lVert \tfrac12 (k_0 + k_0') \rVert \le \tfrac12 (\lVert k_0 \rVert + \lVert k_0' \rVert) = d. $$
Hence, $\lVert \frac12 (k_0 + k_0') \rVert = d$.
Applying the Parallelogram Law,
$$ \left\lVert \frac{k_0 - k_0'}{2} \right\rVert^2 = \frac{\lVert k_0 \rVert^2 + \lVert k_0' \rVert}{2} - \left\lVert \frac{k_0 + k_0'}{2} \right\rVert^2 = \frac{d^2 + d^2}{2} - d^2 = 0. $$
Therefore $k_0' = k_0$.
Step 2: If $\mathscr{M}$ is a closed linear subspace of $\mathscr{H}$, $h \in \mathscr{H}$ and $f_0 \in \mathscr{M}$ is the unique element such that $\lVert h - f_0 \rVert = \operatorname{dist}(h, \mathscr{M}$ then $h - f_0 \perp \mathscr{M}$. Conversely, if $f_0 \in \mathscr{M}$ is such that $h - f_0 \perp \mathscr{M}$ then $\lVert h - f_0 \rVert = \operatorname{dist}(h, \mathscr{M})$.
Proof. Take $f \in \mathscr{M}$. Then $f_0 + f \in \mathscr{M}$ so we have
$$ d^2 = \lVert h - f_0 \rVert^2 \le \lVert h - (f_0 + f) \rVert^2 = \lVert (h - f_0) - f \rVert^2 = \lVert h - f_0 \rVert^2 - 2\operatorname{Re}\langle h - f_0, f \rangle + \lVert f \rVert^2 $$
whence $2\operatorname{Re}\langle h - f_0, f \rangle \le \lVert f \rVert^2$.
Write $\langle h - f_0, f \rangle = re^{i\theta}$. Replacing $f$ with $te^{i\theta}$ in the preceeding inequality, we have
$$ 2\operatorname{Re}\langle h - f_0, te^{i\theta} f \rangle = 2\operatorname{Re} \Big(te^{-i\theta} \langle h - f_0, f \rangle \Big) = 2tr \le \lVert te^{i\theta} f\rVert^2 = t^2 \lVert f \rVert^2. $$
Taking $t \to 0$ we see that $r = 0$. I.e. $h - f_0 \perp f$.
For the converse, suppose $f \in \mathscr{M}$, then $f \perp h - f_0$ means that
$$ \lVert h - f \rVert^2 = \lVert (h - f_0) + (f_0 - f) \rVert^2 = \lVert h - f_0 \rVert^2 + \lVert f_0 - f \rVert^2 \ge \lVert h - f_0 \rVert^2. $$
Thus $\lVert h - f_0 \rVert = \operatorname{dist}(h, \mathscr{M})$.
Step 3: Note that this applies when $\mathscr{M}$ is finite dimensional, because finite dimensional subspaces are closed as a consequence of the Bolzano-Weirstrass Theorem. |
Prove that $C′$ is a subgroup of $G$. | If $a \in C'$, then $axax=xaxa$ for all $x \in G$, so that
$$
a^{-1} xa^{-1}x = (x^{-1}ax^{-1}a)^{-1} = (ax^{-1}ax^{-1})^{-1} = xa^{-1}xa^{-1},
$$
that is, $a^{-1} \in C'$.
Added later: Perhaps this is easier to understand. Start with $ax^{-1}ax^{-1} = x^{-1}ax^{-1}a$. Then taking the inverse of both sides gives $xa^{-1}xa^{-1} = a^{-1}xa^{-1}x$. |
Convergence of Jacobi-Method | We have a more general statement (Householder-John theorem):
Let $A$ and $M+M^*-A$ be Hermitian positive definite and $M$ be an invertible matrix. Then $\rho(I-M^{-1}A)<1$.
Let $\lambda$ and $x\neq 0$ be an eigenvalue and an eigenvector of $I-M^{-1}A$ so
$$
(I-M^{-1}A)x=\lambda x \iff (1-\lambda)Mx=Ax.
$$
We have $\lambda\neq 1$, otherwise $A$ would be singular.
Premultiplying by $x^*$ and taking a conjugate transpose gives
$$
x^*Mx=\frac{1}{1-\lambda}x^*Ax, \quad
x^*M^*x=\frac{1}{1-\bar{\lambda}}x^*Ax.
$$
Hence
$$
x^*(M+M^*-A)x=\left(\frac{1}{1-\lambda}+\frac{1}{1-\bar{\lambda}}-1\right)x^*Ax
=\frac{1-|\lambda|^2}{|1-\lambda|^2}x^*Ax.
$$
Both $A$ and $M+M^*-A$ are HPD, we have $x^*(M+M^*-A)x>0$ and $x^*Ax>0$.
Therefore
$$
1-|\lambda|^2>0\iff |\lambda|<1.
$$
Now to answer the question, set $A:=C$ and $M:=D_C$ to get:
If $C$ and $2D_C-C$ are positive definite then $\rho(I-D_C^{-1}C)<1$ and the Jacobi method for $Cx=b$ is convergent. |
Write a Limit to calculate $f'(0)$ | it is $$\frac{f(x+h)-f(x)}{h}=\frac{\frac{2}{1+(x+h)^2}-\frac{2}{1+x^2}}{h}$$
Can you finish?
Ok, another hint:
The numerator is given by $$-2\,{\frac {h \left( h+2\,x \right) }{ \left( {h}^{2}+2\,xh+{x}^{2}+1
\right) \left( {x}^{2}+1 \right) }}
$$ |
De Rham cohomology of $S^n$ | You are wrong: $S^n$ is not the disjoint union $\mathbb R^n \sqcup \mathbb R^0$ - topologically.
Although $S^n$ is $\mathbb R^n$ with one point at infinity, the topology of this point at infinity is very different from that of $\mathbb R^0$. |
Find the value of K. Use of l Hospital's rule and expansion is not allowed. | (Preamble: Apart from addition and multiplication the only genuine two-ary function allowed in print is the general power function
$$(x,y)\mapsto x^y:=\exp(y\, \log x)\qquad(x>0, y\in{\mathbb C})\ .\tag{1}$$
The function $(x,y)\mapsto \log_x y$ where both $x$ and $y$ are variables is not of this kind. When such a thing should really come up in practice one would have to rewrite it in such a way that only functions of one variable and the "allowed" operations appear. It is true that we sometimes write $\log_a y$. But the $a>1$ here is not a variable; it is a constant scaling parameter chosen in a most suitable way for the problem at hand.)
So after a lot of mind bending I'm interpreting your problem as follows: For $0<|x|\ll1$ one has $0<\cos (3x)<1$. On the other hand
$$\cos(2ix)={e^{-2x}+e^{2x}\over2}=\cosh (2x)\qquad(x\in{\mathbb R})\ .$$
Therefore the equation
$$\left(\cos(3x)\right)^y=\cos(2ix)=\cosh(2x)$$
has a real solution $y$ for such $x$; and according to $(1)$ this $y$ is given by
$$y={\log\bigl(\cosh(2x)\bigr)\over \log\bigl(\cos(3x)\bigr)}=: f(x)\qquad(0<|x|\ll1)\ .\tag{2}$$
The right hand side of $(2)$ is the correct expression of the function $f$ your teacher had in mind.
Now it comes to computing $\lim_{x\to0} f(x)$. Since both de l'Hôpital's rule and expansion are forbidden I have to stop here. It is unclear to me how this limit can be computed without using the analytical properties of the involved functions somehow. |
Non-empty compact subset of an open set? | Theorem 1.4.6 tells you that your $r$ is the distance between some point $z_0\in K$ and some point $w_0\in U^c$. Since $K\subseteq U$, the points $z_0$ and $w_0$ are distinct, so their distance is $>0$. |
How can i calculate $H_i(S^2 \times S^1)$? | Building on ThorbenK's suggestion, set $U=D_1\times S^1$, $V=D_2\times S^1$ where each $D_j$ is a (slightly fattened) hemisphere (i.e. a disc, so, contractible). Obviously $U\cap V= (D_1\cap D_2)\times S^1=(-c,c)\times S^1\times S^1$, which retracts to a $2$-torus.
So
$$H_*(U)=\mathbb{Z},\mathbb{Z}, 0,0$$
$$H_*(V)=\mathbb{Z},\mathbb{Z}, 0,0$$
$$H_*(U\cap V)=\mathbb{Z},\mathbb{Z}\oplus\mathbb{Z}, \mathbb{Z},0$$
Set $X=S^2\times S^1=U\cup V$. By the Mayer-Vietoris sequence
$$0\to H_3(X)\to H_2(U\cap V)\to H_2(U)\oplus H_2(V)\to H_2(X)\to H_1(U\cap V)\to H_1(U)\oplus H_1(V)\to H_1(X)\to H_0(U\cap V)\to H_0(U)\oplus H_0(V)\to H_0(X)\to 0$$
Or, as we already know the homology groups of $U$, $V$ and $U\cap V$,
$$0\to H_3(X)\to \mathbb{Z}\to 0 \to H_2(X)\to \mathbb{Z}^2\to \mathbb{Z}^2\to H_1(X)\to \mathbb{Z}\to \mathbb{Z}^2\to H_0(X)\to 0$$
We immediately see that $H_3(X)=\mathbb{Z}$.
The maps $\rho_j:H_j(U\cap V)\to H_j(U)\oplus H_j(V)$ are given by $((i_1)_*, (i_2)_*)$ where $i_1:U\cap V\to U$ and $i_2:U\cap V\to V$ are the inclusions.
So, at the level $0$, homology is generated by the class of a point; given $[p]\in H_0(U\cap V)$, then obviously $\rho_0([p])=([i_1(p)], [i_2(p)])$, so $\rho_0:\mathbb{Z}\to\mathbb{Z}^2$ is $\rho_0(x)=(x,x)$, hence the cokernel is
$$H_0(X)=\mathbb{Z}$$
As $\rho_0$ is injective, the map $H_1(X)\to H_0(U\cap V)$ has image $0$ and its kernel, which is $H_1(X)$, is the image of $\sigma:H_1(U)\oplus H_1(V)\to H_1(X)$, which is then surjective. Therefore $H_1(X)=(H_1(U)\oplus H_1(V))/\rho_1(H_1(U\cap V))$.
Moreover, consider two generators of $H_1(U\cap V)$: the cycle $[\gamma]$ given by the $S^1$ that is also a factor of $X$ and the cycle $[\gamma']$ given by the intersection between the two disc $D_1$ and $D_2$ (the equator of $S^2$). We have that $(i_j)_*([\gamma])=[i_j\circ\gamma]$, whereas $(i_j)_*[\gamma']=0$. So $\rho_1:\mathbb{Z}^2\to\mathbb{Z}^2$ is $\rho_1(x,y)=(x,x)$.
Its kernel is $\mathbb{Z}$ and it is homomorphic to $H_2(X)$.
On the other hand, its image is $\mathbb{Z}$, therefore $H_1(X)=\mathbb{Z}^2/(1,1)\mathbb{Z}=\mathbb{Z}$.
