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H: Confusion about unjustified argument in solution to IMO 2018 algebra problem
The first algebra question in IMO 2018 is:
Let $\mathbb{Q}_{>0}$ denote the set of all positive rational numbers. Determine all functions
$f\colon \mathbb{Q}_{>0} \to \mathbb{Q}_{>0}$ satisfying
$$f\left(x^2 f\left(y \right)^2 \right) = f\left(x \right)^2f\left(y \right)$$
for all $x, y \in \mathbb{Q}_{>0}.$
In the first part of the solution they say:
Take any $a, b \in \mathbb{Q}_{>0}.$ By substituting $x = f\left(a \right) \dots$
This is where my issue is, the question asks you to prove for all $x \in \mathbb{Q}_{>0}$. Wouldn't substituting $x$ for $f\left(a \right)$ only be justified if $f$ was a bijective function?
Link: https://www.imo-official.org/problems/IMO2018SL.pdf
AI: There is no issue with the substitution.
The identity holds for all rational $x$. Since $f(a) $ is a rational number, we're allowed to set $ x = f(a)$.
We are saying "For all $a$, we have $x = f(a)$ and thus...
We are not saying that "For all $x$, there exists an $a$ such that $ x = f(a)$."
Yes, it might be that the image of $ f(a) $ isn't $\mathbb{Q}$.
(In particular, since the only solution is $f(x) = 1$.)
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H: Probability that exactly $k$ balls in each group are black and rest are white
Assume that we have a total of $n$ balls, $b$ of which are black and the remaining are white. I want to partition them into $g$ groups of size $r$ (i.e., $n=g \times r$) such that for $b/k$ of the groups exactly $k$ balls in each group are black; the rest of groups have only white balls. I want to compute the probability of this happening if one was to pick the balls for the groups at random (without replacement). For the purposes of this problem you can assume that $k$ divides $b$.
My thinking is that we can imagine we have two bags (one with blacks of size $b$ and another with whites) and that we start populating groups one by one and keeping track of the remaining balls of each bag such that the probability is
$$\frac{{b \choose k}{n-b \choose r-k} \times {b-k \choose k}{n-b-(r-k) \choose r-k} \times \cdots \times {b-(b/k-1)k \choose k}{n-b-(b/k-1)(r-k) \choose r-k}}{{n \choose r}}$$
i.e., we stop at the last mixed group since for the last group there are no black balls left.
Is this calculation correct?
I think the denominator is wrong, I think it might be ${n \choose k}$ (which is equal to all possible orderings of the balls in a line) but I am not sure.
AI: Since we are calculating a probability, we can safely assume the $g$ groups are distinguishable, and that the balls are distinguishable. With this formulation, the sample space is simply the set of all assignments of the $n$ balls to the $g$ groups such that each group gets $r$ balls, and this can be done in $\binom{n}{r;r;\cdots;r}$, the multinomial coefficient with $g$ copies of $r$ at bottom.
For the event you are considering, the first thing you need to do is pick $\frac{b}{k}$ of the groups to get the black balls, which can be done in $\binom{g}{b/k}$ ways (this is a key factor missing in your analysis). Then there are $\binom{b}{k;k;\cdots;k}$ ($\frac{b}{k}$ $k$'s in the bottom) ways to distribute the black balls to these bins. Finally you must distribute the white balls. Each of the bins with black balls must get $r-k$ white balls, while the remaining bins must get $r$ white balls; this assignment can be done in $\binom{n-b}{r-k;r-k;\cdots;r-k;r;r;\cdots;r}$ (there are $\frac{b}{k}$ $r-k$'s and $g-\frac{b}{k}$ $r$'s on the bottom). The probability you need is the product of these three quantities, divided by the size of the sample space.
EDIT: In comment, the OP asks a slight extension where the desired condition is that $\lfloor\frac{b}{k}\rfloor$ bins (let's call them "special") contain exactly $k$ black balls, and the remaining black balls can be distributed arbitrarily. In this case the analysis is a bit more complex, but follows the same logic. There are $\binom{g}{\lfloor b/k\rfloor}$ ways to pick the bins that get $k$ black balls, and then $$\binom{b}{k;k;\cdots;k;b-k\lfloor b/k\rfloor}$$ to pick the $k\lfloor b/k\rfloor$ black balls that will go in the special bins and distribute them. The $b-k\lfloor b/k\rfloor$ selects the black balls that will later be distributed to the non-special bins.
