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H: can i have open sets in closed interval $[0,1]$?
I have to show that if $A=[0,1]$ then $(a,1]$ and $[0,b)$ are open in $A$ for $0\le a<1$ and $0<b\le 1$. when i finish my proof a friend tell me that is not true because when $x=1$ in $(a,1]$ i can't find a $r>0$ such that the open ball is complete contain in $(a,1]$ then isn't in $[0,1]$, so, i don't know if true or false that statement because i know in $\Bbb R$ isn't. What do you think?
Edit: My proof consist in find $\epsilon$ but i take $\inf\{1-x,x\}$ for $(a,1]$ and $\inf\{b-x,x\}$ for $[0,b)$. it's the correct choose?
AI: Your friend is wrong and apparently does not understand the subspace topology. A set $U\subseteq A$ is open in $A$ in the subspace topology that $A$ inherits from $\Bbb R$ if and only if there is some ordinary open set $V$ in $\Bbb R$ such that $U=V\cap A$. In your case, for instance, $(a,1]$ is open in $A$ if $0\le a<1$, because $(a,1]=(a,2)\cap A$, and $(a,2)$ is open in the usual topology on $\Bbb R$. Similarly, $[0,b)$ is open in $A$ for $0<b\le 1$ because $[0,b)=(-1,b)\cap A$, and $(-1,b)$ is open in $\Bbb R$.
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H: Non-trivial normal subgroup of $G$ intersects the center $Z(G)$ non-trivially
I've seen a lot of information about this problem when $G$ is a $p$-group. But that need not be the case here.
Let $G$ be a group such that $G/Z(G)$ is abelian. Let $H$ be a non-trivial normal subgroup of $G$. Show $H\cap Z(G)$ is a non-trivial subgroup of $G$.
Clearly $H\cap Z(G)$ is a subgroup of $G$, but I am not sure how to show non-triviality. How does $G/Z(G)$ being abelian help us? Any hints?
AI: We must assume $G$ is nontrivial, of course, though the problem does not mention it (I guess it is implied by the existence of a nontrivial normal subgroup...).
If $G/Z(G)$ is abelian, then $G$ is nilpotent (of class at most $2$). If $G$ is finite, then it is the product of its $p$-parts, and the problem is reduced to the problem for $p$-groups, which is solved here for nilpotent groups of any class.
But here’s an argument that does not require finiteness of $G$ or knowing about the nilpotency: note that because $G/Z(G)$ is abelian, then $[G,G]\subseteq Z(G)$.
Let $h\neq e$ be an element of $H$. If $h$ is central in $G$, there is nothing left to do. If $h$ is not central in $G$, let $g\in G$ be an element such that $gh\neq hg$. Then
$$e\neq [h,g] = h^{-1}g^{-1}hg = h^{-1}(g^{-1}hg)\in H.$$
But we also have $[h,g]\in [G,G]\subseteq Z(G)$, thus exhibiting a non-trivial element of $H\cap Z(G)$.
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H: Find the arclength of the curve defined by $r(t)=i+9t^2j+t^3k$ for $0 \leq t \leq \sqrt28$.
First I found $r'(t)=\langle 1,18t^2,3t^2\rangle$ and so the magnitude of $r'(t)= \sqrt{1+(18t)^2+(3t^2)^2}$ thus the integral from $0$ to $\sqrt{28}$ of $\sqrt{1+324t^2+9t^4} dt$. When I plugged $\sqrt{28}$ in, I get $\sqrt{1+324(28)+9(784)}$ to get $\sqrt{16,129} - \sqrt{1}$ which is $126$. This is not the correct answer.
AI: Observe that
\begin{align*}
r'(t)&=18tj+3t^2k\\[4pt]
\text{length }&=\int_{0}^{\sqrt{28}}\sqrt{(18t)^2+(3t^2)^2}dt=\int_{0}^{\sqrt{28}}3t\sqrt{36+t^2}dt=\left.\frac32\frac{(36+t^2)^{3/2}}{3/2}\right|_{0}^{\sqrt{28}}
\end{align*}
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H: Is this function lebesgue integrable or not?
I'm trying to see if this function is lebesgue integrable.
$$\int_0^1 \frac{(-1)^{\lfloor 1/x \rfloor}}{x^2} dx.$$
How can I prove it?
I try the following:
Let $f(x)=\frac{(-1)^{\lfloor 1/x \rfloor}}{x^2}$.
\begin{align*}
\int_0^1 |f(x)| dx&=\sum_{n=1}^{\infty} \int_{1/(n+1)}^{1/n} |f(x)| dx\\
&=\sum_{n=1}^{\infty} \int_{1/(n+1)}^{1/n} \frac{1}{x^2} dx\\
&=\sum_{n=1}^{\infty} \left(\frac{-1}{n}+\frac{1}{n+1}\right)<\infty.
\end{align*}
Thus $f(x)$ is L.I.
I'm wrong?
AI: No, the integrand here is not Lebesgue integrable. To prove that $f : x↦ \frac{(-1)^{\lfloor 1/x\rfloor}}{x^2}$ is Lebesgue integrable, you have to prove that $f$ is measurable (this is the case here) and that $\int_0^1 |f(x)|\,\mathrm{d}x$ is finite. But $|f(x)| = \frac{1}{x^2}$, therefore,
$$
\int_0^1 |f(x)|\,\mathrm{d}x = \int_0^1 \frac{\mathrm{d}x}{x^2} = \lim_{\varepsilon\to 0} \int_\varepsilon^1 \frac{\mathrm{d}x}{x^2} =\lim_{\varepsilon\to 0} \left(\frac{1}{\varepsilon}-1\right) = +\infty.
$$
Your error comes from the fact that $\frac{1}{1/n} = n$ (actually I had difficulties to find it) so your last sum should be
$$
\sum_{n=1}^\infty\left((n+1)-n\right) = \infty.
$$
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H: Show the following defines a topology on $\mathbb{R}$
For the open sets of $\mathbb{R}$ take those sets that are open in the
usual sense and periodic with period $1$ ( $t \in U \iff t+1 \in U \;
, \forall U \; \; open $) PROVE this is a topology on R
attempt:
I am having some trouble proving this.Trivially $\varnothing $ satisfies this and $\mathbb{R}$ as well.
Now, if we consider a colletion $(U_{\alpha})$ of open sets, then take $x \in \bigcup U_{\alpha}$
then $x+1 \in \bigcup U_{\alpha}$ it follows that $x+1\in U_{\alpha}$ for some $\alpha$ so the union is in the collection
finally, take $U_1$ and $U_2$ and let $x \in U_1$ and $x \in U_2$ so$x+1 \in U_1$ and $x+1 \in U_2$ and so $x + 1 \in U_1 \cap U_2$
so $U_1 \cap U_2$ is in the collection
is this enough to prove this collection is a topology on $\mathbb{R}$?
AI: No, the open sets have the extra condition
$$\forall x \in U: x+1 \in U\tag{1}$$
which is voidly satisfied for $\emptyset$ and trivial for $U=X$.
If $U_i, i \in I$ is a collection of periodic Euclidean open sets, $U:=\bigcup_i U_i$ is Euclidean open (we know that the Euclidean topology is a topology) and to check it is still periodic, take any $x \in U$. Then $x \in U_i$ for some $i$, and as $U_i$ is periodic, $x+1 \in U_i$ and so $x+1 \in U$. So $U$ is periodic (it satisfies $(1)$).
The finite intersection you did OK.
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H: How can we conclude that $\mathbb{P}(A)\leq \mathbb{P}(B)?$
Given two events $A$ and $B$, if the event $A$ happens, then $B$ happens. How can we conclude that
$$\mathbb{P}(A)\leq \mathbb{P}(B)?$$
AI: "if A happens then B happens" means that
$A \subseteq B$, that imply also $\mathbb{P}[A] \leq \mathbb{P}[B]$
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H: Topological difference between the compact interval $I$ and the Cantor set
There is an homeomorphism between the Cantor set $X = 2^\omega$ (with the product topology) and the Cantor ternary set $\mathcal{C}=[0,1] \smallsetminus \bigcup_{n=0}^\infty \bigcup_{k=0}^{3^n-1} \left(\frac{3k+1}{3^{n+1}},\frac{3k+2}{3^{n+1}}\right)$, but this homeomorphism seems to hold as well between $X$ and the compact interval $I = [0,1]$, using the binary expansion, in spite of the ambiguity when the expansion ends in all 1's (e.g. $0.01111\cdots = 0.10000\cdots$), which merely duplicates the neighborhoods near these points, but the union of two open sets is always an open so that does not affect the continuity of the homeomorphism [edit: actually it turns out that it does prevent the continuity of the translation in the direction $[0,1] \rightarrow \{0,1\}^\omega$, see the answer].
I am missing any quirk stemming from this binary expansion ambiguity? Is there any reason at all why the Cantor set is presented first as a ternary set, instead as merely $I$, in topology textbooks?
Edit: Let's note the binary expansion function $f: [0,1] \rightarrow \{0,1\}^\omega$. Taking a neighborhood $V$ of $y=(0,0,1,1,1,\cdots)$ in $\{0,1\}^\omega$, defined as $V=(j_n)_{n\in \mathbb{N}}/(j_0=j_1=0 \wedge j_2=1)$, there seem to be no neighborhood $U$ of $x=0.25 \in I, x \in U$, such that its image $f(U) \subset V$.
So, they are not homeomorphic after all? Is it because I took the topology induced by the real line topology on $I$?
AI: As explained in the comments, the Cantor set $X$ is disconnected and $I$ is connected, and there cannot exist any homeomorphism between connected and disconnected spaces. Such an homeomorphism would allow to build continuous paths in $X$ from continuous paths in $I$ (by composition) and thus $X$ would be connected ; contradiction.
The proposed binary expansion function $f$ is not surjective thus not bijective, and not even continuous (as shown at the end of the question), and so is not an homeomorphism. However the conversion function in the reverse direction, $g: \{0,1\}^\omega \rightarrow [0,1]$ is continuous, but not injective.
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H: A question on a sequence problem
The question:
Given 6 beads which consist of red, blue, yellow, green, white and black. How many ways can a rings of beads be formed?
I'm confused by the question, would this mean this is asking for the combination or permutation of the colors above?
AI: It is hopefully asking for permutations. $6$ distinct objects can be arranged in a circle in $5! $ ways. But, in case of a ring, we don't count clockwise and anti-clockwise as different arrangements. So, the no. of rings that can be made is $$\frac{5! }{2}=60$$
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H: $\text{Q Find the area enclosed by the curves "}y^2+x^2=9\text{" and "}\left|\left(x^2-y\left|x\right|\right)\right|=1$
$\text{Q Find the area enclosed by the curves "}y^2+x^2=9\text{" and "}\left|\left(x^2-y\left|x\right|\right)\right|=1\text{" which contains the origin}.$
I tried to plot the graph on desmos. and got the following graph--
I am unable to approach how to get the area. perhaps definite integration would work, but I am not able to integrate it. Direct integration of the function is not possible, hence breaking it into parts would be required. the graph is symmetric on the y-axis, hence if only half the area is found out, we can deduce the next.
I am also unfortunately unable to figure out what the limits would be,(perhaps I guess they need to be obtained by solving the circle and the second curve.)
Also, there is an issue that a circle cannot be integrated directly (as it is not a function). I think to subtract the small areas from the total area of the circle would be a better idea.
please help. Thanks in advanced.
AI: Hint: you can split $\left|\left(x^2-y\left|x\right|\right)\right|$ into four functions, then, split the area you are trying to find into manageable pieces each one with a well defined function on top and bottom.
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H: express a cosine product as a single cosine
Problem
I have this expression $c\cos(\theta_1)\cos(\theta_2),$
where $c$ is a constant term that eats away any constant factor e.g the following can be considered true:
$$c\cos(\theta_1)\cos(\theta_2) = c\cos(\theta_1)\cos(\theta_2) / \pi$$
Because both expressions differ only by a constant factor.
$\theta_1\theta_2$ Are free variables. The goal is trying to find an expression of the form
$c\cos(f(\theta_1, \theta_2))$ where $f$ is just a function of the two variables.
More context
What I am trying to do is, I have two random points $p_1, p_2$ in the unit sphere. And the function:
$f(v) = (v\cdot p_1)(c\cdot p_2) = |v||p_1|\cos(\theta_1)|v||p_2|\cos(\theta_2) = c\cos(\theta_1)\cos(\theta_2)$ (in this case $c=1$)
The final goal is to find a new point $p_3$ in the sphere such that $v\cdot p_3 = f(v)$
I know there's a whole infinity of such vectors, the nice thing however is that if you find an explicit representation you can just pick one based on nice heuristics, e.g pick one on the line connecting $p_1, p_2$
AI: It seems to me you might be looking for the identity
$$2\cos(\theta_1)\cos(\theta_2) = \cos(\theta_1 + \theta_2) + \cos(\theta_1 - \theta_2)$$
Unfortunately, it does mean that
$$f(\theta_1, \theta_2) = \arccos\bigr(\cos(\theta_1 + \theta_2) + \cos(\theta_1 - \theta_2)\bigr) = \arccos\bigr(\cos(\theta_1)\cos(\theta_2)\bigr)$$
Fortunately, it does not matter much in your case.
To recap, you have two unit vectors, $\hat{c}$ and $\hat{v}$, and (many random pairs of) unit vectors $\hat{p}_1$ and $\hat{p}_2$, and you want to find a third unit vector $\hat{p}_3$ (per such pair), such that
$$\hat{p}_3 \cdot \hat{v} = (\hat{p}_1 \cdot \hat{v})(\hat{p}_2 \cdot \hat{c}) \tag{1}\label{None1}$$
Since $-1 \le \hat{p}_2 \cdot \hat{c} \le 1$, we can express $\hat{p}_3$ as rotated away from $\hat{p}_1$ around the axis $\vec{a} = \hat{v}\times\hat{p}_1$ enough to make the equation true. (In the case of $\hat{p}_2 \perp \hat{c}$, that angle is zero; $\hat{p}_3 = \hat{p}_1$.)
First, find the unit rotation axis $\hat{a}$:
$$\hat{a} = \frac{\hat{v}\times\hat{p}_1}{\left\lVert\hat{v}\times\hat{p}_1\right\rVert}$$
We rotate $\vec{p}_1$ by angle $\theta$ around unit axis vector $\hat{a}$, using Rodrigues' rotation formula, to obtain the desired vector:
$$\vec{p}_3 = \vec{p}_1\cos\theta + (\hat{a}\times\vec{p}_1)\sin\theta + \hat{a}(\hat{a}\cdot\vec{p}_1)(1 - \cos\theta) \tag{2}\label{None2}$$
Substituting $\eqref{None2}$ into $\eqref{None1}$ and writing it in Cartesian coordinate form yields a function of form
$$C_2 \cos\theta + C_1 \sin\theta + C_0 = 0 \tag{3}\label{None3}$$
where $C_2$, $C_1$, and $C_0$ are constants depending only on $\hat{a}$, $\hat{v}$, $\hat{c}$, $\hat{p}_1$, and $\hat{p}_2$.
When you know $\theta$, substituting it into $\eqref{None2}$ yields the desired $\hat{p}_3$.
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H: While using the method of proof by contradiction, are we "assuming" consistency?
I am aware of threads here and here which asks something similar. However, I had something very specific to ask under the same context.
I have a very elementary question about the connection between Godel's second incompleteness thorem and the method of proof by contradiction. From my limited knowledge that I have managed to gather from books and the internet, Godel's second theorem states that "No consistent, recursively
axiomatized theory that includes Peano Arithmetic can prove its own consistency."
Consistency means that an axiomatic system cannot result in some statement, and its negation to be both simultaneously provable from the axioms or in other words, the axiomatic system is free of contradictions.
Proof by contradiction, I suppose, is based on the above fact. Since the axiomatic system is consistent, we should not be able to prove that a statement is both true and false at the same time. In other words, we are basically assuming that the system is free from contradictions based on which, we have proofs by contradiction. Proof by contradiction is routinely used in many mathematical disciplines like real analysis which I presume includes the Peano Arithmetic for the construction of real numbers.
However, Godel's second theorem states that we can never prove formally, within that system, its own consistency.
My question: Are we then just "assuming" consistency of the axiomatic system and then construct proofs based on contradiction? Consistency has to be assumed because Godel's second theorem makes it clear that we can never prove that the axiomatic system is consistent. Please clarify if consistency is assumed or if it is not so, please clarify the error in this thought process.
AI: We do not make that assumption, per se, when we prove things in a given system. Proof by contradiction is a valid rule of inference in (classical) logic... we can use it to prove things in a consistent system and we can use it to prove things in an inconsistent system. There is no assumption here... the rules are the rules.
Of course, if you are using an inconsistent set of axioms, you will be able to, in principle, to prove every statement. That's not a very good state of affairs from a philosophical standpoint.
So we want to know our axioms are consistent, not so that we can use proof by contradiction (again, that's cart-before-horse), but so that we know we're doing something meaningful. And Godel tells us (with a few caveats) that we'll never have a proof of a strong system from a weak (i.e. "safer") system, so tells us we will basically be assuming consistency rather than proving it in a philosophically satisfying way.
(Note also that even without knowledge of Godel's theorem, we would know that we need to take some consistency on faith since we need to assume the consistency of whatever simple "system" we make our most basic arguments about how proofs fit together. The hope that Godel destroys is that within this simple system, we might prove the consistency of a strong system like ZFC, where real heavy-duty mathematics is done)
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H: If $A,B,C$ form a partition of $\Omega$ then describe the smallest $\sigma$-algebra containing the sets $A,B$ and $C$.
Question:
If $A,B,C$ form a partition of $\Omega$ then describe the smallest $\sigma$-algebra containing the sets $A,B$ and $C$.
This seems like a straightforward question but it is giving me a really hard time.
At first I thought the set is $\mathcal{F}=\{ \emptyset , \Omega \}$. However, this obviously doesn't include $A,B$ and $C$. So I thought intuitively that the set should just be $\mathcal{F} = \{ \emptyset , \Omega, A,B,C\}$. Now this makes sense as it satisfies the definition of the $\sigma$-algebra but I don't have a solid idea on how to prove that this is indeed the smallest possible set. I feel as if I'm missing some kind of fine detail that'll just make everything click but at the moment this all is just really confusing.
Any help would be greatly appreciated, thank you.
AI: $\mathcal{F} = \{ \emptyset , \Omega, A,B,C\}$ is a nice try. But unfortunately, it is still not a $\sigma$-algebra, because the complement of $A$, $A^c=\Omega-A= B\cup C$ (here, we used the assumption that $A, B, C$ form a partition of $\Omega$) is not in $\mathcal{F}$. Similarly, $B^c= A\cup C$, and $C^c=A\cup B$ are not in $\mathcal{F}$.
So we have to enlarge $\mathcal{F}$ further by including them.
Therefore, we try $\mathcal{F}=\{ \emptyset , \Omega, A,B,C, B\cup C,A\cup C, A\cup B\}$.
This is indeed a $\sigma$-algebra, which I invite you to prove it rigorously by showing that it satisfies the three properties of the $\sigma$-algebra:
(i) $\Omega\in \mathcal{F}$ (trivial!)