In conclusion
$$H_*(X)=\mathbb{Z}, \mathbb{Z},\mathbb{Z},\mathbb{Z}$$ |
Probability of dependent events. Why is this wrong solution? | First we have to regard that you´re drawing the balls without replacement. That means if a new ball is drawn the probability to draw another new ball is $\frac{9-1}{15-1}=\frac8{14}$
So let us draw two new balls ($n$) first and then one used ball ($u$): $nnu$
The probability is $\frac{9}{15}\cdot \frac{8}{14}\cdot \frac{6}{13}$
But we can also draw the other two orders: $nun$ and $unn$. Both with the same probability like $nnu$. Therefore the probability to draw 2 new balls and one used ball is
$$3\cdot \frac{9}{15}\cdot \frac{8}{14}\cdot \frac{6}{13}\approx 47.47\%$$
This is equal to what you have written with the binomial coefficients: $\large{\frac{\binom{9}{2}\cdot \binom{6}{1}}{\binom{15}{3}}}$ |
Why is the equation $\frac{(z-i)^2}{(z^2+1)^2}=\frac{1}{(z+i)^2} $ in the residue theorem accurate? | Well, $z² + 1 = (z-i)(z+i)$. As the latter is squared in the denominator, you can cancel out $(z-i)^2$. |
There exists a functional in the dual of a Banach space that distinguishes points | Let $S$ be the scalars. That is $S=\mathbb R$ or $S=\mathbb C.$
(1). If $x\ne y$ and $x, y$ are linearly dependent then WLOG $x\ne 0$ and $y=kx$ with $1\ne k\in S.$ Let $<x>$ be the $1$-dimensional vector subspace generated by $\{x\}.$ Then $<x>$ is closed.
Let $f(rx)=r$ for $r\in S$. Then $f\in <x>^*$ with $\|f\|=1.$ By Hahn-Banach there exists an extension $\phi\in B^*$ of $f$. And $\phi (x)=f(x)=1\ne k=f(y)=\phi (y).$
(2). If $x,y$ are linearly independent let $<x,y>$ be the $2$-dimensional vector subspace generated by $\{x,y\}.$ Now $<x,y>$ is closed. If $a,b,a',b'\in S$ then $$ax+by=a'x+b'y\implies (a-a')x+(b-b')y=0\implies (a=a'\land b=b')$$ because $x,y$ are linearly independent. So for $z\in <x,y>$ there are unique scalars $a,b$ such that $z=ax+by.$
Define $f(ax+by)=a$ for $ a,b\in S$. Now $<y>$ is also closed and $x\not \in <y>$ so $$0<c=\inf \{x+yd:d\in S\}.$$ So when $a\ne 0$ we have $$\|ax+by\|=|a|\cdot \|x+a^{-1}by\|\geq |a|\cdot c=|f(ax+by)|c.$$ And when $a=0$ we have $f(ax+by)=0.$ So $\|f\|\leq 1/c<\infty,$ so $f\in <x,y>^*.$
By Hahn-Banach, $f$ extends to $\phi\in B^*.$ And we have $\phi(x)=f(x)=1\ne 0=f(y)=\phi(y).$
Remark. For Banach space $Y$ and $g\in Y^*$, for all $x\in Y$ we have $|g(x)|=\|g\|\cdot d(x,g^{-1}\{0\})$ where $d(x,g^{-1}\{0\})=\inf \{\|x-y\|: g(y)=0\}.$ So in (2) we have $\|f\|=1/c.$ |
Algebra of bounded functions on a completely regular space | I find it easy to prove that it is a normed algebra, so proving that it is complete is all that remains to do;
You need to see that
if $(f_n)$ is a Cauchy sequence in $B_T$, then for every $t\in T$, the sequence $\bigl(f_n(t)\bigr)$ is a Cauchy sequence in $\mathbb{C}$, which follows directly from $\lvert f_n(t) - f_m(t)\rvert \leqslant \lVert f_n - f_m\rVert$.
So the pointwise limit $f(t) = \lim_{n\to\infty} f_n(t)$ exists. Then you need to know that a (locally) uniform limit of continuous functions is continuous, and finally, you need to show that $\lVert f_n - f\rVert \to 0$ (but that is very easy).
the points of $T$ are homeomorphically included in the space $\mathscr{M}$ of the maximal ideals of the algebra $B_T$ and that, in this inclusion, the image of $T$ in $\mathscr{M}$ is an everywhere dense subset (quoting the book). Here I know that $\mathscr{M}$, which I think to be intended with the weak$^\ast$ topology, is the same as the set of all the complex non-null continuous multiplicative linear functionals $B_T\to\mathbb{C}$, but I cannot go further...
Right, it is intended to carry the (subspace topology of) the weak$^\ast$ topology, and $\mathscr{M}$ is identified with the space of unital complex algebra homomorphisms on $B_T$. (The correspondence between a maximal ideal $\mathfrak{M}$ and a unital algebra homomorphism $\varphi$ is $\varphi \mapsto \ker \varphi$ in the one direction, in the other, it is given by $\mathfrak{M} \mapsto (x \mapsto \operatorname{can} (x + \mathfrak{M}))$, where $\operatorname{can}$ denotes the canonical isomorphism between a Banach algebra in which every nonzero element is invertible and $\mathbb{C}$ as implied by the Gelfand-Mazur theorem).
The inclusion map $\varepsilon\colon t \mapsto \varepsilon_t$, where $\varepsilon_t(f) = f(t)$ is clear, and one needs to see that it is an embedding for completely regular $T$.
Fix an arbitrary $t_0 \in T$. First, we show that $\varepsilon$ is continuous at $t_0$. A neighbourhood basis of $\lambda \in B_T^\ast$ in the weak$^\ast$ topology is given by sets of the form
$$V(\lambda;\delta; f_1,\dotsc, f_n) = \left\{\mu\in B_T^\ast : \bigl(\forall 1 \leqslant i \leqslant n\bigr)\bigl(\lvert \mu(f_i) - \lambda(f_i)\rvert < \delta\bigr) \right\}$$
for $\delta > 0$ and finitely many $f_i$. Now
$$\varepsilon^{-1}(V(\varepsilon_{t_0}; \delta; f_1,\dotsc, f_n)) = \left\{ t\in T : \bigl(\forall 1 \leqslant i \leqslant n\bigr)\bigl( \lvert f_i(t) - f_i(t_0)\rvert < \delta\right\} = \bigcap_{i=1}^n f_i^{-1}\left(D_\delta(f_i(t_0))\right)$$
is a finite intersection of (open) neighbourhoods of $t_0$, hence a(n open) neighbourhood of $t_0$. Thus $\varepsilon^{-1}(U)$ is a neighbourhood of $t_0$ for every neighbourhood $U$ of $\varepsilon_{t_0}$, i.e. $\varepsilon$ is continuous at $t_0$.
Next we see that $\varepsilon$ is open at $t_0$, that means that the image of every neighbourhood of $t_0$ is a neighbourhood of $\varepsilon_{t_0}$ [in $\varepsilon(T) \subset \mathscr{M}$]. So, given a neighbourhood $U$ of $t_0$, we need to find a $\delta > 0$ and finitely many $f_i \in B_T$ such that
$$\varepsilon(T) \cap V(\varepsilon_{t_0};\delta; f_1,\dotsc,f_n) \subset \varepsilon(U),$$
which would follow from
$$\varepsilon^{-1}\bigl(V(\varepsilon_{t_0};\delta;f_1,\dotsc,f_n)\bigr) \subset U.$$
Now, by the complete regularity, there is an $f \in B_T$ with $f(t_0) = 1$ and $f(T\setminus U) \subset \{0\}$. Then $V\left(\varepsilon_{t_0};\frac{1}{2};f\right)$ does the job.
So, $\varepsilon$ is continuous at every point, and open at every point, hence it is continuous and open, i.e. an embedding.
To see that $\varepsilon(T)$ is dense in $\mathscr{M}$, we suppose there were a $\varphi \in \mathscr{M}\setminus \overline{\varepsilon(T)}$ and derive a contradiction. By definition, $\varphi \in \mathscr{M}\setminus \overline{\varepsilon(T)}$ means there is a (weak$^\ast$) neighbourhood of $\varphi$ not intersecting $\varepsilon(T)$, i.e. there is a $\delta > 0$ and finitely many $f_1,\dotsc,f_n\in B_T$ with $V(\varphi;\delta;f_1,\dotsc,f_n)\cap \varepsilon(T) = \varnothing$, or, in other words,
$$\bigl(\forall t\in T\bigr)\bigl(\exists \nu \in \{1,\dotsc,n\}\bigr)\bigl(\lvert \varphi(f_\nu) - f_\nu(t)\rvert \geqslant \delta\bigr).\tag{$\ast$}$$
Then consider
$$F\colon t \mapsto \sum_{\nu=1}^n \lvert f_\nu(t) - \varphi(f_\nu)\rvert^2 = \sum_{\nu = 1}^n (f_\nu(t)-\varphi(f_\nu))(\overline{f_\nu(t)} - \overline{\varphi(f_\nu)}).$$
Since $B_T$ is an algebra containing the constant functions and $B_T$ is conjugation-invariant [$f\in B_T \iff \overline{f}\in B_T$], we have $F\in B_T$, and by $(\ast)$, we have $F(t) \geqslant \delta^2$ for all $t\in T$, hence $F$ is a unit in $B_T$, whence $\varphi(F) \neq 0$. On the other hand,
$$\begin{align}
\varphi(F) &= \varphi\left(\sum_{\nu = 1}^n (f_\nu-\varphi(f_\nu)\cdot \mathbb{1})(\overline{f_\nu} - \overline{\varphi(f_\nu)}\cdot \mathbb{1})\right)\\
&= \sum_{\nu = 1}^n \varphi\left(f_\nu - \varphi(f_\nu)\cdot \mathbb{1}\right)\cdot\varphi\left(\overline{f_\nu} - \overline{\varphi(f_\nu)}\cdot \mathbb{1}\right)\\
&= \sum_{\nu=1}^n \left(\varphi(f_\nu) - \varphi(\varphi(f_\nu)\cdot \mathbb{1})\right)\cdot\varphi\left(\overline{f_\nu} - \overline{\varphi(f_\nu)}\cdot \mathbb{1}\right)\\
&= 0,
\end{align}$$
since $\varphi(\mathbb{1}) = 1$, so $\varphi(\varphi(f_\nu)\cdot \mathbb{1}) = \varphi(f_\nu)$.
Finally we treat the last item,
any complex bounded [and continuous] function on this image of $T$ admits a unique continuous extension to $\mathscr{M}$.
Well, $\mathscr{M}$ is a (topological, not linear) subspace of $B_T^\ast$ endowed with the weak$^\ast$ topology. Since $\varepsilon \colon T \to \varepsilon(T)$ is a homeomorphism, every continuous and bounded function $f$ on $\varepsilon(T)$ can be identified with an element $g = f\circ \varepsilon \in B_T$, and with the canonical embedding $J$ of $B_T$ into its bidual, we obtain a continuous map $J(g) = J(f\circ\varepsilon) \colon B_T^\ast \to \mathbb{C}$ (where $B_T^\ast$ is endowed with the weak$^\ast$ topology, since that's the one we're interested in). Then
$$\tilde{f} = J(f\circ\varepsilon)\bigl\lvert_{\mathscr{M}}$$
is a continuous extension of $f$ to $\mathscr{M}$. The continuity is clear, since it's the restriction of a continuous function, and
$$\tilde{f}(\varepsilon_t) = J(f\circ\varepsilon)(\varepsilon_t) = \varepsilon_t(f\circ\varepsilon) =(f\circ\varepsilon)(t) = f(\varepsilon_t)$$
shows it is indeed an extension of $f$.
