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H: Identification of $\mathbb F_2[X]/(X^4+X+1)$
As mentioned in the title I would like to show that we can identify $\mathbb F_2[X]/(X^4+X+1)$ with the set $K$ of polynomials: $p_0+p_1 a+p_2a^2+p_3 a^3$ in a variable $a$ that we assume satisfies $a^4+a+1=0$.
If you have any ideas than can help me with that would be great.
AI: This is a very general construction:
Let $\mathbb F$ be any field and $p(X)\in \mathbb F[X]$. The quotient ring $R=\mathbb F[X]/(p(x))$ consists of elements of the form $q(X)+I$, where $I=(p(X))$.
Now, you can use the division algorithm for polynomials and obtain that $q(X)=p(X)\cdot t(X)+r(X)$ for a unique polynomial $r(X)$ of degree smaller than the degree of $p(X)$. It thus follows that every element in $R$ is equivalent to some element of the form $r(X)+I$, where $\deg(r)<\deg(p)$.
A straightforward verification shows that no two such elements are equivalent, and thus the set $R$ can be identified with the set of polynomials $r(X)\in \mathbb F[X]$ of degree smaller than the degree of $p(X)$. The ring operations are given by addition and multiplication on the representatives, and then taken modulo $I$.
A direct verification shows that the element $\alpha = X+p(X)$ in $R$ satisfies the equation $p(x)=0$. |
H: Evaluate $ 1 + \frac{1}{3}\frac{1}{4}+\frac{1}{5}\frac{1}{4^2}+\frac{1}{7}\frac{1}{4^3}+\dots$
Evaluate $$ 1 + \frac{1}{3}\frac{1}{4}+\frac{1}{5}\frac{1}{4^2}+\frac{1}{7}\frac{1}{4^3}+\dots$$
All i could do was to see that $$\frac{1}{3}=\frac{1}{2.1+1},\frac{1}{5}=\frac{1}{2.2+1},\frac{1}{7}=\frac{1}{2.3+1},\dots$$
SO, we should be able to write
$$ 1 + \frac{1}{3}\frac{1}{4}+\frac{1}{5}\frac{1}{4^2}+\frac{1}{7}\frac{1}{4^3}+\dots=1+\sum_{n=1}^{\infty}\frac{1}{2n+1}.\frac{1}{4^n}$$
This is what i could simplify the given question to...
But, this is doing me no good.. I could not go any further..
I would be thankful if some one can help me to clear this.
I would request users who are trying to help me to just give necessary hints but not post it as answer.
Thank You.
AI: From this, for $|x|<1,$
$$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots$$
$$\text{and }\ln(1-x)=-x-\frac{x^2}2-\frac{x^3}3-\frac{x^4}4+\cdots$$
Now subtract
Observe that $$ 1 + \frac13\frac14+\frac15\frac1{4^2}+\frac17\frac1{4^3}+\cdots$$
$$=2\sum_{0\le r<\infty}\frac1{2r+1}\left(\frac12\right)^{2r+1}$$ |
H: Prove sequence in a complete metric space converges if the series of distances converges.
Assume that X is complete, and let $(p_n)$ be a sequence in X. Assume that $\sum\limits_{n=1}^\infty d(p_n, p_{n+1})$ converges. Prove that $(p_n)$ converges.
(X,d) is a metric space.
Don't quite know how to start. Any help/hints appreciated!
AI: How to start: look at the definition of "complete metric space". Every Cauchy sequence converges. OK, we have to show that $(p_n)$ is a Cauchy sequence. Using, somehow, the fact that the series of distances converges.
Next, consider the definition of a Cauchy sequence. Given a positive $\varepsilon$ we have to show that, if $n$ is big enough, then $d(p_n,p_m)\lt\varepsilon$ for all $m\gt n$. Right? Is that how the Cauchy criterion is stated in your book?
Hmm. How do we get from $d(p_n,p_{n+1})$ to $d(p_n,p_m)$?? The triangle inequality!! If $m=n+k$ then$$d(p_n,p_m)=d(p_n,p_{n+k})\le d(p_n,p_{n+1})+d(p_{n+1},p_{n+2})+\cdots+d(p_{n+k-1},p_{n+k})\le d(p_n,p_{n+1})+d(p_{n+1},p_{n+2})+d(p_{n+2},p_{n+3})+\cdots$$
Hmm. The remainder of a convergent series. Goes to zero, doesn't it? |
H: What is the cardinality of $\omega^\alpha\cdot n+1$?
How could I determine the cardinality of $\omega^\alpha\cdot n+1$? where $\alpha$ is an ordinal.
I think that, if $\alpha$ is infinite then $|\omega^\alpha\cdot n+1|>\aleph_0$, Is that right?
AI: If $\alpha < \omega_1$ then $\omega^{\alpha} < \omega_1$, so certainly $\omega^{\alpha} \cdot n + 1 < \omega_1$ and so $|\omega^{\alpha} \cdot n + 1| < \aleph_1$.
You can prove that $\omega^{\alpha} < \omega_1$ for all $\alpha < \omega_1$ by transfinite induction on $\alpha$.
Clearly it's true when $\alpha=0$, and if it's true for $\omega^{\alpha}$ then it's true for $\omega^{\alpha+1} = \omega^{\alpha} \cdot \omega$.
Now suppose $\alpha < \omega_1$ is a limit and $\omega^{\beta}$ is countable for all $\beta < \alpha$. Then
$$|\omega^{\alpha}| = \left| \bigcup_{\beta < \alpha} \omega^{\beta}\right| \le \sum_{\beta < \alpha} |\omega^{\beta}| \le |\alpha| \cdot \sup_{\beta < \alpha} |\omega^{\beta}| = \aleph_0 \cdot \aleph_0 = \aleph_0$$
However $|\omega^{\omega_1}| > \aleph_0$ because there is an injection $\omega_1 \to \omega^{\omega_1}$ sending $\alpha < \omega_1$ to the function $f_{\alpha} \in \omega^{\omega_1}$ defined by $f_{\alpha}(\alpha) = 1$ and $f_{\alpha}(\beta) = 0$ for $\alpha \ne \beta < \omega_1$.
A generalisation of the above arguments allows you to prove that $|\omega^{\alpha}| = |\alpha|$ for all $\alpha \ge \omega$. |
H: Is there a function that gives the same result for a number and its reciprocal?
Is there a (non-piecewise, non-trivial) function where $f(x) = f(\frac{1}{x})$?
Why?
It would be nice to compare ratios without worrying about the ordering of numerator and denominator. For example, I might want to know whether the "magnitude" of the ratio (maybe the "absolute ratio") of the widths of two objects is greater than $2$, but not care which is larger.
It occurred to me that there's a common solution for this problem when comparing the difference of two numbers: the square of a number is the same as the square of its opposite - $(a-b)^2=(b-a)^2$. This is really useful with Euclidean distances, because you don't have to worry about the order of subtraction or use absolute values. Can we get the same elegance for ratios?
Difference: $g(a-b)=g(b-a) \rightarrow g(x)=x^2$
Ratio: $f(\frac{a}{b})=f(\frac{b}{a}) \rightarrow f(x)=\ ?$
AI: $$ \frac{x}{x^2+1} $$
the inverse of Will Jagy's $x + \frac{1}{x}$.
I like $$x - \frac{1}{x} = \frac{x^2-1}{x}$$
because it contains sign information (input magnitude greater or less than one) that you may choose to ignore, and gives a nice zero for $x = \frac{1}{x} = \pm 1$. If you choose to take the absolute value of it (ignoring the sign) it gives you the desired $f(x) = f(\frac{1}{x})$ |
H: Proving composition of functions
I am trying to prove the following theorems:
Let A, B, and C be nonempty sets and let $f : A \rightarrow B$ and $g : B \rightarrow C$.
If $g \circ f : A \rightarrow C$ is an injection, then $f : A \rightarrow B$ is an injection.
If $g \circ f : A \rightarrow C$ is a surjection, then $g : B \rightarrow C$ is an surjection.
Any suggestions? Thanks!
AI: Suppose that $f(x_1) = f(x_2)$. Then we have $$g(f(x_1)) = g(f(x_2))$$ so the injectivity of $g \circ f$ implies.....
Suppose that $c \in C$, and choose an $a \in A$ such that $g(f(a)) = c$, by surjectivity of $g \circ f$. So can you conclude that $g$ is onto? |
H: Exact sequence of $R$-modules
Let $0\longrightarrow N\overset{f}{\longrightarrow}M\overset{g}{\longrightarrow}L\longrightarrow0$
be a short exact sequence of $R$-modules. Prove that this chain splits iff
$f(N)$ is direct summand of $M$.
Thanks a lot.
AI: Suppose $\;M=f(N)\oplus K\;$ , and define $\;F:M\to N\;$ by $\;F(m) = F(f(n)+k):=n\;$ . Show that $\;F\;$ is a well defined $\;R- $ homomorphism and $\;F\circ f=\text{Id}\,_N\;$ .
The other direction: suppose there exists an $\;R-$ homom. $\;F:M\to N\;$ s.t. $\;F\circ f=\text{Id}\,_N\;$. Show that
$$\begin{align*}(1)&\;\;\ker F\cap f(N)=\{0\}\\
(2)&\;\;M=f(N)\oplus\ker F\end{align*}$$
Note: Point (2) above may be a little tricky but not too much if you understand what's going on here. |
H: Existence of a function from $[0,1]$ to an arbitrary measurable set
For any measurable $E\subset \mathbb{R}$ with measure 1, does there exist a continuous function
$$T:[0,1]\to E$$
Such that $\mu\circ T=\mu$, where $\mu$ is Lebesgue measure on the real line?
AI: Hint: A continuous image of $[0,1]$ is compact and connected. |
H: Let $R$ be an integral domain, $M$ is free $R$-module with finite basis
Let $R$ be an integral domain, $M$ is free $R$-module with finite basis. Prove that two finite bases of $M$ have the same cardinality.
Help me some hints.
AI: If $R$ commutative ring then there is morphism $f:R\rightarrow k$ on to some field $k$. Suppose $R^n=R^m$ for some $n, m\in \mathbb N$. $k^m=(R/\ker f)^m=(R\otimes R/\ker f)^m=R^m\otimes R/\ker f=R^n\otimes R/\ker f=(R\otimes R/\ker f)^n$ $=(R/\ker f)^n=k^n$.
Its easy to see that $R$ module $k\cong R/\ker f$ has $k$ module structure. We reduced our problem to the fact that two vector spaces has bases of the same cardinality and in this case we know that $n=m$. |
H: What does it mean for an ultrafilter to have a limit?
I got this question from the construction of the Stone-Čech compactification using ultrafilters given in Wikipedia. There they say that if $F$ is an ultrafilter in a compact Hausdorff space $K$ then it has a unique limit.
I can see this in the case $F$ is principal. It makes sense to think about a limit of an ultrafilter $F$ as a common limit point to all subsets which are in $F$. If $F$ is principal, clearly there is only one such limit point, namely the generator.
If $F$ is not principal, then all subsets in $F$ are infinite, and by compactness of $K$ one can find subsequence for each of these subsets that has a limit point (unique by Hausdorffness). But how do I see that there is one unique limit point for a subsequence of each subset in $F$?
AI: Let $\mathscr{U}$ be an ultrafilter in the compact space $X$. Let $\mathscr{F}=\{\operatorname{cl}U:U\in\mathscr{U}\}$; then $\mathscr{F}\subseteq\mathscr{U}$, so $\mathscr{F}$ is a family of closed sets with the finite intersection property. Since $X$ is compact, $\bigcap\mathscr{F}\ne\varnothing$. Let $x\in\bigcap\mathscr{F}$; if $V$ is any open nbhd of $x$, and $U\in\mathscr{U}$, then $V\cap\operatorname{cl}U\ne\varnothing$, so $V\cap U\ne\varnothing$. Thus, $\mathscr{U}\cup\{V\}$ is a filter, and since $\mathscr{U}$ is an ultrafilter, this means that $\mathscr{U}\cup\{V\}=\mathscr{U}$, i.e., that $V\in\mathscr{U}$. Thus, every open nbhd of $x$ is in $\mathscr{U}$, which by definition means that $\mathscr{U}\to x$.
If $X$ is also Hausdorff, this limit point is unique, because $\bigcap\mathscr{F}$ is a singleton. To see this, suppose that $x,y\in\bigcap\mathscr{F}$, and $x\ne y$. If $X$ is Hausdorff, there are disjoint open sets $V$ and $W$ such that $x\in W$ and $y\in V$. But then the argument in the previous paragraph shows that $V\in\mathscr{U}$ and $W\in\mathscr{U}$, which is absurd. |
H: Significance of rank of frechet derivative in definition of manifold?
In studying manifolds, the stipulation that the derivative be full rank is confusing to me on an intuitive level. Can anyone please explain how I should think about this intuitively? What does it mean for the derivative to be full rank? What would it mean for the derivative to not be full rank?
AI: In what scenario did you encounter this condition? I think the most famous place you need the full rank assumption for the inverse function theorem in $\mathbb R^n$. Here, full rank is equivalent to the Jacobian (which is an $n \times n$ matrix) having non-zero determinant. This is a generalization of the derivative from $\mathbb R \to \mathbb R$ not being zero, this is usually synonymous to a function being one to one on connected sets (I.e. Intervals). Can you see how this is just a generalized analogy? |
H: Is $\sum_{n=1}^{\infty}\frac{2^nn!}{(n+1)!}$ absolutely convergent?
I'm very uncomfortable with factorials just because I haven't done many of them. But my basic understanding is if I start with (for example) $(n+1)!$ then this is equivalent to $(n+1)*(n)$ and if it were $(n-1)!$ then this is equivalent to $(n-1)*(n-2)*(n-3)...$
Is my understanding correct? If so did I solve this correctly using the ratio test?
$$\sum_{n=1}^{\infty}\frac{2^nn!}{(n+1)!}$$
So I did the following:
$$|\frac{2^{n+1}(n+1)!}{(n+2)!}*\frac{2^nn!}{(n+1)!}|$$
When I simplified this I got:
$$\frac{2(n+1)!}{(n+2)!}$$
And further...
$$\frac{2}{(n+2)!}$$
And then I took the limit of this and found it to be $0$
Is this correct?
AI: Note that, since limit of the summand
$$ \lim_{n\to \infty} \frac{2^n n!}{(n+1)!}=\infty, $$
then the series diverges. See a related problem. |
H: Need help with $\int_0^\infty e^{-x}\ln\ln\left(e^x+\sqrt{e^{2x}-1}\right)\,dx$
I need help with this integral:
$$\int_0^\infty e^{-x}\ln\ln\left(e^x+\sqrt{e^{2x}-1}\right)\,dx\approx0.20597312051214...$$
Is it possible to evaluated it in a closed form?
AI: A useful identity here is ${\rm arccosh\,} z = \ln( z + \sqrt{z^2 -1} )$.
Therefore $$\int_0^{\infty} e^{-x} \ln \ln( e^x + \sqrt{e^{2x}-1}) \;dx= \int_0^{\infty} e^{-x}\ln {\rm arccosh\,} e^x \;dx =\int_1^{\infty} \frac{\ln {\rm arccosh\,} y}{y^2} \;dy\, $$
Change integration variable $z={\rm arccosh\,}y$,
$$= \int_0^{\infty} \frac{\ln z \sinh z}{ \cosh^2 z} \;dz\,.$$
Integrate by parts and use the definition of the Euler's $\gamma$ constant and $\Gamma$ function see here,
$$= -\gamma + \ln\left[\frac{\Gamma\left(\frac{1}{4}\right)\Gamma\left(\frac{5}{4}\right)}{\Gamma^2\left(\frac{3}{4}\right)} \right]\,. $$
We can further simplify this using $\Gamma(1-z)\Gamma(z)=\pi/\sin \pi z$ and $\Gamma(z)\Gamma(z+\frac{1}{2})=2^{1-2z}\sqrt{\pi}\Gamma(2z)$ for $z=3/4$ and $\sin (3\pi/4)=1/\sqrt{2}$. This gives $$=-\gamma - 3\ln 2 - 2 \ln \pi + 4 \ln \Gamma(1/4)\,. $$ |
H: For any real numbers $a,b,c$ show that $\displaystyle \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$
For any real numbers $a,b,c$ show that:
$$ \min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{a^2+b^2+c^2}{2}$$
OK. So, here is my attempt to solve the problem:
We can assume, Without Loss Of Generality, that $a \leq b \leq c$ because of the symmetry.