(ii) If $X\in\mathcal{F}$, then the complement $X^c=\Omega-X$ is still in $\mathcal{F}$ (you just have to check finitely number of times, since $\mathcal{F}$ is finite)
(iii) If $X_1,..., X_k\in\mathcal{F}$, then $X_1\cup \cdots X_k\in\mathcal{F}$
(again you just have to check finitely number of times, since $\mathcal{F}$ is finite)
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H: If 2 dice are rolled what is probability that "product of given values of dices is > 10" | "given the result is a double"
I am new in the field of probabilities. I came accross this problem.
Is my solution correct?
If 2 dices are rolled, what is the probability of P("product of given values of dices is > 10" | "given the result is a double")?
I think that P("values are identical") = 6/36 # (1,1),(2,2)..(6,6)
And all double dices whose product is greater than 10 is (4,4),(5,5),(6,6)
P("product of given values of dices is > 10" | "values are identical") = (3/6) / (6/36)
Is this true? Thank you.
AI: You can calculate the conditional probability of the events like you did:
Let $ X = \text{product of given values of dices is > 10}$ and $Y = \text{values are identical}$
$$P(X| Y) = \frac{P(X \land Y)}{P(Y)} = \frac{\frac{3}{36}}{\frac{6}{36}}$$
Your problem was taking the sample space of $P(X \land Y)$ to be the set of identical throws, where it is the whole set of two dice throws $\Omega$
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H: finding value of $p,q$ such that function $f(x)$ is continuous at $x=-1$
If $$f(x)=\left\{\begin{matrix}
\sin\bigg(\pi(x+p)\bigg)\;\; , &x<-1 \\
q\bigg(\lfloor x \rfloor^2+\lfloor x \rfloor\bigg)+1\;\;,\;\;&
x\geq -1 \end{matrix}\right.$$ where $\lfloor x \rfloor $ represent the integer part of $x,$ . Then what values of $p,q$ function $f(x)$ is continuous at $x=-1$
what i try: If function $f(x)$ is continuous at $x=-1.$ Then
$$f(-1^{-})=f(-1^+)=f(1)$$
Here $f(-1^{-})=\lim_{h\rightarrow 0}f(-1-h)=\lim_{h\rightarrow 0}\sin(\pi(-1-h)+\pi p)=\sin(-\pi +\pi p)=-\sin(\pi p)$
and $f(-1^+)=\lim_{h\rightarrow 0}f(-1+h)=\lim_{h\rightarrow 0}q\bigg(\lfloor h-1\rfloor^2+\lfloor h-1\rfloor\bigg)+1=-q+1$
and $f(-1)=q\bigg(1-1\bigg)+1=1$
so for continuity at $x=-1,$
$-\sin(\pi p)=1-q=1$, we get $q=0$ and $\displaystyle \sin(\pi p)=-1=-\sin\bigg(2n\pi+\frac{3\pi}{2}\bigg)\Longrightarrow p=2n+\frac{3}{2}$
so we have $p=2n+1.5$ and $q=0,$ But answer given as $q\in \mathbb{R}$
please help me How $q\in \mathbb{R}$ and whats wrong in my answer. Thanks
AI: When $\;x\ge-1\;$ but very close to $\;-1\;$ , say $\;x\in[-1,-1+\epsilon)\,,\,\,0<\epsilon<<1\;$, we have $\;\lfloor x\rfloor=-1\;$ ,so
$$q\left(\lfloor x\rfloor^2+\lfloor x\rfloor\right)=q(1-1)=0\implies\lim_{x\to-1^+} f(x)=1$$
whereas
$$\lim_{x\to-1^+}f(x)=\lim_{x\to-1^+}\sin\left(\pi(x+p)\right)=\sin\left(-\pi+p\pi\right)=-\sin p\pi$$
Thus, it must be
$$-\sin p\pi=1\iff\sin p\pi=-1\iff p=\frac n2\,,\,\,n=3\pmod 4\,,\,\,q\in\Bbb R$$
Of course, instead of $\;p=\frac n2,\,n=3\pmod 4\,,\,$ we can simply write $\;p=-\cfrac\pi2+2k\pi\,,\,\,k\in\Bbb Z$
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H: Prove that if $E[X_1^p]<\infty$, then $\frac{\max_{1\le i\le n} X_i}{n^{1/p}} \rightarrow 0$ in probability where $\{X_n\}$ is i.i.d and non-negative
Suppose $\{X_n\}_{n\geq 1}$ are iid and non negative. Define $M_n=\max \limits_{i=1,\ldots,n}\{X_i\}.$
Prove if $E[X_1^p]<\infty$, then $\frac{M_n}{n^{1/p}}\rightarrow 0$ in probability.
As a previous result I have that $P[M_n>x]\leq nP[X_1>x].$ Is this result useful to prove this convergence? I would appreciate any hint.
AI: Let $Y_i=X_i^{p}$. Then $(Y_i)$ is i.i.d with finite mean. By SLLN $\frac {S_n} n \to EY_1$ almost surely where $S_n=Y_1+Y_2+...+Y_n$. Now $\frac {Y_n} n=\frac {S_n-S_{n-1}} n=\frac {S_n} n-\frac {n-1}n\frac {S_{n-1}} {n-1} \to EY_1-EY_1=0$ almost surely.
It is an elementary fact that if $\frac {x_n} n \to 0$ then $\frac {\max {\{x_1,x_2,..,x_n\}}} n \to 0$ also.
We have proved that $\frac {M_n^{p}} n \to 0$ which implies $\frac {M_n} {n^{1/p}} \to 0$ almost surely. Of course almost sure convergence implies convergence in probability.
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H: The difference between the statements for sequences of function $f_n(x)$
Let I be an interval and c ∈ I.
Statement A: For all $\epsilon$ > 0, there is $\delta$ > 0 such that,for all $n ∈ \mathbb{N}$ and for all $x ∈ I$ satisfying $|x−c|≤\delta$, $|f_n(x)−f_n(c)| ≤ \epsilon$.
Statement B: For all $n ∈ \mathbb{N}$ and for all $\epsilon > 0$, there exists $\delta > 0$ such that whenever $x ∈ I$ and $|x−c|≤\delta$, then $|f_n(x)−f_n(c)| ≤ \epsilon$.
In my opinion, the difference between the statement is I think statement A says that all the functions are continuous at a certain point, whereas statement B says that each function is continuous at all points
My group says my answer is wrong but I am not sure why.
AI: Look at what the difference between the statements is:
A: For all $\varepsilon>0$, there is a $\delta>0$ such that for all $n\in \Bbb N$ and [...]
B: For all $n\in \Bbb N$ and for all $\varepsilon>0$, there is a $\delta>0$ such that [...]
The difference here is that in statement A, whatever $\delta$ you choose should work for all $n$ simulatneously. The constant $c$ is still fixed in both cases, so we are only interested in continuity at $c$, and not the whole interval.
As an example to illustrate the difference, consider the family of functions $f_n(x) = nx$ (and, if you'd like to specify it, $c = 0$). It satisfies B, since you can pick $\delta = \frac\varepsilon n$. But it does not satisfy A, because there is no single $\delta$ you can choose that works for all $n$. Higher $n$ demands ever smaller $\delta$, and you're not allowed to pick $0$.
Statement B says each function, by itself, is continuous at $c$. It does not in any way state any kind of relationship between the different functions. Statement A says more. It says that not only are they continuous at $c$, but also that as $n$ grows, their "steepness" (in the sense of $\varepsilon$-$\delta$, not in the sense of derivative) is bounded.
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H: Is there a way to add positives and negatives with the same algorithm?
It is taught to us in grade school that you can add (positive) numbers like this:
5 2 3
4 5 6
-------
5 7 9
But if we change it to 523 + (-456), we cannot use the same algorithm
5 2 3
-4 -5 -6
---------
1 -3 -3
I know that I can evaluate that to 100 + -(30) + (-3) which gets 67, but in my situation, 17 is the maximum integer, and -17 is the minimum integer (excluding the actual numbers being added). Is there a way to do this where the 523 + 456 also works?
AI: The reason why long long data type in c++ represents numbers range from $\ -2^{63}\ $ to $\ 2^{63}-1\ $ is because the representation used is $64$-bit twos complement. An analogous representations in decimal would be $n$-digit "tens complement", for some $\ n\ $, which could represent all the numbers in the range $\ -\frac{10^n}{2}\ $ to $\ \frac{10^n}{2}-1\ $, so it's not at all clear (not, at lest, to me) where your "maximum integer" of $17$ comes from, or what it is meant to be the maximum of.
Nevertheless, if you represent decimal numbers using sequences of positive and negative digits from $-9$ to $9$, as you have done, you can convert any representation to any equivalent one without needing to use any numbers outside the range $-9$ to $9$.
In your example, $\ 1\ -3\ -3\ $, for instance, first replace the rightmost digit by $7$ and decrease the next digit to the left by $1$ to get the equivalent representation $\ 1\ -4\ \ 7\ $. You don't even need to get the $7$ by subtracting $3$ from $10$, since you can get it by subtracting one less than $3$, namely $2$, from $9$.
Next, replace the new middle digit, $-4$, with $6$, and decrease the leftmost digit by $1$ to get the equivalent representation $\ 0\ 6\ 7\ $, from which the new leftmost digit $0$ can be deleted to give you $67$.
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H: Point on line closest to origin given the line's parametric equations
So I'm given the line described by:
$$\begin{cases} x = \frac{2}{3} + \lambda \\ y = \frac{1}{3} + \lambda \\ z = \lambda \end{cases}$$
And I'm asked to determine the point that is closest to the origin. I have the following formula for the distance between a point and a line (using parametric equations) in $\Bbb R^3$:
$$ d(\vec{p}, L) = \frac{\mid (\vec{p} - \vec{q}) \times \vec{r} \mid}{\mid \vec{r} \mid} $$ where $\vec{p}$ is the point and $\vec{x} = \vec{q} + \lambda\vec{r}$ are the parametric equations for the line.
Somehow I think I'm supposed to fill these in and find some sort of function for it, but I'm not really sure. Am I now supposed to calculate this for each $x, y$ and $z$ seperatly? I'm a little confused as I might go on about doing this.
AI: If $(x,y,z)$ is a point on the line, then the distance to the origin is given by
$d(\lambda)= \sqrt{(\frac{2}{3}+\lambda)^2+(\frac{1}{3}+\lambda)^2+ \lambda^2}.$
Since $d$ is minimal $ \iff d^2$ is minimal, you have to determine $t $ such that $f(t)= \min f(\mathbb R)$, where
$$f(t)=(\frac{2}{3}+t)^2+(\frac{1}{3}+t)^2+ t^2.$$
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H: Does a reduced row echolon only have one solution?
In general, when finding the reduced row echolon for a matrix, is there only one solution, even if there is no solution?
$$\begin{pmatrix}2&3&5\\ \:-2&-3&-3\end{pmatrix}$$
I got the following answer by hand (no solution because $1$ does not $= -2$):
$$\begin{pmatrix}1&0&-2\\ 0&1&1\end{pmatrix}$$
But an online calculator gave this answer (no solution because $0$ does not $= 2$):
$$\begin{pmatrix}1&1.5&2.5\\ \:0&0&2\end{pmatrix}$$
Is mine wrong or are they both correct?
Thank you
AI: Your row echelon form is wrong. The second column is a scalar multiple of the first, which means you cannot have a leading one in the second column, because elementary row operation don't change linear relations among columns.
In your matrix, the first two columns form a linearly independent set: this rules out that this matrix can be obtained from the given one by elementary row operations.
A row echelon form from your matrix is obtained by the operations
\begin{align}
\begin{pmatrix}
2 & 3 & 5 \\
-2 & -3 & 2
\end{pmatrix}
&\to
\begin{pmatrix}
1 & 3/2 & 5/2 \\
-2 & -3 & 2
\end{pmatrix}
&& R_1\gets \tfrac{1}{2}R_1
\\&\to
\begin{pmatrix}
1 & 3/2 & 5/2 \\
0 & 0 & 7
\end{pmatrix}
&&R_2\gets R_2+2R_1
\\&\to
\begin{pmatrix}
1 & 3/2 & 5/2 \\
0 & 0 & 1
\end{pmatrix}
&& R_2\gets\tfrac{1}{7}R_2
\\&\to
\begin{pmatrix}
1 & 3/2 & 0 \\
0 & 0 & 1
\end{pmatrix}
&& R_1\gets R_1-\tfrac{5}{2}R_2
\end{align}
The last step produces the reduced row echelon form.
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H: A Doubt regarding Cycloid
If we consider a cycloid made by a wheel. Then will the cycloid intersect the wheel when the wheel touches the topmost point of the cycloid? Thus will the radius of curvature be same to that of the wheel at that point?
AI: From the parametric equations
$$\begin{align}x&=R(\theta-\sin\theta),\\y&=R(1-\cos\theta)\end{align}$$
the radius of curvature is
$$\frac{R((1-\cos\theta)^2+\sin^2\theta)^{3/2}}{|(1-\cos\theta)\cos\theta-\sin^2\theta|}=2^{3/2}R\sqrt{1-\cos\theta}.$$
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H: Kolmogorov's three-series theorem- what can be said about distribution of $X_1$?
We assume that $X_n$ is i.i.d. and we know that $\sum_{n=1}^{\infty} X_n$ converges almost surely.
The question is what can be said about $X_1$ distribution.
I wanted to use Kolmogorov's three-series theorem. Then I know that:
$\sum_{n=1}^{\infty} P(|X_n|>C)$ converges
$\sum_{n=1}^{\infty} Var(Y_n)$ converges
$\sum_{n=1}^{\infty} \mathbb{E} (Y_n)$ converges
where $Y_n=X_n \mathbb{1}_{\{|X_n|<C\}}$
But what can be concluded about distribution of $X_1$ from here?
AI: If $\sum X_n$ converges almost surely then $X_n \to 0$ almost surely. Hence $X_n \to 0$ in probability. But then $P(|X_1| >\epsilon) =P(|X_n| >\epsilon) \to 0$, so $P(|X_1| >\epsilon) =0$ for every $\epsilon >0$. This implies $P(|X_1| >0)=0$ which means $X_1=0$ with probability $1$.
If you are keen on using the Three Series Theorem you can argue as follows:
Let $N>0$ and $Y_n=X_nI_{|X_n| \leq N}$. By the Three Series Theorem $\sum var (Y_n) <\infty$. But $Y_n$'s are identically distributed so all the terms in this sum are the same. This gives $var (Y_1)=0$ which means $Y_1$ is a constant random variable. Now $X_1I_{|X_1| \leq N}$ is a constant for each $N$. I will let you verify that $X_1$ itself must be a constant. But then the convergence of $\sum X_n$ shows that the constant must be $0$.
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H: A statistics problem (normal distribution)
If $X\sim N(0, \sigma^2)$, how can I compute $\operatorname{Var}(X^2)$? Here is my idea... but I cannot get there.
$$\operatorname{Var}(X^2) = E(X^4) - (E(X^2))^2$$
AI: There are many ways to do this, but here's one. The moment-generating function of $X$ is $\exp\frac{\sigma^2t^2}{2}$. Its power series begins $1+\frac{\sigma^2t^2}{2}+\frac{\sigma^4t^4}{8}$. Multiplying the $t^4$ coefficient by $4!$ gives $E(X^4)=3\sigma^4$. I assume you know $E(X^2)=\sigma^2$, so the result is $2\sigma^4$.
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H: Is $dX/dt=X(t)$ the correct ODE for $X(t)=e^t$?
For a school project for chemistry I use systems of ODEs to calculate the concentrations of specific chemicals over time. Now I am wondering if
$$ \frac{dX}{dt} =X(t) $$
the same is as
$$ X(t)=e^t . $$
As far as I know, this should be correct, because the derivative of $ e^t $ is the same as the current value. Can anyone confirm that this is correct (or not)?
I already searched for it on the internet but can't really find any articles about this. Thanks!
AI: The differential equation
$$ \frac{d X}{dt}=X(t)$$
has the general solution
$$X(t)=Ce^t$$
where $C \in \mathbb R.$
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H: why $(3/4,3/4)\notin B(0,1)$?
I have an excerpt from my textbook which explain that elements of the product topology are not all of the from $U_1\times U_2$ for $U_i\in\mathcal{T}_i$ , where the profuct topology is defined as :
Suppose that $(T_1,\mathcal{T}_1)$ and $(T_2,\mathcal{T}_2)$ are two topological spaces. Then the product topology on $T_1\times T_2$ is the topology with basis $$\mathcal{B}=\{U_1\times U_2: U_1\in\mathcal{t}_1,U_2\in\mathcal{T}_2\}$$
We call $(T_1,\times T_2, \mathcal{T})$ the topological product on $T_1$ and $T_2$
So my textbook gives this example:
Consider $B(0,1)\in\Bbb R^2$, which is open in the product topology (it is open for the usual metric $\Bbb R^2$, whicih is the 'product metric', $\mathcal{q}_p$ with $p=2$). Suppose $B(0,1)=U_1\times U_2$. Then $(3/4,0)\in B(0,1)$,so $3/4\in U_1$,and $(0,3/4)\in B(0,1)$,so $3/4\in U_2$: but then $U_1\times U_2 \ni (3/4,3/4)\notin B(0,1)$
So I don't exactly understand why $(3/4,3/4)\notin B(0,1)$?
AI: $(x,y) \in B(0,1) \iff x^2+y^2 < 1.$
But this is not true for $x=y=3/4.$
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H: Is this proof correct? [$\lim_{x\to-\infty}f=\lim_{x\to+\infty}f=+\infty\implies\ f$ has a global minimum]
I'm trying to prove that if $f : \Bbb R\to\Bbb R$ is a continuous function that verifies:
$$\lim\limits_{x\to-\infty}f=\lim_{x\to +\infty}f+\infty$$
Then $f$ has a global minimum
So, since:
$\lim\limits_{x\to-\infty}f=\lim\limits_{x\to+\infty}f=+\infty\to\exists x\in\Bbb R, \exists \delta_1,\delta_2 \gt 0 \phantom{2} / \phantom{2}\forall c\in (x, x + \delta_2) :f(c)\geqslant f(x), \forall c \in (x - \delta_1, x) : f(c) \leqslant f(x) $
Since x may not be unique, letting:
$$m=\min \{f(x_1),\ldots,f(x_i)\}, \phantom{2} i\in\Bbb N$$
We have that there is a global minimum at the $x_i$ of $m$
Is my reasoning correct?
AI: It is not clear how you've used the continuity of $f$ to conclude that $f$ has any local minimum. (What you've written after the $\implies$ is precisely that $x$ is a local minimum.)
Even assuming that that were true, you don't know whether $f$ has finitely many local minima and so, it may not make sense to take $\min$. (In fact, you don't even know if $f$ has countably many local minima. It is actually easy to construct an $f$ which does not.)
The correct idea would be to do something like the following:
Let $y_0 = f(0)$. Since $f(x) \to \infty$ as $x\to\pm\infty$, there exist $M_1, M_2$ with $M_1 < 0 < M_2$ such that
$$f(x) > y_0 \quad \forall x < M_1$$
and
$$f(x) > y_0 \quad \forall x > M_2.$$
Now, $I = [M_1, M_2]$ is compact and so, $f$ achieves its minimum on it. Let $m$ be this minimum. The claim is that this is the global minimum. Proving this is not tough. (Note that you must necessarily have $m \le y_0$ since $0 \in I$.)