Now, $\mathscr{M}$ is a compact (quasicompact and Hausdorff) space, since it is a weak$^\ast$-closed subset of the norm-closed unit ball of $B_T^\ast$, hence normal, and in particular completely regular. Thus the uniqueness of the extension is equivalent to the denseness of $\varepsilon(T)$ in $\mathscr{M}$.
Note: This is a functional-analytic construction of the Stone-Čech compactification of a completely regular space. Since compact spaces are completely regular, and subspaces of completely regular spaces are completely regular, only completely regular spaces can have compactifications (a compactification of a topological space $X$ is a compact space $K$ together with an embedding of $X$ into $K$ as a dense subspace). This construction shows that every completely regular space has a compactification, namely the Stone-Čech compactification - many completely regular spaces also have other compactifications, for example locally compact (but not compact) [Hausdorff] spaces have the Alexandrov compactification aka one-point compactification, which is the smallest compactification a non-compact space can have; the two very rarely coincide. The Stone-Čech compactification is in a sense the largest compactification that a space can have, it is characterised by each of the following criteria:
every continuous map $f \colon X \to K$ where $K$ is compact has a (unique) continuous extension to the compactification.
every bounded continuous real (or complex) valued function has a (unique) continuous extension to the compactification.
The latter criterion directly translates to an isometric isomorphism between the algebra $C_b(X)$ of bounded continuous functions on $X$ (which is called $B_XT$ here) and the algebra $C(\beta X)$ of continuous functions on the Stone-Čech compactification $\beta X$ of $X$. |
Problem about components in a compact Hausdorff space | That $\sim$ is an equivalence relation is shown in freakish's answer.
If $x,y$ belong to the same component $C$ of $X$, then $x \sim y$ simply because each $f(C)$ is a connected subset of $\mathbb{R}$ containing $0,1$.
For the converse you have to consider quasicomponents. The quasicomponent of a point $x$ is the intersection of all closed-open sets containing $x$. For compact Hausdorff spaces quasicomponents agree with components. See e.g. Quasicomponents and components in compact Hausdorff space.
If $x,y$ do not belong to the same component, they do not belong to the same quasicomponent. This means that not all closed-open sets can contain both $x,y$. W.lo.g. we may assume that there exists a closed-open $C$ such that $x \in C$, $y \notin C$. Define $f : X \to \mathbb{R}$ by $f(a) = 0$ for $a \in C$ and $f(a) = 1$ for $a \in X \setminus C$. This is a continuous function not attaining the value $1/2$. Hence $x \not\sim y$.
Note that the above argument does not use the fact that $\sim$ is an equivalence relation. In fact, we have shown that $x \sim y$ iff $x,y$ belong to the same component of $X$ which implies that $\sim$ is an equivalence relation. However, or argument requires a compact Hausdorff $X$. freakish's proof is valid for any $X$. |
inverse trig and an algebra nitpick? | In general, $-f(x) \neq f(-x)$.
However, since $\cot x = \frac{1}{\tan x} = \frac{1}{\frac{\sin x}{\cos x}} = \frac{\cos x}{\sin x}$, it happens that $$\cot -x = \frac{\sin -x}{\cos -x} = \frac{-\sin x}{\cos x} = -\cot x.$$
Functions that share the property $-f(x) = f(-x)$ are called 'odd functions'. Functions such that $f(x) = f(-x)$ are called 'even functions'. Products of even and odd functions are much like products of even and odd numbers. $\sin x$ is odd, $\frac{1}{\cos x}$ is even; hence the product of the two is odd. |
Compute and find 2009th decimal(2009th digit after the point), without automation, the following sum | You can write your series as
$$f(x)=\sum_{1}^{x}\frac{81\times10^k}{(10^k-1)(10^{k+1}-1)}=9\sum_{1}^{x}\frac{1}{10^k-1}-\frac{1}{10^{k+1}-1}$$
Where $x=9$. By telescopy we can show:
$$f(x)=\frac{10^{x+1}-10}{10^{x+1}-1}=1-\frac{9}{10^{x+1}-1}$$
So your sum is $S=1-1111111111^{-1}=0.\overline{9999999990}$. So since $2009\equiv 9\pmod{10}$, the digit is $9$. |
Showing rational numbers are algebraic | The rational number $5/7$ is a zero of the polynomial $7x+(-5)$. We have $n=1$, $a_1=7$, $a_0=-5$.
So try showing that works with every rational number. |
Can a partially ordered set always be partitioned by its chains? | Yes, the transfinite induction in your comment works!
Let $f$ be a function that maps every nonempty subset of $P$ to a maximal chain in that subset (we get $f$ using Choice).
Now we use transfinite recursion to define a sequence of subsets of $P$:
$S_0 = P$
If $S_\alpha = \emptyset$, then $S_{\alpha + 1} = \emptyset$. Otherwise, we have $S_{\alpha + 1} = S_\alpha - f(S_\alpha)$.
For every limit ordinal $\alpha$ we have $S_\alpha = \bigcap_{\beta < \alpha} S_\beta$.
Let $S = \bigcap_{\alpha} S_\alpha$. Now, we have some smallest ordinal $\gamma$ such that $S_\gamma = S$: For every element $x$ of $P - S$, let $\gamma_x$ be the smallest ordinal such that $x \notin S_{\gamma_x}$, and then let $\gamma$ be the limit ordinal of the set $\{\gamma_x : x \in P - S\}$.
Now, we know that $S = \emptyset$, since otherwise, we would have $S \subseteq S_{\gamma+1} \subset S_\gamma = S$.
So the set $\{f(S_\alpha) : \alpha < \gamma\}$ will be a partition! This is because for some $\alpha < \beta$, we have $S_\beta \subseteq S_\alpha - f(S_\alpha)$, so the ses $f(S_\alpha)$ and $f(S_\beta)$ are disjoint, and because $\bigcup_{\alpha < \gamma} f(S_\alpha) = P - S_\gamma = P$. |
Show that $o(\frac{a_n}{n})+o(\frac{1}{n})+O(\frac{a_n^2}{n})=O(\frac{1}{na_n})$ | The problem is equivalent to proving that $$o (a_n^2) + o (a_n) + O (a_n^3) = O (1).$$ Since $a_n = o (1),$ the RHS will be $= o (1) = O (1)$. |
How many $0$'s are at the end of $(38!)^{20}$? | $160$ is the correct answer!
Proof)
$$
\newcommand{\floor}[2]{\left\lfloor\frac{#1}{#2}\right\rfloor}
\floor{38}{5}
+\floor{38}{5^2}
= 7 + 1
$$
Hence, $8 \times 20 = 160$. |
Characterizations of bounded real zero sets | Let us call the solution set to be $V$. We can consider this set in $\mathbb{RP}^n$ where $V \subset \lbrace x_0 \neq 0 \rbrace$. Let us call the homogenized polynomial $\tilde{p}$. Then $V(\tilde{p}) = \overline{V}$ in Zariski topology.
If the set $V$ is unbounded, then the closure of $V$ in real topology has non-empty intersection with hyperplane at infinity that is the set $\lbrace x_0 = 0 \rbrace$. This implies that $\overline{V} \cap \lbrace x_0 = 0 \rbrace \neq \emptyset$.
Assume that $V$ is irreducible and $\overline{V} \cap \lbrace x_0 = 0 \rbrace \neq \emptyset$, then we get a sequence of points $p_i V$ such that $p_i$ converges to $p$ in hyperplane at infinity. Let $p_i = [p_{0i} : p_{1i} : \dots : p_{ni}]$, then one of the ratios $p_{ji}/p_{0i}$ is unbounded(since the point $[0 : 0 : \dots : 0]$ is not a point). Thus we get that the solution set is unbounded. Now if $V$ is not irreducible, then we can argue similarily for each of the irreducible components.
Thus the $V$ is bounded iff $\overline{V} \cap \lbrace x_0 = 0 \rbrace = \emptyset$. This holds iff the highest degree homogeneous term of the polynomial $p$ has no zeros except $(0, \dots , 0)$.
For example : Let $p(X_1,X_2) = X_1^2 + X_2^2 - 1$. Then $\tilde{p}(X_0,X_1,X_2) = X_1^2+X_2^2 - X_0^2$. The zeros of this polynomial with $X_0 = 0$ is $(0,0)$ and hence $V(p)$ must be compact, which clearly is true. |
Asymptotics for convolutions of arithmetic functions | Those things don't have a simple asymptotic because
$$\sum_{d |n} \frac{\mu(d)^r}{ d^2}n =n\prod_{p^k \| n} \sum_{d |p^k}\frac{\mu(d)^r}{ d^2}=n\prod_{p | n} (1+ \frac{(-1)^r}{p^2})$$
With $n=m!$ and $m$ large then $\prod_{p | n} (1+ \frac{(-1)^r}{p^2})$ will be close to $1/\zeta(2)$ if $r$ is odd or $\zeta(2)/\zeta(4)$ if $r$ is even, whereas for $n$ prime $\prod_{p | n} (1+ \frac{(-1)^r}{p^2})$ will be close to $1$. |
Rotation matrices vs quaternions? | The relation is as follows: Given the rotation angle $\theta$ and the unit vector (axis) $\mathbf{u}$, you have to form the quaternion
$$
\mathbf{q}=\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\mathbf{u}.
$$
Then the double-sided action
$$
R(\mathbf{v})=\mathbf{q}\mathbf{v}\mathbf{q^*}
$$
(where $\mathbf{q^*}$ is the conjugate quaternion and the operation is quaternion multiplication) gives you the rotated vector. |
Lie algebra for SO(3) as a skew symmetric matrix | $$
\begin{split}
\forall A\in SO(3), AA^T=I &\Rightarrow \frac{d}{dt}(AA^T)=0=\dot AA^T+A\dot A^T=\dot AA^T+(\dot AA^T)^T\\
&\Rightarrow (\dot AA^T)^T=-\dot AA^T
\end{split}
$$
and therefore $\dot AA^T$ is a $3\times 3$ skew symmetric matrix. Therefore, the Lie algebra, which is identified with the tangent space at identity:
$$
so(3):=T_eSO(3)=R_{A*}^{-1}T_ASO(3)=\{\dot A(t)A^{-1}(t)|_0|A(0)=A\}
$$
is the vector space of all $3\times 3$ skew symmetric matrices ($R_{A*}^{-1}$ is the right pull back). |
Prove $(I-J_n)^{-1}=I-\frac{1}{n-1}J_n$ | $J_n^2=nJ_n$, so $(I-J_n)\left(I-\dfrac1{n-1}J_n\right)=I-\dfrac1{n-1}J_n-J_n+\dfrac1{n-1}J_n^2=I$ |
How to prove that a polynomial has one exactly one root using Banach's fixed point theorem? | Yes the first part is (mostly) fine, you have by the mean value theorem
$$d(f(x),f(y)) = |f(x) - f(y)| \leq \sup_{z\in\mathbb{R}}|f'(z)||x-y|\leq \frac{1}{5}|x-y| = \frac{1}{5}d(x,y)$$
Therefore it is a contraction. For the next part, consider
$$g(x) := \frac{1}{3+x^4}$$
If $g$ has a unique fixed point $x_0$ (by the Banach fixed point theorem), then
$$g(x_0) = x_0 = \frac{1}{3+x_0^4} \iff x_0^5 + 3x_0 - 1 =0$$
as in the proof of the Banach fixed point theorem, you may take any point, say $x=0$, then
$$x_0 = \lim_{n\rightarrow\infty}g^n(x)$$
Where I am using the notation $g^{n}(x) = g^{n-1}(g(x))$. |
Adjoint of projection onto direct sum of Hilbert spaces | You are right, the adjoint of $\pi_n : K \to K_n$ is the canonical inclusion $\iota_n : K_n \to K$.