$a \leq b \leq c$ implies that $0 \leq b-a \leq c-a$. Since both sides are positive and $y=x^2$ is an increasing function for positive numbers we conclude that $(b-a)^2 \leq (c-a)^2$ and since $(a-b)^2=(b-a)^2$ we obtain $(a-b)^2 \leq (c-a)^2$.
Therefore:
$\min\{(a-b)^2,(b-c)^2,(c-a)^2\}=$ $$\min\{(b-c)^2,\min\{(a-b)^2,(c-a)^2\}\}=
\min\{(a-b)^2,(b-c)^2\}.$$
So, we have to prove the following inequality instead:
$$\min\{(a-b)^2,(b-c)^2\} \leq \frac{a^2+b^2+c^2}{2}$$
Now two cases can happen:
$$(a-b)^2 \leq (b-c)^2 \implies |a-b| \leq |b-c|=c-b \implies b-c \leq a-b \leq c-b \implies 2b-c \leq a \implies b \leq \frac{a+c}{2}.$$
$$(b-c)^2 \leq (a-b)^2 \implies |c-b| \leq |a-b|=b-a \implies a-b \leq c-b \leq b-a \implies c \leq 2b-a \implies b \geq \frac{a+c}{2}.$$
So, this all boils down to whether $b$ is greater than the arithmetic mean of $a$ and $b$ are not.
Now I'm stuck. If $a,b,c$ had been assumed to be positive real numbers it would've been a lot easier to go forward from this step. But since we have made no assumptions on the signs of $a,b$ and $c$ I have no idea what I should do next.
Maybe I shouldn't care about what $\displaystyle \min\{(a-b)^2,(b-c)^2\}$ is equal to and I should continue my argument by dealing with $(a-b)^2$ and $(b-c)^2$ instead. I doubt that using the formula $\displaystyle \min\{x,y\}=\frac{x+y - |x-y|}{2}$ would simplify this any further. Any ideas on how to go further are appreciated.
AI: Note that LHS does not change if you replace $(a,b,c)$ by $(a-t,b-t,c-t)$. Thus we can first minimize RHS with respect to $t$,after we replace $(a,b,c)$ by $(a-t,b-t,c-t)$. It turns out that RHS is minimized when $\displaystyle t = \frac{a+b+c}{3}$, with minimum value being
$$\frac{1}{6} ((a-b)^2 + (b-c)^2 + (c-a)^2)$$
So it suffices to show that $$\min\{(a-b)^2,(b-c)^2,(c-a)^2\} \leq \frac{1}{6} ((a-b)^2 + (b-c)^2 + (c-a)^2)$$
Without loss of generality, assume that $$a \leq b \leq c$$Then clearly, $\min\{(c-b),(b-a), (c-a)\} = \min\{(c-b),(b-a)\}$, so
$$\min\{(c-b)^2,(b-a)^2, (c-a)^2\} = \min\{(c-b)^2,(b-a)^2\}$$
Moreover,
$$(c-a) = (c-b)+(b-a) \ge 2 \min\{(c-b),(b-a)\}$$
and
$$\begin{eqnarray}
\frac{1}{6} ((a-b)^2 + (b-c)^2 + (c-a)^2) &\ge& \frac{1}{6} (\min\{(c-b),(b-a)\})^2 + (\min\{(c-b),(b-a)\})^2 + (2\min\{(c-b),(b-a)\})^2 \\
&=& \min\{(c-b)^2,(b-a)^2\} \\
&=& \min\{(c-b)^2,(b-a)^2, (c-a)^2\}
\end{eqnarray}$$ |
H: Hartshorne Theorem III.5.2 (finite generation of cohomology for coherent sheaves on projective schemes over a noetherian ring)
Hartshorne, Algebraic Geometry, Theorem III.5.2, reads (in part)
Theorem 5.2
Let $X$ be a projective scheme over a Noetherian ring, and let $\mathcal{O}_X(1)$ be a very ample invertible sheaf on $X$ over $\operatorname{Spec} A$. Let $\mathscr{F}$ be a coherent sheaf on $X$. Then:
(1) for each $i \geq 0$, $H^u(X, \mathscr{F})$ is a finitely-generated $A$-module;
[...]
The proof proceeds by reducing to the case $X = \mathbb{P}^r_A$ and then checking things for $\mathcal{O}_X(q)$, which is fine. Then we need to establish things for arbitrary coherent sheaves. The proof here reads:
[...]
In general, given a coherent sheaf $\mathscr{F}$ on $X$, we can write $\mathscr{F}$ as a quotient of a sheaf $\mathscr{E}$, which is a finite direct sum of sheaves $\mathcal{O}(q_i)$ for various integers $q_i$. Let $\mathscr{R}$ be the kernel,
$$0 \to \mathscr{R} \to \mathscr{E} \to \mathscr{F} \to 0,$$
Then $\mathscr{R}$ is also coherent. We get an exact sequence of $A$-modules
$$\cdots \to H^i(X, \mathscr{E}) \to H^i(X, \mathscr{F}) \to H^{i+1}(X, \mathscr{R}) \to \cdots$$
Now the module on the left is finitely generated because $\mathscr{E}$ is a sum of $\mathcal{O}(q_i)$, as remarked above. The module on the right is finitely generated by the induction hypothesis.
[...]
This last sentence I don't understand. It doesn't explicitly state what the inductive hypothesis is, but I assume that it's that $H^j(X, \mathscr{F})$ is a finitely-generated $A$-module for $j > i$. In this case, what we know is that we have an exact sequence
$$\cdots \to H^i(X, \mathscr{F}) \to H^{i+1}(X, \mathscr{R}) \to H^{i+1}(X, \mathscr{E}) \to H^{i+1}(X, \mathscr{F}) \to \cdots$$
where the two rightmost terms are finitely generated as $A$-modules. Since we don't seem to know anything about $H^i(X, \mathscr{F})$ on the left side yet, I don't see how this helps us establish anything about $H^{i+1}(X, \mathscr{R})$.
What have I missed?
AI: Your induction hypothesis shouldn't be on a specific $\mathcal{F}$, but all coherent sheaves at the same time. If you assume that $H^j$ is finitely generated for all $j > i$ for all coherent sheaves (in particular $\mathcal{F}$ and $\mathcal{R}$), then you can say something about $H^i(X,\mathcal{F})$, assuming that you already know the cohomology of $H^i(X,\mathcal{E})$, which amounts to cohomology of $O(n)$ on projective spaces. |
H: Binary relations: transitivity and symmetry
I've been looking at some examples for transitivity and symmetry.
Suppose $A=\{0,1,2 \} $ and the relation $R=\{ (0,0),(1,1),(2,2),(1,2),(2,1) \}$
Well for starters this is clearly reflixe since $\forall x \in A ,xRx$.
As for symmetry, we define it as : $\forall x,y \in A, xRy \rightarrow yRx$. However, when looking at the relation above, well $(1,2) \rightarrow (2,1)$ is there, but what about zero? Aren't you suppose to have a relation for zero as well? What about if $x=y$, does symmetry hold? For example: $(0,0) \rightarrow (0,0)$. Does symmetry hold for when both $x$ and $y$ are the same values?
For transitivity, it needs to satisfy: $\forall x,y,z, (xRy \land yRz)\rightarrow xRz$. Well in the above case, zero is not related to anything...
The reason I'm questioning this is that the question itself claims that this relation is reflexive (yes), symmetric and transitive and therefore an equivalence relation. Though the confusion remains as to why the transitivity is there and why the symmetry is there.
AI: $R$ is indeed both symmetric and transitive. These properties are a bit different in character from reflexivity. Reflexivity requires certain specific ordered pairs to be in the relation; these don’t. These are only conditional requirements: if some ordered pairs are present, then certain others must also be present.
Symmetry says that if $\langle x,y\rangle\in R$, then $\langle y,x\rangle\in R$. Can you find any $x,y\in A$ such that $\langle x,y\rangle\in R$, but $\langle y,x\rangle\notin R$? No, so $R$ is symmetric. Symmetry of $R$ just says that if $x$ and $y$ are distinct elements of $A$, then either both of the pairs $\langle x,y\rangle$ and $\langle y,x\rangle$ belong to $R$, or neither of them belongs to $R$. In this case $A=\{0,1,2\}$, so the only pairs of elements to be considered are $0$ and $1$, $0$ and $2$, and $1$ and $2$.
$0$ and $1$: Neither $\langle 0,1\rangle$ nor $\langle 1,0\rangle$ is in $R$; this is fine.
$0$ and $2$: Neither $\langle 0,2\rangle$ nor $\langle 2,0\rangle$ is in $R$; this is fine.
$1$ and $2$: Both $\langle 1,2\rangle$ and $\langle 2,1\rangle$ are in $R$; this is fine.
Conclusion: $R$ is symmetric.
Transitivity says that if $\langle x,y\rangle\in R$ and $\langle y,z\rangle\in R$, then $\langle x,z\rangle\in R$. it’s a conditional requirement: if certain ordered pairs are in $R$, then certain other ordered pairs have to be in $R$ as well. If the conditions of the if part aren’t met, it doesn’t impose any requirement. In the particular case at hand we have $\langle x,y\rangle\in R$ and $\langle y,z\rangle\in R$ when:
$x=y=z=0$;
$x=y=z=1$;
$x=y=z=2$;
$x=1,y=2$, and $z=1$; and
$x=2,y=1$, and $z=2$.
For any other choice of values for $x,y$, and $z$, the condition that $\langle x,y\rangle\in R$ and $\langle y,z\rangle\in R$ is not satisfied, so transitivity says nothing about those choices. And in every one of these five cases the ordered pair $\langle x,z\rangle$ is in $R$, as required for transitivity, so $R$ is transitive. |
H: Drawing Planar Graphs
Is it possible to draw a planar graph on 11 vertices in which each face (country) has 3 neighbours?
And is there some method after to draw it to confirm that it is in fact indeed planar?
I drew some figures, but then I'm not entirely sure if it is actually planar...
Thanks.
AI: First, the sum of the degrees must be an even number, since $\Sigma Deg(v_i)= 2|E(G)|$. If your graph is "simple", i.e., it is not bipartite, then it is planar iff it does not contain a $K_5$ as a subgraph. Since in your problem, $Deg(v_i)\leq 3$ , it cannot have a $K_5$ as a subgraph, since every vertex of a $K_5$ has degree $4$. If your graph is bipartite, then you have to avoid a $K_{3,3}$ as a subgraph. |
H: Are there a link between series convergence and countability of sets?
Could you please help me understand this question:
Suppose $E \subset [0,1]$ and
for each sequence $(a_n)$ , $a_n \in E$ and there are no duplicate members at $a_n$ , the series $\sum_{n=1}^\infty a_n$ converges.
Prove $E$ is countable set.
I tried this way:
If $x_n$ is sequence of all members of E than $a_n$ are sub sequences of $x_n$.
Series $\sum_{n=1}^\infty x_n$ converges, therefore $x_n\to 0$.
and series $\sum_{n=1}^\infty a_n$ also converges, therefore $a_n\to 0$.
So we have sequence converging to zero. What next ?
Thanks.
AI: Note that the condition of $\sum a_n$ converging is much stronger than the condition of $I$ being countable. In fact the convergence of $\sum a_n$ implies that for every $\varepsilon$, the set $I \cap [\varepsilon, 1]$ is finite. Do you see why? Use this fact to enumerate $I$. |
H: Boundedness of a real sequence.
Let $\{a_n\}$ be a sequence in $\mathbb{R}$ such that $\sum |a_n||x_n| < \infty$ whenever $\sum |x_n| < \infty$. Prove that $\{a_n\}$ is bounded.
My Attempt :
We have to show $\exists$ $B \in \mathbb{R}^+$ s.t. $|a_n| < B$.
$\sum |x_n| < \infty$ gives $\lim |x_n| = 0$ i.e. for any $\epsilon > 0 $ $\exists$ $m_1 \in \mathbb{N}$ s.t. $|x_n| < \epsilon$ when $n > m_1$
$\sum |a_n||x_n| < \infty$ gives $\lim |a_n x_n| = 0$ i.e. for any $\epsilon > 0 $ $\exists$ $m_2 \in \mathbb{N}$ s.t. $|a_nx_n| < \epsilon$ when $n > m_2$.
I can not approach any more. I require $|x_n| > C$ $\forall n > m$ for some real $C$ and natural number $m$ to prove $a_n < B$.
One more question I have in this context. What will be if we do not consider absolute convergence of the series $\sum x_n$ and $\sum a_n x_n$ ?
Thank you for your help.
AI: It might be easiest to approach this by contradiction : Suppose $(a_n)$ is unbounded, then for each $k \in \mathbb{N}, \exists a_{n_k}$ such that
$$
|a_{n_k}| > 2^k
$$
So consider the sequence $(x_n)$ such that
$$
x_{n_k} = \frac{1}{2^k}, \text{ and } x_n = 0 \text{ otherwise}
$$
Then
$$
\sum|x_n| < \infty, \quad\text{ but }\quad \sum |a_n||x_n| = +\infty
$$
This is a contradiction, so $(a_n)$ must be bounded. |
H: Group action with finite stabilizer.
Let $G$ be a group generated by $\{g_1,g_2,\ldots , g_n\}$. Let $X$ be a space with a $G$-action on it, i.e. $G$ is acting on $X$. Suppose for each $x\in X$, the set $\{g_i;g_i(x)=x\}$ is trivial. Does it imply that the stabilizer of $x$, i.e. $\{g\in G;g(x)=x\}$ is finite? In other words to calculate finite stabilizer subgruops is it enough to consider generators only?
AI: Let $G = (\mathbb{Z}, +)$. It is generated by one element $g_1 = 1$.
Let $G$ act on the set $X = \{0, 1\}$ like this: if $g \in G$ is even, then $g$ acts trivially. Otherwise, it swaps $0$ and $1$.
Now you can see that $g_1$ doesn't fix any elements of $X$, but both elements in $X$ have infinite stabilizers. |
H: Convergence of events in a probability space with respect to $L^2$
Define for events $X, Y$ that $d(X,Y) = P((X-Y) \cup (Y-X)) $ = $ P(X \bigtriangleup Y) $, show that $d(X_n,X) \rightarrow 0$ if and only if $\chi_{X_n}$ converges in $L^2$ to $\chi_X$ (these are indicator functions which take on values of 1 when in the set and 0 outside of the set).
I know how $L^2$ convergence works, but just applying the definitions here I wasn't able to get much far. Using the hint given below, we can see that:
if $d(X_n,X) \rightarrow 0$ then $ P(X_n \bigtriangleup Y) \rightarrow 0 $ as $n \rightarrow \infty$, now we want to show $||\chi_{X_n}-\chi_{X}||_2^2 \rightarrow 0$, we notice $||\chi_{X_n}-\chi_{X}||_2^2 = \int(\chi_{X_n}-\chi_{X})^2 = \int\chi_{X_n \bigtriangleup X} = P(X_n \bigtriangleup X) \rightarrow 0$. Is that how on direction would go? Would the other direction be similar?
AI: Hint: $(\chi_A - \chi_B)^2 = \chi_{A \triangle B}$ |
H: Question having multiple answers?
i am studying for a test and i seem to of stumbled across a question which i found the answer to, but there seem to be others answers aswell. I am confused and would appreciate any help.