Additional note:
It is possible that this minimum is achieved at uncountably many points. For example, consider $f(x) = |x-1| + |x+1|$. $f$ achieves its global minimum at all points in $[-1, 1]$. However, you do have that the function values at all these points is the same.
It is also possible that the function has infinitely many minima of different values. For example, consider
$$f(x) = \begin{cases}
0 & x = 0\\
x^2\sin\left(\dfrac1x\right) & x \in \left[-\dfrac1\pi,\dfrac1\pi\right]\setminus\{0\}\\
\left|x^2 - \dfrac{1}{\pi^2}\right| & \text{otherwise}
\end{cases}$$
Here, $f$ has infinitely many distinct local minima. Thus, your original construction (which assumes that you have only finitely many mimima) will not work.
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H: What is the intuition behind the formula $p \leftrightarrow (q \leftrightarrow (r \leftrightarrow ...))$?
I cooked up the formula $p \leftrightarrow (q \leftrightarrow (r \leftrightarrow ...))$ and naively thought it is a sort of "equivalence" relation. It turns out I am wrong. Suppose you have four variables in total, setting everything to false will make the formula true. However, if you have five variables, setting all variables to false will make the formula false! Having looked at a few experiments on truth tables, I nevertheless failed to find an easy intuition, but I am intrigued by the gap between the formula's simple construction and illusive patterns. I feel something is there but I couldn't grok it.
One thing I noticed include the fact that $\leftrightarrow$ is associative and commutative, so you can shift the variables around. Maybe this helps?
AI: Let one have the finite sequence of boolean values $p_1, \cdots, p_n$. S/he takes any two among them (say, $p, q$), remove two of them, and add $p \iff q$. This decreases the length of sequence by 1, so by $n-1$ step s/he gets to the final result.
On each step, since $T \iff T$ is true, $F \iff F$ is true and $T \iff F$ is false, the parity of Falses among $p_1, \cdots, p_n$ does not changes. So, by any choice of each steps, the parity of False after final consequence is the parity of number of Falses among initial state.
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H: Find coordinates of the center of the mass - line integral
Find the coordinates of the center of the mass of the curve
$$ x^2+y^2=1, x+2y+3z=12 $$
I find calculating line integrals in 3D problematic and really don't know how to approach this one. I think that the curve we get is an elipse but how to find its parameters?
AI: Parameterise the circle $x^2+y^2=1$ as
$$(x,y) = (\cos(t),\sin(t)) \qquad t \in [0,2\pi]$$
as usual. Plug this into the other equation:
$$x+2y+3z=12 \implies \cos(t)+2\sin(t)+3z=12 \implies z = 4 - \frac{\cos(t)+2\sin(t)}{3}$$
So the parameterisation is
$$(x,y,z) = \bigg(\cos(t),\sin(t),4 - \frac{\cos(t)+2\sin(t)}{3}\bigg) \qquad t \in [0,2\pi]$$
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H: A question about regularity of parameterizations of a surface
Can someone give an example of two permetrazations (1-1) of a surface that satisfy: At a same point (on the surface) one permetrazation is regular but the other is not regular
AI: Consider the $xy$-plane $S$. Then we can give a 1-1 parametrization $\overline{x}:\mathbb{R}^2\to S$ given by
$$\overline{x}(u,v)=(u,v,0),$$
which is regular (which you can check yourself).
Now, consider another parametrization $\overline{y}:\mathbb{R}^2\to S$ given by
$$\overline{y}(u,v)=(u^3,v^3,0),$$
Then it is still 1-1 parametrization of $S$, but it is not regular, since the differential of $\overline{y}$ at $(u,v)=(0,0)$ is given by
$$d\overline{y}_{(0,0)}=\left.\left[
\begin{array}{cc}
3u^2 & 0 \\
0 & 3v^2 \\
0 & 0 \\
\end{array}
\right]\right|_{(u,v)=(0,0)}=\left[
\begin{array}{cc}
0 & 0 \\
0 & 0 \\
0 & 0 \\
\end{array}
\right]$$
which does not have full column rank.
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H: Set notation: $U\otimes V\simeq V\otimes U$ with $V,U$ vector spaces
Let $U,V$ be finite-dimensional vector spaces over a common field and show that $U\otimes V\simeq V\otimes U$. Will someone explain what the notation $\simeq$ means in this case?
AI: It means that there exists a vector space isomorphism between $U \otimes V$ and $V \otimes U$.
In this case, this is very easy: $U$ and $V$ are finitedimensional and both $U \otimes V$ and $V \otimes U$ have dimension $\dim V \dim U$, so they are isomorphic.
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H: Find $\sigma$ for which $\sum _{i=1}^{n} \frac{1}{a_i\cdot a_{\sigma(i)}}$ is maximal
$$\text{Let: } 0 < a_1 \lt a_2 \lt \dots \lt a_n$$
$$\text{Find } \sigma \in S_n \text{ for which :}$$
$$\sum _{i=1}^{n} \frac{1}{a_i\cdot a_{\sigma(i)}} \text{ is maximal}$$
I think the maximum value of the sum will be reached if $\sigma = e$ but i don't know how to prove that. Any idea is really appreciated! Thanks for help!
Note: $e = \begin{pmatrix}
1 & 2 & ... & n\\
1 & 2 & ... & n
\end{pmatrix}$
AI: By Cauchy Schwarz inequality
$$\sum_{i=1}^n \frac{1}{a_ia_{\sigma(i)}} \leq \left(\sum_{i=1}^n\frac{1}{a_i^2}\right)^{1/2}\left(\sum_{i=1}^n\frac{1}{a_{\sigma(i)}^2}\right)^{1/2}=\sum_{i=1}^n\frac{1}{a_i^2}$$
The equality occurs when $a_i=\lambda a_{\sigma(i)}$ for all $i$. This would mean that $$\prod_ia_i=\lambda^n \prod_{i}a_{\sigma(i)}=\lambda^n\prod_ia_i \implies \lambda=1$$
Thus $\sigma$ is the identity permutation.
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H: To prove that $p$ is a prime number
I'm reading a book about proofs and fundamentals on my own and, currently, I'm having trouble proving this result.
Theorem: Let $p$ be a positive integer bigger than or equal to $2$ and such that, for any integers $a$ and $b$, if $p|ab$, then $p|a$ or $p|b$. Show that $p$ is a prime number.
Sketch Work: My work so long has been assuming that $p$, in this condition, is a composite number. Hence, there exists some integers $m$ and $n$, such that $p = mn$ and $1 < m,n < p$. If $p|ab$, then there is some integer $k$ such that $kp = ab$. How do I show from here that $p$ will not divide $a$ and $b$ and hence we have a contradiction?
Is my strategy good? What should be the next step? Thank you!
AI: If $p$ is composite, then $p=mn$ where $1<m,n<p.$ We have $p \mid mn,$ so by assumption $p\mid m$ or $p \mid n$ which is a contradiction since $m,n<p.$
Edit: The hypothesis says that if $p \mid ab,$ then $p \mid a$ or $p \mid b.$ In order to use this, you need to find integers $a$ and $b$ such that $p \mid ab.$ The hypothesis does not say that there definitely exist such integers. Assuming that the conclusion does not hold guarantees such existence and then puts you in a position where you can use the hypothesis.
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H: EDIT: Can cdf have a set of continuity points which is a.s. different from $\emptyset$?
It can be easily proven that a cumulative distribution function (cdf)
$$F(x)=\mathbb{P}\left(X\le x\right)\hspace{0.3cm}\text{for }-\infty<x<+\infty$$
is right-continuous $\left(\text{that is }F(x)=F(x^+)\right)$, but NOT left-continuous $\left(\text{that is }F(x)\neq F(x^-)\right)$.
Given this, could it be even possible that there exists a set of continuity points of $F$
$$D=\{x: F(x^-)=F(x)\}$$
which is a.s. different from $\emptyset$? (That is, considering $\mathbb{P}$ as the probability measure corresponding to the cdf $F$, $\mathbb{P}(D\neq\emptyset)=0$). Clearly, for $D$ it would be sufficient to specify the left-continuity condition, since, as above-stated, right-continuity is already granted by definition of cdf. However, the question is: could one have that such a D is a.s. different from $\emptyset$ if cumulative distribution functions are known to be not left-continous at any points at all?
AI: The set $D^{\complement}$ of discontinuity points is a countable set.
This becomes clear if we write $D^{\complement}=\bigcup_{n=1}^{\infty}E_n$ where $E_n=\{x\in\mathbb R\mid F(x)-F(x^-)>\frac1{n}\}$.
Observe that for every $n$ the fact $P(E_n)\leq1$ tells us that the set $E_n$ has less than $n$ elements.
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H: Direction Derviative, Different Results, Possible?
Given:
\begin{equation*}
f(x,y) = e^{xy}, \quad P=(1,1),\ \mathbf{v} = (-\frac{\sqrt{3}}{2}, \frac{1}{2})
\end{equation*}
I am trying to calculate the directional derivative of $f$ in direction of $\mathbf{v}$.
My problem is, that when I apply two different ways of caculating the derivative I get different results. However I do not find a mistake in my calculation:
1) First approach:
\begin{align*}
f(P) &= e^{1*1} = e \\
P + t \mathbf{v} &= (1,1)^T + (-\frac{t\sqrt{3}}{2}, \frac{t}{2})^T = (1 -\frac{t\sqrt{3}}{2}, 1 + \frac{t}{2})^T \\
\end{align*}
\begin{equation*}
(1 -\frac{t\sqrt{3}}{2})(1 + \frac{t}{2}) = 1 + \frac{t}{2} - \frac{t \sqrt{3}}{2} - \frac{t^2 \sqrt{3}}{4}
\end{equation*}
\begin{align*}
f(P + t \mathbf{v}) = e^{1 + \frac{t}{2} - \frac{t \sqrt{3}}{2} - \frac{t^2 \sqrt{3}}{4} } \\
\lim_{t \rightarrow 0} \frac{f(P + t \mathbf{v}) - f(P)}{t} = \lim_{t \rightarrow 0 } \frac{e^{1 + \frac{t}{2} - \frac{t \sqrt{3}}{2} - \frac{t^2 \sqrt{3}}{4} } - e}{t} = \lim_{t \rightarrow 0 } \frac{0}{t} = 0
\end{align*}
2) Second approach:
The directional derivative should be equal to the Jacobian of $f$ at $P$ times $\mathbf{v}$:
\begin{align*}
J(f) &= [ye^{xy}, xe^{xy}] \\
J(f, P) &= [e, e] \\
J(f, P) \mathbf{v} &= [e, e] (-\frac{\sqrt{3}}{2}, \frac{1}{2})^T \\
&= \frac{e}{2}(1 - \sqrt{3})
\end{align*}
However that is not equal to zero. I am inclined to believe that there is a mistake in my first approach, but the calculations seem correct.
AI: First notice that you've made a mistake here:
$$(1 -\frac{t\sqrt{3}}{2})(1 + \frac{t}{2}) = 1 + \frac{t}{2} - \frac{t \sqrt{3}}{2} - \frac{t^2 \sqrt{3}}{4} = \frac{2 +t(1 - \frac{\sqrt{3}(1+t)}{2})}{2}$$
If you distribute the $t$ in the last equation you get $1 + t/2 -t(\sqrt 3/4)-t^2(\sqrt 3/4)$ regardless I don't see the point of your last equality since $1 + \frac{t}{2} - \frac{t \sqrt{3}}{2} - \frac{t^2 \sqrt{3}}{4}$ is perfectly fine to work with when taking a limit as $t \to 0.$
Your biggest mistake however comes from when you calculate the limit as $t \to 0$ of
$$Q_t = \frac{f(P+tv)-f(P)}{t} = \frac{\exp(1+ \frac t 2 - \frac{\sqrt3}{2}t - t^2 \frac{\sqrt 3}{4} ) - e}{t}.$$
It is correct that the numerator converges to $e- e = 0$ but this doesn't mean that $Q_t \to 0$ since the numerator also converges to $0$.
You can use l'hospital's rule to find this limit. Taking the derivative of the numerator and denominator you get
$$ \frac{(1/2 - \sqrt 3/2 - t(\sqrt 3/2) )\exp(1+ \frac t 2 - \frac{\sqrt3}{2}t - t^2 \frac{\sqrt 3}{4})}{1}$$
which converges to $$\left(\frac{1}{2} - \frac{\sqrt 3}{2} \right) e = \frac{e}{2}(1 - \sqrt 3).$$
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H: Why is that if $a\in\overline{A}$ then for every $n>0$ we have $B(a,1/n)\bigcap A \neq \emptyset$.
I don't quite understand one thing here:
Why is that if $a\in\overline{A}$ then for every $n>0$ we have $B(a,1/n)\bigcap A \neq \emptyset$.
I drew a picture below to visualize my understanding of this:.
Well, as you see it's not quite clear why the ball has some intersecting points with $A$.
I know the defintion of closure, which is:
The closure of $A,\overline{A}$, is the set $$\begin{align*}\overline{A}&:=\{x\in T: U\bigcap A\neq\emptyset \text{ for every open set $U$ that contains $x$}\}\\&=\{x\in T : \text{ every (open) neighbourhood of $x$ interects $A$}\}\end{align*}$$
Is my second visualization more accurate?
AI: If you look closely at your definition of $\overline{A},$ you would realize your picture is incorrect. In fact, since $\overline{A}$ consists of those $x$ whose every neighbourhood intersects $A,$ therefore in particular the neighbourhood $B(a,\frac1n)$ of $a$ needs to intersect $A.$ In other words $$B\left(a,\frac1n\right) \cap A \neq \emptyset.$$
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H: Value of m for which the function will give integers as an output.
$F(m)=(2m^3+2m)/(m^2+1)$ and $g(m)=(m^4+1)/(m^2+1)$ What are the values of $m$ other than $1$ for which solution of both function will be integers. Please tell if there is any formula to find so or any technique?
AI: Whenever $m^2+1 \ne 0$, you find that $F(m)=2m$ and $g(m)=m^2-1+2/(m^2+1)$, so $F(m)$ will be an integer for all integers $m$.
Assumig that only integers are allowed for $m$, $g(m)$ will be an integer just for $m \in \{-1,0,1\}$.
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H: Evaluate $\lim_{n\to \infty} \sum_{k=0}^n \frac{\sqrt {kn}}{n}$
I'm not sure which would be the best way to compute this limit. As you might have observed, if you expand the infinite sum and rearrange some terms you get:
$$\lim_{n\to \infty}\frac{\sqrt n}{n}\!\cdot\!\sum_{k=0}^n \sqrt k$$
it is easy to see that when you take the limit it yields to an indetermination of the type: $$\mathrm{0}\!\cdot\!\mathrm{\infty}$$ which you can easily transform into one of the type: $$\frac{\infty}{\infty}$$ by dividing by the opposite:
$$\lim_{n\to \infty}\frac{\sum_{k=0}^n \sqrt k}{\frac{n}{\sqrt n}}=\lim_{n\to \infty}\frac{\sum_{k=0}^n \sqrt k}{\sqrt n} $$
I don't know if using L'Hôpital is valid here (as there is an infinite sum in the numerator), how ever, if you differentiate both numerator and denominator, all the square roots in the numerator will go away (except for $\sqrt n$) because they are just constants and you will be left with:
$$\lim_{n\to \infty}\frac{\frac{1}{\sqrt n}}{\frac{1}{\sqrt n}} = \lim_{n\to \infty}\frac{\sqrt n}{\sqrt n} =\lim_{n\to \infty} 1 = 1$$
Is this result correct? I don't even know if it is possible to take the limit of an infinite sum which is not an integral as you are taking discreet steps instead continuous ones.
Thank you for everything.
AI: By the trapezoidal rule, you have\begin{align}\sum_{k=1}^n\sqrt k&\approx\frac12(\sqrt1+\sqrt n)+\int_1^n \sqrt x\,dx\\&=\frac12+\frac12\sqrt n+\frac23n^{3/2}-\frac23\\&=\frac12\sqrt n+\frac23n^{3/2}-\frac16\end{align}and therefore\begin{align}\lim_{n\to\infty}\frac1{\sqrt n}\sum_{k=1}^n\sqrt k&=\lim_{n\to\infty}\frac12+\frac23n-\frac1{6\sqrt n}\\&=\infty.\end{align}
|
H: What does it mean when a system is made dimensionless and what is the exact technique for that?
For school research I'm working on a system of ODEs to describe a chemical oscillator (the Oregonator). This system is described with the following system:
$$ \frac {dX}{dt}=k_1AY-k_2XY+k_3AX-2k_4X^2 $$
$$ \frac{dY}{dt}=-k_1AY-k_2XY+\frac{1}{2}k_cfBZ $$
$$ \frac{dZ}{dt}=2k_3AX-k_cBZ $$
The variables $k_- $ are all constants and so are $f$, A and B.
Most papers however go on to make it dimensionless. What does it exactly mean to make a system dimensionless (in chemistry) and what is the technique I should use to make a (or the above) system dimensionless?
AI: Begin by writing each variable as a constant of the same dimension, times a dimensionless variable. Then simplify your equations to cancel such constants. For example, let's write $X=X_0x$ with $X_0$ constant and $x$ dimensionless, and handle your other capitalized variables the same way. We also need to nondimensionalize time, e.g. with $t=t_0\tau$ where $t_0$ is a constant with the dimension of time. So$$\begin{align}\frac{dx}{d\tau}&=\frac{t_0}{X_0}\frac{dX}{dt}\\&=\frac{t_0}{X_0}\left(k_1A_0Y_0ay-k_2X_0Y_0xy+k_3A_0X_0ax-2k_4X_0^2x^2\right)\\&=K_1ay-K_2xy+K_3ax-2K_4x^2\end{align}$$where I've defined$$K_1:=\frac{t_0k_1A_0Y_0}{X_0},\,K_2:=t_0k_2Y_0,\,K_3:=t_0k_3A_0,\,K_4:=t_0k_4X_0.$$These constants are also dimensionless: for example, $K_1$ has the same dimension as the ratio of the dimensionless $dx/d\tau$ to the dimensionless $ay$. So if you go through all the equations this way, only dimensionless quantities will remain.
|
H: What is the radius of the small circle inscribed in a square?
If we are given a square with sides of length 4cm. The smaller circle is tangent to the larger circle and the two sides of the square as shown in the photo below. How can i find the length of the radius of the smaller circle?
My approach :
The radius of the large circle is 2cm ( 1/2 length of the square side ).
there is a right triangle( 45-45-90) where each leg is equal 2cm , and the hypotenuse( which is the segment from the center of the big circle passing through the center of the small one until the vertex of the square ) is 2$ \sqrt 2 $
the segment that joins the radius of the small circle until the vertex of the square is equal -2+2 $\sqrt 2 $.
I think in order to know the radius i have to subtract ( -2+2$\sqrt2 $- x ) where x is the small space between the vertex of the square and the arc of the small circle.
Thanks in advance.