Let $(x_1, x_2, \ldots) \in K$ and $y_n \in K_n$. We have:
\begin{align}
\Big\langle \pi_n(x_1, x_2, \ldots), y_n\Big\rangle_K &= \langle x_n, y_n\rangle_{K_n} \\
&= \Big\langle (x_1, x_2, \ldots), (\underbrace{0, \ldots, 0}_{n-1}, y_n, 0, \ldots)\Big\rangle_{K}\\
&= \Big\langle (x_1, x_2, \ldots), \iota_n(y_n)\Big\rangle_{K}\\
\end{align}
Therefore $\pi_n^* = \iota_n$. |
Differentiability properties of $\psi(x)\cos(\phi(x)),\,\, \psi(x)\sin(\phi(x))$ at $x=0$ | Yes, these properties can hold simultaneously.
For example, let
$$\psi(x) = x, \quad \phi(x) = 0$$
These are both smooth on $\mathbb{R}$, so it suffices to drop any right-only or range-restriction considerations. For the conditions on $\psi$, it is clearly strictly increasing, and satisfies $\psi(0) = 0$, $\psi'(0) = 1 > 0$. Moreover, $\phi$ is smooth, and $\phi' \psi = 0$, so certainly $\lim_{x \to 0^+} \phi'(x) \psi(x) = 0$. Hence these satisfy your requirements on $\psi$ and $\phi$.
It follows that
$$f_1(x) = x, \quad f_2(x) = 0$$
These are smooth, and all derivatives of order $n \geq 2$ (and hence all even ones) are identically zero. Lastly, we have that $f_1'(x) = 1$, so $f_1'(1) = 1 \neq 0$. |
Find all $n \in \mathbb{N}$ such that ${{2n}\choose{n}} \equiv (-1)^n\pmod{2n+1}$ | The solutions are those $n \in \mathbb N$ for which $2n+1$ is either prime or a Catalan pseudoprime.
We say that $2n+1$ is a Catalan pseudoprime if it is a composite number and
$$(-1)^n\, C_n \equiv 2 \pmod{2n+1}$$
where $C_n$ is the $n$-th Catalan number, that is,
$$C_n = \frac 1 {n+1} \binom {2n} n.$$
Rewriting the definition, we see that this means
$$(-1)^n \frac 1 {n+1} \binom {2n} n \equiv 2 \pmod {2n+1}$$
and multiplying by $(-1)^n (n+1)$,
$$\binom {2n} n \equiv (-1)^n (2n + 2) \equiv (-1)^n \pmod {2n+1}$$
which is the original congruence.
The only known Catalan pseudoprimes are $5907$, $1194649$ and $12327121$. So the only other solutions to the congruence that we know of are $2953$, $597324$ and $6163560$. |
How Do I Find My Car | First, I'll ignore the fact that a complete search takes an infinite amount of time, so that your question really depends on the maximum length you theoretically could cover on the real line. Lets codify this by saying that you're going to accept an error rate of 1% (so in 1% of the cases, you stop before you find your car).
Lets look at a few special cases-:
You flip a fair coin. If it is heads, you go towards $\infty$, otherwise you go towards $-\infty$. This will have an average success rate of 50%, so it is not acceptable at the 1% level.
Walk in one direction until the tail probability equals 0.5%, then turn around and head in the other direction until that tail probability is 0.5%. Now, your expected success rate is 99%.
No 2 is optimal because it maximizes the "density per unit length" of the overall search path, which means you spend more time searching in higher probability regions, without "deadheading" as much as you would if you did multiple "turnarounds" while searching. We want to minimize turnarounds, but still get the highest probability regions in our search........ |
proving a group is soluble (solvable if you are american) | The theorem you need is
Every finite $p$-group is solvable.
and
If $N\lhd G$, $N$ and $G/N$ are solvable, then $N$ is solvable.
So from part (a), you have proven that $n_7=1$ or $n_2=1$.
Let $P\in Syl_7(G)$.
Then $P\lhd G$. Also note that $P$ is $7$-group and $G/P$ is $2$-group.
Hence $P$ and $G/P$ are solvable, so is $G$.
If you let $Q\in Syl_2(G)$, the same argument follows as well. |
product of $f(x)f(y)f(z)\geq1$ for $f(x)=ax^2+bx+c$ with $xyz=1$ and $a+b+c=1$. | By Holder:
$$(ax^2+bx+c)(ay^2+by+c)(az^2+bz+c)\geq\left(a\sqrt[3]{x^2y^2z^2}+b\sqrt[3]{xyz}+c\right)^3=1.$$ |
Proving a function to be continuous in Topology | Let $U$ be an open subset of $Y$. You want to prove that $f^{-1}(U)$ is an open subset of $X$. But, by the definition of the topology on $X$, this is true indeed, since $A\in\mathcal T$ if and only if $A=f^{-1}(B)$ for some $B\in\mathcal U$. |
Prove that $(\sup\{y\in\Bbb R^{\ge0}:y^2\le x\})^2=x$ | Since you have the l.u.b. axiom, you can use that, for any two bounded sets $A,B$ of nonnegative reals, we have
$$(\sup A) \cdot (\sup B)=\sup( \{a\cdot b:a \in A,b \in B\}).$$
Applying this to $A=B=S(x)$ we want to find the sup of the set of products $a\cdot b$ where $a^2\le x$ and $b^2 \le x.$ First note any such product is at most $x$: without loss assume $a \le b$, then we have $ab \le b^2 \le x.$ This much shows $\sqrt{x}\cdot \sqrt{x} \le x$ for your definition of $\sqrt{x}$ as a sup.
Second, since we may use any (independent) choices of $a,b \in S(x)$ we may in particular take them both equal, and obtain the products $t\cdot t=t^2$ which, given the definition of $S(x)$, have supremum $x$, which shows that $\sqrt{x}\cdot \sqrt{x} \ge x$. |
How to prove that $W=U_1 \oplus(U_2 \cap W)$ if $U_1 \subseteq W$, given that $V=U_1 \oplus U_2$ and $U_1,U_2,W$ are subspaces of $V$? | It is better to use the definition of the direct sum.
The assumptions may be rewritten as:
a)
any element of $V$ is the sum of an element of $U_1$ and an element of $U_2$.
b) The intersection of $U_1$ and $U_2$ is the null vector.
Now you want to prove to things:
1) any element of $W$ is the sum of an element of $U_1$ and an element of $U_2\cap W$.
2) The intersection of $U_1$ and $U_2\cap W$ is the null vector.
It seems pretty clear how to proceed now. You will need the assumption $U_1\subset W$ to get 1) from a). |
Define a canonical homomorphism from G/G' to H/H' where K' is the commutator of K. | Hint: $f: G \to H$ is a group homomorphism. Consider the canonical homomorphism $\pi :H \to H/H'.$ Let $\phi : G \xrightarrow{f}H \xrightarrow{\pi} H/H'$ be the combined group homomorphism. Show that every for $a, b \in G, \phi ([a, b])$ is the identity element of $H/H'$ where $[a, b] := aba^{-1}b^{-1}.$ |
Justify $\zeta(3)=2\int_0^1 \left(Li_2(e^{-2\pi i x})+Li_2(e^{2\pi i x}\right))\log \Gamma(x)dx$ | So we have to prove that
$$ \frac{\zeta(3)}{4} = \int_{0}^{1}\sum_{k\geq 1}\frac{\cos(2\pi k x)}{k^2}\log\Gamma(x)\,dx $$
but over the iterval $(0,1)$ the graph of the Fourier cosine series $\sum_{k\geq 1}\frac{\cos(2\pi k x)}{k^2}$ is just the graph of a parabola, since $\sum_{k\geq 1}\frac{\sin(2\pi k x)}{k}$ is a sawtooth wave. Moreover, such a parabola is symmetric with respect to the point $x=\frac{1}{2}$. The claim boils down to:
$$ \frac{\zeta(3)}{4} = \int_{0}^{1/2}\left(\frac{\pi^2}{6}-\pi^2(x-x^2)\right)\log\left(\Gamma(x)\Gamma(1-x)\right)\,dx $$
or, through the reflection formula for the $\Gamma$ function and the substitution $x\to\frac{1}{2}-x$:
$$ \frac{\zeta(3)}{4}=\frac{\pi^2}{12}\int_{0}^{1/2}(12x^2-1)\log\left(\frac{\pi}{\cos(\pi x)}\right)\,dx$$
that is equivalent to:
$$ 3\zeta(3)=\pi^2\int_{0}^{1/2}(1-12x^2)\log(\cos(\pi x))\,dx \tag{1a}$$
or to:
$$ \frac{3\zeta(3)}{\pi^3} = \int_{0}^{1/2}\tan(\pi x)(x-4x^3)\,dx,\tag{1b}$$
$$ \frac{\zeta(3)}{\zeta(2)} = \int_{0}^{1}\frac{\pi}{2}\cot(\pi x/2)x(x-1)(x-2)\,dx.\tag{1c}$$
Since the Weierstrass product of the cosine function gives
$$ \cos(\pi x)=\prod_{n\geq 0}\left(1-\frac{4 x^2}{(2n+1)^2}\right)\tag{2}$$
and
$$ \int_{0}^{1/2}(1-12x^2)\log\left(1-\frac{4 x^2}{(2n+1)^2}\right)\,dx \\= \frac{1}{3}+4 n (1+n)+2 n (1+n) (1+2 n) (\log(n)-\log(n+1)) \tag{3}$$
the whole question boils down to an exercise in summation by parts involving the derivative of the $\zeta$ function (in particular, $\zeta'(-2)$, that is related with $\zeta(3)$ via the reflection formula for the $\zeta$ function). As an alternative, $(1)$ can be checked by expanding both $12x^2-1$ and $\log\cos(\pi x)$ as Fourier series. On the other hand, Chen Wang's solution to the linked question just relies on Kummer's Fourier series for $\log\Gamma$. The Weierstrass product also provides the identities
$$ \frac{\pi}{2}\cot\left(\frac{\pi x}{2}\right)=\frac{1}{x}+\sum_{n\geq 1}\left(\frac{1}{x-2n}+\frac{1}{x+2n}\right) \tag{4}$$
$$ \tan\left(\frac{\pi x}{2}\right)=\frac{4}{\pi}\sum_{n\geq 0}\frac{x}{(2n+1)^2-x^2} \tag{5} $$
that can be used to directly tackle $(1b)$ or $(1c)$.