The question: The slope of a line is double of the slope of another line. If Tan. of the angle between them is 1/3, find the slopes of the line.
In order to solve this, i used a formule m2-m1/1+m1m2 = Tan Theta = 1/3
Solving the quadratic equation by taking m2 = 2m and m1 = m,i got 1 or 2, but the answer in my book is different and strange.
The answer: "1 and 2, or 1/2 and 1, or -1 and -2, or -1/2 and -1"
If possible, can anybody please help me how that answer was reached?:/
AI: HINT:
We have $$\left|\frac{m_2-m_1}{1+m_1m_2}\right|=\frac13$$
If $m_2=2m_1,$ we get $$\left|\frac{m_1}{1+2m_1^2}\right|=\frac13$$
$$\implies 1+2m_1^2=3|m_1|$$
Now for real $b,$ $$|b|=\begin{cases} +b &\mbox{if } b\ge0 \\
-b & \mbox{if } b<0 \end{cases} $$ |
H: solving without a calculator: 2-4*x*arctan(x) = 0.
I was doing some review problems and one of them is find the points of inflection of $\arctan (x)$. I found the second derivative, but was not able solve the problem. The part of the problem I could not solve is in the title.
$2 -4x\arctan (x) = 0$
I get $\frac{1}{2} = x\arctan(x)$
This is where I get lost.
AI: Given $f(x) = \arctan(x)$, we have that
$$f'(x) = \frac{1}{1 + x^2} = (1 + x^2)^{-1}$$
Hence the second derivative is $$f''(x) = -(1 + x^2)^{-2} \cdot 2x = \frac{-2x}{(1 + x^2)^2}$$
This is zero if and only if $x = 0$.
Actually solving $2x\arctan{x} = 1$ by hand is difficult, since the solution is not very pleasant. I doubt there's any meaningful closed form. |
H: How to prove that minimum of two exponential random variables is another exponential random variable?
How can I prove that the minimum of two exponential random variables is another exponential random variable, i.e. Z = min(X,Y)
AI: Note that you must assume that $X$ and $Y$ are independent,
otherwise the result is easily seen to be false.
There is a constant $\lambda$ such that
$P(X \geq t)=e^{-\lambda t}$ for every $t>0$.
There is a constant $\mu$ such that
$P(Y \geq t)=e^{-\mu t}$ for every $t>0$.
Then for every $t>0$ we have
$$
P(Z \geq t)=P(X\geq t,Y\geq t)=P(X\geq t)P(Y\geq t)=e^{-(\lambda+\mu)t}
$$
So $Z$ is an exponential random variable with
parameter $\lambda+\mu$. |
H: Proving Path-Connectedness
I am working on a homework question that gives $A \subseteq X$ and three points $x, y, z \in A$. Suppose also that there are continuous functions $f, g : [0,1] \to X$ with $f([0,1]) \subseteq A$, $f(0)=x$, $f(1)=y$ and $g([0,1]) \subseteq A$, $g(0)=y$, $g(1)=z$. I am asked to show that there is a function $h:[0,1] \to X$ with $h([0,1]) \subseteq A$, $h(0)=x$, and $h(1)=z$.
I think I have proved that $h(0)=x$ and $h(1)=z$, but I am not sure how to prove $h([0,1]) \subseteq A$. So far, I have attempted to have $h=f+g$ and use the triangle inequality to show that $d(x,z) < d(f(0),g(1))$ and say that this implies $h([0,1]) \subseteq A$. To be honest, I am not sure that this is the correct way to do this. Any help or advice would be appreciated.
AI: Imagine that the parameter in each of the paths measures time, so for instance, from time $t = 0$ to time $t = 1$, a point moves from $x$ to $y$ according to the function $f$, all the while staying within $A$.
The function $g$ parametrized a path from $y$ to $z$ during that same time interval. But what if we delayed the start of $g$ so that it ran from $t = 1$ to $t = 2$? Then, we would have a point moving from $x$ to $y$ during $[0, 1]$ and from $y$ to $z$ during $[1, 2]$. Here's a piecewise formula for such a path, where I have deliberately chosen to call the variable $s$:
$$
k(s) = \begin{cases}
f(s) & 0 \le s \le 1 \\
g(s - 1) & 1 \le s \le 2
\end{cases}
$$
This function almost satisfies all your criteria. Do you see why $k([0, 2]) \subseteq A$, for example? The one issue with $k$ is that although it gets the point from $x$ to $z$, it takes too long! There's a simple solution: run the filmstrip of $k$ at double speed.
Let $s = 2t$. The resulting function $h: [0, 1] \to A$ has formula
$$
h(t) = \begin{cases}
f(2t) & 0 \le t \le \frac{1}{2}\\
g(2t - 1) & \frac{1}{2} \le t \le 1
\end{cases}
$$
does the job nicely. |
H: Vertices of a median through R?
i am studying for a test and seem to have hit yet another obstacle. Hint/Answers are highly appreciated.
The question: The vertices of a Triangle PQR are P (2,1), Q(-2,3), R(4,5). Find equation of the median through the vertex R.
I don't know how to solve this problem. At first i thought about taking the intercept, but still ended up with with a extra unknown variable. In my book the answer is given as :"3x-4y+8=0".
Note: Please provide a answer in layman's terms, i only have a beginner level education on the subject.
AI: The equation you're looking for is the equation for the line through $R$ and through the midpoint betwenn $Q$ and $P$. The midpoint between $Q$ and $P$ is $$\frac 12 (Q+P) = (0,2).$$
So you're looking for the equation of a line through $(4,5)$ and $(0,2)$. You know that any line has the form $ax+by+c=0$ for some constants $a,b$. Plugging in $(4,5)$ and $(0,2)$ for $x,y$, we get two equations:
$$ 4a+5b+c=0$$ and $$2b + c= 0 $$
Now comes the point where we have only two equations, but three unknowns! The solution is that the equation is only unique up to multiplication by non-zero numbers. That is, if your equation is $ax+by+c=0$, then for $d \neq 0$, also $dax+dby+dc=0$ will also describe the same line.
Since this is not a line through the origin, you can assume that $c \neq 0$, so you can divide by $c$. That way, you get the equation $a^\prime x + b^\prime y + 1 = 0$, describing the same line (here $a^\prime = a/c$ and $b^\prime = b/c$). Now you have two unknowns and two equations! This can be solved! |
H: Let $N$ be a submodule of $R$-module $M$, $M/N$ is free $R$-module. Prove that $N$ is direct summand of $M$.
Let $N$ be a submodule of $R$-module $M$, $M/N$ is free $R$-module.
Prove that $N$ is direct summand of $M$.
Thanks for any insight.
AI: Hint: The following exact sequence splits (as $M/N $ is free), so $N$ is a direct summand of $M$,
$0\rightarrow N\rightarrow M\rightarrow M/N\rightarrow 0$ |
H: $(\sum_{n=1}^{100} n!)^7$ is of the form $7k+5$
Could any one just check for me that $(\sum_{n=1}^{100} n!)^7$ is of the form $7k+5$ or not? I got it that but they asked me to show it is of the form $7k+4$
Thank you!
AI: Observe that $\displaystyle r!\equiv0\pmod7 $ for $r\ge7$
$\displaystyle\implies \sum_{n=1}^{100} n!\equiv\sum_{n=1}^6 n!\pmod7 $
Now, using Wilson's Theorem, $\displaystyle6!\equiv-1\pmod 7$
As $\displaystyle 6\equiv-1\pmod7, $
$\displaystyle5!\equiv (-6)\cdot5!\pmod7\equiv-(6!)\equiv-(-1)\equiv1$
The rest terms are small enough to be handled directly |
H: Does there exist a sequence such that $\lim\limits_{n\to\infty}\frac{n(a_{n+1}-a_{n})+1}{a_{n}}=0$?
Question:
Does there exist a positive sequence $\{a_{n}\}$ such
$$\lim_{n\to\infty}\dfrac{n(a_{n+1}-a_{n})+1}{a_{n}}=0?$$
If it exists, can you make an example? if not, why not?
My try: we consider this sequence
$$a_{n}=\dfrac{1}{n}$$
then the
$$\lim_{n\to\infty}\dfrac{n(a_{n+1}-a_{n})+1}{a_{n}}=\lim_{n\to\infty}\dfrac{\dfrac{n}{n(n+1)}+1}{\dfrac{1}{n}}\to+\infty$$
But I can't take a example such condition? Thank you
AI: Take $a_n=\ln(n)$:
$$\lim_{n\to\infty}\frac {n\cdot\ln\left(\frac{n+1}{n}\right)+1}{\ln n}$$
Can you take it from here?
EDIT: the intuition behind taking a logarithmic function is that you need a slowly growing function $f$ such that the difference between the values of $f$ for consecutive integers is small, so it will cancel the linear factor, but as $n$ goes to infinity $f(n)$ goes to infinity as well.
Notice that taking $a_n =\sqrt{n}$ will yield limit of a constant, so we do need a "slower" growingfunction. |
H: Limit $\lim_{n\to \infty} \frac{1}{2n} \log{2n\choose n}$
$\lim_{n\to \infty} \frac{1}{2n} \log{2n\choose n}$
I could not approach it beyond these simple steps,
$\lim_{n\to \infty} \frac{1}{2n} \log(\frac{2n!}{(n!)^2})$
$=\lim_{n\to \infty} \frac{1}{2n} [\log(2n)+\cdots +\log(n+1)-\log(n)-\cdots-\log1]$
$=\lim_{n\to \infty} (\log(2n)^{1/2n}+\cdots+\log(n+1)^{1/2n}-\log(n)^{1/2n}-\cdots-\log1^{1/2n})$
Now,I understand that I have to create a sum of limit and produce an integration or use the formula $\lim_{n\to \infty} \log(1+\frac1x)^x=e$ but I cannot do it. Please help!
AI: HINT:
As $\displaystyle \binom{2n}n=\frac{(2n)!}{n! n!}=\prod_{1\le r\le n}\frac{n+r}r$
$$\ln\binom{2n}n=\sum_{1\le r\le n}\ln\left(\frac{n+r}r\right)=\sum_{1\le r\le n}\ln\left(\frac{1+\frac rn}{\frac rn}\right)$$
As
$$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
$$\lim_{n\to\infty}\frac1n\ln\binom{2n}n=\int_0^1\ln\left(\frac{1+x}x\right)dx$$ |
H: What is the name of this factor-algebra?
In the polynomial algebra $k[x_1,x_2,\ldots, x_n]$ consider an ideal $I$ generated by the polynomials of the form $x_i^k-x_i$, $i=1 \ldots n$ and $k=2,3,\ldots.$
Consider the quotient algebra $A=k[x_1,x_2,\ldots, x_n]/I$.
What is the name of the algebra $A$? Any reference?
AI: It suffices to mod $x_i^2-x_i = x_i(x_i-1)$. We have $k[x_i]/(x_i(x_i-1)) \cong k \times k$ by the Chinese Remainder Theorem. It follows that $A \cong \otimes_{i=1}^{n} (k \times k) \cong k^{2^n}$ as $k$-algebras. It is the universal $k$-algebra generated by $n$ idempotent elements. I don't know a special "name". |
H: A polynomial is zero if it zero on infinite subsets
Let $f(t_1, ... , t_n)$ be a polynomial over a field $F$. Suppose there exist infinite subsets $X_1, ... , X_n$ of $F$ such that $f(x_1, ... , x_n) = 0$ for all $(x_1, ... , x_n) ∈ X_1× \cdots ×X_n$. Prove that $f$ is the zero polynomial.
Not sure how to start on this one!
AI: Hints
Use the fact that a non-zero polynomial in a single variable over a field has only finitely many zeros.
Induction on $n$. |
H: Spivak Calculus chapter 1, problem 1v
this is my first question here. I am self-studying Spivak's Calculus Fourth Edition and am stuck on Chapter 1 Problem 1v.
The question is to prove that: $x^n - y^n = (x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$
So I expand the right side out to:
$x(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})-[y(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})]$
Which led me to:
$x^n + x^{n-1}y+...+x^2y^{n-2}+xy^{n-1}-(x^{n-1}y+x^{n-2}y^2+...+xy^{n-1}+y^n)$
I can now cancel out so that I end up with:
$x^n+x^2y^{n-2}-x^{n-2}y^2-y^n$
So the question is, how do I get rid of those two remaining components? I'm assuming it had something to do with the expansion '...' but am not sure how to go about it.
Any help is greatly appreciated!
AI: Writing Prahlad's answer in similar notation to your question, with a few more terms explicit,
$x(x^{n-1}+x^{n-2}y+x^{n-3}y^2+...+x^2y^{n-3}+xy^{n-2}+y^{n-1})-[y(x^{n-1}+x^{n-2}y+x^{n-3}y^2+...+x^2y^{n-3}+xy^{n-2}+y^{n-1})]$
Which leads to:
$x^n + x^{n-1}y+x^{n-2}y^2...+x^2y^{n-2}+xy^{n-1}-(x^{n-1}y+x^{n-2}y^2+...+x^2y^{n-2}+xy^{n-1}+y^n)$
You can now cancel and end up with:
$x^n-y^n.$ |
H: Ideals in a real/complex number field?
Considering a real or complex number field (with traditional addition and multiplication) I see no ideals besides $\mathbb{R}$ and $\{ 0\}$ or $\mathbb{C}$ and $\{ 0 + 0i\}$.
Quick web search gave no satisfactory results. Yet I believe this couldn't be considered trivial, could it?
I am contemplating quite a few ideas for example looking for possible kernels of homomorphisms. Maybe you could give some references? Ideally I'd love to figure out even multiple different proofs for this.
AI: Fields only have the two obvious ideals. That is because in a field, every non-zero element is invertible, and invertible elements generate the entire ring as ideals. |
H: The trapezoid rule and the integral: $\displaystyle\int_1^4(x-1)(x-4)\,\mathrm dx$
I've tried this many times, and the result I get is always $-4$, I don't know why..
I first get $h$, $ (4-1)/6 $ , then I put $h/2(x)$, $x = 2(-2) + 2(-2)$, which is equal to $1/2(-8) = -4$, what am I doing wrong?
Thanks in advance!
$\displaystyle\int_1^4(x-1)(x-4)\,\mathrm dx\qquad$Use $6$ strips.$\qquad$Area $\approx\color{grey}{\boxed{\color{black}{-4}\,}}$ sq. un.
AI: Since you found that $h=\dfrac{4-1}{6} = \dfrac{1}{2}$, then for $\color{blue}{n=6}$ strips, the trapezoid rule tells you that
$$A \approx \frac{1}{4}\left[f(1) + 2f\left(\frac{3}{2}\right) + 2f(2) + 2f\left(\frac{5}{2}\right) + 2f(3) + 2f\left(\frac{7}{2}\right) + f(4)\right]$$
where $f(x)=(x-1)(x-4)$. Evaluating this gives you
$$\begin{aligned} A &\approx \frac{1}{4}\left[0+ 2\left(\frac{1}{2}\right)\left(-\frac{5}{2}\right) + 2(1)(-2) + 2\left(\frac{3}{2}\right)\left(-\frac{3}{2}\right)+2(2)(-1) + 2\left(\frac{5}{2}\right)\left(-\frac{1}{2}\right)+0\right] \\ &= \ldots\end{aligned}$$
Can you take things from here? |
H: There exist $n \in \mathbb{N}$ such that $\mathrm{Im} f^{n} + \mathrm{Ker} f^{n} = M$ if $M$ satisfies DCC
Let $M$ be an $R$-module, and let $f$ be an $R$-automorphism on $M$. Prove that if $M$ satisfies the descending chain condition, then there exist $n \in \mathbb{N}$ such that $\operatorname{Im} f^{n} + \operatorname{Ker} f^{n} = M$.