AI: Let $A$ be the upper right corner of the square. The smaller circle is the image of the larger circle by a homothetic transformation centered at $A$. The ratio of this transformation is $\frac{\sqrt{2}-1}{\sqrt{2}+1}$ because the farthest point from $A$ in the large circle is at a distance $2(\sqrt{2} + 1)$ and its image is the nearest point from $A$ which is at distance $2(\sqrt{2}-1)$. Hence the radius of the smaller circle (assuming the larger circle has radius 2) is $2 \frac{\sqrt{2}-1}{\sqrt{2}+1} = 6 - 4 \sqrt{2}$
|
H: Poisson or exponential distribution
I have to complete the question for a homework task and I'm confused if it is a Poisson or exponential distribution. any insight would be appreciated.
cars arrive at a car wash in a town at an average rate of 50
per hour.
Q: If you arrive within a three minutes of another car , you must wait outside. What is the probability that, when you turn up to vote, must wait outside?
AI: The time between consecutive rivals is exponentially distributed, with a mean of $2$ minutes. It couldn't be Poisson-distributed, as it's continuous. It's at most a minute with probability $1-e^{-1/2}$. What is Poisson-distributed is the number of arrivals per hour, or minute or whatever.
|
H: Covariant and contravariant components of vectors
I am struggling with the covariance and contravariance of vectors. In my physics classes, the professor explained that if covariant components transform with a certain matrix, then contravariant components transform with its inverse. However, I find the latter to be the transpose of the inverse.
Here is my line reasoning:
Let $ V $ be an $ n $-dimensional real vector space and let $ \langle , \rangle $ be a positive definite scalar product on $ V $.
Let $ \mathcal{B} = \{e_i, \dots , e_n\} $ and $ \mathcal{B}' = \{e_i', \dots , e_n'\} $ be bases of $ V $ and $ v \in V $. We denote by $ X $ the components of $ v $ with respect to $ \mathcal{B} $ and with $ X' $ its components with respect to $ \mathcal{B}' $. We have that $ X = NX' $, where $ N = ((e_1')_{\mathcal{B}}, \dots, (e_n')_{\mathcal{B}}) $, i.e. $ N $'s columns are the coordinate vectors of the $ e_i' $s with respect to $ \mathcal{B} $.
We define $ \varphi : V \to V, \; \varphi (e_i) = e_i' $. Then, if we fix the base $ \mathcal{B} $ on both the domain and the codomain, $ \varphi $ is represented by the matrix $ N $.
We define $ X $ to be the contravariant components of $ v $ (with respect to $ \mathcal{B} $).
Let $ C $ be the $ n \times n $ matrix that represents $ \langle, \rangle $ with respect to $ \mathcal{B} $.
We have that the covariant components of $ v $ in the base $ \mathcal{B} $ are
\begin{equation}\label{}
v_i = \langle v, e_i \rangle = \langle e_i, v \rangle = ((e_i)_{\mathcal{B}} )^TCX
\end{equation}
while those in the base $ \mathcal{B}' $ are
\begin{equation}\label{}
v_i' = \langle v, e_i' \rangle = \langle e_i', v \rangle = ((e_i')_{\mathcal{B}})^TCX = (N(e_i)_{\mathcal{B}} )^TCX = ((e_i)_{\mathcal{B}} )^TN^TCX
\end{equation}
Then we have
\begin{equation}\label{}
\begin{pmatrix}
v_1' \\
\vdots \\
v_n' \\
\end{pmatrix}
= N^TCX = N^T
\begin{pmatrix}
v_1 \\
\vdots \\
v_n
\end{pmatrix}
\end{equation}
This shows that to get from $ X $ to $ X' $ I use the matrix $ N^{-1} $, while for the covariant components I use the matrix $ N^T $.
What am I doing wrong?
AI: I think the problem is in the formalism. You are using the intrinsic one, but my experience is that physicists like using indices.
Assume that the $e_i$s are the vectors of a basis of $V$ and that the $\varepsilon^i$s are their duals. You can take different bases $e_{k'}$ and $\varepsilon^{k'}$ so that $e_{i'} = E_{i'}^k e_k$ and $\varepsilon^{l'} = P_j^{l'}\varepsilon^j$ for some coefficients $E_{i'}^k$ and $P_j^{l'}$ giving matrices $E$ and $P$ (I am using the Einstein convention here, my $E$ is your $N$). But then
\begin{align*}
\varepsilon^{i'}(e_{k'}) & = \delta_{k'}^{i'} \\
& = P^{i'}_j\varepsilon^j(E_{k'}^le_l) \\
& = P_j^{i'}E_{k'}^l \delta_l^j = P_l^{i'}E_{k'}^l.
\end{align*}
Without using indices, this means that $PE = \mathrm{id}$. But $E$ is invertible, so $P=E^{-1}$.
You can also define an action of a matrix on a covector saying $P$ acts on $\varepsilon^k$ as $P\varepsilon^k := \varepsilon^k(P^{-1}\cdot{})$, i.e. as the inverse transpose. Take $E$ to be the change of basis from $e_i$ to $e_{k'}$ and $P$ its inverse. Then $P\varepsilon^k(Ee_i) = \varepsilon^k(P^{-1}Ee_i) = \varepsilon^k(e_i) = \delta_i^k$, which means that after a change of coordinates $P\varepsilon^k$ is the dual of $Ee_k$. You can then rephrase this by saying that the covector $\varepsilon^k$ changes by the inverse transpose of $P$.
|
H: If $A_t = cos(X_t)$ and $B_t = sin(X_t)$ find the infinitesimal increment for $Y_t = A_t^2 + B_t^2$
If $X_t$ is Brownian motion, I'm not sure how to apply Ito's lemma to get $d Y_t$ for $ Y_t = A_t^2 + B_t^2$ where $A_t = cos(X_t)$ and $B_t = sin(X_t)$ in particular, I get confused because $sin^2(x) + cos^2(x) = 1$ so shouldn't $d Y_t$ be just 0 and I'm not sure how to handle the correlation between $A_t$ and $B_t$ which are clearly not independent.
AI: Your reasoning using the trigonometric identity is correct and so applying Ito's lemma is unnecessary here. However it is possible and gives the same result.
In the case of $f(x) = x^2$, Ito's lemma tells you that
$$dA_t^2 = 2 A_t dA_t + d \langle A \rangle_t$$
This means we want to find an SDE satisfied by $A_t$. Ito's lemma applied with the function $\cos$ tells us that
$$dA_t = - \sin(X_t) dX_t - \frac{1}{2} \cos(X_t) dt$$
where we use the fact that $\langle X \rangle_t = t$. Using this, we also find that $d \langle A \rangle_t = \sin(X_t)^2 dt$.
Hence substituting into the first equation gives $$dA_t^2 = - 2 \cos(X_t) \sin(X_t) dX_t - \cos(X_t)^2 dt + \sin(X_t)^2 dt$$
Similar calculations show that
$$dB_t^2 = 2 \cos(X_t) \sin(X_t) dX_t - \sin(X_t)^2 dt + \cos(X_t)^2 dt.$$
Hence,
$$dY_t = dA_t^2 + dB_t^2 = 0$$
as you predicted.
|
H: Second order ODE solution - help me spot a mistake
Solve following ODE:
$$
(1-x)x''+2(x')^2=0; x(0)=2, x'(0)=-1
$$
$$
x''=\frac{-2(x')^2}{1-x}
$$
substitute $x'=u(x)$ and assume $u \neq 0$
$$
uu'=\frac{-2(u)^2}{1-x} \\
\frac{dx}{1-x}=-\frac{du}{2u} \\
-\ln{|1-x|}=-\frac{1}{2}\ln{|u|}+c \\
|1-x|=\sqrt(|u|)e^c
$$
undo substitution $x'=u(x)$ and use $x(0)=2, x'(0)=-1$:
$$
|1-2|=\sqrt(|-1|)e^c \\
1 = e^c \\
c = 0
$$
so
$$
|1-x|=\sqrt{|x'|}\\
x' = (1-x)^2 \\
\frac{dx}{dt}=(1-x)^2 \\
dt = \frac{dx}{(1-x)^2} \\
t + c = \frac{1}{1-x}
$$
use $x(0)=2$
$$
0 + c = \frac{1}{1-2} \\
c = -1
$$
finally:
$$
t - 1 = \frac{1}{1-x} \\
\frac{1}{t-1} = 1 - x \\
x = 1 - \frac{1}{t-1} \\
x(t) = \frac{t-2}{t-1}
$$
Wolfram Alpha however, says the ODE is $x(t)=\frac{t+2}{t+1}$. Where is my mistake?
AI: The first stage of the solution should be kept with an undetermined constant up to the point
$$
u(x)=C(x-1)^2.
$$
Then using $u(2)=-1$ gives $C=-1$ and thus
$$
x'(t)=-(x-1)^2,
$$
which has a different sign than the equation you got. The sign difference in the solution follows from here.
|
H: $\exp ^r z=\exp rz$ for all $z\in\mathbb{C}$ and $r\in\mathbb{R}$
Let $z\in\mathbb{C}$ and $r\in\mathbb{R}$. Assuming that $\exp ^r z$ can be a multi-valued function (and $\exp rz$ cannot), there always exists (for any given $z$ and $r$) some value of $\exp ^r z$ such that
$$\exp ^r z=\exp rz.$$
How I could I prove this?
As a starting point, I'm using the fact that $\exp z=\sum_{n\ge 0}\frac{z^n}{n!}$ for all $z\in\mathbb{C}$.
I'm motivated by the following proof that, for all $w,z\in\mathbb{C}$, $\exp w\exp z=\exp (w+z)$:
$$\begin{align}\exp w\exp z&=\sum_{n\ge 0}\frac{w^n}{n!}\sum_{k\ge 0}\frac{z^k}{k!}=\sum_{n\ge 0}\sum_{0\le k\le n}\frac{w^k}{k!}\frac{z^{n-k}}{(n-k)!}\\&=\sum_{n\ge 0}\frac{1}{n!}\sum_{0\le k\le n}\frac{n!}{k!\, (n-k)!}w^k z^{n-k}=\sum_{n\ge 0}\frac{(w+z)^n}{n!}\\&=\exp (w+z).\end{align}$$
Maybe the proof of the relation in my question could be done in a similar fashion.
AI: Since $\exp w\exp z=\exp(w+z)$, the cases $r\in\Bbb N$ follow by induction, and $r=0$ is trivial, and $\exp-w\exp w=1$ extends to $r\in\Bbb Z$. We get to $\in\Bbb Q$ by writing $r=p/q$ with $p\in\Bbb Z,\,q\in\Bbb N$ so$$(\exp^rz)^q=\exp^pz=\exp pz=\exp(qrz)=(\exp rz)^q.$$We extend to $r\in\Bbb R$ by continuity in $r$: for any sequence of rationals $r_n$ with $\lim_{n\to\infty}r_n=r$,$$\exp^rz=\lim_{n\to\infty}\exp^{r_n}z=\lim_{n\to\infty}\exp(r_nz)=\exp rz.$$
|
H: Leftside limit of the Distribution function
Let $(\Omega,\mathrm{P})$ be a probabilty space and $X$ be a random variable.
Why is $\lim\limits_{s \uparrow t}\mathrm{P}(X < s)= \mathrm{P}(X \leq t)$?
I first thought that this follows from the continuity from below, but this doesn't works. If I consider $x_n \to t $ with $x_n < t$ then $x_n$ can also decreasing in some points...
AI: For a monotonically increasing function $F$ we can write $\sup \{F(y): y<x\}=\lim_{ y \uparrow x}F(x)=\lim F(x_n)$ where $x_n$ is any sequence incresaing to $x$. Hence $\lim_{s\uparrow t} P(X<s)$ is same as $\lim P(X<t-\frac 1 n)=P(X \leq t)$.
|
H: finding a power series for $f(z)=\frac{1}{1+z^2}$ centered at $0$
In an exercise I am asked to find a power series for $f(z)=\frac{1}{1+z^2}$ centered at $0$.
My approach was the following:
$f(z)=\frac{1}{1+z^2}=\frac{1-z^2}{(1+z^2)(1-z^2)}=(1-z^2)\frac{1}{1-z^4}=(1-z^2)\sum_{n\geq0}z^{4n}$
But this does not seem right because a power series is defined as $\sum a_n(z-a)^n$ and I have something like: $\sum a_n(z-a)^{kn}$, plus I don't know how to pull that $(1-z^2)$ inside the sum.
So how can I write this function as a power series?
Usually what is the main approach when trying to write a function as a power or as a Laurent series? Because I allays find myself having some trouble doing so.
Edit:
I just realized that we can continue to manipulate this expression: $(1-z^2)\sum_{n\geq0}z^{4n}$
$$(1-z^2)\sum_{n\geq0}z^{4n}=\sum_{n\geq0}z^{4n} - z^2\sum_{n\geq0}z^{4n}=$$
$$\sum_{n\geq0}z^{4n} -\sum_{n\geq0}z^{4n+2}=\sum_{n\geq0}(-1)^n z^{2n}$$
AI: $\frac 1 {1+z^{2}}=1-z^{2}+z^{4}-z^{6}+...$ for $|z| <1$. RHS is a geometric sum and this is special case of the sum of a geometric series.
|
H: A proof question using Riemann Lemma
Let $l>0$, $f(x)$ is continuous on $[-l,l]$ and differentiable at $x=0$. Please use Riemann lemma to prove that
$$\begin{equation}
\lim_{n\rightarrow\infty}{\frac{1}{\pi}\int_{-l}^{l}{f(x)\frac{\sin{nx}}{x}}dx}=f(0)
\end{equation}$$
I am sorry for being stupid but I am stuck at this question for 2 hours with no clue of how to even start. Please help me......
AI: Hint: $\frac 1 {\pi} \int_{-l} ^{l} \frac {sin (nx)} x dx\to 1$ as $n \to \infty$ (as seen by making the substitution $y=nx$). Hence it is enough to show that $\int_{-l}^{l} (f(x)-f(0)) \frac {sin (nx)} x dx \to 0$. Let $g(x)= \frac {f(x)-f(0)} x$ for $x \neq 0$ and $g(0)=f'(0)$. Apply Riemann Lebesgue Lemma to this function to finish the proof.
|
H: Show that the function $f(x)=\frac{x-1}{2(x+1)}$ is continuous in $a=3$
What I've done is the following.
$$\biggl|\frac{x-1}{2(x+1)}-\frac{3-1}{2(3+1)} \biggr|<\epsilon$$
By some calculation, I got
$$\biggl|\frac{3x-5}{2(x+1)} \biggr|<\epsilon.$$
This is greater than zero when $x>5/3$ or $x<-1.$
So, then, we do the calculations to get two deltas for these relations and the complement of these relations. We get $\delta_1$ and $\delta_2,$ and for delta, we pick the minimum of these two.
The thing is that when I do these calculations, there is a lot of relations that come into play (like when dividing, multiplying, and so on). I have a feeling that my approach isn't correct. Is that true?
AI: You have made some mistakes in calculation. You should get $|f(x)-f(3)|$ becomes $|\frac {x-3} {4(x+1)}|$ which is $\leq \frac {|x-3|} {4( 4-|x-3|)}$ since $|x+1|=|(x-3)+4 | \geq 4-|x-3|$. Hence we we want $\delta$ such that $\frac {\delta} {4(4-\delta)} <\epsilon$ which is same as $\delta <\frac {16 \epsilon} {1+4\epsilon}$.
|
H: Longest Rubik's cube algorithm - maximization problem
For a given Rubik's cube algorithm $A$ let $\mathfrak C(A)$ be the number of times, we have to repeat algorithm $A$ to get back to where we've started. For example if $A=RUR'U'$ then $\mathfrak C(A)=6$ The question is:
What is the greatest value of $\mathfrak C(A)$, we can achieve?
This question is equivalent to following maximization problem:
Maximize $LCM(a_1,a_2,...a_i,b_1,b_2,...b_j)$ under the following conditions:
$$~a_1,a_2,...a_i,b_1,b_2,...,b_j \in \mathbb N_+~$$
$$a_1+a_2+...+a_i=8$$
$$b_1+b_2+...+b_j=12$$
$$2 \ | \ (a_1+a_2+...+a_i+b_1+b_2+...+b_j)$$
Where LCM is Least Common Multiple function.
Thanks for all the help.
AI: Wikipedia's article on the Rubik's cube group says that the largest order of any element in the group is 1260. For instance, $RU^2D'BD'$ is one such move.
|
H: Not every matrix on V ⊗ W can be written as a tensor product of a matrix on V and another on W.
I am reading some notes about tensor product of vector spaces (those in here) where the following sentence can be found:
Note that not every matrix on V ⊗ W can be written as a tensor product
of a matrix on V and another on W.
Matrices in V ⊗ W are defined as a Kronecker product. So, how could this claim be possible? Has it to do with the problem of splitting the Kronecker product of two matrices?
More generally: does the general picture changes if we consider V and W infinite-dimensional vector spaces?
AI: Two elements:
If you read the sentence following the one you quote, there is an explanation based on dimensions.
Consider a $4 \times 4$ matrix $C$ with only one zero entry. If you try to write $C = A \otimes B$, either none of the $A,B$ entries is equal to zero and then $C$ will have no zero entry. Or they have at least one, and then $C$ will have at least four zero entries.
The second argument can be extended for infinite dimensions.
|
H: Uniform convergence of sequence of functions $\frac{2+nx^2}{2+nx}$ on [0,1]?
I have recently been trying some questions related to the uniform convergence of a sequence of functions. And meanwhile, I got stuck in one of the problems in which I have been supposed to discuss the point-wise and uniform convergence of the sequence of functions defined as
$$f_n(x)=\frac{2+nx^2}{2+nx}$$
on the interval $[0,1]$
I have found out its point-wise limit that is given by
$$f(x)=
\begin{cases}
1, \text {if } x = 0,1\\
x, \text {if } 0 < x < 1
\end{cases}$$
So the first half has been done.
In the second half, let $\epsilon>0$ be given.
Now I need to find an $m$(if possible) such that $|f_n(x)-f(x)| <\epsilon$ for all $x$ in $[0,1]$ and for all $n\geq m.$
So I see that if $x$ is $0$ or $1$, then any natural number $m$ will work.
But the problem is when $x$ is neither of them.
If I assume that we get a natural number $m$ such that the definition of uniform convergence holds then or further calculations, we get that $n>\dfrac{2(1-x-\epsilon)}{x\epsilon}$ for all $n\geq m.$
Now I am not understanding how to proceed.
Help, please!