Now we have a situation that often occurs in mathematics: when proving something, we go through some lemma that is crucial for other situations. In this case we have that the RHS of $(3)$ is summable over $n\geq 1$, so the "smallness" of such term gives us that
$$ \frac{4n(n+1)+\frac{1}{3}}{2n(2n+1)(n+1)} $$
is an excellent approximation of $\log\left(1+\frac{1}{n}\right)$ for any $n\geq 1$, and we may even estimate the error by approximating the associated integral with the Cauchy-Schwarz inequality, integration by parts or other techniques. In our case we get that the approximation error is less than $1.3\cdot 10^{-3}$ for any $n\geq 1$, and
$$ \log(2)\approx\left(\frac{5}{6}\right)^2. $$ |
Singleton in a complete metric space is a complete metric space, but has no interior point. Baire? | Yes, it has an interior point. If $p\in M$, then $p$ is an interior point of $\{p\}$, if we see $\{p\}$ as a metric subspace of $M$. Of course, in $M$, the set $\{p\}$ has, in general, no interior point (it has only if $p$ is an isolated point of $M$). |
a question about a complex integral, I am struggling with it! | This is an improper integral so it should be viewed as a limiting procedure of the form
$$\lim_{b\rightarrow 1^-}\lim_{a\rightarrow 0^+} \int_a^b \frac{\ln x}{1-x^2}\,dx.$$
The series for $\dfrac{1}{1-x^2}$ converges uniformly (which is the key) over such intervals so you can interchange series and integral without trouble. This is a standard theorem in real analysis. |
Left cosets of $H$ in $G$. Symmetric group of order $3$ | "The cosets are displayed as (1)H,(12)H,(13)H,(23)H".
This is definitely wrong. But you are right, there are 3 cosets, since the group has order 6 and the subgroup has order 2.
The cosets are
$H = (1)H = (13)H$,
$(123)H =\{(123)(12)\}$ and
$(132)H = \{(132),(23)\}$. |
Question about integrable functions | Hint: let $A_n = \{x: |f(x)| > 1/n\}$. |
Baby Rudin exercise 3.3 solution, possible typo in solutions manual? | You're right in both cases; both guides are wrong. (Here's one that seems to get it right.)
Note that the first manual you mentioned (the one you linked to in a comment) makes the same mistake again two lines further down, where it says that the limit satisfies $s^2-s-2=0$, which should be $s^2-\sqrt s-2=0$. |
show using probability axioms $P[A\bigcap B] \ge P[A] + P[B] - 1$ | $A = (A \cap B) \cup (A \setminus B)$
$B = (A \cap B) \cup (B \setminus A)$
$p A + p B = 2 p (A \cap B) + p (A \setminus B) + p (B \setminus A) = p (A \cap B) + p (A \cup B) \le p (A \cap B) + 1$. |
Show that $(X^*)^*=\epsilon\epsilon 'X$ | First, note that $(X^{*})^*$ means that it satisfies the equation:
$$
(u,(X^{*})^*v) = (X^*u,v)
$$
for $u \in U$, $v \in V$, but
$$
(X^* u,v) = \epsilon'(v, X^{*}u) = \epsilon' (Xv,u) = \epsilon' \epsilon(u,Xv)
$$
by definition of "quadratic space $\epsilon'$", by definition of $X^*$, and by definition of "quadratic space $\epsilon$", respectively. Therefore,
$$
(u,(X^{*})^*v) = \epsilon' \epsilon(u,Xv)
$$
for any $u \in U$ and $v \in V$, which implies $(X^{*})^* = \epsilon \epsilon' X$ since the form is non-degenerate. |
What is the probability of winning the prize | There are $\ {6\choose 4}\ $ subsets of the player's $6$ cards which could be the one exactly matched, and the matching set could occur in any of $\ {6\choose 4}\ $ sets of positions of the $6$-ball draw. There are thus $\ {6\choose 4}^2\ $ different ways in which the match could occur, each with the same probability of
$$
\frac{4\cdot3\cdot2\cdot1\cdot42\cdot41}{48\cdot47\cdot46\cdot45\cdot44\cdot43}\ .
$$
Thus, the probability of such an exact match occurring is
$$
{6\choose 4}^2\frac{4\cdot3\cdot2\cdot1\cdot42\cdot41}{48\cdot47\cdot46\cdot45\cdot44\cdot43}=\frac{4305}{4090504}\approx0.00105
$$ |
$\lim_{h \to 0} \int_{x}^{x+h} \ln(t) dt$ | Yes, you're correct. You can argue it that way, or even if you go as far as integrating first you'll find the same result:
\begin{eqnarray*}
\lim_{h\to 0} \int_x^{x+h}\ln(t)dt & = & \lim_{h\to 0} \left . t\ln(t) - t \right |_x^{x+h} \\
& = & \lim_{h\to 0}(x+h)\ln(x+h) - (x+h) - x\ln(x) + x \\
& = & \lim_{h\to 0} x \ln\left (\frac{x+h}{x}\right ) + h\ln(x+h) - h \\
& = & x \ln(1) + 0 - 0 \;\; =\;\; 0.
\end{eqnarray*} |
Find $x$: $\sqrt[3]{1+\sqrt{x}}+\sqrt[3]{1-\sqrt{x}}=\sqrt[3]{5}$ | Cubing both sides $$5=1+\sqrt x+1-\sqrt x+3\sqrt[3]{5(1-x)}$$
$$\iff\sqrt[3]{5(1-x)}=1$$
Cube both sides |
How do I find the time of an scheduled event if it is given between a range? | Apparently this is a matter of getting the right interpretation, having that ah-ha moment and also a bit of playing around like Where's Wally?
In the passage the key as mentioned in the comments is what to do with but in $25$ minutes. After the edit, it is more likely that it meant the time after the show started.
To better illustrate this situation below are the variables:
$\textrm{Time when the show started = x}$
$\textrm{But in 25 minutes =} x + \frac{25}{60}$
Time elapsed between 5 AM and 25 minutes before the show started:
$\left(x-\frac{25}{60}\right)-5$
Two thirds of the time which will be to 8 AM, but in 25 minutes.
$\frac{2}{3}\left(8-\left(x+\frac{25}{60}\right)\right)$
Finally all that is left to do is to equate both expressions:
$\left(x-\frac{25}{60}\right)-5=\frac{2}{3}\left(8-\left(x+\frac{25}{60}\right)\right)$
Simplifying:
$\left(x-\frac{5}{12}\right)-5=\frac{2}{3}\left(8-\left(x+\frac{5}{12}\right)\right)$
$\frac{12x-5-60}{12}=\frac{2}{3}\left(\frac{96-12x-5}{12}\right)$
$12x-65=\frac{2}{3}\left(91-12x\right)$
$36x-195=182-24x$
$60x=377$
$x= 6 \frac{17}{60}$
Hence the time when the show began was $\textrm{6:17 AM}$
This checks with the alternatives. Needless to say that probably this was the intended meaning. |
Prove there exists no uniform distribution on a countable and infinite set. | Let $X$ be a random variable which assumes values in a countable infinite set $Q$. We can prove there is no uniform distribution on $Q$.
Assume there exists such a uniform distribution, that is, there exists $a \geq 0$ such that $P(X=q) = a$ for every $q \in Q$.
Observe that, since $Q$ is countable, by countable additivity of $P$,
$1 = P(X \in Q) = \sum_{q \in Q}{P(X = q)} = \sum_{q \in Q}{a}$
Observe that if $a=0$, $\sum_{q \in Q}{a}=0$. Similarly, if $a>0$, $\sum_{q \in Q}{a} = \infty$. Contradiction. |
ideal sheaf modulo square of ideal sheaf equals restriction? | If you tensor the standard exact sequence
$$
0 \to I \to O_X \to O_Y \to 0
$$
with $I$, you get
$$
I \otimes I \to I \to I \otimes O_Y \to 0.
$$
Note that the first map takes $f \otimes g$ to $fg$, hence its image is $I^2$. This proves that
$$
I/I^2 \cong I \otimes O_Y \cong I\vert_Y.
$$ |
Existence of at least one positive root | The general direction is good, but you may want to make the statement that $f(L)\approx c_n L^n$ more precise. To be specific, for sufficiently large $x$, the absolute value of each term in the expression $\frac{c_{n-1}}{c_nx} + ...+ \frac{c_0}{c_nx^n}$ is less than $\frac1n$. Therefore,
$$f(x) = c_n x^n \left(1+\frac{c_{n-1}}{c_nx} + ...+ \frac{c_0}{c_nx^n}\right)>c_n x^n \left(1+n\left(-\frac1n\right)\right)=0$$
You can then proceed as you did. |
Prove that for $\cos (\alpha ) = \frac{1}{3}$, $\alpha < \frac{\pi}{2} - \frac{1}{3}$ | You mean that: if $0<\alpha<\frac{\pi}{2}$ and $\cos\alpha=\frac{1}{3}$, then $\alpha<\frac{\pi}{2}-\frac{1}{3}$?
Let $F(x)=\cos x-\frac{1}{3}$, then $F(0)=\frac{2}{3}$ and $F(\frac{\pi}{2}-\frac{1}{2})=\sin\frac{1}{3}-\frac{1}{3}<0$ since $\sin x<x (0<x<\frac{\pi}{2})$.
This shows that $\alpha<\frac{\pi}{2}-\frac{1}{3}$. |
Maximizing the number of nodes in a graph | This is the degree diameter problem. There is a known bound on the number of vertices (the Moore bound). There are very few graphs that actually attain this bound for specific values of the diameter, your $d$, and degree, your $e$. The Moore bound is $M_{2,d} = 2d+1$ and, for $e>2$, $M_{e,d} = 1+e \frac{(e-1)^d-1}{e-2}$.
The current state of the art (maximal graphs with given degree and diameter) is summarized here. As I write this, diameter $= 2$ is well studied, after that, degree $\leq 16$ is well studied with degree $=2$ very well studied. After that, diameter $\leq 6$ is well studied and thereis much detailed knowledge for degree $\leq 16$ and diameter $\leq 4$. We have complete knowledge for diameter $=2$ and degree $\leq 7$ and for diameter $=3$ and degree $\leq 4$. For other diameter/degree pairs, there is a gap between vertex counts of explicitly constructed graphs and the provable maximum. |
Structure of a subspace of $X = X_1 \oplus \cdots \oplus X_r$ | Basically you are composing projections with restrictions.
Let f be in Hom(V,V).
My TEXing is weak but this is the idea;
f = (projection of X composed with the restriction of f to X) $ \oplus$ (projection of X composed with the restriction to X) $\oplus$ (projection of X composed with the restriction of Y) $\oplus $(projection of Y composed with the restriction of f to Y). |
Minimum distance between the curves of two inverse functions? | Consider the orthogonal projection onto the line $x+y=0$. The distance between orthogonal projections of two points does not exceed the distance between points themselves (for the same reason that a leg of a right triangle is not longer than its hypotenuse: Pythagorean theorem). Therefore, if we can find the distance between projections of two curves, that will be a lower bound on the distance between the curves themselves. With luck and foresight, the lower bound will be sharp, i.e., it will be the actual distance.