AI: Note that $ \operatorname{im}(f) \supseteq \operatorname{im}(f^2) \supseteq \operatorname{im}(f^3) \supseteq \cdots$ so by the DCC, there is an integer $n$ such that $\operatorname{im}(f^n)=\operatorname{im}(f^{2n}).$ Let $m\in M.$ Since $\operatorname{im}(f^n)=\operatorname{im}(f^{2n}),$ there is some $t\in M$ such that $f^n(m)=f^{2n}(t).$ Write $m=(m-f^n(t)) + f^n(t)$ and we're done. |
H: Expectation of a squared random variable and of its absolute value
Is it true that if $\mathbb{E}[X^2]<\infty$ then $\mathbb{E}[|X|]<\infty$? If so, why?
AI: The most direct way to see this, without referring to a single theorem, is to consider the set $E=\{|X| \leq 1\}$. Then if $1_E$ is the indicator function of $E$,
$$\mathbb{E}[|X|] = \mathbb{E}[|X|\cdot 1_E] + \mathbb{E}[|X| \cdot 1_{E^c}] \leq \mathbb{E}[1 \cdot1_E] + \mathbb{E}[X^2 \cdot 1_{E^c}] \leq1+\mathbb{E}[X^2] < \infty.$$
The more clever arguments using Jenson's or Holder's inequalities provide a much sharper bound, however. |
H: How to calculate percentage for two different values?
I have two data one is like$=75$ and second data is dislike$=127$, both values are counted separately. But i need to show percentage of this.
For example $40\%$ likes and $60\%$ dislikes for the above something like.
I know to calculate percentage for single value $(\%/100)*x$ but for the above i don't have idea.
Can anyone help me on this?
AI: Likes: $\quad\dfrac{\text{total number of likes}}{\text{total number of those answering}}\;=\;\dfrac{75}{75+127}\times 100\% \approx 37.13\%$.
Dislikes: $\dfrac{\text{total number of dislikes}}{\text{total number of those answering}} \;=\;\dfrac{127}{75 + 127} \times 100\%\approx 62.87\%$ |
H: Suppose $a_n+b_n$ converges. Does $a_n*b_n$ converges also?
$a_n, b_n$ - sequences
Suppose $a_n+b_n$ converges. Does $a_n b_n$ converge also?
I tried thinking if I can learn something about $a_n$ and $b_n$ by the assumption $a_n b_n$ converges.
I also tried to develop this equation $|a_n b_n - L| < \epsilon$ assuming it is converging.
I didn't get any bright conclusions.
Will be glad help.
AI: $a_n =n, b_n =-n$
$a_n+b_n=0$ but then...
$a_nb_n=-n^2\rightarrow -\infty$ |
H: Suppose $\mu(X)<\infty$. Show that if $\ f\in\ L^q(X,\mu)$ for some $1\le q<\infty$ then $\ f\in\ L^r(X,\mu)$ for $1\le r
Suppose $\mu(X)<\infty$. Show that if $\ f\in\ L^q(X,\mu)$ for some $1\le q<\infty$ then $\ f\in\ L^r(X,\mu)$ for $1\le r<q$
In the solution the assistant wrote this:
If $1\le r<q$ then $q/r\ge 1$. Let p be the dual exponent to $q/r$ (i.e. $\frac{1}{q/r}+\frac{1}{p}=1)$
then by Hölder Inequality
$\int_{X}|f|^rd\mu\le\underbrace{(\int_{X}|f|^qd\mu)^{r/q}}_{??????}(\int_X1^pd\mu)^{1/p}<\infty$
for me the middle part doesn't make sense (it is not the hölder inequality) can somebody show it ?
instead of this can't we say (only in this case) that
$\int_{X}|f|^rd\mu\le (\int_{X}|f|d\mu)^r$ then
$(\int_{X}|f|d\mu)^r\le ((\int_{X}|f|^qd\mu)^{1/q}(\int_X1^pd\mu)^{1/p})^r<\infty$ ?
AI: A direct approach: since $r\leqslant q$, $|f|^r\leqslant|f|^q+1$ pointwise, and $1$ is integrable because $\mu$ has finite mass.
Note: The proof in your post is Hölder's inequality, applied to the functions $u=|f|^r$ and $v=1$, and to the conjugate exponents $q/r$ and $p$. Then $|u|^{q/r}=|f|^q$ and $|v|^p=1$. |
H: Characterizations of convexity relying only on gradient
Suppose $f:\mathbb{R}^n\rightarrow \mathbb{R}$ is once (and only once) continuously differentiable.
Are there any characterizations of convexity that rely only on the gradient $\nabla f$?
In the one-dimensional case this would be that $f'$ is non-decreasing.
AI: For $a\neq b$ consider $g_{a,b}(t) = f((1-t)a + tb)$. You have
$$g_{a,b}'(t) = \langle\nabla f((1-t)a+tb),b-a\rangle.$$
If $f$ is convex, so is $g_{a,b}$ (and $f$ is convex if $g_{a,b}$ is for all $a,b$), and from the monotonicity of $g_{a,b}'$ we obtain
$$\langle \nabla f(b) - \nabla f(a), b-a\rangle \geqslant 0.\tag{1}$$
$(1)$ holds for all $a,b$, and that means that $\nabla f$ is monotonic. Convince yourself that this is a necessary and sufficient condition for the convexity of $f$. |
H: Integrability of functions
Briefly justify the following facts:
a) $|x|$ is integrable on $[-1,2]$
b) $x^{\frac{1}{4}}$ is integrable on $[0,9] $.
c) The function $h(x)=\begin{cases}
x^2& x\in[0,1] \\
2x+3 &x\in(1,2]
\end{cases} \quad$
is integrable on $[0,3]$.
d) if $f, g$ are integrable on $[a,b]$ the so is $f-g$.
I would really appreciated if some one can give me at least directions how to to this? Thank you so much! And i am not sure should I just use theorems or I need to prove those things..
AI: (a), (b), and (c) are just direct applications of the fact that continuous functions are integrable ((c) is piecewise continuous, from which we split the integral; more generally, there is a theorem that states that if the set of discontinuities of the function has measure zero and $f$ is bounded over a compact subset, then it is integrable over that subset). Also, in (c) there is the issue of the fact that the function is not defined over $(2,3]$, and thus the integral is not even well-defined.
(d) is true because of linearity; $\int_a^b{(f-g)d x}=\int_a^b{fdx} -\int_a^b{gdx}$. Thus, since $\int_a^b{fdx}$ and $\int_a^b{gdx}$ are assumed to exist, $\int_a^b{(f-g)dx}$ exists and is equal to their difference.
Let me know if you want me to show the continuity/piece-wise continuity of the functions in (a), (b), (c).
Here is proof that (a) is continuous. We write this as the piece-wise function $$|x|=\begin{cases} x & x\in (0,\infty) \\ -x & x\in(-\infty,0] \end{cases}.$$ On the interval $(0,\infty)$, it is clear that $|x|$ is continuous because $x$ is continuous (for each $\epsilon>0$ we choose $\delta=\epsilon$ to so that $|x-y|<\delta|$ implies $|x-y|<\epsilon$ for every $x\in (0,\infty)$.) Similarly, on $(-\infty,0)$ it is continuous because $-x$ is continuous on this domain (proof similar to above). Thus, it remains to see that $|x|$ is continuous at $x=0$. It suffices to show that the limit from the right and left is equal $0$. Since $\lim_{x\rightarrow 0^-}{-x}=-\lim_{x\rightarrow 0^-}=-0=0=\lim_{x\rightarrow 0^+}{x}$, we see that this is the case. Thus, $|x|$ is continuous.
(b) is a rather ugly computation.
(c) is not continuous at $x=1$ because the left-hand limit is 1 and the right-hand limit is 5. The integral is not defined, however, because we can't integrate over the region $(2,3]$ since $h$ isn't defined here. |
H: Diophantine Equation.
How many solutions are there in $\mathbb{N} \times \mathbb{N}$ to the equation $\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{1995}?$
How would you solve this? I have tried but am not sure how I should proceed with this.
AI: HINT:
$$\frac1x+\frac1y=\frac1m\iff (m-x)(m-y)=m^2$$ |
H: Normalizing data to a given value.
I would like compare the data by normalizing it to a given value.
for example:
[20,30,40,50,60,70]
How do I normalize the given set of elements to its first value, that is, 20.
Please also let me know if did not understand it correctly.
I understand that I need to take the mean and the standard deviation; then subtract mean, then divide by the standard deviation.
Is that correct?
AI: Normalizing as in your third paragraph is not tied to a particular value. It uses the whole data set and changes it to one with zero mean and unit standard deviation. Do you understand how to calculate the mean and standard deviation of your data?
There are various normalizations that you could use that do depend on the first value. You could just subtract the first value from each data point. That will make the first zero and show you how much each data point differs from the first. You could divide by the first point instead (assuming it is not zero). Unless your data is ordered in some way, it seems strange to single out one point for special treatment. |
H: Evaluating $\lim\limits_{x\to \infty} \sqrt{x^2-3x+5} - \sqrt{x^2+2x+1}$
I cannot solve the following limit with radicals:
$$\lim_{x\to \infty} \sqrt{x^2-3x+5} - \sqrt{x^2+2x+1}. $$
AI: Multiply by $$\dfrac{\sqrt{x^2 - 3x + 5} + \sqrt{x^2 + 2x + 1}}{\sqrt{x^2 - 3x + 5} + \sqrt{x^2 + 2x + 1}}$$
This gives us:
$$\lim_{x\to \infty} \Big(\sqrt{x^2-3x+5} - \sqrt{x^2+2x+1}\Big)\cdot \dfrac{\sqrt{x^2 - 3x + 5} + \sqrt{x^2 + 2x + 1}}{\sqrt{x^2 - 3x + 5} + \sqrt{x^2 + 2x + 1}} $$
$$ = \lim_{x\to \infty} \dfrac{(x^2 - 3x + 5) - (x^2 + 2x + 1)}{\sqrt{x^2 - 3x + 5} + \sqrt{x^2 + 2x + 1}}$$
$$ = \lim_{x\to \infty} \dfrac{-5x + 4}{\sqrt{x^2 - 3x + 5} + \sqrt{x^2 + 2x + 1}}$$
Now, try to evaluate.
You should obtain a limit of $$\dfrac{-5}{2}$$ |
H: How find this $a_{n}$ such $\lim_{n\to+\infty}\left(\dfrac{a_{n+1}+1}{a_{n}}\right)^n=e$
if postive sequence $\{a_{n}\}$ such
$$\lim_{n\to+\infty}\left(\dfrac{a_{n+1}+1}{a_{n}}\right)^n=e$$
find a example the $a_{n}$ such this condition?
This problem is from analysis book : show that
$$\lim_{n\to\infty}\left(\dfrac{a_{n+1}+1}{a_{n}}\right)^n\ge e$$
and I can't prove this.But I can't find this inequality when :$=?$
AI: Since the limit when $a_n=n^{c}$ with $c\gt1$ is $\mathrm e^c$, the first idea that comes to mind is to try $a_n=n\log n$. Behold! This works.
Proof: $a_{n+1}=(n+1)\log n+O(1)$ hence $\frac{a_{n+1}+1}{a_n}=1+\frac1n+o\left(\frac1{n}\right)$. |
H: Let $G$ be a finite group and $\phi:G \to K$ be a surjective homomorphism and $n \in \mathbb{N}. $ If $K$ has an element of order $n,$ so does $G.$
Let $G$ be a finite group$\ ,\phi:G \to K$ be a surjective homomorphism and $n \in \mathbb{N}. $ If $K$ has an element of order $n,$ so does $G.$
May I know if my proof is correct? Thank you.
Proof: Let $w \in K$ such that $o(w)=n.$
Also, $\ \exists g \in G$ such that $w = \phi(g), $whence $w^n = \phi(g^n)=e_K.$
Since $|G| < \infty, o(g^n) \leq |G|.$ Let $m = \min\{n' \in \mathbb{N}:n' \leq |G| \ \wedge g^{nn'} = e_G\}.$
Given any $s \in \mathbb{N}$ such that $g^s=e_G, $ then $\phi(g^s)=w^s=e_K, $whence $n|s.$
Since $(g^n)^{\frac{s}{n}} = e_G$ and $m$ is the minimum element, $\frac{s}{n} \geq m, $ whence $s \geq mn$
Hence, $o(g) =mn$ and $o(g^m)=n.$
AI: The proof is OK. But you might use the well known and generally useful fact that a cyclic group of order $m$ has a cyclic subgroup (and hence elements) of any order $d$ dividing $m$, to reduce this to a one-liner. With $o(w)=n$, $\phi(g)=w$ and $m=o(g)$ one has $e_K=\phi(e_G)=\phi(g^m)=w^m$, so $n$ divides $m$, and the cyclic subgroup $\langle g\rangle$ of $G$, which has order $m$, contains an element of order $n$. |
H: Ideals generated by a set in a ring without multiplicative unity
It is well known that in a ring $R$ with $1\neq 0$, the ideal generated by $S$ is $$I(S)=\{a_1 s_1 b_1+a_2 s_2 b_2+\dots+a_n s_n b_n: n\in\mathbb{N}, a_i, b_i\in R, s_i\in S\}.$$
Is there a similar expression for the ideal generated by a set in a ring without $1$?
AI: For $A,B\subseteq R$ define $AB$ as the set consisting of products of elements in $A$ and $B$ and sums thereof. Also let $\mathbb{Z}S$ denote the set of elements of the form $ns$ with $n\in\mathbb{Z}$ and $s\in S$ and sums thereof.
Then, for a ring without unit,
$$I(S)=\mathbb{Z}S+RS+SR+RSR$$
It's clear that this set is an ideal. Moreover it is clear that any ideal containing $S$ must contain this set.
Note that if a unit exists, then $\mathbb{Z}S\subseteq RS$; moreover $RS\subseteq RSR$ and $SR\subseteq RSR$. So $\mathbb{Z}S+RS+SR+RSR\subseteq RSR$ and $I(S)=RSR$. |
H: Definition of the probability distribution of a random variable?
I have a vocabulary problem. I understand that a "probability distribution" is a function from a sigma algebra to the reals. But then what is a "probability distribution" in "the probability distribution of a random variable?"
AI: A random variable is a measurable function $X:(\Omega,\mathcal F,P)\to(S,\mathcal S)$ between a probability space $(\Omega,\mathcal F,P)$ and a measurable space $(S,\mathcal S)$. The probability distribution of $X$ is the unique measure $\mu$ on $\mathcal S$ such that $$\mu(B)=P(X^{-1}(B)),
$$ for every $B$ in $\mathcal S$. The measurability of the function $X$ is exactly the condition one needs to ensure that this formula makes sense. |
H: write the trasition map $\phi$ between $\sigma_1$ and $\sigma_2$. Verify $\det( J(\phi))$ and find $T_p(S)$.
Sphere $$S=\{(x,y,z) \mid x^2+y^2+z^2=R^2\}$$
$$
\sigma_1(u,v)=(u,v, \sqrt{1-u^2-v^2}) \\
\sigma_2(u,v)=(\tilde u, \sqrt{1-\tilde{u}^2 -\tilde{v}^2}, \tilde v)
$$
I guess $\{\sigma_1, \sigma_2\}$ forms an atlas for $S$.
Question: write the trasition map $\phi$ between $\sigma_1$ and $\sigma_2$. Verify $\det( J(\phi))\gt 0$ and find $T_p(S)$.
AI: $\sigma_1$ and $\sigma_2$ do not form an atlas. (The lower right portion of the sphere has no coordinate chart, for instance). But they're part of an atlas.
The first question is "What's the domain for the transition function? In other words, for what pairs $(u, v)$ are $\bar{u}$ and $\bar{v}$ defined? {I'm using bars because I don't know how to make tildes over the symbols). Well, it sure looks as if the second coordinate needs to be positive, because it's the result of taking a square root. So we know $v > 0$. (In your drawing, it looks as if you've got the positive-$y$ coordinate increasing to the LEFT, which is a bit weird, but I can work with that). So let's take a point $(u, v)$ with $u^2 + v^2 < 1$ and $v > 0$, and apply $\sigma_1$ to it. We get
$ \sigma_1(u, v) = (u, v, \sqrt{1 - u^2 - v^2} )$.