AI: To prove using just the definition that that the convergence is not uniform note that $|f_n(x)-f(x)|=|\frac {2-2x} {2+nx}|$ for $0<x<1$. Put $x=\frac 1 n$ to get $\frac {2-2/n} {2+1} \to \frac 2 3$. This shows that $\sup_x |f_n(x)-f(x)|$ does not tend to $0$.
|
H: Verification of proof: $f(x) = e^x$ is continuous at $a = 2$
$$|e^x-e^2|<\epsilon$$
So if $x<2$ then $2-x<\delta $
Calculations:
$$e^2-e^x<\epsilon$$
$$-e^x<\epsilon -e^2 $$
$$e^x>e^2-\epsilon$$
$$x>\ln(e^2-\epsilon)$$
$$-x<-\ln(e^2-\epsilon)$$
$$2-x<2-\ln(e^2-\epsilon)$$
Then we get:
$$2-x<2-\ln(e²-\epsilon) = \delta_1 $$
And then for $x>2$:
$x-2<\delta$
$$e^x-e^2<\epsilon $$
Calculations:
$$e^x<\epsilon+e^2 $$
$$x<\ln(\epsilon+e^2)$$
$$x-2<\ln(\epsilon+e^2)-2$$
Then we get:
$$x-2<\ln(\epsilon+e^2)-2=\delta_2$$
So that means $\delta=\min{(\delta_1,\delta_2)}$
AI: Overall, your proof is working. There is however some important pitfalls:
You write a succession of inequalities without stating if those are equivalent. This is however critical if you want to go reverse, i.e. if $\vert x-2 \vert < \delta$ then $\vert e^x - e^2 \vert < \epsilon$ which is what is at the end required.
You can't take the logarithm of negative numbers. This is the case for example when $\epsilon > e^2$. You have to deal with those cases.
As you know the Mean value theorem, a prove can be:
It exists $c \in (x, 2)$ such that:
$$\vert e^x - e^2\vert = e^c \vert x - 2 \vert \le e^3 \vert x - 2 \vert$$ providing that $\vert x - 2 \vert \lt 1$. And also
$$\vert e^x - e^2\vert = e^c \vert x - 2 \vert \le e^3 \vert x - 2 \vert \le \epsilon$$ if $\vert x - 2 \vert \lt \delta$ for $\delta \lt \min(\epsilon / e^3, 1)$.
|
H: General solution of $tx''-x'+4t^3x=4t^3$
The task is to find general solution of:
$tx''-x'+4t^3x=4t^3$
The hint is to substitute $s=t^2$
My attempt: First I guessed that $x=1$ satisfies the equation and that is our particular solution. Now we have to find the solution to homogeneous eqaution:
$tx''-x'+4t^3x=0$
My quess was to use $x=\exp{(at^2)}$ and after plugging it into equation we get that a=+/- i so our solutions would be $\cos (t^2)$ and $\sin (t^2)$
So the finall answer is that general solution is $x(t)= C_1 \cos (t^2) + C_2 \sin (t^2) +1 $
But the problem is that I didn't use the hint in any part of it and I don't think that's the right way to solve this problem. What's the right way to do it?
AI: If you substitute $s = t^2$
$$\frac{dx}{dt} = \frac{dx}{ds}.(2t)$$
$$\frac{d^2x}{dt^2} = 4t^2\frac{d^2x}{ds^2} + 2\frac{dx}{ds}$$
Now, substituting those into our original DE
$$4t^3\frac{d^2x}{ds^2} + 2t\frac{dx}{ds} - 2t\frac{dx}{ds} + 4t^3x = 4t^3$$
$$\frac{d^2x}{ds^2} + x = 1$$
The simplification basically transforms this into a linear DE which can be easily factorised and solved
|
H: Integral $\int_{1/\sqrt{2}}^1 r^3 \sqrt{1/r^{2} -1}dr$
I have integral $\int_{1/\sqrt{2}}^1 r^3 \sqrt{\frac{1}{r^2}-1}dr$.
I calculated:
$$
\int_{1/\sqrt{2}}^1 r^3 \sqrt{\frac{1}{r^2}-1}dr= \int_{1/\sqrt{2}}^1 r^3 \sqrt{\frac{1-r^2}{r^2}}dr=\int_{1/\sqrt{2}}^1 r^2\sqrt{1-r^2} dr.
$$
But then I dont know how to proceed.
AI: Let $r=\sin{\theta}$:
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin^2{\theta}\cos^2{\theta} \; d\theta\overset{1}{=}\frac{1}{4} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin^2{2\theta} \; d\theta\overset{2}{=}\frac{1}{8} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} (1-\cos{4\theta}) \; d\theta=\left(\frac{1}{8} \theta-\frac{1}{32}\sin{4\theta}\right) \bigg \rvert_{\frac{\pi}{4}}^{\frac{\pi}{2}}=\boxed{\frac{\pi}{32}}$$
$1$: Sine double angle ($2\sin{\theta}\cos{\theta}=\sin{2\theta}$)
$2$: Power reducing identity for sine ($\sin^2{\theta}=\frac{1}{2} \left(1-\cos{2\theta}\right)$)
|
H: How to get every possible combination to choose k object from a set of n elements
How can I get the actual combinations to choose k objects of n elements.
The binomial coefficient only tells me how many possiblities exist, but I actually don't know the combinations. What methods are there?
AI: Brute force method: to obtain all possible subsets represent every number from $0$ to $2^n-1$ in binary form and pick out from the set the elements which binary position is filled with $1$. If you are interested only in subsets with $k$ elements, check if the sum of binary digits is equal to $k$.
An algorithm which does not require generation of wrong-sized subsets is the following.
Set $k$ counters with indices $1,2,\dots k$ to the lowest possible values (from $1$ to $k$).
Increment the admissible counter. The counter with index $\ell$ is admissible if its current value $m_\ell$ is less than its largest possible value ($n-k+\ell$) and all counters with higher indices have the largest possible values (from $n-k+\ell+1$ to $n$).
If the value $m_\ell$ of the incremented counter is not the largest possible ($n-k+\ell$), reset the indices of all counters with higher indices to the values from $m_\ell+1$ to $m_\ell+k-l$.
goto 2.
An example of this algorithm can be found in a comment for $n=5$, $k=2$.
More effective algorithms are much more complicated. See for example https://epubs.siam.org/doi/abs/10.1137/1.9781611973068.107
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H: Solve this differential equation $x^2y''-5xy'+6y=0$
Solve this equation
$$
\begin{cases}
x^2y''-5xy'+6y=0 \\
y(-1)=3 \\
y'(-1)=2
\end{cases}
$$
I got
$$y=c_1x^{3+\sqrt3}+c_2x^{3-\sqrt3}$$
I have three little questions.
Could I solve the problem by substituting $(-1)^{\sqrt3}=\cos((\sqrt 3) \pi)+i\sin((\sqrt 3)\pi)$?
I need to substitute $t=-x$?
If i have to use 2. , isn't the problem wrong because the problem itself contains $y(-1)=3$?
AI: This is a Cauchy-Euler ODE. In order to extend your solution for $x<0$ replace $x$ by $|x|$ (or from the beginning use the substitution $t=\ln|x|$). So we have
$$y(x)=c_1|x|^{3+\sqrt3}+c_2|x|^{3-\sqrt3}.$$
Now solve
$$\begin{cases}
y(-1)=c_1|-1|^{3+\sqrt3}+c_2|-1|^{3-\sqrt3}=3 \\
y'(-1)=-(3+\sqrt{3})c_1|-1|^{2+\sqrt3}-(3-\sqrt{3})c_2|-1|^{2-\sqrt3}=2
\end{cases}$$
and find $c_1,c_2$:
$$c_1= \frac{3-3\sqrt{3}}{2},\quad c_2 = \frac{3+3\sqrt{3}}{2}.$$
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H: Guidance requested for vector dot product question.
Helping my child out with their year 11 exam preparation, specifically vectors and dot products, I think I may have figured out the answer but I'd like to get some confirmation or, more likely, a short sharp shock of education :-)
Keep in mind it's some thirty-plus years since I've had to tackle this stuff. The question is phrased thus:
If vector a is perpendicular to vector b-a, which of the following are necessarily true?
1) a.(b-a) = 0
2) a.b = a.a
3) a = b
4) a.b = |a|2
The ones they stated as necessarily true were all but 3.
So here is my reasoning. Consider the vectors as follows. If b-a is perpendicular, then the b vector must be like this (although the triangle could of course be oriented in other ways):
/|
/ |
/ |
b / | b-a
/ |
/ |
/______|
a
Now, obviously, item 1 is true because the dot product is |a||b-a|cosθ, where θ = 90 hence cosθ = 0.
In terms of the other three statements, I used Pythagoras on the magnitudes to work out:
$${b^2} = {a^2} + {(b-a)^2}$$
$${b^2} = {a^2} + {b^2 -2ab + a^2}$$
$${b^2} = {2a^2} + {b^2 -2ab}$$
$${2a^2} = {2ab}$$
$${a} = {b}$$
So it appears the magnitude of a and b is the same. That would, of course, mean the triangle is not so much a triangle as two congruent lines. This would explain why items 2 and 4 were true - b-a becomes a zero-length vector which I suppose could be considered perpendicular to a.
But the only reason why I can think that item 3 could be false is if that zero-length vector may corrupt things. The other three statements deal with magnitudes only but it may be that that a zero-length vector may be rewritten as zero units north or zero units west, and they may be considered different.
Other than that, I'm not sure why item 3 would not be true as well. Of course, it's quite possible that I've made some mistake in the reasoning above, in which case I'd appreciate some guidance so I can once again become a hero to my son :-)
AI: I don't know what your son is supposed to know about the dot product... which is critical for the answer.
However if he knows that dot product is distributive vs. addition then $$a \cdot (b-a) = a \cdot b - a \cdot a=0.$$
Therefore 2. is clear and 4. also as $a \cdot a = \vert a \vert^2$.
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H: A question obout the sum of series in $[-\infty,+\infty]$
Suppose that the work set is $[-\infty+\infty]$, we suppose that $$\sum_{n=0}^{+\infty}a_n<+\infty.$$ Now can we says that $$\sum_{n=0}^{+\infty}(a_n-b_n)=\sum_{n=0}^{+\infty}a_n-\sum_{n=0}^{+\infty} b_n\quad$$
In the $[-\infty,+\infty]$ each series(well defined) converges, therefore for me is yes. It's true?
AI: If $\displaystyle\sum_{n=1}^\infty a_n = -\infty$, then the result need not hold. Consider the counterexample:
$$a_n = -1, b_n = -1.$$
If we assume further that $-\infty < \displaystyle\sum_{n=1}^\infty a_n$, then it does hold (in a reasonable sense) as I shall show now.
Let $S_N = \displaystyle\sum_{n=1}^N a_n - b_n,$ $A_N = \displaystyle\sum_{n=1}^N a_n,$ and $B_N = \displaystyle\sum_{n=1}^N b_n.$
You wish to answer whether the equality
$$\lim_{N \to \infty}S_N = \lim_{N \to \infty}A_N - \lim_{N \to \infty}B_N$$
holds.
Clearly, we have $S_n = A_n - B_n$ for all $n \in \Bbb N$.
The convergence of $(A_n)$ (to a real number) is part of the hypothesis.
If $(B_n)$ converges to a real number, then it follows that your desired equality holds.
If $(B_n)$ converges to $-\infty$, then the RHS is $\infty$. Moreover, we can bound $A_n$ and thus, it is easy to show that $S_n \to \infty$ as well. Thus, the equality holds again.
The same consideration applies to the case $B_n \to \infty$.
Now, let us consider the case that $(B_n)$ diverges (that is, not converges). In this case, we have that the RHS is not defined.
We show that the LHS isn't defined either. That is, show that $(S_n)$ diverges as well.
Note that
$$B_n = A_n - S_n \quad \forall n \in \Bbb N.$$
If $(S_n)$ were convergent, then $(A_n - S_n)$ would be as well and in turn, so would $B_n$. This gives us a contradiction and thus, we see that $(S_n)$ is divergent.
Thus, we conclude as follows:
Equality does hold in the following sense:
One side of the equality is defined if and only if the other is. Moreover, whenever both sides are defined, they are equal.
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H: Find accumulation points of $a_n=\alpha(3-\frac{1}{n})^{n^{(-1)^{n}}}+\sqrt[n]{2^{n(-1)^n}+6^{n(-1)^{n+1}}}$ regarding $\alpha$
I solved it for $\alpha<0$ and $\alpha>0$. I can't find limit of $\alpha(3-\frac{1}{n})^n$ (the $n$ is even) when $\alpha = 0$. Solution says that this limit is equal to zero but i don't know why it says so because non alpha part goes to infinity and $0*\infty$ is undefined
AI: Well, if $\alpha=0$ then:
$$\alpha\left(3-\frac{1}{n}\right)^n=0\left(3-\frac{1}{n}\right)^n=0 \quad\forall n \in \mathbb{N}$$
so it's not a $0 \cdot \infty$-type limit.
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H: Find the value of function with given conditions
Let $f(x)$ be a fifth degree polynomial with leading coefficient unity.
If $f(1)=5, f(2)=4, f(3)=3, f(4)=2 , f(5)=1$ find $f(6)$
I know I can solve this by assuming a polynomial equation and then finding the coefficients of every term and finding the value of $f(6)$ but I wanted to know if there is any other method of solving this problem. Any hints to solve this kind of problem is much appreciated.
AI: Consider $g(x) = f(x) - (6 - x)$. It is clear that $g(x) = 0$ for $x \in \{1, \ldots, 5\}$.
Moreover, $g$ is also a degree $5$ polynomial with leading coefficient unity. Thus, we may factorise it as
$$f(x) - (6 - x) = g(x) = (x - 1)\cdots(x - 5).$$
Subsituting $x = 6$ in the above gives
$$f(6) = 5! = 120.$$
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H: "Sequentially compactness" of compact Riemann surface
I'm in trouble show that for any sequence $\{x_n\}_n$ in a compact Riemann surface, there exist a subsequence $\{ x_{n_k}\}_k$ of $\{x_n\}_n$, a chart $(U,\varphi)$ and an $x\in U$ such that the sequence $\{\varphi (x_{n_k})\}_k$ converges to $\varphi(x)$.
Any help?
AI: Since the Riemann surface is compact, you can cover it by finitely many charts $(U_i,\varphi_i)$, each of whose closures is compact. By the pigeonhole principle, find a chart $(U_{i_0},\varphi_{i_0})$ such that for infinitely many $n_k$, $x_{n_k}\in U_{i_0}$. Can you finish the proof from here?
The proof should make it clear that this is really not a proper statement about Riemann surfaces per se, since we have no use for the fact that the transition maps $\varphi_{UV}:\varphi_U(U)\longrightarrow\varphi_V(V)$ are holomorphic. It is at best a statement about compact and locally Euclidean topological spaces as this answer suggests.
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H: How to calculate $\int \frac{\sinh \left(x\right)}{\cosh \left(x\right)-\cos \left(y\right)}dy$?
How to calculate $$\int \frac{\sinh \left(x\right)}{\cosh \left(x\right)-\cos \left(y\right)}dy$$ I know that the answer is $2\arctan\left(\coth\frac{x}{2}\cdot\tan\frac{y}{2}\right)$ but don't have idea how to get it...
AI: hint
Put $$t=\tan(\frac y2)$$
with
$$\cos(y)=\frac{1-t^2}{1+t^2}$$
and
$$dy=2\frac{dt}{1+t^2}$$
it becomes
$$\int \frac{2\sinh(x)}{\cosh(x)-\frac{1-t^2}{1+t^2}}\frac{dt}{1+t^2}=$$
$$\int \frac{2\sinh(x)dt}{\cosh(x)-1+t^2(\cosh(x)+1)}$$
Remember that
$$\cosh(x)-1=2\sinh^2(\frac x2),$$
$$\cosh(x)+1=2\cosh^2(\frac x2)$$
and
$$\sinh(x)=2\sinh(\frac x2)\cosh(\frac x2)$$
Now, with the substitition $$u=t\cosh(\frac x2)$$, you will get the result.
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H: Choosing berries
You have 2 blue berries, 2 red berries and 2 black berries
In how many ways can you pick 2 berries?
My work:
you can choose two red berries or you can choose two blackberries or you can choose two blueberries or you can choose 1 blue and 1 red berry or you can choose 1 black and 1 blue berries or you can choose one black and one redd berrie
Extra condition:
Order don't matter and berries identifies only by colour
HEnce:
$ \binom{2}{2} + \binom{2}{2} + \binom{2}{2} + \binom{2}{1} \cdot \binom{2}{1}+ \binom{2}{1} \cdot \binom{2}{1}+ \binom{2}{1} \cdot \binom{2}{1}$=15
AI: This is a simple application of the "stars and bars" method.
You are asked to choose $2$ objects (two berries) from a choice of $3$ (three types of berries), where repitition is allowed but order does not matter. Hence we have
$$H^3_2 = C^{3+2-1}_2=C^4_2=6$$
Edit:
$H$ is probably not commonly seen, so I shall explain below.
In general, you are asked to choose $r$ objects from a set of $n$ objects. Repetition may or may not be allowed, and Order may or may not matter.
$\color{red}{\text{Repetition not allowed, }}\color{red}{\text{Order does not matter}}$
$$C^n_r = \frac{n!}{r!(n-r)!}$$
$\color{green}{\text{Repetition allowed, }}\color{red}{\text{Order does not matter}}$
$$H^n_r = C^{n+r-1}_r = \frac{(n+r-1)!}{r!(n-1)!}$$
$\color{red}{\text{Repetition not allowed, }}\color{green}{\text{Order matters}}$
$$P^n_r = \frac{n!}{(n-r)!}$$
$\color{green}{\text{Repetition allowed, }}\color{green}{\text{Order matters}}$
$$n^r$$
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H: At what value does this integral reach its minimum?
The problem is to find the value of $a>1$ at which this integral $$\int_{a}^{a^2} \frac{1}{x}\ln\Big(\frac{x-1}{32}\Big)dx$$
reaches its minimum value. I don't want to know the number, I just want feedback on the ideas I'm trying. Considering that $$f(x)=\frac{1}{x}\ln\Big(\frac{x-1}{32}\Big)<0$$
whenever $x<33$ and the fact that $f(x)=0$ at $x_0=33$, we have that all the area associated with the graph of this function is negative until the point $x_0$. So we want to find the value of $a$ which yields the greatest portion of this "negative area". The fundamental theorem of calculus states that $$\int_a^bf(x) = F(b) - F(a)$$
which gives us an expression which can be differentiated. Particularly, the first derivative is $$f' =\frac{1}{a^2}\ln\Big(\frac{a^2-1}{32}\Big)-\frac{1}{a}\ln\Big(\frac{a-1}{32}\Big)$$
or $$f'=\frac{1}{a^2}\ln\Big(\frac{a^2-1}{32}\Big)\cdot 2a-\frac{1}{a}\ln\Big(\frac{a-1}{32}\Big)\cdot 1$$
I think the second is correct since, per the fundamental theorem, we are applying the primitive $F$ to the upper and lower bound functions $a^2$, and $a$ - so this is a sum of two composite functions differentiated by the chain rule. It depends on the correct interpretation of $a$ I think.
If we differentiate again, we can find the values of $a$ for which $f(a)''>0$. This will be value of $a$ where the integral reaches its minimum value. However, applying the second derivative test to an integral doesn't seem proper since it is a tool for studying the concavity of $f$ - what would be the interpretation here?
AI: Your second expression for $f'(a)$ is correct. Also, you shouldn't differentiate it again, but rather solve for the critical points of $f'(a)$ to find the minimum value of the integral because the integral is $f(a)$:
$$f'(a)=0=\frac{1}{a}\left(2\ln{\left(a^2-1\right)}-2\ln{32}-\ln{\left(a-1\right)}+\ln{32}\right)$$
$$ \ln{32}=\ln{\left(a-1\right)}+2\ln{\left(a+1\right)}$$
Now, solve for $a$ then test if its a minimum or maximum by using the first derivative test.