So, what is the projection of $\{(x,y):y=e^x\}$ onto the line $x+y=0$? It's a closed half-line. Its endpoint comes from $(x,e^x)$ such that $e^x-x$ is minimal. Minimizing $e^x-x$ over $x$, we see that the minimum is at $x=0$. The point $(0,1)$ projects to $(-1/2,1/2)$.
By symmetry, the projection of $\{(x,y):y=\ln x\}$ onto the line $x+y=0$ is a closed half-line with endpoint $(1/{2},-1/2)$.
The distance between projections is $\sqrt{1^2+1^2} = \sqrt{2}$. Therefore, the distance between curves is at least $\sqrt{2}$. This is the lower bound.
And it's sharp: the distance between the aforementioned point $(0,1)$ and its counterpart $(1,0)$ is $\sqrt{2}$. |
Which of the Following Sets are compact (C.S.I.R 2015) | Your solutions seem correct to me. For the last one you could like to be able to show a direct contradiction of the definition of compactness, that is, an open cover which does not admit finite subcover.
This counterexample could be easily provided by a family of open subsets of $K$ such as
$$U_n=K\cap\{z\in\mathbb{C} : n-1<Imz<n+1\}, ~ n\in\mathbb{N}$$
I guess no further explanation is required for the first three questions.
A note about question 3: $A_n$ is not required to be discrete in order to prove compactness. A finite set will always be compact, in any topology. |
Is there a non-random variable with Borel set preimage? | Take $(\Omega,\mathcal{F})=([0,1],\mathcal{B}_{[0,1]})$ and $\xi(\omega)=\omega+2\cdot1_N(\omega)$, where $N\subset [0,1]$ is a non-measurable set. Then for each $x\in \mathbb{R}$, $\xi^{-1}(\{x\})$ is either empty or a singleton so that $\xi^{-1}(\{x\})\in \mathcal{B}_{[0,1]}$. However, $\xi^{-1}([2,3])=N\notin \mathcal{B}_{[0,1]}$. |
Is this a valid proof of the parallel postulate? | Right from the beginning you speak about lines with particular slopes. But this assumes that you have a rectangular coordinate system to measure the slopes with, and in order to construct such a coordinate system, you need the parallel postulate already.
So your argument does not work as a "proof of the parallel postulate" in the usual sense of a proof that start from the other axioms of Euclidean plane geometry.
Something like your calculations (which I haven't checked in detail) will prove that if we start by defining the plane to be $\mathbb R^2$ with certain operations, then that plane happens to satisfy the parallel postulate. This can definitely be a relevant thing to prove when you want to argue that analytic geometry works in the first place, or that Euclidean geometry is consistent relative to our usual foundations (such as ZFC).
It's just not the problem that "proof of the parallel postulate" is usually taken to be about. |
Proving that the characters of an infinite Abelian group is a basis for the space of functions from the group to $\mathbb{C}$ | The classical proof of this uses a measure and hence integral (Fourier transform) that can be defined because of the topological properties of the underlying group. In case of finite Abelian groups, this boils down to the discrete topology, in case of infinite ones you need the local compactness to ensure what is called the Pontryagin duality. See e.g. http://en.wikipedia.org/wiki/Pontryagin_duality |
Limit theorems, prove function has a limit at every point | The posted question assumes $\lim_{x\to0}f(x)=0$, but that was not given. What was given is that $\lim_{x\to0}f(x)$ exists. That the limit is $0$ must be proved.
The functional equation $f(x+y)=f(x)+f(y)$ entails that $f(0)$ $=f(x+(-x))$ $=f(x)+f(-x)$; hence we must have $f(-x)=-f(x)$, i.e. this is an odd function. If an odd function $f$ has a limit $L$ at $0$, then
$$
\lim_{x\downarrow0}f(x) =\lim_{x\to0} f(x) = \lim_{x\uparrow0} f(x),
$$
but the last one is
$$
\lim_{x\uparrow0} f(x) = \lim_{x\downarrow0} f(-x) = -\lim_{x\downarrow0} f(x) = -L.
$$
So $L=-L$, and therefore $L=0$.
As to using $f(1)=1$, notice that the function $g(x)=3x$ also satisfies the equation $g(x+y)=g(x)+g(y)$ (and similarly if we had used any other number besides $3$), so certainly that condition must be used.
What the two equalities $f(x+y)=f(x)+f(y)$ and $f(1)=1$ imply is that $f(x)=x$ whenever $x$ is rational. Let's see that first for positive rational numbers.
\begin{align}
1 =f(1) & = f\left(\frac15+\frac15+\frac15+\frac15+\frac15\right) \\[10pt]
& = f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) = 5f\left(\frac 15 \right)
\end{align}
so $f\left(\dfrac15\right)= \dfrac15$. And
$$
f\left(\frac35\right)=f\left(\frac15\right) + f\left(\frac15\right) + f\left(\frac15\right) = \frac15+\frac15+\frac15 = \frac 3 5
$$
and similarly for all other positive rational numbers.
If $x$ is irrational, look at a sequence of rational numbers approaching $x$ and then use continuity. |
What is the right way of generating character+digit permutations? | The first method is clearly correct. You take the total number of strings of the proper lengths and subtract those that do not have at least one digit leaving those that do.
I believe the second method is counting possibilities more than once. Take ABCD59 for example. Did you make sure the 5th spot was a digit before filling in the others or was it the 6th? |
Help with Proof in Cohn's Measure Theory: Approximation Using Regularity of Measure | RHS of (3) is $\geq \mu (L-W)>\mu (A) -\epsilon$ since $K=L-H$ is a compact subset of $A$. [ Here we used the fact that supremum of a set of real numbers is greater than or equal to any member of that set]. Since $\epsilon$ is arbitrary it follows that RHS $\geq \mu(A)$. The reverse inequality follows from the fact that $\mu(K) \leq \mu(A)$ for every compact set $K$ contained in $A$ |
Simplify $\lim_{n \to \infty}\frac{(1-A)(1-B)}{(2-A-B)(A-B)}=\frac{\gamma(\gamma-1)}{2\gamma-1}?$ | Hint
I suppose that you need to use (and reuse) the Taylor expansion (have a look at formula 35 here)
$$\Gamma \left(\frac{1}{n}\right)=n-\gamma +\frac{6 \gamma ^2+\pi ^2}{12 n}+O\left(\frac{1}{n^2}\right)$$
$$\frac{1}{\Gamma \left(\frac{1}{n}\right)}=\frac{1}{n}+\frac{\gamma }{n^2}+O\left(\frac{1}{n^3}\right)$$ and then
$$\Gamma(1+\epsilon)= 1-\gamma \epsilon +\frac{1}{12} \left(6 \gamma ^2+\pi ^2\right) \epsilon
^2+O\left(\epsilon ^3\right)$$
It works (a bit tedious). |
How can I find the integral of this function using trig substitution? | Factor out a $\sin\theta$ from the $\sin^3\theta$. Write the remaining $\sin^2\theta$ as $1-\cos^2\theta$. Now let $u=\cos\theta$.
$$
\sin^3\theta\, \cos^2\theta \,d\theta= (1- {\cos^2\theta} )\cos^2\theta\sin\theta \,d\theta
= (\cos^2\theta- {\cos^4\theta} ) \sin\theta \,d\theta\ \
\buildrel{u=\cos\theta}\over = \ \ -(u^2-u^4)\,du
$$ |
Find value for $k$ such that $(x^2-k)$ is a factor for $f(x)=2x^4+(3k-4)x^3+(2k^2-5k-5)x^2+(2k^3-2k^2-3k-6)x+6$ | Your approach is good. Just add $f(\sqrt k)+f(-\sqrt k)=0$ to get rid of the terms with $\sqrt k$. |
Find the value which satisfies these equations | Not really familiar with DDMs, but algebra says:
$$
\begin{align}
Y &= C + I \\
\implies Y &= 0.6Y + 50 + 10 \\
\implies 0.4 Y &= 60 \\
Y &= \dfrac{60*10}{4} \\
&= 150
\end{align}
$$ |
Divisibility in $\mathbb F_p[x]$ | Well, note that $$x^{p^n} - x + n a = \sum_{i=1}^n(x^{p^i}-x^{p^{i-1}} + a)$$
I claim that every term of the sum is divisible by your polynomial. In fact, the first term of the sum is divisible by the second term, and so on. Better than that, the first term is the $p$-th power of the second term, and so on. Can you prove this? |
Determine continuity and differentiability of the real function $f(x)=\sum\limits_{n\geq1}\frac{1}{n^x}$ | This is the Riemann zeta function. Here are some useful things you can read: Click here ! |
show that $ \ a_n \leq x \leq a_n+\frac{1}{b^n} \ $ for any base $b \neq 10 $ holds for any real number in $[0,1]$. | Take the difference
$$0\leq x - \sum\limits_{i=1}^{n} \frac{a_i}{b^i} = \sum\limits_{i=n+1}^{\infty} \frac{a_i}{b^i} \leq ...$$
and because $0\leq a_i \leq b-1$ and assuming an integer base $b>1$, we have
$$ ... \leq \sum\limits_{i=n+1}^{\infty} \frac{b-1}{b^i}=
\frac{b-1}{b^{n+1}} \left(\color{red}{\sum\limits_{i=0}^{\infty} \frac{1}{b^{i}}}\right)=
\frac{b-1}{b^{n+1}} \left(\frac{1}{1-\frac{1}{b}}\right)=\frac{1}{b^n}$$
the sum in red is an infinite geometric progression with ratio $\left|\frac{1}{b}\right|<1$. |
Show that $f({x_1+x_2+\cdots x_n\over n})\le \frac 1n [f(x_1)+f(x_2)+\cdots f(x_n)]$ | This is a special caes of Jensen's inequality. Standard proofs are given on this wiki page.
It is also possible to prove this special case by a slightly unusual induction.
We let $P(n)$ denote the proposition for $n$. You have already shown that $P(2)$ is true.
Now the proof goes like this:
We show that, for $n \geq 2$, if $P(n)$ is true, then $P(2n)$ is true.
We show that, for $n \geq 2$, if $P(n)$ is true, then $P(n - 1)$ is true.
These two statements combined together will show that $P(n)$ is true for all $n$.
For 1, we have $$f(\frac{x_1 + \dots x_{2n}}{2n}) \leq \frac 1 2[f(\frac{x_1 + \dots + x_n}n) + f(\frac{x_{n + 1} + \dots + x_{2n}}n)] \leq \frac 1 {2n}[f(x_1) + \dots + f(x_{2n})]$$ where the first inequality uses $P(2)$ and the second inequality uses $P(n)$.
For 2, we put $x_n = \frac {x_1 + \dots + x_{n - 1}}{n - 1}$ in $P(n)$ and get $P(n - 1)$. |
Question on proof from Eisenbud's Commutative Algebra. | The key point is to understand the $R$-submodule $IN\subset N$.