Now we want to apply $\sigma_2^{-1}$ to that, i.e., we want to find $(\bar{u}, \bar{v})$ such that $\sigma_2(\bar{u}, \bar{v})$ is the same as the displayed expression above. Well,
$\sigma_2(\bar{u}, \bar{v}) = (\bar{u}, \sqrt{1 - \bar{u}^2 - \bar{v}^2}, \bar{v})$
and for this to be equal to the previous expression requires that they be term-by term equal, i.e., that
$\bar{u} = u$
$\sqrt{1 - \bar{u}^2 - \bar{v}^2} = v$, and
$ \bar{v} = \sqrt{1 - u^2 - v^2}$.
Working with just the first two, we get
$\bar{u} = u$ and
$\sqrt{1 - \bar{u}^2 - \bar{v}^2} = \sqrt{1 - u^2 - \bar{v}^2} = v$
which we can solve to get
$1 - u^2 - \bar{v}^2 = v^2$, or
$\bar{v}^2 = 1 - v^2 - u^2$, or
$\bar{v} = \sqrt{1 - v^2 - u^2}$.
Why did I take the positive square root in that last formula? Because on the domain where the two coordinate charts overlap, we also have $\bar{v} > 0$.
OK: so if you know $u$ and $v$, then the formulas for $\bar{u}$ and $\bar{v}$ are just
$(\bar{u}, \bar{v}) = (u, \sqrt{1 - v^2 - u^2})$
and that's your transition function:
$\phi(u, v) = (u, \sqrt{1 - v^2 - u^2})$.
(It's possible that your text defines the transition to go in the other direction, but with this example in front of you, you should be able to do that as well).
Computing the $2 x 2$ derivative matrix for the function $\phi$ should be just an exercise in calculus for you. And showing the det of that matrix is positive...well, it'll be straightforward, I promise.
To describe $T_P(S)$, we need to specify the point $P$. I'm going to start with some $(u, v)$-pair satisfying $u^2 + v^2 < 1$ (i.e., in the domain of $\sigma_1$, which I'll call $U$), and say that $P = \sigma_1(u, v)$.
Next, I'm going to observe that for any smooth curve $\gamma : [-a, a] \rightarrow S$
with $\gamma(0) = P$, there's a number $0 < b \le a$ and a curve $\beta : [-b, b] \rightarrow U$ such that $\gamma(t) = \sigma_1(b(t))$ for $t \in (-b, b)$. Why? Let $Q = \sigma_1(U)$. That's an open set on $S$. The preimage $\gamma^{-1}(Q)$ is therefore an open set in the reals, and it contains $0$. It therefore contains some open interval $(c, d)$ around $0$. Let $b = \min(|c|, |d|)$ and you've got what you need: the image of $\gamma$ on the interval $(-b, b)$ lies entirely in the image of $\sigma_1$. $\sigma_1$ is 1-1 onto its image. So for any $t \in (-b, b)$, there point $\gamma(t)$ in the image of $\sigma_1$ corresponds to a unique point $(u, v)$ in $U$. Call this point $\beta(t)$. Whew!
(This, by the way, is why we often work with coordinate charts, which go from $S$ to ${\mathbf R}^2$, rather than parametrizations, like your $\sigma_1$ and $\sigma_2$. If we had a coordinate chart, I'd just say $\beta = chart \circ \gamma$.
Anyhow, the point is that every smooth curve on $S$ passing through $P$ corresponds to a smooth curve in $\mathbf{R}^2$ passing through the point $(u, v)$. Let's look at just two of those:
$\beta_1(t) = (u + t, v)$
$\beta_2(t) = (u, v+t)$
When we map these to $S$ by applying $\sigma_1$, we get
$\gamma_1(t) = \sigma_1(\beta_1(t)) = (u+t, v, \sqrt{1 - (u+t)^2 - v^2})$ and
$\gamma_2(t) = \sigma_1(\beta_2(t)) = (u, v+t, \sqrt{1 - u^2 - (v+t)^2})$.
Let's compute their derivatives:
$\gamma_1'(t) = (1, 0, \frac{-1}{\sqrt{1 - (u+t)^2 - v^2}} (u+t))$ and
$\gamma_2'(t) = (0, 1, \frac{-1}{\sqrt{1 - u^2 - (v+t)^2}} (v+t))$.
These derivatives, at $t = 0$, give two tangent vectors:
$\gamma_1'(0) = (1, 0, \frac{-u}{\sqrt{1 - (u+t)^2 - v^2}})$ and
$\gamma_2'(t) = (0, 1, \frac{-v}{\sqrt{1 - u^2 - (v+t)^2}})$.
Your tangent plane must contain both of these. So you need an equation of the form
$ Ax + By + Cz = 0$
that they both satisfy. Here's one:
$A = \frac{-u}{\sqrt{1 - (u+t)^2 - v^2}}$
$B = \frac{-v}{\sqrt{1 - (u+t)^2 - v^2}}$
$C = 1$.
If you correctly computed the Jacobian, then these two tangent vectors should have been the first and second columns of your Jacobian matrix, and the coefficient vector $[A, B, C]$ for the plane turns out to be just a multiple of the cross-product of those two columns. This works in general for surfaces, but for higher dimensions you need a generalized cross-product, and besides, it only works in the case where your $n$-manifold is embedded or immersed in dimension $n+1$. |
H: Evaluate the following using Simpson's rule
I was asked to evaluate the following using Simpson's rule (by using 2 strips). Below is the function and my answer to it. What am I doing wrong?
$$\begin{align}
\color{red}{\int_1^{1.6}\dfrac{\sin2t}{t}\,\mathrm dt} & \color{blue}{=\int_1^{1.6}\dfrac{\cos2t^2-\sin2t}{t^2}=\dfrac{0.6}2\big[f(1)+4(1.3)+f(3)\big]} \\\,\\
&\color{blue}{=\big[-1.32-5.1+0.6\big]}
\end{align}$$
When I plugged in the values I put my calculator in Radian mode but the answer I get is negative. The correct answer is 0.2460
Thank you.
AI: Your formula is wrong. With $f(x) = \sin(2x)/x\;$ you get using the simple Simpson rule
$$\frac{0.6}{6}\left(f(1.0) + 4f(1.3)+f(1.6)\right)\approx 0.245897.$$
and if you take two strips, you get
$$\frac{0.6}{6}\left(f(1.0) + 4 f(1.15)+ 2 f(1.3)+ 4 f(1.45)+f(1.6)\right) \approx 0.245982.$$
What is purpose of the $\cos\;$ expression? |
H: Evaluate $\lim\limits_{n \to\infty} \left(\frac{n-1} {2n+2}\right)^n$
What is the easiest way to evaluate this limit?
$\displaystyle{\lim_{n \to\infty} \left(n-1 \over 2n+2\right)^n}$
$$
\text{Is this possible ?$\,$:}\quad
\lim_{n \to\infty}\left(n/n - 1/n \over 2n/n + 2/n\right)^n
=
\lim_{n \to\infty}\left(1 - 1/n \over 2 + 2/n\right)^{n}
=
\lim_{n \to\infty} \left(1 \over 2\right)^{n} = 0
$$
AI: For simple limits like this, the easiest way is often to sandwich the sequence between simpler sequences for which the limit is known.
Notice that $n-1 \leq n$ and $2n+2 \geq 2n$ entail
$$
\frac{n-1}{2n+2} \leq \frac{1}{2}
$$
so that for $n \geq 1$ one has
$$0 \leq \left(\frac{n-1}{2n+2}\right)^n \leq \left(\frac{1}{2}\right)^n \xrightarrow[n\to\infty]{} 0.$$
Conclude with the squeeze theorem.
Remark. This method gives more than just the limit: you get a good estimate of how quickly the sequence converges. For example, it is now easy to give a $n$ such that $\left(\frac{n-1}{2n+2}\right)^n \leq 10^{-100}$. |
H: References on Ring and Module Theory
Next semester I will take a course about Ring and Module Theory. Can anyone tell me the best texbooks and problems books about it. I only know some famous problems books such as Exercises in Classical Ring Theory, Exercises in Modules and Rings, Exercises in Basic Ring Theory.
AI: Joseph Rotman - An Introduction to Homological Algebra (2nd
edition)
is a great book about homological algebra, but it contains many sections about modules and rings. I advise you to give it a look since it is a wonderfully written book! |
H: Converting CFG to CNF
I have the following problem of converting CFG to CNF:
$$
\begin{aligned}
S \Rightarrow\,& bA \mid aB\\
A \Rightarrow\,& bAA \mid as \mid a\\
B \Rightarrow\,& BB\mid bs\mid b
\end{aligned}
$$
I know that Chomsky normal form only has productions of the type $A\Rightarrow BC$ and $A \Rightarrow a$.
In the case of $A\Rightarrow bAA$, can I replace $b$ by $C_1$, i.e., $A\Rightarrow C_1AA$ and $C_1\Rightarrow b$?
Is this the right way to do it? Can some one guide me?
AI: Hint:
Suppose we have $N \to PQR$. We replace it by two productions $N \to N_1R$ and $N_1 \to PQ$. If $N \to ab$, we take $N \to P_1P_2$, $P_1 \to a$ and $P_2 \to b$.
For example, let's consider $A \to bAA$ (it is, obviously, not in CNF). You could replace it by productions $A \to A_1A_2$, $A_1 \to b$, $A_2 \to AA$ (all in CNF), or you could use another set of productions. |
H: If $P$ is a prime ideal of $R$, $\sqrt{P^{n}}=P\ \forall n\in\mathbb{N}$?
Let $R$ be a commutative ring. If $P$ is a prime ideal of $R$, $\sqrt{P^{n}}=P\ \forall n\in\mathbb{N}$?
AI: Yes that is true. Note that $P^n\subseteq P$ and taking radicals of both sides gives the $\subseteq$ direction of the equality. So we are left to prove that $P\subseteq \sqrt{P^n}.$ Suppose $x\in P.$ Then $x^n\in P^n$ and by definition of the radical, this implies that $x\in \sqrt{P^n},$ which solves the problem. |
H: Isomorphism of rings under different operations
i am new to this site. i have been reading through its posts and question and they are really amazing. however, i found a link to a question asked about one year ago and a question i don't know how to tackle came to my mind. this is the question.
Let R be a ring with a 1. Define R¯ to have the same elements of R with addition ⊕:a⊕b=a+b+1 and multiplication ⊗:a⊗b=ab+a+b Prove that R¯ is isomorphic to R.
how do i go about answering this, because clearly, there is no direct way of proving the conditions for an isomorphism, that is:
1. f(x + y) = f(x) + f(y)
2. f(xy) = f(x)f(y)
i thought about trying to find the inverse of the definition of ⊕ but i couldnt figure out exactly how to get it.
by the way, i have confirmed that both R and R¯ are indeed rings. if i am repeating a certain question or similar question please just provide the link, i guess it might suffice. but so far i haven't found anything close. i have only found trivial examples of isomorphisms.
thanks in advance
AI: Note that the additive identity of $R^-$ is the element that $R$ calls $-1$ and the multiplicative identity is the element that $R$ calls $0$. This suggests that
$$
x \mapsto x-1
$$
will give the desired ring homomorphism $R \to R^-$; this map is clearly bijective.
We check:
$$
x + y \mapsto x + y - 1 = (x-1)\oplus(y-1)
$$
$$
xy \mapsto xy - 1 = (x - 1)(y -1 ) + (x - 1) + (y - 1) = (x-1)\otimes (y-1)
$$ |
H: Measurable with Respect to a Complete Space
Let $f:(X, A,\mu) \to [0,\infty]$ have a Lebesgue integral. Show that $f$ is measurable with respect to the completion of the sigma algebra $A$.
I know that I have to show that the pre-image of every subset of $X$ is contained in the completion of $A$, but I'm having trouble figuring out how that follows from the function being Lebesgue integrable. Here's the solution that I've put together:
Suppose $f:(X,A,\mu) \to [0,\infty]$ has a Lebesgue integral. Fix an integrable set E. Then there exists an increasing sequence of simple functions $u_n$ with $u_n\leq f$ and $u_n\to u$. Moreover, $\int_Efd\mu=\lim\int_E u_nd\mu=\int_E\lim u_nd\mu=\int_Eud\mu
\implies\int_E(f-u)d\mu$.
So let g=f-u. Take $E_+=\{x\in E:g(x)>0\}$ and $E_-=\{x\in E:g(x)<0\}$. If m($E_+$)>$0$, then there is some closed set K$\subset E_+$ with m(K)>$0$. But
$\int_Kgd\mu=\int_Egd\mu-\int_{E-K}gd\mu=0-\sum_{n=1}^{\infty}\int_{E_n-K}gd\mu=0$
So it follows that g=$0$ almost everywhere on K, contradicting m(K)>$0$. Thus, m($E_+$)=$0$. Similarly, m($E_-$)=$0$.
$\implies g=0$ almost everywhere $\implies f=u$ almost everywhere. But $u_n$ are measurable by design and therefore u is measurable. Since f and u differ only on sets of measure $0$, this means that f is measurable not with respect to A, but with respect to the completion of A with respect to $\mu$.
AI: Here's a little something to get you started: There are simple functions $u_n\le f$ and $v_n\ge f$ so that $\int(v_n-u_n)\,d\mu\to0$. Replacing $u_n$ by $u_1\vee u_2\vee\ldots\vee u_n$ and $v_n$ by $v_1\wedge v_w\wedge\ldots\wedge v_n$, you may assume that $u_n$ form an increasing sequence, and $v_n$ a decreasing one. So they both converge pointwise, to limits $u$ and $v$ with $u\le f\le v$. $u$ and $v$ are both measurable (why?), and $u=f=v$ a.e. (again, why?). |
H: Finding the volume between two concentrical hemispheres
I have a sphere with the equation $x^2 + y^2 +z^2 = b^2$ with a another sphere $x^2 + y^2 +z^2 = a^2$ located inside of it so that $0<a<b$. I am trying to find the volume between these two spheres above the $xy$ plane, so as I understand they would both be hemispheres. I am having trouble when finding how to set up this integral. To for example get the limits for $z$ for both spheres I was going to set $z^2 = a^2 - x^2 - y^2$ equal to $z^2 = b^2 - x^2 - y^2$, but then I end up with $a^2 = b^2$. I imagine this is wrong or I should be switching to spherical coordinates.
AI: Setting the equations equal to each other is finding where the spheres intersect, which doesn't work (since they don't!).
Switch to spherical coordinates -- both rectangular and cylindrical coordinates are going to require you to break up the integral in ways you don't want.
In spherical coordinates $\rho$ is going from $a$ to $b$ and $\phi$ from $0$ to $\pi/2$ (since you're only above the $xy$ plane) while $\theta$ goes from $0$ to $2\pi$. |
H: $\frac{x_1^2}{\cos^2 \left( \arctan \frac{x_2}{x_1} \right)} =x_1^2 +x_2^2 $?
Is the relationship true? I don't seem able to prove it myself. I know that $$\cos(\arctan(x)) =\frac{1}{\sqrt{1+x^2}}$$ but I am at a loss here.
Thank you.
AI: So, $$\cos\left(\arctan \frac{x_2}{x_1}\right)=\frac1{\sqrt{1+\left(\frac{x_2}{x_1}\right)^2}}$$
$$\implies\cos^2\left(\arctan \frac{x_2}{x_1}\right)=\frac1{1+\left(\frac{x_2}{x_1}\right)^2}=\frac{x_1^2}{x_1^2+x_2^2}$$ |
H: Lebesgue measure nested sequences
I am asked to prove the following:
Let $E \subset \mathbb R$ be Lebesgue measurable. Then there is a sequence of open sets $(O_n)$ and a sequence of closed sets $(F_n)$ such that $F_n \subset E \subset O_n$, $F_n \subset F_{n+1}$ and $O_{n+1} \subset O_n$ for all $n$. Furthermore $\lambda((\bigcap_{n=1}^\infty O_n) \setminus (\bigcup_{n=1}^\infty F_n))=0$.