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H: Another proof (check) that the set of isolated points of a set in $\mathbb R^n$ is countable
Just thinking if this proof works, and i have some "not so explicit" ending so i would love if you suggest a way for refining this.
Theorem : $S \subset \mathbb R^n$ is a set. Then the set of isolated points of $S$ is countable.
Proof: Let $S_{\lambda}$ denote the set of isolated points of $S$.
Thus $\forall ~ \bf{x}$ $ \in S_{\lambda}$, $\exists ~ \varepsilon_{\bf{x}} \in \mathbb R^+$ such that $B(\bf{x}$ ,$~\varepsilon_{\bf{x}})\cap S =\phi$
We know that the set of open balls of rational coordinates is countable.
Say $\bf{x}$ $=(x_1,x_2,\cdots,x_n)$ and chooae rationals $r_j$ such that $||x_j -r_j||<\frac{\varepsilon_{\bf{x}}}{4n}$.
Set $r_{\phi_x}=(r_1,r_2,\dots, r_n)$
Then $||~ \bf{x}$ $-r_{\phi_x}~||< \frac{\varepsilon_{\bf{x}}}{4}$
Thus $B(r_{\phi_x,\frac{\varepsilon_{\bf{x}}}{4}}) \subseteq B(\bf{x}$ $,\varepsilon_{\bf{x}})$.
Like this we can generate a ball with rational coordinate which is a subset of the ball with center at $\bf{x}$ with radius $\varepsilon_{\bf{x}}$ for all $x \in S_{\lambda}$.
To end, we define the map $\Gamma:B_{\mathbb Q}\to B_{S_{\lambda}}$ (where $B_{\mathbb Q}$ is the set kf balls with rational coordinates (and is countable) and $B_{S_{\lambda}}$ is the set of balls $B(\bf{y}$ $,\varepsilon_{\bf{y}})$ for all $\bf{y}$ in $S_{\lambda}$) such that $\Gamma (B(r_{\phi_x},\frac{\varepsilon_{\bf{x}}}{4}))=B(\bf{x}$ $,\varepsilon)$.
This map is injective when the $r_{\phi_y}$'s are kept fixed and hence $B_{S_{\lambda}}$ is countable which gives (please suggest a good way, i mean intuitively it makes me think its correct but, here i am in need of much help) $S_{\lambda}$
Much thanks!
AI: Around each isolated point $x ∈ S$ is an open $n-ball$ $B(x)$ such that $B(x)\cap S=\phi$.Then there is an $n-ball$ $A_x$ with rational radius and rational center coordinates such that $x ∈ A_x ⊂ B(x)$. The map $x\longmapsto A_x$ is a one-to-one correspondence between the isolated points of
$S$ and a subset of the countable set of all open $n-balls$ with rational center and radius.
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H: $ \sup_{x\in C}\|x\|=\sup_{x^*\in B^*}\sup_{x\in C}\langle x^*,x\rangle $
Let $X$ be a separable Banach space, the associated dual space is denoted by $X^*$ and the usual duality between $X$ and $X^*$ by $\langle , \rangle$. Let $B^*$ the closed unit ball of $X^*$.
Take $C$ nonempty weakly compact convex subsets of $X$. Can we say that:
$$
\sup_{x\in C}\|x\|=\sup_{x^*\in B^*}\sup_{x\in C}\langle x^*,x\rangle
$$
It is obvious that
$$
\sup_{x^*\in B^*}\sup_{x\in C}\langle x^*,x\rangle\leq \sup_{x\in C}\|x\|
$$
But for inverse inequality. $C$ should it be bounded?
AI: The reverse inequality holds without any assumptions on $C$.
Let $x \in C$. By the Hahn-Banach theorem, there exists $x^* \in B^*$ such that $\lVert x \rVert = \langle x^*,x\rangle$. Thus
$$\lVert x \rVert \leq \sup_{y \in C} \,\langle x^*,y\rangle \leq \sup_{y^* \in B^*}\sup_{y\in C} \,\langle y^*,y\rangle.$$
Taking the supremum over $x \in C$ yields the result.
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H: Is there an easier way to solve the given problem?
If $x + 2 + \sqrt{3}i=0$ then find the value of $2x^4+3x^3-x^2-15x+36$
If you try to find the values of $x^2, x^3 $and $ x^4$, and then put the values in, we can find the solution to be 1. But is there a better way to solve this problem? Where you can probably break the second polynomial into a simple one and then solve it, saving a lot of time and making the calculations less prone to errors?
AI: Real coefficients in the quartic inspired me to multiply $x + 2 + \sqrt{3}i\;$ by the conjugate, which gives
$$(x + 2 + \sqrt{3}i)(x + 2 - \sqrt{3}i)=x^2+4x+7.$$
Dividing the given expression by $x^2+4x+7$ (just to see if we get a simple form) leads to $$2x^4+3x^3-x^2-15x+36=\underbrace{(x^2+4x+7)}_{0}\cdot(2x^2-5x+5)+1$$
Yes, if $x + 2 + \sqrt{3}i=0\;$ then $2x^4+3x^3-x^2-15x+36=1.$
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H: Is $x \approx x$?
If I write $x \approx y$, does this mean (a) $x$ is sufficiently close to $y$ for some practical purpose, or (b) $x$ is sufficiently close to $y$ for some practical purpose, but is not equal to $y$?
If (a) is true, then it appears $x \approx x$.
This question appears to have more importance when considering something like the small angle approximations. Is the statement $\sin(x) \approx x $ true when $0 \leq x \leq 1$, or true when $0<\sin(x) \leq 1$?
AI: Keeping in mind that "$\approx$" isn't actually a formal notion at the outset, any reasonable approach to it (e.g. via nonstandard analysis, where "$\approx$" is interpreted as "is in the halo of") will say that it does extend equality: everything is approximately equal to itself.
It's worth noting that the way approximation is cropping up here is in questions of asymptotics: when we say "$x\approx \sin(x)$ near $0$," we don't really care about how close $\sin(0.1)$ is to $0.1$, what we're really interested in is $\lim_{x\rightarrow 0}{\sin(x)\over x}$. There is a formal framework for discussing this sort of analysis, namely big-O notation and its relatives, and in this framework everything is indeed close to itself.
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H: Average of a tensor with respect to a group action
Consider a smooth manifold $M$ (assume it boundary-free and orientable) and a tensor field $\mathcal{G}\in\Gamma(\otimes^hTM\otimes^kT^*M)$. Let $\Phi:\mathbb{T}^p\times M \rightarrow M$ be a torus action on the manifold. What is the definition of the average of the tensor $\mathcal{G}$ with respect to the action $\Phi$?
I think it should be defined as
$$ \bar{\mathcal{G}} = \frac{1}{\int_M d\alpha}\int_M (\Phi_{\alpha})^*\mathcal{G}\;d\alpha_1\wedge ... \wedge d\alpha_p. $$
Is my intuition correct? It seems at least to be coherent with the fact that when $\mathcal{G}$ is invariant with respect to the torus action, then $(\Phi_\alpha)^*\mathcal{G}=\mathcal{G}$ and hence $\bar{\mathcal{G}}=\mathcal{G}$ as expected.
AI: You should probably average it with respect to the action (but your integral is over $M$). If $G \circlearrowright M$ is a smooth action, $G$ is compact Lie group, $\nu$ the Haar (probability) measure on $G$ and we also denote by $g$ the map $M \ni p \mapsto g\cdot p \in M$, it seems that $$\overline{\mathcal{G}}_x(\theta^1,\ldots, \theta^h, X_1,\ldots, X_k) = \int_{G} (g^*\mathcal{G})_x(\theta^1,\ldots, \theta^h, X_1,\ldots, X_k)\,{\rm d}\nu(g)$$makes more sense, where $x \in M$, $\theta^i \in T_x^*M$ and $X_j \in T_xM$ for all $1 \leq i \leq h$ and $1 \leq j \leq k$ are all fixed. Then we have that $g^*\overline{\mathcal{G}} = \overline{\mathcal{G}}$ for all $g \in G$, by construction.
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H: Can you take the (Euclidean) norm of a scalar quantity (non-vector)?
Context: This question
My question is whether or not you can take the (Euclidean) norm of a scalar non-vector quantity. My intuition is no, but I just wanted to confirm it.
I understand that a norm in essence gives you a 'length' or 'magnitude' which is a scalar quantity, so if you already had a scalar quantity $k$, does $\Vert k \Vert = k$ where $\Vert . \Vert$ is a norm?
Thanks.
AI: Every norm $\|\cdot\|$ on $\Bbb R$ is related to the absolute value $|\cdot|$ via the silly computation $\|x\| = \|x1\| = |x|\|1\| = \|1\||x|$, since you can regard the "vector" $x$ as the number it actually is. Corollary: the only norm $\|\cdot\|$ in the vector space $\Bbb R$ with $\|1\| = 1$ is $\|\cdot\| = |\cdot|$.
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H: Showing $\mathbb{R}$ with the standard topology union irrational subsets is normal.
We are given that $\mathcal{T}$ is the standard topology on $\mathbb{R}$ and $\mathcal{O}=\{O\in\mathcal{P}(\mathbb{R}):O=U\cup A, U\in\mathcal{T}, A\subseteq\mathbb{R}\backslash\mathbb{Q}\}$. It can easily be shown that $(\mathbb{R},\mathcal{O})$ is a topology. Now we want to show that this space is normal, and as a hint, we are given that every metric space is normal.
So I assume that the idea is to show that $(\mathbb{R},\mathcal{O})$ is metrizable, by finding a suitable metric $d$ that induces the same topology $\mathcal{O}$. However, I have no idea where to start to find such a metric. Is this the correct approach?
AI: That space is the Michael line. It is not metrizable, but it is paracompact and therefore normal. There is a proof of paracompactness at the link, but the result is quite easy using the fact that the usual topology on $\Bbb R$ is paracompact, so it’s worth trying on your own if you’ve done anything with paracompactness.
I gave a very different proof that the Michael line is normal in this answer to an earlier question.
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H: Affine function is a diffeomorphism?
Given an affine function $f:\mathbb{R}^n \to \mathbb{R}^n$ defined for all $x\in \mathbb{R}^n$ by $$f(x)=T(x)+a$$ such that $T$ is an invertible Linear map and $a\in \mathbb{R}^n$, is $f$ a diffeomorphism?
AI: It is a diffeomorphism iff $T$ is invertible.
It is easy to see that $f$ is invertible iff $T$ is invertible with inverse
$$f^{-1}(x) = T^{-1}(x - a).$$
As $T^{-1}$ is a linear transformation, it is differentiable everywhere.
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H: Why is a countable set closed?
If $T$ is an uncountable set, show that $$\mathcal{T}=\{\emptyset,T,\text{ all sets whose complements is at most countable}\}$$ forms a topology on $T$.
Answer:
We have $∅$ and $T$ in T . To check the other two axioms, it is easier to work with
complements, i.e. at most countable sets, and show that they satisfy the ‘closed set’ axioms:
finite unions are closed (clear, since they are at most countable), and arbitrary intersections
are closed (again, they are at most countable, or empty)
I don't exactly get the things in bold, in particular:
Why is the finite union of at most countable sets closed? It is definitely not clear for me. Could somebody explain?
In particular, why is an at most countable set closed?
AI: What do you mean by "open" and "closed" in the first place?
In topology there is no such thing as an "open set" or "closed set" as such. Rather, there are many different notions of "open-ness" and "closed-ness" depending on which topology we're looking at. The more precise statement here is:
Every at-most-countable set is closed in the sense of $\mathcal{T}$,
but the "in the sense of $\mathcal{T}$"-bit is often dropped when there is only one topology being considered.
Remember that the closed sets are by definition the complements of the open sets. The open sets, in turn, are exactly the elements of the topology $\mathcal{T}$ - again by definition: "open set in the sense of $\mathcal{T}$" just means "element of $\mathcal{T}$."
In this case, all sets with at-most-countable complements are open in the sense of $\mathcal{T}$; this is what the definition of $\mathcal{T}$ explicitly says (incidentally, note that including $T$ in $\mathcal{T}$ is redundant since $T\setminus T$ is already at most countable). Consequently, all at-most-countable sets are closed in the sense of $\mathcal{T}$. (And so since the union of finitely many countable sets is countable, we have that $\mathcal{T}$ is closed under finite intersections as desired.)
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H: Probability distribution and "inclusion"
Is there such thing as a probability distribution "included" in another?
For example, take two random variables X and Y, where Y takes it values in a set Sy included in Sx. How do you formalize this?
In practice, given observations on variables X and Y, can you test for such an inclusion?
AI: The concept of absolute continuity might interest you. Suppose $X, Y$ are random elements taking values in a space $\mathcal{X}$, with distributions $\mu$ and $\nu$ respectively (these are probability measures on $\mathcal{X}$). Then the inclusion you describe corresponds to $\nu$ being absolutely continuous with respect to $\mu$. That is, for any measurable subset $A\subset \mathcal{X}$ such that $\mu(A)=0$, we also have $\nu(A)=0.$ Or in other words: if $\mathbb{P}(X\in A)=0$, then also $\mathbb{P}(Y\in A)=0.$
This intuitively means that if $X$ cannot take value in some (measurable) subset $A\subset \mathcal{X}$, then $Y$ cannot take value in that subset either.
I know of no straightforward way to test this in general.
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H: How to show that $H=\langle a,b\rangle=\{a^m\cdot{b^n}\mid m,n\in{\mathbb{Z}}\}$ is a subgroup of $\langle G,\cdot\rangle$?
I have $a,b\in{G}$ and also $G$ is Abelian.
I thought that first I should show that $H\subset{G}$ and then show that $\forall{h_1,h_2}\in{H}$ we will have $ h_1\cdot{h_2}\in{H}$.
First to show that $H$ is a subset of $G$, I have taken $a^m\cdot{b^n}\in{H}$.
Since $a,b\in{G}$ we have $a^m\cdot{b^n}\in{G}$ from the closure axiom.
Secondly to show that $H$ is subgroup I think I need to show first that $H$ is also Abelian but I'm having difficulties approaching this proof.
If I prove that $H$ is Abelian then if I take $h_1=a^m\cdot{b^n},h_2=a^x\cdot{b^y}\in{H}$. Then $$h_1\cdot{h_2}=a^m\cdot{b^n}\cdot{a^x}\cdot{b^y}=a^{m+x\in{\mathbb{Z}}}\cdot{b^{n+y\in{\mathbb{Z}}}}\in{H}.$$
Is this the correct approach? If yes the how can I show that $H$ is Abelian ?
AI: One way to do this is the One-step Subgroup Test.
You have shown that $H\subseteq G$.
We have $e=a^0b^0\in H$, so $H\neq\varnothing$.
Let $g=a^nb^m, h=a^rb^s\in H$ for $n,m,r,s\in\Bbb Z$. Then
$$\begin{align}
gh^{-1}&=(a^nb^m)(a^rb^s)^{-1}\\
&=(a^nb^m)(b^{-s}a^{-r})\\
&=a^n(b^m(b^{-s}a^{-r}))\\
&=a^n((b^mb^{-s})a^{-r})\\
&=a^n(b^{m-s}a^{-r})\\
&=a^n(a^{-r}b^{m-s})\quad (G\, (\text{ and hence } H)\text{ is abelian})\\
&=(a^na^{-r})b^{m-s}\\
&=a^{n-r}b^{m-s},
\end{align}$$
but $n-r, m-s\in\Bbb Z$. Hence $gh^{-1}\in H$.
Hence $H\le G$.
(It is not necessary to check whether $H$ is abelian to verify that it is a subgroup. However, since I made use of commutativity, consider that for any $p,q\in H\subseteq G$, we have $pq=qp$ since $G$ is abelian. Hence $H$ is abelian. )
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H: "generalised" gamma-like integral $\int_0^\infty x^ne^{-f(n)x}dx$
I have noticed that if we have an integral of the form:
$$I[f]=\int_0^\infty x^ne^{-f(n)x}dx=\frac{1}{f^{n+1}(n)}\int_0^\infty x^ne^{-x}dx=\frac{n!}{f^{n+1}(n)}$$
I was wondering what kind of restrictions would need to be applied to $f$ in order for this integral to converge for all values of $n$. An obvious one to me is that $f(n)>0$ and for it to converge for $n\to\infty$ we would require $$f(n)>(n!)^{\frac 1{n+1}}\tag{1}$$
are there any other requirements or functions people can think of that fit these requirements? Thanks
some obvious ones to me are:
$f(n)=n!,(an)!$ but could an exponential work?
In terms of satisfying $(1)$ using sterlings approximation as suggested we get
$$(n!)^{\frac{1}{n+1}}\approx (2\pi)^{1/2(n+1)}\frac{e^{1/(n+1)}n}{n^{1/2(n+1)}}$$
and as this approaches infinity I believe it is approx to $n$ so I think:
$$f(n)=|n|^\alpha+a$$
would converge for all values $n$ where $\alpha>1,a>0$
AI: The integral converges as long as $n > -1$ and $f(n) > 0$. It diverges at $0$ if $n \le -1$, and diverges at $\infty$ if $f(n) \le 0$ and $n \ge -1$.
If you want $\lim_{n \to \infty} \frac{n!}{f(n)^{n+1}} = 0$, noting that $n! \sim \sqrt{2\pi} n^{n+1/2} e^{-n}$ by Stirling's approximation, $f(n) = n/e$ would work, while
$f(n) = t n $ for $0 < t < 1/e$ would not.
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H: Why is the dimension of a crystallographic group unique?
The algebraic definition of a crystallographic group goes as follows:
If a group $\Gamma$ fits into a short exact sequence
$$0 \to \mathbb{Z}^n \overset{i}{\to} \Gamma \overset{p}{\to} G \to 1$$
such that $i \left(\mathbb{Z}^n \right)$ is maximal abelian in $\Gamma$ and $G$ is finite, then $\Gamma$ is a crystallographic group. In this case, $n$ is called the dimension of $\Gamma$ and $G$ is called the holonomy group of $\Gamma$.
Why is the dimension of a crystallographic group unique? I guess it has something to do with $i \left(\mathbb{Z}^n \right)$ being maximal abelian, but I can't figure it out.
AI: Hint. Your question is equivalent to proving that $\mathbb{Z}^n$ cannot contain $\mathbb{Z}^m$ with finite index, unless $m=n$.
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H: Limit of p th norm of function as p tends to infinity
Let $f\colon[0,1]\to\mathbb R$ be continuous. Let $c_p=\left(\int_0^1|f(x)|^pdx\right)^{\frac{1}{p}}$. Then the limit $\lim_{p\to \infty}c_p$ is?
I know $\inf f(x) \leq c_p\leq \sup f(x)$ for $x\in [0,1]$. But no idea whether the sequence is increasing or decreasing and how to proceed to find the limit.
AI: If $f$ is identically $0$, it's clear. Otherwise, consider $g$ defined by $g= f/\lVert f\rVert_\infty$ (as the supremum of $f$ on $[0,1]$, $\lVert f\rVert_\infty$ exists, is positive, and attained at some $x^\ast \in[0,1]$). We have that $g$ is continuous, and $|g| \leq 1$.