(For an element $m\in M$, I'll denote by $\bar m$ its image in $N$ in what follows.)
i) How do you write a typical element of that submodule $IN$ ?
Answer : $\Sigma i_k\bar p_k $ with $i_k\in I$ and $p_k\in M$.
ii) And how do you write a typical element of $N$ ?
Answer : $\bar m $ with $m\in M$ .
But the hypothesis of (b) says that you can write $m=\Sigma r_l m_l+\Sigma j_tm_t'$ (with $r_l\in R, m'_t\in M , j_t\in I$), so that $\bar m =\Sigma j_t\overline {m_t'}$, an element in $IN$ according to i).
So indeed $N=IN $ , as claimed.
By the way, you can find a mnemonic for Nakayama's lemma here. |
let $x > -1$ prove for all $n \geq 1$, $(1+x)^n \geq 1+nx$ | The idea here is correct, but the way it is written, it's not clear that this is a proper proof of the inductive step. When proving by induction, you start with the $n=k$ case, and then prove the $n=k+1$ case. If we are to assume the implications are top implies bottom, then what you have done is started with the $n=k+1$ case and then worked down. However, this is precisely what you are trying to prove to begin with.
An example of when this kind of logic doesn't work:
Suppose we want to prove $1=0$. We have
$$\underset{\color{red}X}{(1=0)}\Rightarrow\underset{\color{green}\checkmark}{(0\cdot1=0\cdot0)}\Rightarrow\underset{\color{green}\checkmark}{(0=0)}$$
While the conclusion is in fact true, it does not imply $1=0$.
Instead, start with
$$(1+x)^k\ge1+kx$$
and then work towards proving $(1+x)^{k+1}\ge1+(k+1)x$ from there. |
theoretical question regarding deduction and relation between $\vdash$ and $\vDash$ | You have to prove a sort of soundness theorem for your calculus.
Hint about soundness :
if $\Gamma \vdash a$, the $\Gamma \vDash a$.
Assume that $\Gamma \vdash a$ and consider the ususal cases :
(i) $a \in \Gamma$; then obviously $\Gamma \vDash a$, because every assignment $v$ that satisfies every formula in $\Gamma$ will satisfies also $a$.
(ii) $a$ is an axiom; this means that $a$ is not a tautology. Consider the case $a := p$.
According to the definition of derivation, we have $\{ q \lor r \} \vdash p$, but obviously : $\{ q \lor r \} \nvDash p$.
(iii) $a$ is derived with the inference rule from a previous formula in the derivation. |
Solving for vector from equations. Can Dot product product be bought down as the denominator | You essentially have two equations in three unknowns, for which there is not a unique solution. If $p_3=(x,y,z)$ then since $\bar{p}$ and $b_2$ are known, even if both expressions involving 2d vectors on the left sides compute to known numbers, you can only get so far as
$$[1] \ \ p_3 \cdot b_1=k_1,\ \ p_3 \cdot b_2=k_2,$$
where $k_1=p_2.x+\bar{p}\cdot b_1$ and $k_2=p_2.y + \bar{p}\cdot b_2.$ It is unclear (to me) just what $p_2.x$ and $p_2.y$ mean, and whether $x,y$ denote two dimensional vectors which are known. But the point is, whatever they denote, if their values are known real numbers then $k_1,k_2$ are just two real numbers, and then replacing $p_3$ by its unknown coordinates $(x,y,z)$ and replacing $b_1,b_2$ by whatever numerical coordinates they have, the system [1] ends up as two linear equations in the three unknowns $x,y,z$. But two linear equations in three unknowns are never enough to determine the values of those unknowns. |
Partial Derivative of a Dot Product with Respect to one of its Vectors | $\frac {\partial f}{\partial \textbf v}$ is a shorthand for $\left(\frac {\partial f}{\partial v_1},...,\frac {\partial f}{\partial v_n}\right)$, in other words it is the gradient of $f$. In this case, if you expand the dot product notation in terms of the coefficients, you obtain $\frac {\partial f}{\partial v_i}=u_i$, so $\frac {\partial f}{\partial \textbf v} = \textbf u$.
In general, most rules for taking derivatives generalise well to taking derivatives with respect to vectors, as is done here, or even matrices. For a useful reference, I recommend the matrix cookbook, which has a list of identities. Proving a few might help you in your understanding. |
Prove for all $x \in \mathbb{R}$ at least one of $\sqrt{3}-x$ and $\sqrt{3}+x$ is irrational | The sum of two rational numbers is rational, but
$$(\sqrt 3 -x)+(\sqrt 3+x)=2\sqrt 3$$
which is irrational. Therefore they cannot both be rational. |
What is $\displaystyle\lim_{x\to\infty}(\left \lfloor{-\dfrac{1}{x}}\right \rfloor )$? | As $x\to+\infty$, $-\frac1x$ approaches $0$ from the negative side; it is negative for any sufficiently large finite $x$. Thus we must conclude that the limit of its floor is $-1$, this being the floor of an arbitrarily small (in magnitude) negative number. |
$\overline{V(I)-V(J)}=V(\bigcup_{n=1}^{\infty}I\colon J^n)$ | I assume that $I,J$ are ideals of a commutative ring $R$?
After replacing $R$ by $R/I$ we may clearly assume that $I=0$, so that $0:J^n=\mathrm{Ann}(J^n)$.
Let $I \subseteq R$ be an ideal. Then the following are equivalent:
$\overline{\mathrm{Spec}(R) \setminus V(J)} \subseteq V(I)\\
\mathrm{Spec}(R) \setminus V(J) \subseteq V(I)\\
V(J) \cup V(I)=\mathrm{Spec}(R)\\
IJ \subseteq \sqrt{0}\\I \subseteq (\sqrt{0}:J) \\
V(\sqrt{0}:J) \subseteq V(I)$
We have used here that $(\sqrt{0}:J)$ is a radical ideal, which is easy to prove. This proves
$$\overline{\mathrm{Spec}(R) \setminus V(J)} = V\bigl((\sqrt{0}:J)\bigr)$$
We always have $(\sqrt{0}:J) \supseteq \sqrt{\bigcup_{n \geq 1} \mathrm{Ann}(J^n)}$, and the converse inclusion holds when $J$ is finitely generated (this is easy to check). In general we won't have equality. |
Quaternion to Euler angles conversion | Be careful when you convert between quaternions and euler angles. One of the main source of confusions are the conventions adopted to represent angles. The same quaternion can represent a rotation or it's inverse based on the adopted convention.
Your equations seems to be correct at first glance. As a comparison write some code to compute the conversion to the rotation vector. I do not know what the website does and I believe that when it comes to quaternions it is safe not to use software from different sources because you do not know the conventions that are adopted.
By the way, quaternions are provided by matlab in the aerosym toolbox (included in the complete version), but they follow the JPL convention and not the Hamiltonian. |
If $a=a'$ mod $n$ and $b=b'$ mod $n$ does $ab=a'b'$ mod $n$? | We can write $a' = a + n k$ for some $k$, and likewise $b' = b + n m$; then
$$a'b' = (a + nk)(b + nm) = ab + n(\text{stuff}) \equiv ab \pmod{n}$$
upon distribution the multiplication. |
Use intermediate value theorem to prove $\frac{1}{\sqrt{x-5}} = 2^x -8$ has a solution | Hint. For $x>5$, consider the continuous function,
$$f(x)=2^x - 8-\frac{1}{\sqrt{x-5}}.$$
Note that $f(6)=2^6-8-1=55>0$.
Now the limit of $f(x)$ as $x\to 5^+$ is $-\infty$ and, by the definition of limit, there is $5<x_0<6$ such that $f(x_0)<0$ (you don't have to find it explicitly).
Therefore by IVT, $f$ has a zero in the interval $(x_0,6)\subset (5,6)$. |
End - extension in the complete theory that isn't elementary extension | Let $L = \{\leq\}$, and let $T$ be the complete theory of $N = (\mathbb{N}^*,\leq)$, the reverse order on the natural numbers. Let $M = ((\mathbb{N}\cup \{-1\})^*,\leq)$, so $M$ adds a new maximal element $-1$ on top of $N$.
Then $M$ is an end extension of $N$, and it is also a model of $T$ since it is isomorphic to $N$, but it is not an elementary extension: let $\varphi(x)$ be the formula $\forall y(y\leq x)$ which expresses that $x$ is a maximal element. Then $N\models \varphi(0)$ but $M\not\models \varphi(0)$. |
Integrating this gamma kernel-like function | Making a change of variable $t=b/x$ and assuming that $b>0$ we have that
$$
\int_{0}^{\infty }x^{1-a}e^{-b/x}\,\mathrm d x=\int_0^{\infty } (b/t)^{1-a}e^{-t}bt^{-2}\,\mathrm d t\\
=b^{2-a}\int_{0}^{\infty }t^{a-3}e^{-t}\,\mathrm d t=\frac{\Gamma (a-2)}{b^{a-2}},\quad \text{ when }a>2
$$
what coincides with your result. |
question about a line in Hatcher about long exact sequence in cohomology | Consider the map $p_0:I\times Y\to\partial I\times Y, (t,y)\mapsto (0,y)$, then if $i:\partial I\times Y\hookrightarrow I\times Y$ is the inclusion, then $i\circ p_0$ is the map $(t,y)\mapsto (0,y)$ and is clearly homotopic to $\mathrm{id}_{I\times Y}$, hence $p_0^*i^*=\mathrm{id}$ in cohomology. This shows that $i^*$ is injective, and by exactness, if $j$ is the inclusion of pairs $j:(I\times Y,\emptyset)\hookrightarrow(I\times Y,\partial I\times Y)$, then $\mathrm{Ker}~i^*=\mathrm{Im}~j^*=0$, so $j^*=0$ and the connecting homomorphism $\delta$ is onto because $\mathrm{Im}~\delta=\mathrm{Ker}~j^*$. You get short exact sequences
$$0\to H^{n}(I\times Y)\xrightarrow{\;i^*\;} H^{n}(\partial I\times Y)\xrightarrow{\;\delta\;}H^{n+1}(I\times Y,\partial I\times Y)\to 0$$
and the first arrow is split (through $p_0^*$). |
Spiral of Theodorus - Discussion | From the Wikipedia entry for Wilbur Knorr, on one of his books:
The Evolution of the Euclidean
Elements: A Study of the Theory of
Incommensurable Magnitudes and Its
Significance for Early Greek
Geometry (Dordrecht: D. Reidel Publishing Co., 1975).
This work incorporates Knorr's Ph.D.
thesis. It traces the early history of
irrational numbers from their first
discovery (in Thebes between 430 and
410 BC, Knorr speculates), through the
work of Theodorus of Cyrene, who
showed the irrationality of the square
roots of the integers up to 17, and
Theodorus' student Theaetetus, who
showed that all non-square integers
have irrational square roots. Knorr
reconstructs an argument based on
Pythagorean triples and parity that
matches the story in Plato's
Theaetetus of Theodorus' difficulties
with the number 17, and shows that
switching from parity to a different
dichotomy in terms of whether a number
is square or not was the key to
Theaetetus' success.