My attempt goes along the lines of this:
Since $E$ is measurable there exists an open set $O_k$ and a closed set $F_k$ such that $F_k \subset E \subset O_k$ where $\lambda(O_k \setminus E)$ and $\lambda(E \setminus F_k)$ can be made "controllably" small. Now I set $O = \limsup O_k$ and $F \liminf F_k$ to get the nestedness. However I am not able to "connect the dots" and get to the last equality (the measure result). Am I onto something here?
Thank you for your time.
AI: Write $O = \cap_n O_n$ and $F = \cup_n F_n$. Let $k \ge 1$. Then
$F_k \subset F \subset E \subset O \subset O_k$,
$O \setminus F \subset O_k \setminus F_k$, and
$O_k \setminus F_k = (O_k \setminus E) \cup (E \setminus F_k)$.
Thus $\lambda(O \setminus F) \le \lambda(O_k \setminus F_k) \le \lambda(O_k \setminus E) + \lambda(E \setminus F_k)$. Now let $k \to \infty$. |
H: How to prove Trigonometry equation?
how to solve following equation
$$
\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{1}{9}\right)
=
\cos^{-1}\left(\frac{3}{5}\right)
$$
How to prove the above equation?
AI: $\newcommand{\+}{^{\dagger}}%
\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
\newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
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\begin{align}
&\tan\pars{\overbrace{\cos^{-1}\left(\frac{3}{5}\right)}^{\ds{\equiv\ x}}}
=
\tan\pars{\tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{1}{9}\right)}
=
{1/4 + 1/9 \over 1 - \pars{1/4}\pars{1/9}} = {13 \over 35}
\end{align}
$$
\tan\pars{x} = {\root{1 - \cos^{2}\pars{x}} \over \cos\pars{x}}
=
{\root{1 - \pars{3/5}^{2}} \over 3/5} = {4/5 \over 3/5} = {4 \over 3}
\color{#0000ff}{\Huge\not=} {13 \over 35}
$$ |
H: Determine if the number $ \sqrt{8+2\sqrt{10+2\sqrt{5}}} - \sqrt{8-2\sqrt{10+2\sqrt{5}}} $ is rational
$ \sqrt{8+2\sqrt{10+2\sqrt{5}}} - \sqrt{8-2\sqrt{10+2\sqrt{5}}} $
I have tried to raising it to the square, but I can't obtain the result.
$ \sqrt{8+2\sqrt{10+2\sqrt{5}}} - \sqrt{8-2\sqrt{10+2\sqrt{5}}}= k $
$ 2\sqrt{10+2\sqrt{5}} -2\sqrt{10+2\sqrt{5}} = k^2 $
$2(\sqrt{(10+2\sqrt{5})(10-2\sqrt{5}})=2\sqrt{80}=8\sqrt{5}=k^2$
Is this a good lead?
@EDIT one more thing, how to show that $ \sqrt{8+2\sqrt{10+2\sqrt{5}}} + \sqrt{8-2\sqrt{10+2\sqrt{5}}} $ equals to $\sqrt{10}+\sqrt{2}$
AI: So, your steps aren't quite valid; if you have a statement of the form
$$
\sqrt{a} - \sqrt{b} = c
$$
Then, while it's true that
$$
\left(\sqrt a - \sqrt b\right)^2 = c^2
$$
this unfortunately doesn't simplify to
$$
a - b = c^2
$$
rather, you get
$$
a - 2\sqrt a\sqrt b + b = c^2
$$
(check this by "foil"-ing).
What is true however, is that
$$
(\sqrt a - \sqrt b)(\sqrt a + \sqrt b) = a - b
$$
(again, check by "foil"-ing), and that $\sqrt a - \sqrt b$ is rational if and only if $\sqrt a + \sqrt b$ is (and so, as a consequence of this assumption, $a - b$ would be rational as well).
So
$$\left(\sqrt{8+2\sqrt{10+2\sqrt{5}}} - \sqrt{8-2\sqrt{10+2\sqrt{5}}}\right)\left(\sqrt{8+2\sqrt{10+2\sqrt{5}}} + \sqrt{8-2\sqrt{10+2\sqrt{5}}}\right)\\
= 8+2\sqrt{10+2\sqrt{5}} - (8-2\sqrt{10+2\sqrt{5}})\\
= 4\sqrt{10+2\sqrt{5}}$$
Now, you might try squaring things. |
H: If $X_j$ is iid Unif$(-1,1)$ and $Y_n=\frac{\sum X_j}{\sum X_j^2+\sum X_j^3}$, show that $\sqrt{n}Y_n\rightarrow N(0,3)$ in distribution.
If $X_j$ is iid Unif$(-1,1)$ and $\displaystyle Y_n=\frac{\sum X_j}{\sum X_j^2+\sum X_j^3}$, show that $\sqrt{n}Y_n\rightarrow N(0,3)$ in distribution.
I know from central limit theorem that $\sqrt{n} \frac{1}{n}\sum X_j \rightarrow N(0,\frac{1}{3})$ but I'm not sure how to show the denominator of $Y_n$ goes to some constant in probability or something else in distribution.
AI: I convert my comments into an answer.
First remark that: $$
\sqrt{n}\cdot Y_n = \frac{\frac{1}{\sqrt{n}}\sum X_i}{\frac{1}{n}\sum X_i^2 + \frac{1}{n}\sum X_i^3}.
$$
The numerator converges in law to a normal random variable by the Central Limit Theorem.
The denominator converges almost surely (in probability is sufficient) to some constant by the Law of Large Numbers.
Determine the variance of the normal random variable, the limit constant for the denominator, and finally conclude with Slutsky's lemma. |
H: Prove by induction that $a-b|a^n-b^n$
Given $a,b,n \in \mathbb N$, prove that $a-b|a^n-b^n$. I think about induction. The assertion is obviously true for $n=1$. If I assume that assertive is true for a given $k \in \mathbb N$, i.e.: $a-b|a^k-b^k$, I should be able to find that $a-b|a^{k+1}-b^{k+1}$, but I can't do it. Any help is welcome. Thanks!
AI: To complete the induction, note that
$a^{k + 1} - b^{k + 1} = a^{k + 1} - a^kb + a^kb - b^{k + 1} = a^k(a - b) + b(a^k - b^k), \tag{1}$
then simply observe that
$(a - b) \mid a^k(a - b), \tag{2}$
which is obvious, and that
$(a - b) \mid (a^k -b^k) \tag{3}$
by the induction hypothesis
$(a - b) \mid (a^k - b^k). \tag{4}$
Since $a - b$ divides both summands, it divides their sum.QED
Hope this helps. Cheerio,
and as always,
Fiat Lux!!! |
H: How to show the metric space is complete?
the space is the Real line with bounded metric (i.e. $d/(1+d)$, $d$: euclidean).
We thought that since nd the space real line with euclidean metric is complete and the bounded metric is smaller than d, then any cauchy sequence that converges with eucliedan metric will also converge with bounded metric. but there could be more cauchy seq. with bounded metric, so we have to show that these also converge. this is the part we couldn't do.
AI: Write $e(x,y) = d(x,y)/(1+d(x,y))$. Let $\{x_n\}$ be a Cauchy sequence in the metric $e$.
Let $\epsilon > 0$ be given. Let $f(t) = \dfrac{t}{1+t}$. Since $f$ is increasing on $[0,\infty)$ and $\lim_{t \to 0^+} f(t) = 0$, there exists $\delta$ with the property that $e(x,y) < \delta$ implies $d(x,y) < \epsilon$. Let $N$ be an index with the property that $n,m \ge N$ implies $e(x_n,x_m) < \delta$. We conclude that $n,m \ge N$ implies $d(x_m,x_n) < \epsilon$. It follows that $\{x_n\}$ is Cauchy in the metric $d$ and converges to a number $x$.
Since $e(x_n,n) \le d(x_n,x)$ you get $x_n \to x$ in the metric $e$. Thus $e$ is a complete metric. |
H: Convergence of sum and multiplication
Iv'e got a question and I find it difficult to me.
Let $ a_n,b_n$ be sequences. Assume that $a_n+b_n$ converges.
Is $a_n\cdot{b_n}$ essentially converges?
Prove.
Thank you!
AI: No. Consider $a_n = n$, $b_n = -n$. |
H: $A = \left\{ (1,x) \in \mathbb{R}^2 : x \in [2,4] \right\} \subseteq \mathbb{R}^2$ is bounded and closed but not compact?
Is it true that set $A = \left\{ (1,x) \in \mathbb{R}^2 : x \in [2,4] \right\} \subseteq \mathbb{R}^2$ is bounded and closed but is not compact. We consider space $(\mathbb{R}^2, d_C)$ where $$d_C(x,y) = \begin{cases} d_E(x,y) \quad \text{if x, y, 0 in one line} \\ d_E(x,0)+d_E(0,y) \quad \text{in any other case} \end{cases}$$
Where of course $d_E$ is euclidean metric.
AI: Yes, it’s true. In fact, $A$ is a closed, discrete set in $\Bbb R^2$ with the metric $d_C$, and an infinite, discrete set is never compact.
For each $a\in[2,4]$ let
$$U_a=\left\{\langle x,ax\rangle:\frac12<x<\frac32\right\}\;;$$
$U_a$ is an open interval on the line through $\langle 0,0\rangle$ and $\langle 1,a\rangle\in A$, so it’s an open set in the space; in fact, it’s the open $d_C$-ball of radius $\frac{\sqrt{a^2+1}}2$ centred at $\langle 1,a\rangle$. $U_a\cap A=\langle 1,a\rangle$ for each $a\in[2,4]$, so $A$ is discrete. In fact, the open sets $U_a$ for $a\in[2,4]$ are even pairwise disjoint. Thus, $\{U_a:a\in[2,4]\}$ is an open cover of $A$ that has no finite subcover. (It even has no proper subcover: every one of the sets $U_a$ is actually needed in order to cover $A$.)
The point of $A$ furthest from the origin is $\langle 1,4\rangle$, and $d_C(\langle 0,0\rangle,\langle 1,4\rangle)=\sqrt{17}$, so $A$ lies within a ball of finite radius centred at the origin and is therefore bounded. I’ll leave it to you to verify that $A$ is closed in this topology; the ideas that I used to construct the sets $U_a$ should help, if you don’t already see how. |
H: $\max\{\chi(G):G$ embeds on projective plane$\}=6$
My lecture notes in Discrete Mathematics state that
$$
\max\{\chi(G) \; : \; G \text{ embeds on projective plane} \}=6,
$$
but I have no idea where this comes from. $\chi(G)$ is the chromatic number of $G$. How does such an emebedding looks like?
AI: The complete graph $K_6$ has chromatic number $6$ and embeds in the
projective plane. A nice way to see the embedding is to start from
the regular icosahedron in the sphere $S$, which has six opposite pairs
of vertices. For each pair $\pm P, \pm Q$, each of $\pm P$ is
adjacent to $Q$ or $-Q$ but not both. So the image of the icosahedron
in the projective plane $S / \{\pm1\}$ is an graph with six vertices,
each adjacent to the other five, i.e. an embedding of $K_6$. |
H: How to prove surjectivity part of Short Five Lemma for short exact sequences.
Suppose we have a homomorphism $\alpha, \beta, \gamma$ of short exact sequences:
$$
\begin{matrix}
0 & \to & A & \xrightarrow{\psi} & B & \xrightarrow{\phi} & C & \to & 0 \\
\ & \ & \downarrow^{\alpha} & \ & \downarrow^{\beta} \ & \ & \downarrow^{\gamma} \\
0 & \to & A' & \xrightarrow{\psi'} & B' & \xrightarrow{\phi'} & C' & \to & 0
\end{matrix}
$$
If both $\alpha, \gamma$ are surjective then so is $\beta$. This can be proved using the properties of the diagram somehow.
I've tried several things.
AI: It should be straightforward.
We want to prove $\beta$ is surjective, so start out from an arbitrary element $b'\in B'$. We can do one thing: consider $c':=\phi'(b')\in C'$.
Since $\gamma$ is surjective, we get $c$ with $\gamma(c)=c'$.
That the pair $\phi,0$ of maps is exact means nothing else but that $\phi$ is surjective. It yields an element $b\in B$, such that $\gamma\phi(b)=c'=\phi'(b')$.
Now we might not get $\beta(b)=b'$ with this element $b$, nevertheless, we have that $\beta(b)$ and $b'$ has the same image under $\phi'$, so by exactness, this gives $a'\in A'$ such that $\psi'(a')=b' - \beta(b)$.
Can you take it on from here? |
H: Permutation and combination of letters
I need help with the following question:
"Given that a computer can only type letters A,B,C,D and E, how many ways can I type in 6 letters such that they must contain at least all of the different letters?
ie. AABCDE ABCDEA ABCDEB"
My professor said that the number of ways is:
6C4 * 5!
because,
5! - There are 5! ways to arrange the letters
6C4 - Since 4 letters are going to be unique, out of the 6 letter positions, choose 4 positions to have these unique letters.
However, I had a different answer(which was wrong). Mine is:
6C5 * 5! * 5
I chose 5 out of 6 positions to include all ABCDE, and there are 5! ways of arranging them. As for the last position, it can include any of the 5 letters.
Could anyone explain to me why is my answer wrong?
Major thanks to all :3
AI: Choose the first five positions to include ABCDE, and an ordering of them, say $$
\text{EBDCA_},$$ and then pick one of our five options for the last space, say A, so we get $$\text{EBDCAA}.$$
Alternately, we choose the first four and the last position to include ABCDE, and an ordering of them, say $$\text{EBDC_A},$$ and then pick one of our five options for the last space, say A, so we get $$\text{EBDCAA}.$$ Clearly, these are the same result, but your method counts them as though they are not.
It turns out that your answer is exactly double the correct answer. That is because, for any final arrangement of letters, your method counts two "different" ways to obtain it (as we saw above for one particular arrangement). |
H: set of linear and continuous functions
Let E,F be normed vector spaces
$ \mathcal L(E,F) = \{ f \in Hom(E,F) | f-continuous\} $
Why
1) $Hom(\Bbb K^n,\Bbb K) = \mathcal L (\Bbb K^n,\Bbb K) = \Bbb K^n $
2) $\mathcal L (\Bbb K^n, F) = F^n$
AI: First part of 1) says that any linear map $\Bbb K^n\to \Bbb K$ is continuous (or equivalently, bounded). If you want to prove it, show that a sequence $v_k$ in $\Bbb K^n$ is convergent iff all the coordinate sequences $(v_k)_i$ are convergent (for the particular norm considered in $\Bbb K^n$). Then any linear $f:\Bbb K^n\to\Bbb K$ will be uniquely determined by the $n$ values $f(e_i)$ where $(e_i)$ is the standard basis of $\Bbb K^n$, so if you have a sequence $v_k$ converging to $w$ of $\Bbb K^n$ then write $f(v_k)=f\left(\sum_i (v_k)_i\,e_i\right) = \sum_i (v_k)_i\,f(e_i)$ by linearity, which is a finite sum of numbers, and each coordinate sequence $(v_k)_i$ converges to $w_i$, so we have $\ f(v_k)\ \to\ \sum_i(w_i)\,f(e_i)=f(w)$.
2) Exactly the same works for any normed space $F$ as codomain, with the difference that $f(e_i)$ will be a vector in $F$ instead of a scalar. The main observation is that any linear map $f:\Bbb K^n\to F$ is uniquely determined by the $n$ vectors $f(e_1),\, \dots,\ f(e_n)$. |
H: i.i.d. random variables with continuous distribution function are equal with probability 0
Let $X_1, X_2, ...$ be i.i.d. (real) random variables with continuous distribution function (in particular, we're not assuming absolute continuity).