Then
$$
\frac{c_p}{\lVert f\rVert_\infty} = \left(\int_{[0,1]} |g(x)|^p dx \right)^{1/p} \leq \left(\int_{[0,1]} 1 dx \right)^{1/p} = 1 \tag{1}
$$
Further, for every $\varepsilon>0$, there exists some neighborhood $V_\varepsilon$ of $x^\ast$ of some size $\delta>0$ such that $g(x^\ast)\geq 1-\varepsilon$. Therefore, for all $p$,
$$
\frac{c_p}{\lVert f\rVert_\infty} \geq \left(\int_{V_\varepsilon} |g(x)|^p dx \right)^{1/p} \geq (1-\varepsilon)|V_\varepsilon|^{1/p}=
(1-\varepsilon)\delta^{1/p} \xrightarrow[p\to\infty]{} 1-\varepsilon \tag{2}
$$
Since this holds for all $\varepsilon>0$, we get
$$
\lim_{p\to\infty}\frac{c_p}{\lVert f\rVert_\infty} = 1 \tag{3}
$$
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H: general approach to comment on bijection of these functions whose graph can't be drawn
consider the following class of functions defined as:
If $f : \Bbb R \to \Bbb R$ be the function such that
$$ f(x)=x|x|-4 :x \in \Bbb Q$$
$$ f(x)=x|x|-\sqrt{3} :x \notin \Bbb Q$$
then $f(x)$ is there a general approach to comment on bijection of this function?
my approach:
generally for finding whether a function is one -one/many-one and into/onto I draw graphs and match range with co-domain while also checking if $f'(x)>0$ or $f'(x)<0$ to get answer but here the only method I can think of here is finding an example which makes the function many-one (here
$f(2)=f(\sqrt{\sqrt{3}})=0$) and into (clearly $f(x) \neq \sqrt{3}$)
Is there any way which does not require finding counterexamples.kindly help.
AI: To disprove a function being one-one, you do have to come up with a counterexample. (Or at least, show the existence of one.)
While graphing and $f'$ are tools that can help for nice enough functions, you always have the definition to turn back to.
In this case, you could try by assuming $f(x_1) = f(x_2)$ and see where it leads you. This can be done systematically.
Since the function is defined piece-wise, it would make sense to take the following cases:
$x_1, x_2 \in \Bbb Q$
$x_1 \in \Bbb Q$ and $x_2 \notin \Bbb Q$
$x_1, x_2 \notin \Bbb Q$
It is easy to see that both cases 1. and 3. will give you $x_1|x_1| = x_2|x_2|$ which would force $x_1 = x_2$.
So only 2. is to be checked. Here you set up the equation
$$x_1|x_1| - 4 = x_2|x_2| - \sqrt3$$
or
$$x_2|x_2| = x_1|x_1| - 4 + \sqrt 3.$$
Now, one notes that if the RHS is irrational, then one can find an irrational $x_2$ that satisfies the equation. (Note that the RHS being irrational is not necessary, though.)
An easy way to do that if by putting $x_1 = 0$ or by putting $x_1 = 2$, the latter gives you the counterexample you created.
I agree that this last part did involve some level of "observation" but given how random functions can be, this should be reasonable.
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H: Tell me what it would look like to factor $x^2-1 = (x+1)(x-1)$ a different way
During the factoring of $x^2-1$ I saw a $+x$ and $-x$ were introduced but I wonder how the factoring would go if the $+x-x$ were added in reverse, like so $-x+x$.
I was shown $x^2-1$ can be factored to $(x+1)(x-1)$ thusly...
$x^2-1 =$
$x^2+x-x-1 =$
$(x^2+x)-(x+1) =$
$(x*x+x*1)+(-1)*(x+1) =$
$x*(x+1)+(-1)*(x+1) =$
$(x+(-1))(x+1) =$
$(x-1)(x+1) =$
I want to know how $x^2-1$ can be factored to $(x+1)(x-1)$ if instead of $x^2+x-x-1 =$ the $+x$ and $-x$ were brought in the other way around like so $x^2-x+x-1 =$
I tried to factor this to $(x-1)(x+1)$ and got lost along the way.
I'll start the equation again.
$x^2-1 =$
$x^2-x+x-1 =$
... what happens next?
AI: I don't understand what you don't understand,
First of all, Addition is a commutative operation so, it won't matter how you "add"
You're lost at this point..?
$x^2-x+x-1$
Hint: Just Take the common factor out, like you did first, earlier you took $+1$ common. Something else this time. and obviously, $+x$ from the first two terms.
Just remember that the result would be the same, however you factor out.
If you still couldn't understand,
$$x^2-x+x-1$$
$$=(x)(x-1)+(+1)(x-1)$$
$$ (x-1)[(x)+(+1)]$$
$$\implies (x-1)(x+1)$$
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H: Calculate area using double integral
I'm trying to calculate the area defined by the following curves: $y=x^2, 4x=y^2, y=6$ using double integrals.
I'm wondering whether my solution is correct:
Area = $\int^{6}_{0}\int^{\sqrt{4x}}_{0}1dydx - \int^{6}_{0}\int^{x^2}_{0}1dydx$
Is it correct?
Thanks!
AI: No, that is not correct.
First, you should compute the intersection points of the curves $y=x^2$ and $4x=y^2$, which are $(0,0)$ and $\left(\sqrt[3]4,2\sqrt[3]2\right)$. When $y\in\left[0,2\sqrt[3]2\right)$, the curve $y=x^2$ is to the right of the curve $4x=y^2$; after that, it is located to the left.
So, you should compute$$\int_0^{2\sqrt[3]2}\int_{y^2/4}^{\sqrt y}1\,\mathrm dx\,\mathrm dy+\int_{2\sqrt[3]2}^6\int^{y^2/4}_{\sqrt y}1\,\mathrm dx\,\mathrm dy.$$
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H: self-adjoint operator intuition
can someone please explain self-adjoint operator intuition to me.
And why when $T^* = T^{-1}$, $T$ preserves the inner product and therefore preserves the the orthonormal basis and the length and distance?
thank you
AI: Think about a symmetric matrix, it’s a simpler case. I think that the term arise from an observation: matrices that are diagonal if we describe them with orthonormal bases have a symmetric structure if we change coordinates with an orthogonal transformation. This is a different way to interpret the def immediatley using the specrtal theorem.
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H: Simulate stars and bars problem
Suppose that I have $n$ balls to be divided among $k$ buckets so that each bucket has a non-negative number of balls. This is a classic stars and bars problem: the number of ways to do this division is $\binom{n+k-1}{k-1}$. (In my application, $k>>n$.)
I would like to run a simulation where, for each repetition, I can randomly select one of the $\binom{n+k-1}{k-1}$ possible divisions. But I do not know how to make this random selection.
Could you please help with an algorithm?
I have a feeling that I just need to find the right discrete distribution and simulate realizations of it, but I do not know how to arrive at this discrete distribution.
AI: If you want to simulate choosing a bucket at random for each ball, you can simply generate $n$ random integers in the range $[1,k]$. If you want to generate each possible distribution of balls with equal probability, you could generate $n$ different random integers in the range $[1,n+k-1]$ and use them in conjunction with the stars-and-bars model to generate the associated distribution.
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H: To find elements of truth set
STATEMENT : $x$ is a real number and $5$ $\in$ {$y$ is real number | $x^2+y^2 <50$ }.
Since $5$ is element of above set. so we have $x^2 + 25 < 50$. so we have $x^2 < 25$. So Truth set is {$-4,-3,-2,-1,0,1,2,3,4$}. Is my work correct? Thank you
AI: You assumed $x$ is an integer. But $4.9$ is an allowed value for $x$.
$x^2<25$ is equivalent to $|x|<5$ or $x\in (-5,5).$
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H: $\epsilon$-$\delta$ Proof of Limit in $\mathbb{R}^4$
Question: Determine $\lim_{(a,b,c,d) \to (0, 0, 0, 0)} \frac{a^2d^2 -2abcd + b^2c^2}{a^2 + b^2 + c^2 + d^2}$ and prove your result using the $\epsilon$-$\delta$ definition of a limit.
Attempt: I have completed the first part of the question using polar coordinates and found that the limit is $0$, but am having trouble proving my result.
Using $z=(a,b,c,d)$ and $f(z)=\frac{a^2d^2 -2abcd + b^2c^2}{a^2 + b^2 + c^2 + d^2}$.
RTP: $\forall \epsilon>0, \exists \delta >0$ s.t. ($z \in \mathbb{R}^4$ and $0 < |z| < \delta$) $\Rightarrow$ $(|f(z)|< \epsilon)$
However, this is at a level (particularly being in $\mathbb{R}^4$) I am not at yet.
Any help would be greatly appreciated.
Update: I've realised I'm not sure what $0 < |z| < \delta$ is, since $|z| = |(a,b,c,d)|$. Do I evaluate $|(a,b,c,d)|$ using the Euclidean norm?
AI: The limit is $0$. To see this, start by writing
$$\left| \frac{a^2d^2 -2abcd + b^2c^2}{a^2 + b^2 + c^2 + d^2}-0\right|=\frac{|a^2d^2 -2abcd + b^2c^2|}{a^2 + b^2 + c^2 + d^2}$$
$$\le \frac{a^2d^2}{a^2 + b^2 + c^2 + d^2}+\frac{2|abcd|}{a^2 + b^2 + c^2 + d^2}+\frac{b^2c^2}{a^2 + b^2 + c^2 + d^2}$$
$$=a^2\frac{d^2}{a^2 + b^2 + c^2 + d^2}+|cd|\frac{2|ab|}{a^2 + b^2 + c^2 + d^2}+b^2\frac{c^2}{a^2 + b^2 + c^2 + d^2}$$
$$\le a^2+ |cd|+b^2$$
Here, we have used the inequality $2|ab|\le a^2+b^2$ to bound the middle term. Then, note that the remaining fractions are all bounded above by $1$.
Applying $|cd|\le c^2+d^2$, we get that
$$
\left| \frac{a^2d^2 -2abcd + b^2c^2}{a^2 + b^2 + c^2 + d^2}-0\right|\le a^2+b^2+c^2+d^2
$$
So, given $\varepsilon>0$, taking $\delta=\sqrt{\varepsilon}$ does the trick.
In more detail, suppose $\varepsilon>0$ is given. Write $z=(a,b,c,d)$ and put $\delta:=\sqrt{\varepsilon}$. Obviously, $\delta>0$. Now, if $0<|z|<\delta$, then
$$ a^2+b^2+c^2+d^2=|z|^2<\delta^2=\varepsilon $$
Combining this with the previous inequalities gives the desired result.
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H: Find the volume between the surface $x^2+y^2+z=1$ and $ z=x^2+(y-1)^2$
I'm trying to find the volume between the surface $x^2+y^2+z=1$ and $ z=x^2+(y-1)^2$ but nothing works for me.
I made the plot and it looks like this:
How could you start? Any recommendation?
AI: Try checking where the two surfaces intersect. Solve the given equations for $z$ and set them equal:
$$\begin{align*}
1-x^2-y^2&=x^2+(y-1)^2\\
0&=2x^2+2y^2-2y\\
0&=x^2+y^2-y\\
0&=x^2+\left(y-\frac12\right)^2-\frac14\\
x^2+\left(y-\frac12\right)^2&=\frac1{2^2}
\end{align*}$$
which corresponds to the cylinder of radius $\frac12$ with cross sections parallel to the $(x,y)$ plane and centered over the point $\left(0,\frac12,0\right)$.
With this in mind, a change to cylindrical coordinates is the "obvious" move. Take
$$\begin{cases}x=r\cos\theta\\y=\frac12+r\sin\theta\\z=z\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=r\,\mathrm dr\,\mathrm d\theta\,\mathrm dz$$
The solid (call it $S$) is then given by the set
$$S=\left\{(r,\theta,z)\mid0\le r\le\frac12,0\le\theta\le2\pi,r^2-r\sin\theta+\frac14\le z\le\frac34-r^2-r\sin\theta\right\}$$
where the last inequality follows from
$$\begin{align*}
&x^2+(y-1)^2\le z\le1-x^2-y^2\\
&\implies r^2\cos^2\theta+\left(r\sin\theta-\frac12\right)^2\le z\le1-r^2\cos^2\theta-\left(\frac12+r\sin\theta\right)^2\\
&\implies r^2-r\sin\theta+\frac14\le z\le\frac34-r^2-r\sin\theta
\end{align*}$$
The volume is then
$$\iiint_S\mathrm dx\,\mathrm dy\,\mathrm dz=\int_0^{2\pi}\int_0^{\frac12}\int_{r^2-r\sin\theta+\frac14}^{\frac34-r^2-r\sin\theta}r\,\mathrm dz\,\mathrm dr\,\mathrm d\theta$$
$$=\frac\pi{16}$$
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H: What is the order of the subgroup of G (rubiks cube group) generated by ?
I got:
=1,FF,RR,FFRR,RRFF
But in my text book the answer is 12? Does any one else know the other elements?
I assumed since FFFFRRRR=1 there were no more?
AI: Let $f=FF$ and $g=RR$. Note that $f$ and $g$ have order $2$, and $fg$ and $gf$ have order $6$, i.e., we have $(fg)^6=e$ and no smaller power of it. This still is not enough to show that the group generated by $f$ and $g$ has order $12$. We also have
$$
g(fg)^2=(fg)^3f.
$$
So the $12$ elements are
$$
e,f,fg,(fg)f,(fg)^2,(fg)^2f,g,gf,(gf)g,(gf)^2,(gf)^2g,(gf)^3=(fg)^3.
$$
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H: Calculating distribution of items among cells with limited capacity
The questioning is:
Given 13 identical balls and 6 different cells, in how many ways can the balls be distributed between the cells so that there is not a single cell that has 3 balls?
My approach to the question was:
Calculate the amount of distributions with no additional terms, which is D(6,13) and then subtract the amount of distribiotions that has at least one cell that has 3 balls.
To calculate the amount of distribiotions that has at least one cell that has 3 balls, I need to calculate the amount of distribiotions that has 1 cell with 3 balls, 2 cells with 3 balls etc... Until 4 cells as 5 cells with 3 balls is impossible given there are only 13 balls. Then apply the inclusion exclusion principle on the solution to find the total amount of distribiotions in which there are n cells, 0 < n < 5 that has 3 balls
To calculate for n 3-balls-cells, I calculated in how many ways the balls can be distributed among the free capacity cells (e.g 6-n cells) which is D(13 - n * 3, 6-n), then multiplied by all the possible positions in which the 3-balls-cells can be in which is C(6,n), and finally subtracted the duplicated calculations (such as the 3,5 cells and the 5,3 cells counted twice) by subtracting P(6, n).
By this calculation, since all the groups are contained within the group of one 3-balls-cell the inclusion exclusion principle will only be to calculate the amount distributions of one 3-balls-cell:
D(6,13) - C(6,1) * D(10,5) =
8568 - 6 * 2002 = -3444
Which doesn't make any sense.
What have I done wrong and what is the way to solve this problem?
AI: You seem to have started the inclusion-exclusion calculation but not finished it. Here’s a complete working out:
Let $A_i$ be the set of distributions in which cell $i$ has $3$ balls. Then
$$\begin{align*}
\left|\bigcup_{i=1}^6A_i\right|&=\sum_{\varnothing\ne I\subseteq[6]}(-1)^{|I|+1}\left|\bigcap_{i\in I}A_i\right|\\
&=\sum_{k=1}^4(-1)^{k+1}\binom6k\binom{(13-3k)+(6-k-1)}{6-k-1}\\
&=\sum_{k=1}^4(-1)^{k+1}\binom6k\binom{18-4k}{5-k}\;,
\end{align*}$$
since it is impossible for more than $4$ cells to contain $3$ balls, so the number of good distributions is
$$\begin{align*}
\binom{18}5&-\sum_{k=1}^4(-1)^{k+1}\binom6k\binom{18-4k}{5-k}\\
&=\binom{18}5+\sum_{k=1}^4(-1)^k\binom6k\binom{18-4k}{5-k}\\
&=\sum_{k=0}^4(-1)^k\binom6k\binom{18-4k}{5-k}\\
&=\binom{18}5-\binom61\binom{14}4+\binom62\binom{10}3-\binom63\binom62+\binom62\binom21\\
&=8568-6\cdot1001+15\cdot120-20\cdot15+15\cdot2\\
&=4092\;.
\end{align*}$$
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H: Use the Cauchy theorem to check whether a function is analytic or not in some region
Let $f $ be complex value defined in some bounded closed region $\gamma $ (rectangle) but we don't know whether it has poles or not in $\gamma$.
Let us consider the integral $I$ given by
$$I=\int_{\gamma} f(z)dz $$
Now we have two cases:
If $I=c$ where $c \in \mathbb{C}$ and $c$ different from $0$ so in this case, we deduce that $f$ is not analytic in the region $\gamma$ because if it is analytic the Cauchy theorem says that $I$ should be zero but we find it different from $0$.
Now the second case when $I=0 $, in this case, we can't say that $f$ is analytic or not in $\gamma$ (because there are some functions which are not analytic but they integral is zero).
So the Cauchy theorem works in one direction if $I$ is different from $0$ then $f$ it is not analytic in $\gamma$.
My question is what is the condition that makes the Cauchy theorem work in the other direction that means
If $I=0$ $\Longrightarrow$ $f$ is analytic and has no poles in $\gamma$
I suggest the condition for any
$$z_{0} \in \gamma \hspace{0.2cm}\& \hspace{0.2cm} \lim\limits_{z \to z_0} \frac{1}{(k-1)!} \left( \frac{d^{k-1}}{dz^{k-1}} (z-z_{0})^{k}f(z) \right) \neq 0 $$
Because the residue theorem tells us that $I=\sum \text{res} (f;z_{0})$
Is this condition is enough to say that theorem work in the other direction?
AI: Nope. This won't work. Consider $\displaystyle\int_\gamma \bar z{}^2\,dz$ with $\gamma$ the unit circle.
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H: How to find the PDF of a function of two random variables
Y is a uniform continuous random variable between [0,L] and X is a uniform continuous random variable given Y=y between [0,y]. What is the PDF of Z = X/Y.
AI: First of all you have to find the joint density $f_{XY}(x,y)$
You know:
Marginal $$f_Y(y)=\frac{1}{L}\mathbb{1}_{[0;L]}(y)$$
Conditional $$f_{X|Y}(x|y)=\frac{1}{y}\mathbb{1}_{[0;y]}(x)$$
So your joint density is
$$f_{XY}(x,y)=\frac{1}{yL}\mathbb{1}_{[0;L]}(y)\mathbb{1}_{[0;y]}(x)$$
....now you can proceed as you explained...please post you efforts
EDIT:
First observe that $z \in [0;1]$
Theb, using the definition of CDF you get
$$F_Z(z)=\mathbb{P}[Z \leq z]=\mathbb{P}[Y >\frac{X}{z}]=$$
$$=\int_0^{Lz}dx\int_{\frac{X}{z}}^L \frac{1}{Ly}dy...$$
Some intermediate results:
$$\frac{1}{L}\int_0^{Lz}dx[ln y]_{\frac{x}{z}}^L=\frac{1}{L}\int_0^{Lz}[lnL-ln(x/z)]dx=...=zln L-(lnL-1)z=z$$
Note: the integral $\int_0^{Lz}ln(x/z)dx$ is solved by parts.