(Note: This is copied from an answer I gave at MathOverflow a year ago.) |
definitition of projective resolution of $R$-modules (with homology) | You are going well, but I think you're just mistaking degrees of boundary maps. A homological complex is usually of the form
$$\cdots \to M_{n}\overset{f_n}{\to}M_{n-1}\to\cdots \to M_0\overset{f_0}\to 0$$
so it's straightforward to have $H_0(M^\bullet)=M_0/\mathrm{im}(f_1)$. |
Is the tangent-cotangent isomorphism orientation preserving? | We can use $\phi$ to pull $\eta$ back to a one-form $\alpha = \phi^* \eta$ on $TM$, which (since $\phi$ is an isomorphism) still has the property that $(d\alpha)^n$ is a volume form on $TM$. It's also not hard to write down a formula for $\alpha$; as a remark before doing so, note that if $X$ is a vector tangent to the manifold $TM$ at $(p, v)$ then
$$
\pi_* d\phi(X) = \pi_* X
$$
since obviously $d\phi$ doesn't affect the horizontal part of vectors in $TTM$. Then one computes:
$$
\alpha_{p,v}(X) = \eta_{\phi(p, v)}(d\phi(X)) = \phi(v) \Big( \pi_* d\phi(X)\Big) = \langle v, \pi_* X \rangle,
$$
which is a useful formula.
It's also not hard to write down an explicit formula for $d\alpha$. We let $K$ be the natural map $T_{p,v} TM \to T_pM$ taking a vector to its vertical part; then one can show
$$
d\alpha(Y,Z) = \langle KY, \pi_* Z \rangle - \langle \pi_* Y, KZ \rangle.
$$
(Two easy ways to check this: one, check it for coordinate fields using an explicit formula for $d\alpha$ in terms of $\alpha$ and our formula above; two, if you already believe that $d\eta$ has the coordinate representation $\sum d p_i \wedge d x_j$, then it should be clear that $d\alpha$ must have the coordinate representation $\sum d q_i \wedge d x_j$, where $q_i$ is the coordinate associated with $\partial / \partial x_i$; it's easy to see that the above formula follows.)
The point is that this gives a simple way to compute $(d\alpha)^n$ applied to a nice set of coordinate fields for $TM$. In particular, if $x_1, \dots, x_n$ are coordinates on some chart of $M$, and as above $q_i$ are the natural extensions to $TM$, then our formula gives
$$
(d\alpha)^n\Big(\frac{\partial}{\partial q_1}, \frac{\partial}{\partial x_1}, \dots, \frac{\partial}{\partial q_n}, \frac{\partial}{\partial x_n} \Big) = n!.
$$
(This is assuming I didn't screw up a sign somewhere which is unfortunately quite possible; I don't have good references in front of me.)
As Dan Fox points out in his comment, whether you consider this to be the "natural" orientation on $TM$ is entirely up to you. |
challenging integral involving $\zeta(5)$ | The purported identity is false, as GEdgar already indicated in the comment, but remarkably accurate:
$$ \begin{eqnarray}
\int_0^1 \frac{\sqrt{1-x^2}}{1-x^2 \sin^2(x)} \mathrm{d} x &\approx& \color{red}{ 0.91392913}60302011781728596 \\
\frac{5 \pi^{8/5}}{32 \zeta(5)^{9/5}} &\approx& \color{red}{0.91392913}77247633495515212
\end{eqnarray}
$$
Here is the Mathematica code used:
In[19]:= N[
NIntegrate[Sqrt[1 - x^2]/(1 - x^2 Sin[x]^2), {x, 0, 1},
WorkingPrecision -> 60], 25]
Out[19]= 0.9139291360302011781728596
In[20]:= N[(5 Pi^(8/5))/(32 Zeta[5]^(9/5)), 25]
Out[20]= 0.9139291377247633495515212
In[21]:= % - %%
Out[21]= 1.694562171378662*10^-9 |
Existence and uniqueness of the minimizer of Moreau-Yosida approximation | This is not an answer, but it won't fit in the comments.
Set $t=1$ in the above, it has no effect on the following.
The fact that $f$ is bounded below allows us to conclude that $F_1$ is bounded below. Other conditions are possible, for example, if $f$ majorises some linear functional, or if $f$ is continuous at some point.
The proof here relies on the epigraph (of $F_1$ in this case) being closed
and bounded below. Then weak compactness of a 'clipped' version of the epigraph is used to show existence of a minimiser.
For uniqueness:
The polarisation identity can be used to get
$\|x-{1 \over 2} (y_1+y_2)\|^2 = {1 \over 2} \|x-y_1\|^2 + {1 \over 2} \|x-y_2\|^2 - {1 \over 4} \|y_1-y_2\|^2$, combining this with convexity of $f$
gives
$F_1({1 \over 2} (y_1+y_2)) \le {1 \over 2} F_1(y_1) + {1 \over 2} F_1(y_2)- {1 \over 4} \|y_1-y_2\|^2$.
Hence if $y_1 \neq y_2$ were two minimisers, we would have an immediate contradiction. |
Convergence of series of bounded linear operators | Hint. Notice that
$$
\left\|I-\sum_{n=0}^\infty \frac{1}{(n+1)!}A^nh^n\right\| \le \sum_{n=1}^\infty \left\|\frac{1}{(n+1)!}A^nh^n\right\| = \sum_{n=1}^\infty \frac{h^n}{(n+1)!}\left\|A^n\right\| \le \sum_{n=1}^\infty \frac{h^n}{(n+1)!}\left\|A\right\|^n
$$
Thus if $\sum_{n=1}^\infty \frac{h^n}{(n+1)!}\left\|A\right\|^n \to 0$ as $h\to 0$ then $\sum_{n=0}^\infty \frac{1}{(n+1)!}A^nh^n \to I$ as $h\to0$. You should be able to prove $\sum_{n=1}^\infty \frac{h^n}{(n+1)!}\left\|A\right\|^n \to 0$ as $h\to 0$ relatively easily. |
Find value of $\prod(1+\alpha^2)$ | Your polynomial is irreducible over $\mathbb{Q}$ since it is irreducible over $\mathbb{F}_2$, hence it is the minimal polynomial of $\alpha$. The minimal polynomial of $\alpha^2$ is given by
$$ \left[\underbrace{\left(x^4-6x^2-9\right)^2}_{\text{even part}}-\underbrace{\left(4x^3+7x\right)^2}_{\text{odd part}}\right]_{x\mapsto \sqrt{x}} = x^4-28x^3-38x^2+59x+81=g(x)$$
and
$$ \prod_\text{cyc}(1+\alpha^2) = g(-1) =\color{red}{13}.$$
As an alternative, the LHS is the determinant of $M^2+I$ where $M$ is the companion matrix of the original polynomial. |
Legendre symbol question | Let the primes $\le N$ be $p_0$ to $p_k$.
Consider the system of congruences
$d\equiv 1^2\pmod{p_i}$, $i=0$ to $k$. By the Chinese Remainder Theorem, this system has infinitely many solutions. Each is a quadratic residue modulo all the $p_i$. |
What does the notation $\mathbb{P}V$ mean for a vectorspace $V$? | I would expect this to mean the projective space defined by $V$: the space nonzero vectors modulo the relation $r\vec{v} \sim \vec{v}$ whenever $r \neq 0$.
So, for example, $\mathbb{P} \mathbb{R}^3$ is the usual real projective plane. |
Convex function defined by an inner product | The only difficulty here is that you are dealing with the second derivative of $g$.
Let $f(x) = g(x)-g(b)$ and $\phi(x) = {1 \over 2} \|f(x)\|^2$.
Then $D \phi(x)(h) = f^T(x) Df(x)(h)$ and
$D^2\phi(x)(h,\eta) = (Df(x) (\eta))^T Df(x)(h) + f^T(x) D^2 f(x)(h,\eta)$,
or
$D^2\phi(x)(h,\eta) = \eta^T Df(x)^T Df(x)h + f^T(x) D^2 f(x)(h,\eta)$.
Since $g'(b)$ is injective, we see that $Df(b)$ is injective and hence so is
$Df(b)^T Df(b)$. Furthermore, by continuity, there is some $\epsilon>0$ and $\delta >0$
such that for $x \in B(b,\delta) $ we have $Df(x)^T Df(x) \ge \epsilon I$.
Since $g$ is $C^2$, and $f(b) = 0$, there is some $\delta_1 \le \delta$ such that
for $x \in B(b,\delta_1)$ we have
$f^T(x) D^2 \phi(x)(h,h) \ge -{1 \over 2} \epsilon \|h\|^2$ and so
$D^2 \phi(x) \ge {1 \over 2} \epsilon I$.
Hence $\phi$ is strictly convex in $B(b,\delta_1)$. |
What is the order of the alternating group $A_4$? | You are missing the identity element. It can be written as an "even" permutation:
$$(12)(12)$$ |
Finding an entire function $f$ | This is impossible. The assumptions imply that $f^{-1}(c) \subset U$ for all $c$ with $|c| < 1$, and the level set of an entire function (of several variables) can never be compact. |
How can I add a curve to my otherwise linear values? | It seems you have already solved the issue of moving around the global unaltered of wrap-around and uneven steps in longitude and lattitude (if not quaternions and quaternion interpolation will solve this).
What you are looking to do requires interpolation of a higher degree (than linear). The curve you are looking for here is probably best suited to be a quadratic curve on the speed of your flight not the position. An equation of the form $speed = \alpha t^2 + 2\beta t(1-t) + \gamma (1-t)^2$ is called a Bézier curve. You could perhaps calculate the time required for the flight and then use interpolation to actually move and the time will stay the same (as the linear flight) but will be smoothed at the start and the end.
Check out this page for helping on implementation for a quadratic bezier curve. |
Fast binomial sampling | Sampling from the binomial distribution over a wide range of values of $n$ and $p$ can't really be done in the same way in all cases. Instead we should use the normal or Poisson approximations to handle large $n$, and the exact distribution to handle small $n$.
Assuming $p \leq 1/2$, one can use a normal approximation for $np \gg 1$ and a Poisson approximation for $n \gg 1$ such that $np \sim 1$. (In practice, "$np \gg 1$" means essentially "$np>20$".) For $p>1/2$ one can draw a sample with $p$ replaced by $1-p$ using one of the above methods, and then subtract the result from $n$.
Note that although it might seem that a Poisson is at least as hard to sample as a binomial, you can write the CDF in terms of the incomplete Gamma function to make it cheap to evaluate. |
$f$ is contained in no codimension $0,1$ primes $\implies f$ is invertible. Why do we need codim 1 primes as well? | In general maximal ideals do not have codimension zero. The codimension of an ideal is its height: the largest chain of prime ideals it contains. In a reasonable ring, say integral and finitely generated over a field, the codimension of a maximal ideal is equal to the dimension of the ring, not zero.
For example, for integral domains, the question asks you to prove that if an element does not lie in any minimal prime ideal (that is a prime ideal containing no other prime ideals other than zero), than this element is invertible.
By the way: it is indeed clear that if an element does not lie any maximal ideal, then it is invertible. The geometric analogue is that if a function does not vanish at any point, then it is an invertible function. But the assertion here is that, for example, if a function on a 3 dimensional space does not vanish on any surface, then it is invertible. This is also obvious when stated in this geometric form, but not so clear for "non-geometric" rings. |
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