How can I show that $P[X_n = X_m] = 0$ for $n \ne m$?
It's pretty clear for an absolutely continuous distribution function, where one can even drop the assumption that the $X_i$ are identically distributed. I'm not sure how to handle the only continuous case though.
AI: It suffices to consider $m=1,n=2$. Given $\epsilon > 0$, take $N$ so that
$P[|X_2| \le N] > 1 - \epsilon$. Now the event $(X_1 = X_2 \in [-N,N])$ is contained in the union of rectangles $(x_i \le X_1, X_2 \le x_{i+1})$ where $x_{i+1} - x_i = \delta$. Take $\delta$ small enough that ... |
H: Proof of convexity of $f(x)=x^2$
I know that a function is convex if the following inequality is true:
$$\lambda f(x_1) + (1-\lambda)f(x_2) \ge f(\lambda x_1 + (1-\lambda)x_2)$$
for $\lambda \in [0, 1]$ and $f(\cdot)$ is defined on positive real numbers.
If $f(x)=x^2$, I can write the following:
$$\lambda x_1^2 + (1-\lambda)x_2^2 \ge (\lambda x_1 + (1-\lambda)x_2)^2$$
$$0 \ge (\lambda ^2 - \lambda) (x_1^2 - x_2 ^ 2) $$
But I am not sure if this is true or not. How can I prove this?
AI: You made a mistake in your rearranging. The following are equivalent: $$\begin{eqnarray*}\lambda x_1^2+(1-\lambda)x_2^2 & \ge & \bigl(\lambda x_1+(1-\lambda)x_2\bigr)^2\\(1-\lambda)x_2^2 + \lambda x_1^2 & \ge & \lambda^2 x_1^2+2\lambda(1-\lambda)x_1x_2+(1-\lambda)^2x_2^2\\\lambda x_1^2 & \ge & \lambda^2 x_1^2+2\lambda(1-\lambda)x_1x_2+\bigl[(1-\lambda)^2-(1-\lambda)\bigr]x_2^2\\\lambda x_1^2 & \ge & \lambda^2 x_1^2+2\lambda(1-\lambda)x_1x_2+(1-\lambda)\bigl[(1-\lambda)-1\bigr]x_2\\\lambda x_1^2 & \ge & \lambda^2x_1^2+2\lambda(1-\lambda)x_1x_2+(1-\lambda)(-\lambda) x_2^2\\0 & \ge & (\lambda^2-\lambda)x_1^2+2\lambda(1-\lambda)x_1x_2+(1-\lambda)(-\lambda) x_2^2\\0 & \ge & \lambda(\lambda-1)x_1^2-2\lambda(\lambda-1)x_1x_2+\lambda(\lambda-1)x_2^2\\0 & \ge & \lambda(\lambda-1)(x_1-x_2)^2\end{eqnarray*}$$
The final inequality is true for all $\lambda$ if $x_1=x_2.$ If $x_1\ne x_2,$ then since $\lambda\ge\lambda-1,$ the final inequality holds exactly when $\lambda\geq 0\geq \lambda-1,$ or equivalently, when $\lambda\in[0,1].$ |
H: Finding number of solutions.
How many solutions does this equation have
$$2 \cos^2\left(\frac12 x \right) \sin^2 x = x^2+x-2$$
where $0 \lt x \le \displaystyle\frac \pi9?$
I observed that $2 \cos^2\left(\frac12x\right)$ can be written as $1+\cos x$. Simplifying $\sin^2 x,$ we get
$$(1+\cos x)^2(1-\cos x)=x^2+x-2$$
But I don't understand what to do after that.
AI: HINT:
In the given range, the left hand side is positive. How about the right side? |
H: Intro analysis - contraction mappings
A function $ f: \mathbb{R} \rightarrow \mathbb{R} $ is called a contraction mapping if there exists a positive constant K < 1 such that
$ |f(x) - f(y)| \leq K |x-y| $
d) Suppose $f:\mathbb{R} \rightarrow \mathbb{R} $ is a contraction mpaping and for any $ x_0 \in \mathbb{R} $, consider the sequence $(x_n) $ defined recursively by $ x_n = f(x_{n-1}) $ for $ n \in \mathbb{N} $
define also $a_n = x_n - x_{n-1}$ so $x_ 0 + a_1 + a_2 + ... + a_n = x_n $ show that $\displaystyle \sum_{n=1}^\infty a_n$ converges absolutely and use this to conclude that the sequence $(x_n)$ converges
part d
Here is my attempt thus far
$ \displaystyle \sum_{n=1}^N |a_n| = |x_n - x_0| $
note that $ |x_2 - x_1| = |x_2 - x_0 -(x_1 -x_0)| \geq |x_2 -x_0| - |x_1 - x_0| $ so $ |x_2 - x_0| \leq |x_2 - x_1| + |x_1 - x_0| $ by a similar approach $ |x_3 - x_0| \leq |x_3 - x_2| + |x_2 - x_1| + |x_1 - x_0| $
then $ |x_n - x_0| \leq |x_1 - x_0| + |x_2 - x_1| + |x_3 - x_2| + ... + |x_{n} - x_{n-1}| $
now since $ |f(x_{n-1} - f(x_{n-2})| = |x_n - x_{n-1}| \leq K|x_{n-1} - x_{n-2}| \leq K^2|x_{n-2} - x_{n-3}| \leq ... $ we get that $ |x_n - x_0| \leq |x_1-x_0|(1 + K + K^2 +....+K^{n-1}) $
$ \displaystyle \lim_{N\rightarrow \infty} \sum_{n=1}^N|a_n| = \lim_{n\rightarrow \infty} |x_n - x_0| \leq \dfrac{|x_1 - x_0|}{1-K} $
now I don't know what to do - I feel as though I should conclude, as the limit is bounded so obviously converges to something.
I'm also not sure on how to go about to prove that the seqn x_n converges
any help thanks
AI: You're basically done. You've shown that for $S_N = \sum_{n=1}^N|a_n|$, the sequence $\{S_N\}$ is bounded. Since the terms in the series $\sum |a_n|$ are positive, you also know $\{S_N\}$ is monotone, and so $\lim_{N\to\infty} S_N$ exists. Thus $\sum_{n=1}^\infty |a_n|$ converges, which in turn implies $\sum_{n=1}^\infty a_n$ converges. Now conclude that $\{x_n\}$ converges by noting $x_N = x_0 + \sum_{n=1}^N a_n$ as described in the prompt (and as such $\{x_n\}$ converges to $x_0 + \sum_{n=1}^\infty a_n$ as $n \to \infty$). |
H: How to find the derivative of $(2x+5)^3(3x-1)^4$
How to find a derivative of the following function?
$$\ f(x)=(2x+5)^{3} (3x-1)^{4}$$
So I used:
$$(fg)'= f'g + fg'$$ and $$(f(g(x)))'= f'(g(x)) + g'(x)$$
Then I got:
$$ f(x)= 6(2x+5)^{2} + 12(3x-1)^{3}$$
and I don't know how to get the solution from this, which is: $$ 6(2x+5)^{2}(3x-1)^{3}(7x+9)$$
AI: I assume that you are supposed to have $$f(x)=(2x+5)^3(3x-1)^4,$$ instead of what you've written. By product rule, we have $$f'(x)=(2x+5)^3\bigl[(3x-1)^4\bigr]'+\bigl[(2x+5)^3\bigr]'(3x-1)^4.$$ Applying chain rule (which you've incorrectly stated, by the way) gives us $$f'(x)=(2x+5)^3\cdot 4(3x-1)\cdot(3x-1)'+(3x-1)^4\cdot3(2x+5)^2\cdot(2x+5)',$$ or $$f'(x)=12(2x+5)^3(3x-1)^3+6(2x+5)^2(3x-1)^4.$$ Now, pull out the common factor $6(2x+5)^2(3x-1)^3,$ and see what happens. |
H: Lagrange multipliers - perturbation of constraints
I have been spending some time learning about Lagrange multipliers lately.
Something is puzzling me though. Reading around (also on Wikipedia) I saw multiple time the interpretation that lagrange multipliers represent the rate of change of the optimal value of the function when a constraint is perturbed.
More formally, when solving the problem: $ min f(x)$ such that $h_1(x) = 0, h_2(x)=0.. h_i(x) = r, ... h_m(x) = 0 $ and when denoting the optimal solution by $x(r)$ we get that $\frac{df}{dr}x(r) = - \lambda_i$.
Firstly, should it not be $\frac{df(x(r))}{dr}$ ?
Secondly, which lambdas?? We are varying the $r$, and from what I see there is no guarantee that the Lagrange multipliers will be the same for all $r$ - even though the Lagrange condition is the same, the constraints are different). Is it for $r$ = 0? In that case, the proof I have seen so far would not work (it just uses the chain rule on the function, and then uses the Lagrange condition being satisfied at $x(r)$ with the multipliers being set to $\lambda_i$.)
Can someone please explain this to me? It is not clear where the Lagrange multipliers come from in this case.
AI: You are right that it should be ${d\over dr}f(x(r))$.
The result will be true for any $r$. When you solve the Lagrange problem, you will find $x=x(r)$ and $\lambda=\lambda(r)$. Substitute $x=x(r)$ into the objective function, and you get the general formula
$${d\over dr}f(x(r))=-\lambda_i(r)$$ |
H: Thinning a Poisson Process
Suppose that events are produced according to a Poisson process with an average of lambda events per minute.
Each event has a probability $p$ of being Type A event, independent of other events.
Let the random variable $Y$ represent the number of Type A events that occur in a one-minute period. Prove that $Y$ has a Poisson distribution with mean $\lambda p$.
I read over it and I feel like I'm missing something because I still see it as having the mean as lambda not $\lambda p$.
AI: Let $X \sim \mathcal{P}(\lambda)$ denote the number of events during a give one-minute period. Each of these events being of Type A with probability $p$ (independently with respect to the others), if $Z_1,Z_2,Z_3,\dots$ is a sequence of independent Bernoulli variables with $P(Z_i = 1)=p$, then the following identity holds in distribution:
$$
Y \overset{(d)}{=} \sum_{i=1}^X Z_i = \sum_{i=1}^\infty Z_i\,1_{i \leq X}.
$$
By independence of $Z_i$ and $X$ for every $i$, one has
$$
E(Y) = \sum_{i=1}^\infty E(Z_i)E(1_{i \leq X}) = p \sum_{i=1}^\infty E(1_{i \leq X}) = pE(X) = p\lambda.
$$
To prove that $Y$ is Poisson with parameter $p\lambda$, we use the generating function:
$$
E[s^Y] = \sum_{n=0}^\infty E[s^Y \mid X = n] \frac{\lambda^n}{n!}e^{-\lambda}
$$
where $E[s^Y \mid X = n] = E[s^{Z_1}]^n = (1+p(s-1))^n$. So finally, one has
$$
E[s^Y] = e^{\lambda(1+p(s-1))}e^{-\lambda} = e^{p\lambda(s-1)}.
$$ |
H: Does Wolfram Alpha fail for $x^x$?
WolframAlpha generates the following graphic for $f_{(x)} = x^x$:
$f_{(x)} = x^x$
Can anyone explain me why this graphic looks like above? I mean why the real part for negative, non-integer, but rational numbers looks like that?for $0^0$?
To clarify my question:
We know that:
$$a^{m/n} = sqrt(a^m, n)$$
So, for example:
$${-1.1}^{-1.1} = 1 / {sqrt({-1.1}^{11}, 10)}$$ that is a complex number, not a real one
AI: By definition $x^x = \exp(x \log(x))$. Presumably Mathematica is using the principal branch of the logarithm. Then for $x < 0$, $\log(x) = \log(|x|) + i \pi$, and so $x^x = \exp(x \log |x|) (\cos(\pi x) + i \sin(\pi x))$. |
H: Probability mean is less than 5 given that poisson distribution states it is 6
I want to find the probability that mean is less than 5 given that poisson distribution states it is 6
ie find p(x<5|x~po(6))
Here is the actual question:
Two grocers agree that the daily demand for a particular item has Poisson distribution. However, grocer A
claims that the mean demand is 3 items per day, while grocer B claims that the mean demand is 6 items per
day. They agree to resolve the disagreement by observing the demand on one particular day: B agrees to
accept A’s claim if the observed demand is 4 or less, and A agrees to accept B’s claim if the observed demand
is 5 or more.
(a) Calculate the probability that A’s claim is accepted when, in fact, B’s claim is correct
AI: I think you meant finding the probability of a Poisson rv with mean 6 taking value less than 5. That is the same as the probability of a Poisson rv taking values less or equal to 4 since Poisson is a discrete prob. distribution. Also, since Poisson variate can only take non-negative discrete values, the required probability will be the sum of the probabilities of a Poisson variate taking values 0,1,2,3,4 and you can calculate individual probabilities from the Poisson pmf. You can pick it from here. |
H: find the limit of ${(n + 1)^{{1 \over {\sqrt n }}}}$
I really tried to think of a way to find the limit of this sequence but couldn't think of something useful. I tried to use Bernoulli's inequality to squeeze the sequence, but it didn't come in handy.
$$\mathop {\lim }\limits_{n \to \infty } {(n + 1)^{{1 \over {\sqrt n }}}} = ?$$
AI: You can do this without logarithms. First' a lemma:
Lemma: For an arbitrary $\varepsilon > 0$ there exists an $n_0 \in \mathbb{N}$ such that for every $n \geq n_0$ we have $$n^\frac{1}{n} < 1 + \varepsilon.$$
To prove the lemma, note that the inequality is equivalent to $$ n < (1 + \varepsilon)^n,$$ and this is true when $n$ is large enough by binomial expansion.
Now, all the terms in your sequence are greater than $1$. You can use this, the lemma above, and the definition of limits to prove that the sequence converges to $1$. |
H: Norm of operator $T_x(f) = f(x)$
Let $X$ be a normed vectorspace and $X'$ be the dual space of $X$. For $x \in X$ we can define $T_x: X' \to \mathbb F$ by $T_x(f) := f(x)$. This is indeed an operator in $X''$. I read that $\| T_x \| = \| x \|$ but I could not figure out how to prove this. Can this be done by means of a Hahn-Banach theorem ?
AI: We have, for every $f \in X'$,
$$\lvert T_x(f)\rvert = \lvert f(x)\rvert \leqslant \lVert f\rVert\cdot \lVert x\rVert$$
by the definition of the norm on $X'$. That yields $\lVert T_x\rVert \leqslant \lVert x\rVert$.
To see the equality, we need, for every $x\neq 0$, to find an $f_x \in X'\setminus\{0\}$ with
$$\lvert f_x(x)\rvert = \lVert f_x\rVert \cdot \lVert x\rVert.$$
The existence of such an $f_x$ follows from the Hahn-Banach theorem, for $x\neq 0$ consider the subspace $M = \mathbb{F}\cdot x$, and the linear functional $\varphi_x \colon M \to \mathbb{F}; \varphi_x(c\cdot x) = c\cdot \lVert x\rVert$. By Hahn-Banach, $\varphi_x$ has an extension $f_x$ of norm $1$. |
H: Question about second condition of pumping lemma
I don't think that I fully understand how to use the pumping lemma to prove that a given language is not regular. I'm reading Sipser and according to him the definition of the pumping lemma is:
"If A is a regular language, then there is a number p ( the pumping length) where, if s is any string of A of length at least p, then s may be divided into three pieces, s = xyz, satisfying the following conditions:
1. for each $i \ge 0, xy^iz\in A$
2. $\left| y \right| \gt 0$
3. $\left|xy\right| \le p$
Now given the exercise:
Show that $A_1 = \{0^n1^n2^n | n\ge 0\}$ is not regular.