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H: What mistake am I making while deriving the expansion of $\cos(\alpha + \beta)$
I was trying to derive the formula for expansion of $\cos (\alpha + \beta)$ by equating the ratio of lengths of two specific chords to the ratio of angles opposite to them but I'm not getting the correct results. Here's how I'm doing it :
In the above diagram,$\angle AOB = \alpha$, $\angle BOC = \beta$, $\angle AOC = (\alpha + \beta)$, $a = \cos{\alpha}$, $b = \sin{\alpha}$, $x = \cos{(\alpha + \beta)}$ and $y = \sin {(\alpha + \beta)}$
And as $a$, $b$, $x$ and $y$ are sines and cosines of $\alpha$ and $(\alpha+\beta)$ respectively, so : $a^2+b^2=x^2+y^2=1$
Now, using the distance formula for coordinate geometry, which states that the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ on the Cartesian Plane is : $\sqrt{(x_1-x_2)^2 - (y_1-y_2)^2}$ units, we obtain :
$$AB = \sqrt{(a-1)^2+(b-0)^2}=\sqrt{a^2+1-2a+b^2}=\sqrt{(a^2+b^2)+1-2a}=\sqrt{1+1-2a}$$
$$\therefore AB = \sqrt{2-2a}$$
$$AC = \sqrt{(x-1)^2+(y-0)^2}=\sqrt{a^2+1-2x+y^2}=\sqrt{(x^2+y^2)+1-2x}=\sqrt{1+1-2x}$$
$$\therefore AC = \sqrt{2-2x}$$
Now, the ratio of lengths of $AB$ and $AC$ would be equal to the ratio of angles opposite to them, that are $\alpha$ and $(\alpha + \beta)$ respectively (this is the part where I think I might be wrong but don't see how).
So, according to me,
$$\dfrac{AB}{AC}=\dfrac{\alpha}{\alpha + \beta} \implies \dfrac{AC}{AB} = \dfrac{\alpha + \beta}{\alpha} = 1 + \dfrac{\beta}{\alpha}$$
$$\implies \dfrac{\sqrt{2-2x}}{\sqrt{2-2a}} = 1 + \dfrac{\beta}{\alpha} \implies \dfrac{2-2x}{2-2a} = \Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2$$
$$\implies \dfrac{1-x}{1-a} = \Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2 \implies 1-x = (1-a)\Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2$$
This leads us to the conclusion that :
$$\cos(\alpha + \beta) = x = 1-(1-a)\Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2 = 1-(1-\cos{\alpha})\Bigg ( 1 + \dfrac{\beta}{\alpha} \Bigg )^2$$
which is not true...
So, where am I going wrong in this?
Thanks!
PS : I am really grateful to those people who are giving alternative methods of derivation but what I really want to know is the mistake in my derivation. Thanks!
AI: You make a mistake exactly where you suspected. Note the ratio of the lengths of the chords is not equal to the ratio of the angles subtended by them. You could see this by applying the law of cosines on $\triangle AOB$ and $\triangle AOC$: $$AB^2=OB^2 + OA^2 -2OA\cdot OB\cos\alpha \\ =1+1-2\cos\alpha\\=2(1-\cos\alpha)\\ \implies AB=2\sin\frac \alpha 2$$ Similarly, $$AC= 2\sin\frac{\alpha+\beta}{2}$$ and $$\frac{AB}{AC} \ne \frac{\alpha}{\alpha+\beta}$$ As for your question in the comments, recall that the arc length is really $r\theta$ and the chord length is $2r\sin\frac{\theta}{2}$. Two quantities will be directly proportional iff their ratio is a constant. But $$\frac{2r\sin\frac{\theta}{2}}{r\theta}=\frac{\sin\frac{\theta}{2}}{\frac{\theta}{2}}$$ which is clearly not constant, except maybe for the case when $\theta\approx 0$.
|
H: An application of Rouché's theorem
i need help to proof the next:
Let $f$ be analytic at $D$ minus a finite numbers of interior points where $f$ has poles. Show that if $0<|f(z)|<1$ over $\partial D$, then the number of poles of $f$ in $D$ is equal to the number of roots of the equation $f(z)=1$ in $D$.
I have tried to prove it using Rouché’s theorem, but i cannot conclude anything , I would appreciate your help.
AI: HINT: Use the argument principle for $g(z)=1-f(z)$. What is the image of $\partial D$ under $g$?
|
H: Evaluate the sum $\sum_{n=0}^{\infty} \frac{2n}{8^n}{{2n}\choose{n}}$
Evaluate the sum $\sum_{n=0}^{\infty} \frac{2n}{8^n}{{2n}\choose{n}}$
I am unable to find a way to solve this sum. I have never seen sums involving binomial coefficients multiplied by $n$. Help will be appreciated
AI: Note that
\begin{align}\binom{2i}{i} \frac1{2^{2i}}&=\frac{\left(\prod\limits_{j=1}^i2j\right)\left(\prod\limits_{j=0}^{i-1}(2j+1)\right)}{2^i 2^i(i!)^2}\\ &=(-1)^i\frac{(-1/2)(-1-1/2)\ldots (-1/2-i+1)}{i!}=(-1)^i\binom{-1/2}i\end{align}
and therefore one can rewrite the power series (for $x\in\Bbb C$ with $|x| <1$)
$$\tag{1}\sum_{n=0}^\infty \binom{2n}n \frac{x^{n}}{4^{n}}=\sum_{n=0}^\infty \binom{-1/2}n(-x)^n=(1-x)^{-1/2},$$
where we have used the Binomial series.
To establish a formula for your series, just differentiate (1) to find that
$$\sum_{n=1}^\infty 2n l\binom{2n}n \frac{x^{n-1}}{4^n}=(1-x)^{-3/2}$$
and hence
$$\sum_{n=0}^\infty 2n \binom{2n}n \frac{x^{n}}{4^{n}}=x(1-x)^{-3/2}.$$
Thus, taking $x= 1/4$ one gets
$$\sum_{n=0}^\infty\frac{2n}{8^n} \binom{2n}n=\frac{1}{4}(1-1/4)^{-3/2}.$$
|
H: Birthday Problem: Probability that at least two also share the same weekday, if weekday uniformly distributed and independent of birth date?
Consider the following problem extension for the birthday problem: We now want know the probability that out of $n$ persons, at least two people were born on the same date and the same weekday (for example, both were born on the 02.01 and on a Tuesday). We also want to know for which $n$ that probability is great than $\frac{1}{2}$.
We can assume that birthdays and weekdays follow a uniform distribution (also, we can assume the absence of leap-years) and are independent events. That is, the probability to be born on a certain date is $\frac{1}{365}$ and to be born on a weekday is $\frac{1}{7}$.
AI: You are looking for $$1 - \dfrac{(2555)_n}{2555^n} \ge \dfrac{1}{2}$$
where $$(2555)_n = (2555)(2554)\cdots (2555-n+1)$$
Plugging this into Wolframalpha gives $n \ge 59.7852$.
Testing this out:
$$1 - \dfrac{(2555)_{60}}{2555^{60}} \approx 0.50252$$
$$1- \dfrac{(2555)_{59}}{2555^{59}} \approx 0.49076$$
If the notation is confusing, another notation would be, the probability that every person has a distinct birthday and birth weekday is:
$$\left(\dfrac{2555}{2555}\right)\left(\dfrac{2554}{2555}\right)\cdots \left(\dfrac{2555-n+1}{2555}\right)$$
So, the probability that at least two people share both a birthday and a birth weekday is the complement of that:
$$1-\left(\dfrac{2555}{2555}\right)\left(\dfrac{2554}{2555}\right)\cdots \left(\dfrac{2555-n+1}{2555}\right) = 1-\dfrac{(2555)_n}{2555^n}$$
In Excel, you can use the following formulas:
$$\begin{array}{rl}\text{Cell A1:} & \text{Number of people} \\ \text{Cell B1:} & \text{Prob All Distinct} \\ \text{Cell C1:} & \text{Prob at least two share}\end{array}$$
This gives you column headers. Next, we want calculations:
$$\begin{array}{rl}\text{Cell A2:} & \text{=ROW()} \\ \text{Cell B2:} & \text{=(2555-ROW()+1)/2555} \\ \text{Cell C2:} & \text{=1-B2} \\ \text{Cell A3:} & \text{=ROW()} \\ \text{Cell B3:} & \text{=B2*(2555-ROW()+1)/2555} \\ \text{Cell C3:} & \text{=1-B3}\end{array}$$
Copy the formulas in cells $\text{A3:C3}$ to $\text{A3:A100}$. You will see that at 60 people, the probability in column $C$ is greater than $0.5$ for the first time.
|
H: Find a function satisfy certain condition
Find function $f:\mathbb{R}\to\mathbb{R}$ such that $f(2)=2$ and
$$\sum_{i=1}^nc_i\frac{\partial(x_1^2+\dots+x_n^2)^{\frac{f(y)}{2}}}{\partial x_i}=\frac{\partial(x_1^2+\dots+x^2_n)^{\frac{f(y)}{2}}}{\partial y}$$
This is what I got so far, firstly I tried to simplify the expression
$$\sum_{i=1}^n2c_ix_i\frac{f(y)}{2}(x_1^2+\dots+x_n^2)^{\frac{f(y)}{2}-1}
=\frac{1}{2}\ln(x_1^2+\dots+x_n^2)(x_1^2+\dots+x_n^2)^{\frac{f(y)}{2}}f'(y)$$
Not sure it's right, but still can't find $f$ satisfy this condition.
AI: There is no such function.
Writing $r^2=\sum_i x_i^2$ and using the fact that $\frac{\partial r}{\partial x_i}=\frac{x_i}{r}$ we can rewrite the above equation equivalently as
$$f(y)r^{f(y)-2}\sum_{i}c_i x_i=f'(y)\ln r~ r^{f(y)}\iff f(y)\sum_{i}c_i x_i=f'(y)r^2\ln r$$
but clearly the last equation cannot be satisfied for a function $f(y)$. If we allowed other dependences maybe this could be done. This can also be done as long as we restrict the domain on a surface
$$A~r^2\ln r=\sum_{i}{c_ix_i}~,~ A\in \mathbb{R}$$
in which case the solution is
$$f(y)=2e^{A(y-2)}$$
but whether this is useful or not depends on the nature of the original problem posed.
|
H: On the codomain of a complex valued function
I want to denote a complex function that its outputs are real. So I must write:
$$f: \mathbb{C} \to \mathbb{R}$$
However, what happens to those $z$ that make $f(z)$ complex? Are they considered and then I cut off the imaginary output. For instance, let:
$$f(z) = z^2 + x_0, \,\,\, x_0 \in \mathbb{R}$$
If $z=i$, it is fine. But what if $z=1+i$, should I cut off the imaginary part $(2i)$ from the output $2i+x_0$?
AI: If you write $f: \mathbb C \to \mathbb R$, you are stating that all the values $f(z)$ are real, for all $z \in \mathbb C$. So if your function is $f(z) = z^2 + x_0$ and you write $f: \mathbb C \to \mathbb R$, you have just made
a mistake.
If not all the values are real, but you "cut off" the imaginary part, you have a different function, namely the real part of your original function.
The original function $f$ does not map $
\mathbb C$ into $\mathbb R$, but $\text{Re} f$ does.
|
H: matrix least square optimization with constraint
I want to Minimize the following equation (the error matrix) by
formulating a cost function and calculating the point where
its gradient equals zero.
\begin{equation}
\hat{X} = \arg \min_{X} \frac{1}{2} {\left\| AX - B \right\|}_{F}^{2}
\end{equation}
where $$X= {\rm Diag}(x)$$
I did this optimization problem without constraint:
$$\eqalign{
E &= AX-B\\
\phi &= \tfrac{1}{2}\|E\|^2 = \tfrac{1}{2}E:E \\
d\phi &= E:dE = E:A\,dX = A^TE:dX \\
&= A^TE:{\rm Diag}(dx) \\
&= {\rm diag}\Big(A^T\big(AX-B\big)\Big):dx \\
\frac{\partial\phi}{\partial x}
&= {\rm diag}(A^TAX) - {\rm diag}(A^TB) \\
}$$
Set the gradient to zero and solve for the optimal $x$.
$$\eqalign{
{\rm diag}(A^TB) &= {\rm diag}(A^TAX) \\
&= \left(I\odot A^TA\right)x \\
x &= (I\odot A^TA)^{-1}\operatorname{diag}\left(A^TB\right) \\
}$$
now, I want to minimize above optimization problem with constraint:\begin{equation}
\hat{X} = \arg \min_{X} \frac{1}{2} {\left\| AX - B \right\|}_{F}^{2} , X= {\rm Diag}(x)
\end{equation}
\begin{align}
\text{subject to } & x_{i}^{min} \leq x_{i} \leq x_{i}^{max} \\
\end{align}
I recently learnt about Lagrange multipliers but I'm not sure how to apply that to my problem. I'll appreciate if any one can help me how to apply that.
AI: Let $a_i$ and $b_i$ be the $i$th columns of $A$ and $B$, respectively. Let $\tilde A$ be the block diagonal matrix whose $i$th diagonal block is $a_i$, and let $b$ be the column vector obtained by vertically concatenating the vectors $b_i$. Notice that
$$\| AX - B \|_F^2 = \| \tilde A x - b \|_2^2.
$$ So your optimization problem is
\begin{align}
\text{minimize} & \quad \frac12 \| \tilde A x - b \|_2^2 \\
\text{subject to} & \quad c \leq x \leq d
\end{align}
where $c$ is the vector whose $i$th component is $x_i^\min$ and $d$ is the vector whose $i$th component is $x_i^\max$. The optimization variable is $x$. Projecting onto the constraint set is easy, so you could solve this problem using the projected gradient method or an accelerated projected gradient method. (Other methods such as interior point methods are also possible.)
When implementing the projected gradient method, it will help to know that the gradient of your objective function
$$
f(x) = \frac12 \| \tilde A x - b \|_2^2
$$
is
$$
\nabla f(x) = \tilde A^T(\tilde A x - b).
$$
|
H: Volume using double integrals
Calculate the volume of the solid bounded by the following surfaces:
$y=x^2, y=1, x+y+z=4, z=0$.
How does on set up the integral?
AI: $y=x^2$ is "parabola shifted infinitely" accordingly z-axis.
Your volume can be calculate by 2 and/or 3 dimensional integral:
$$\int_{-1}^{1}\int_{x^2}^{1}\int_{0}^{4-x-y}dxdydz= \int_{-1}^{1}\int_{x^2}^{1}(4-x-y)dxdy$$
Hope, you can finish it.
|
H: Verify the statements for Riemann-integrable function $f_n(x)$.
For each $n = 1, 2, \cdots$ a function $f_n(x)$ is defined so that it is Riemann-integrable on $[a, b]$ and the series $\sum_{n=1}^{\infty}f_n(x)$ converges $\forall \space x \in [a,b]$.
Which of the following statements are true?
$$\lim_{n\rightarrow\infty}\sup_{x\in[a,b]}|f_n(x)| = 0$$
$$\int_{a}^{b}\bigg(\lim_{n\rightarrow \infty} |f_n(x)|\bigg) dx = 0$$
$$\lim_{n\rightarrow\infty}\int_{a}^{b}|f_n(x)|dx = 0$$
ATTEMPT
I. $f_n$ is integrable, so it is bounded. Therefore, $\sup |f_n(x)|$ exists and equals to $|f_n(c)|, c\in [a, b]$. Since the series converges, $\lim f_n(x) = 0 \space \forall x \in [a, b]$. Hence $\lim \sup _{x \in [a, b]}|f_n(x)| =\lim|f_n(c)| = |0| = 0.$ Answers indicate that this is false, but I don't see where I am mistaken.
II. Recall that $\lim f_n(x) = 0 \space \forall x \in [a, b]$. Then $\lim |f_n(x)| = |0| = 0 \space \forall x \in [a, b]$. Therefore, the integrand is 0 on $[a, b]$, and so is the integral.
III. No idea.
It would be great if there was a counterexample.
AI: Let $ f_n$ defined at $ [0,1]$ by
$$f_n(x)=x^n \; if \; x\ne 1 \; and \; f_n(1)=0$$
$$f_n \; is \; Riemann\; integrable\; at \; [0,1],$$
$$\sum f_n(x) \; converges \; \forall x\in[0,1]$$
but
$$\sup_{x\in [0,1]}|f_n(x)|=1$$
the first statement is false.
|
H: Suppose $[a], [b] \in \mathbb{Z}_n$ and $[a]\cdot[b] = [0]$. Is it necessarily true that either $[a] = [0]$ or $[b] = [0]$?
Let $n \in \mathbb{N}$. Let $R$ be the equivalence relation $\equiv \pmod{n}$. Suppose $[a], [b] \in \mathbb{Z}_n$ and $[a]\cdot[b] = [0]$. Is it necessarily true that either $[a] = [0]$ or $[b] = [0]$?
Proof:
Premises:
$
1.\, [a]\cdot[b] = [0]\\
2. \forall x\forall y\forall k(x \equiv y \pmod{n} \to x \equiv ky \pmod{n})
$
Assume $[a] \neq [0]$:
$
\begin{align*}
&\langle\text{Assumption} \rangle\\
&c \in [b]\\
\iff&\langle\text{Def. of $[b]$} \rangle\\
& cRb\\
\iff&\langle\text{Def. of $R$} \rangle\\
& c \equiv b \pmod{n}\\
\iff&\langle\text{By Premise 2} \rangle\\
& c \equiv ab \pmod{n}\\
\iff&\langle\text{Def. of $R$} \rangle\\
& cRab\\
\iff&\langle\text{Def. of $[ab]$} \rangle\\
& c \in [ab]\\
\iff&\langle\text{$[a][b]=[0]$} \rangle\\
&c \in [0]\\
\end{align*}
$
$[b] \subseteq [0]$
Is my proof heading in the right direction ?
Is the addition of Premise 2 correct? How can I prove it to use it in my proof ?
AI: As pointed out in the comments, the statement is false as well as the "Premise 2".
But you can try to show this: if $p$ is prime, then $[a]_p[b]_p=[0]_p$ implies $[a]_p=[0]_p$ or $[b]_p=[0]_p$. There is more or less nothing to prove apart from correctly using the definition of congruence.
In particular in $\mathbb{Z}/p\mathbb{Z}$ there are no zero-divisors. On the contrary, you can also show that if $n$ is composite then there exist zero-divisors in $\mathbb{Z}/n\mathbb{Z}$.
|
H: Find the distribution of numbers of arrivals of the Poisson process $N(t)$ in time interval $[t, t+\tau)$, $\tau \sim Exp(a)$.
Poisson process has rate $\lambda$ and $\tau \perp \!\!\! \perp N(t)$.
To find distribution i've started with $P(N(t+\tau)-N(t)=k) = P(N(\tau) = k)$. I know that $N(t) \sim Poiss(\lambda t)$, but i don't know what to do next to find distribution.
AI: Hint: use the formula
$$ P(N(\tau) = k) = \int_0^\infty P(N(s)=k|\tau = s) ae^{-as}\, ds.$$
|
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