I assume that $A_1$ is regular and therefore can be divided into a form $s=xyz$ for $s\in A_1$. But now I could simply say that $s=0^p1^p2^p$ where p is the pumping length and divide s such that $x=\epsilon, y= 0^p1^p, z = 2^p$ and claim that because $\left|xy\right| \gt p$ it follows that $A_3$ is not regular. Similarly for the language $L=\{a^nb^la^k | k \neq n+l\}$ I could use the same logic. But all the solutions to those exercises come up with fairly complicated solutions compared to this logic, so I assume something is wrong here.
I think I misunderstand something about the theorem and how to use it to prove some language is not regular, what am I getting wrong?
AI: When you use the lemma, you don't get to pick $x$, $y$, and $z$ for a given string $s$. All you know is that if the language is regular, they must exist, but not necessarily what they are. To show that a language is not regular, you need to show that for any choice of $x$, $y$, and $z$, at least one of the conditions (1), (2), and (3) fails. (Here the easiest way is probably to take a division satisfying (2) and (3) and show that (1) fails.)
Here's a hint: let $n>p$. If (2) and (3) hold, what can you say about $y$? Can you then get a string of the form $xy^kz$ that is not in the language? |
H: PDE using Fourier Transform
Using the Fourier Transform, solve:
$u_t=u_{xx}+\alpha u$ with $\alpha>0$, for $x \in \mathbb{R}, t>0$
with initial data $u(x,0)=f(x)$, with $f$ continuous in $\mathbb{R}$
Apllying Fourier transform in equation and initial data, we obtain
$\partial_t^2 \hat u(\xi)=-\xi^2\hat u + \alpha \hat u $
$\hat u (\xi,0)=\hat f(\xi)$
Solving, we obtain $\hat u(\xi,t)=\hat f(\xi)e^{(\alpha-\xi^2)t}$.
To get $u$, we have to apply the inverse Fourier transform, but I'm not getting no useful result.
AI: You need to make use of the fact that if $f(x)=e^{-ax^2}$, $a>0$ constant, then its Fourier transform is $$\hat{f}(\xi)=\sqrt{{\pi\over a}}\exp\left({-\xi^2\over 4a}\right)$$ which is commonly found in textbooks on the subject. (Note: I am using $\hat{f}(\xi):=\int_{-\infty}^{\infty} f(x)e^{-i\xi x}dx$.)
That, together with the Convolution Theorem, will get you the solution you seek:
$$
u(x,t)=\sqrt{\pi}\exp\left(-{1\over 4}+\alpha t\right)f(x)*e^{-x^2}.
$$ |
H: Organic firm sales drop
Assume a grocery company grows from having 180 to 210 shops and at the same time experiences a 7% drop in sale. How much would their sale have dropped if they had not opened the extra stores.
I can see that to be status quo they should have grown by (210-180)/180=16.66 %, but I can't figure out how to get the final result.
Thanks.
p.s. Couldn't find an appropriate tag.
AI: The only thing that makes sense is to assume all stores have the same sales and that if they had not opened those stores, each store would still have sold that amount. If a store sold one unit last year, they sold $180$ units. This year they sold $0.93 \cdot 180$ units from $210$ stores, so each store sold $\frac {0.93 \cdot 180}{210}$ and the total with $180$ stores would have been $\frac {0.93 \cdot 180}{210}\cdot 180$ |
H: Existence of inaccessible cardinals implies the consistency of ZFC
I wonder if any can sketch for me in very broad lines the proof of the fact that the existence of inaccessible cardinals implies the consistency of ZFC? I don´t know much about set theory, but I find it extremely interesting that this should be the case.
AI: In $\sf ZFC$ one can define a "rank" function on sets. This means, in rough words, how many times we need to iterate the power set function before we can generate a set (taking union at limit stages).
If $\kappa$ is inaccessible then the sets whose rank is smaller than $\kappa$ form a model of $\sf ZFC$. Therefore by the completeness theorem one has the $\sf ZFC$ is consistent, if there exists an inaccessible cardinal.
I'm not getting into details of what exactly are the sets of rank less than $\kappa$, or how to show that all the axioms of $\sf ZFC$ hold in that set. But if one is familiar with these basics definitions then one can easily show that it is the case.
Finally, even if we assume that $\sf ZFC$ is consistent, and therefore has a model, it is still far from implying that there is an inaccessible cardinals. There is a long and curious hierarchy of stronger and stronger assertions regarding the consistency of $\sf ZFC$ (we can require just the consistency of $\sf ZFC$, or the consistency of the theory $\sf ZFC+\rm Con(\sf ZFC)$, and so on; we can require the there are "nice" models of $\sf ZFC$; and more and more). |
H: Invertible matrices
Please help with the next question.
Let A and B be two invertible matrices such that $A+B \neq 0$.
Prove or disprove that A+B is invertible.
Thanks!
AI: What you can conclude with these matrices?
$$A=I_2\ \text{and}\ B=\mathrm{diag}(-1,1)$$ |
H: Symmetric and Transitive closures
Given a relation $R$, is the symmetric closure of the transitive
closure of $R$ equal to the transitive closure of the symmetric
closure of $R$? If yes, prove it. If not, give a counterexample.
Well for this, the only reasoning I can come up with is if I have an arbitrary transitive closure set $R_t = (2,3),(1,3)$ then the symmetric closure of $R_t$ is $R_s= \{ (3,2),(3,1) \}$. Therefore:
$$R_s \cup R_t=\{ (2,3),(1,3),(3,2),(3,1) \}$$
Now, if we have the symmetric closure $R_s= \{ (3,2),(3,1) \}$ the the transitive closure $R_s$ would be $R_t= \{(2,1)\}$
Thus making:
$$R_t \cup R_s =\{ (3,2),(3,1),(2,1) \}$$
This means that they're not equal. First off, is this counterexample valid in this case? If not, then how can I prove otherwise?
AI: if a relation on $\mathbb{N}$ consists of the single element (1,2) then the symmetric closure adds (2,1) and then transitive closure adds the further elements (1,1) and (2,2).
the other way round we only get (2,1) |
H: If a "group" has two identities then is not a group
The story goes like this: A friend and I found this old exercise:
Let $G=\Bbb R-\{-1\}$ and $a*b:=a+b+ab$, is $(G,*)$ a group?
I say that $(G,*)$ is not a group because for any $a\in G$ follows that $0*a=0+a+0a=a+0+a0=a*0$ and also $1*a=1+a+1a=a+1+a1=a*1$, hence it has two identities. My friend says that it doesn't matter that it has two indentities since the definition we've got says that "there exists an element such that blah blah blah". But I say it does matters because then what happens with the inverse, something like this happens: $0=a*a^{-1}=1 \rightarrow 0=1$ (ugly!!!). Am I right? or his evil trolling mind is right?
Or in an (absurd) way is the definition not correct?
(There should be a tag called [settle-argument])
I'm of course joking
AI: $1*a=2a+1\not\equiv a$ and so $1$ is not an identity. Instead, what you demonstrated is that $1$ is a non-trivial center, which all elements other than $0$ will be since this operation is abelian. It is not possible for a set to have two distinct left-right identities, since if $e_1$ and $e_2$ are identities, then $e_1*e_2=e_1$ and $e_1*e_2=e_2$ and so $e_1=e_2$. |
H: Integral of $\int \frac{\sin^22x}{\cos2x}\,\mathrm dx.$
I would like to solve the following integral
$$\int \frac{\sin^22x}{\cos2x}\,\mathrm dx.$$
I know that the derivative of $\cos$ its $\sin$ but how its help me?
Any suggestions?
Thanks.
AI: Hint:
$$\int \frac{\sin^2 2x}{\cos 2x}~dx ~ = ~ \int \sec 2x - \cos 2x~dx ~ = ~ \frac{1}{2} \int \sec u - \cos u ~ du \quad (u = 2x)$$ |
H: On finding a limit by dividing by the highest exponent
Sometimes it’s easy to divide by the highest exponent to find a limit.
$${{{n^3} + 4{n^2}} \over {\root 3 \of n + \root 4 \of {{n^3}} }}$$
So, in the above example (which I just made up; there’s nothing special about it), you should divide by $n^3$. How do you handle the denominator which involves roots?
AI: You can use the same method for fraction exponent:
$${{{n^3} + 4{n^2}} \over {\root 3 \of n + \root 4 \of {{n^3}} }}\sim_\infty\frac{n^3}{n^{3/4}}\ \text{since} \ n^{1/3}=_\infty o(n^{3/4})$$ |
H: Jacobson radical subset of maximal ideal
If we define the jacobson radical $J$ to be the intersection of all maximal right ideals then I am trying to show that if we have a maximal two sided ideal $M$ we must have $J\subseteq M$?
Any ideas
AI: The Jacobson radical of $R$ is the intersection of all right primitive ideals (annihilators of simple right $R$ modules), and every maximal ideal is right primitive.
The equivalence of this definition of the Jacobson radical with the one you gave is pretty easy to prove, and you can find it, for example, in Martin Isaacs' Graduate Algebra starting page 177.
(If you are wondering why every maximal ideal is right primitive, look at it this way: every nonzero right $R/M$ module is faithful (why?) and of course every ring with identity has simple right modules via Zorn's Lemma. That would mean $R/M$ is right primitive, and hence $M$ is a right primitive ideal.) |
H: Did I solve this problem correctly? $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n^2}{n^3+4}$
For the following series:
$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n^2}{n^3+4}$
I found $b_n$ to be $b_n=\frac{n^2}{n^3+4} \gt 0$ and then I took the limit of this and found it to be zero.
AI: You can prove that $b_n$ is decreasing by using the first derivative test (for the function to be decreasing $f'(x)<0$) and considering the function
$$ f(x)=\frac{x^2}{x^3+4}, \quad x\geq 2 . $$ |
H: How to show that two groups makes $S_n$
I need to show that:
$S=\left\{(12),(13),...,(1n)\right\}$ generates $S_n$
$S=\left\{(12),(123\cdots n)\right\}$ generates $S_n$
How do I show that each one of them generates $S_n$?
Thank you!
AI: Can you prove that every element of $S_n$ is equal to a product of transpositions? If so, you just need to show that each of those generating sets contains all transpositions.
Edit: To make this more direct, this task becomes straight forward once you understand how conjugation works in $S_n$. |
H: Find roots of a trigonometric equation
I've been struggling on find the roots of this equation for a while.
$$
f(x)=\cos(2x) - 2\sin(x)\cos(x).
$$
I've already ended transforming all $\cos$ relation to $\sin$ ones, but I'm now stuck on:
$$
f(x)=1-2\sin^2(x) - \sin(2x).
$$
What should I do now?
AI: You have
$$f(x)=\cos(2x)-2\sin(x)\cos(x).$$
The roots will satisfy $f(x)=0$, i.e. $$\cos(2x)-2\sin(x)\cos(x)=0$$ and so $\cos(2x)=2\sin(x)\cos(x)$.
Now, $2\sin(x)\cos(x)=\sin(2x)$, so you have $$\cos(2x)=\sin(2x),$$ and from here, hopefully you can see what to do. |
H: On differentiating an integral with respect to a function
Let $f,g:\mathbb{R}^n \rightarrow \mathbb{R}$, and let
$$
Q = \int \! g(\mathbf{x})f(\mathbf{x}) \, \mathrm{d}\mathbf{x}
$$
What is the result of the following differentiation?
$$
\frac{\partial}{\partial g}Q = \frac{\partial}{\partial g} \int \! g(\mathbf{x})f(\mathbf{x}) \, \mathrm{d}\mathbf{x}
$$
I think that the answer is just $\int \! f(\mathbf{x}) \, \mathrm{d}\mathbf{x}$ [EDIT: I'm Wrong!], but I am afraid that there's something tricky around.
Besides, what if I have to find $ \partial Q/\partial g$ but the integral is defined over a region of $\mathbb{R}^n$? Is everything as before?
Thanks in advance!
AI: The answer is $f$. Or more correctly $h \mapsto \int f(x)h(x) dx $. Its a linear map, and the derivative of a linear map is itself. |
H: On the definition of the direct sum in vector spaces
We say that if $V_1 , V_2, \ldots, V_n$ are vector subspaces, the sum is direct if and only if the morphism $u$ from $V_1 \times \cdots \times V_n$ to $V_1 + \cdots + V_n$ which maps $(x_1, \ldots, x_n)$ to $x_1 + \cdots + x_n$ is an isomorphism.
Looking at the definition of a direct sum in categories in Wikipedia, it is clear that $V_1 \times \cdots \times V_n$ can be given canonical injections so that it is a categorical sum. Thus, we see that if the $u$ is an isomorphism, $V_1 + \cdots + V_n$ with $f_i(v_i)=v_i$ with i between 1 and n as canonical injections is also a sum.
But what if there is another isomorphism between $V_1\times\cdots\times V_n$ and $V_1 +\cdots+V_n$, is $u$ always an isomorphism (true if all $V_i$ are of finite dimension) ? Is $(V_1 + \cdots +V_n,(f_i)_{i=1 \ldots n})$ still a sum (again true if all $V_i$ are of finite dimension) ?
AI: No. It is possible that $V_1$ and $V_2$ are subspaces of some vectorspace $W$ such that the subspace $V_1 + V_2$ of $W$ is isomorphic to $V_1 \times V_2$, but the canonical map $V_1 \times V_2 \to V_1 + V_2$ is not an isomorphism.
Example. Take $W$ the vectorspace of infinite sequences of real numbers (with or without finite carrier; doesn't really matter) and take $V_1$ and $V_2$ to be equal to $W$. Then $V_1 \times V_2 \cong W = V_1 + V_2$ (by interleaving the coordinates of $V_1$ and $V_2$), but the canonical map $V_1 \times V_2 \to V_1 + V_2 = W$ is not injective. |
H: algorithm question how prove that $(n+a)^b = \Theta(n^b)$
this question doctor in college give us as home work
but I don't know how approve it
AI: Recall that $f \in \Theta(g)$ whenever $k_1g(x) \leq f(x)$ an $f(x) \leq k_2g(x)$ for some constants $k_1$, $k_2$ and all $x$ greater than or equal to some $N$. In this case, you must show that there are constants $k_1$ and $k_2$ such that $k_1n^b \leq (n + a)^b$ and $(n + a)^b \leq k_2n^b$ for all $n$ greater than or equal to some $N$. If we choose $k_1$ so that $k_1^{1/b} \leq 1 + a/n$ for all $n \geq N_1$, then we obtain $k_1n^b \leq (n + a)^b$. Similarly, if we choose $k_2$ so that $1 + a/n \leq k_2^{1/b}$ for all $n \geq N_2$, then we obtain $(n + a)^b \leq k^2n^b$. Is it always possible to choose such $k_1$ and $k_2$? Recall what happens to $a/n$ as $n$ becomes arbitrarily large. |
H: Decomposition of polynomials
It is a very simple question but I'm stuck in decomposing this: $x^3+2x^2-2$. I can't find the $x-c$ (Ruffini's rule) form that can enable me to decompose it. Is it possible to decompose?
If I can solve it, I will be able to resolve an entire math problem!. It is an elementary question I know but I can't find a way to continue after several attemps. Please help me, thanks in advance!
AI: By Eisenstein, the polynomial $x^3 + 2x^2-2$ is $\mathbb Q$-irreducible, so there is no rational number $c$ such that $x-c$ divides your polynomial. It does have a real root $c$, though, which is easiest to find by various approximative processes. |
H: When is $c v^\top y y^\top v \ge ||v||^2$
Given $c\in R$ being some constant, $v, y \in R^n$, I want to find conditions for which the following inequality holds true:
$$c v^\top y y^\top v \ge ||v||^2$$
EDIT:
Note that $y y^\top$ is an $n \times n$ matrix.
AI: The way the product is written, $v^\top y^\top y v$ must be parsed as $v^\top (y^\top y) v$ since the products $v^\top y^\top$ and $yv$ don't make sense. Since $y^\top y = \|y\|^2$ this leads to $c \|v\|^2 \|y\|^2 \ge \|v\|^2$. The condition is $\|v\|= 0$ or $c \|y\|^2 \ge 1$. |
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