The next task is to fill up the special bins with white balls, so we can make sure we don't get any additional black balls in. This can be done in $$\binom{n-b}{r-k;r-k;\cdots;r-k;n-b-(r-k)\lfloor b/k\rfloor}$$ ways. As with the black balls, we are both picking which white balls will go in the special bins and distributing them...the distribution of the white balls to non-special bins is left until later.
Finally, we need to distribute the remaining balls to the non-special bins, regardless of whether they are black or white. This can be done in $$\binom{n-r\lfloor b/k\rfloor}{r;r;\cdots;r}$$ where there are $g-\lfloor b/k\rfloor$ $r$'s on the bottom (correction provided by OP).
Note: you pretty much have to do the counts in this order, because you don't have any control on the black/white distribution in the non-special bins.
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H: Finding $|\!\operatorname{Aut}(L(K_4))|$ using Orbit-Stabiliser Theorem
I know that you can find the size of an automorphism group of a simple graph $G$ by using the Orbit-Stabiliser theorem as follows: let $\DeclareMathOperator{Aut}{Aut}A = \Aut(G)$, and $v$ be a vertex of $G$, where $Av$ denotes the orbit of $v$, and $A_v$ is the stabiliser of $v$, then $|A| = |Av||A_v|$. You can find the orbit of a vertex $v$, then fix it, and consider the orbit of another vertex $w$ when $v$ is fixed, e.g. $B = A_v, |A| = |Av||Bw||B_w|$ et cetera, until all non-trivial orbits and stabilisers have been found.
With that being said, I'm not entirely sure how to apply this method to $L(K_4)$, where $L(G)$ denotes the line graph of a simple graph $G$. This is a 4-regular graph of order 6 - so does any vertex have an orbit of size 6, since each vertex has the same degree? Or does its orbit vary and have something to do with its neighbouring vertices? How can we find the orbit of a vertex by inspection?
AI: Does any vertex have an orbit of size 6, since each vertex has the same degree?
There is indeed only one vertex orbit, containing all six orbits of $L(K_4)$. However in general vertices having the same degree is not sufficient for them to be in the same orbit under automorphisms. For example suppose you add a leaf node to the path graph $P_5$ incident at one of the two noncentral nonleaf nodes; this will leave two vertices of degree two which are obviously not in similar positions in the graph. Of course two vertices having the same degree is a necessary condition.
View $K_4$'s vertex set as $\{1,2,3,4\}$, in which case two vertices $ij$ and $k\ell$ of $L(K_4)$ are adjacent if and only if the sets $\{i,j\},\{k,\ell\}$ contain a common term, e.g. $12$ and $23$ are incident. Since permutations of $\{1,2,3,4\}$ induce automorphisms of $K_4$, they also induce automorphisms of $L(K_4)$, and it's easy to see there is a permutation turning $ij$ into $k\ell$ for any $i,j,k,\ell$. This means the action is "transitive" (it's possible to transition from any vertex to any other vertex via an automorphism) so there's only one orbit.
Some authors may summarize this argument simply declaring there is one orbit "by symmetry."
Note that while we know permutations of $\{1,2,3,4\}$ induce automorphisms of $L(K_4)$, we don't a priori know whether or not there are more automorphisms of $L(K_4)$ than just these kinds.
Pick any vertex, say $12$. Note $34$ is not incident to it, but all four other vertices are. Therefore the stabilizer $A_{12}$ must also stabilize $34$. (Why?) This leaves four vertices $13,14,23,24$ for any other automorphism (in the stabilizer) to permute. Can you check the orbit of $13$ (under the action of $A_{12}$ not $A$) includes $14,24,24$ using permutations of $\{1,2,3,4\}$ which fix $12$? (Note these permutations can swap $1$ and $2$.)
Now pick one of these four vertices, say $13$. What is the stabilizer $(A_{12})_{13}$? Argue any element of it must fix not only $12,13$ but also $34,24$, leaving only $14,23$ to be permuted. Note they can be permuted, since all four other vertices besides $14,23$ are connected to both of them, so it is possible to swap these two vertices and drag the edges along with and that's a valid graph automorphism.
In conclusion, $|A|=6\cdot4\cdot2=48$.
Here is a geometric interpretation. The $K_4$ graph can be interpreted as a tetrahedron, whose full symmetry group $T_h$ has $12$ rotations (two $120^{\circ}$ rotations for each of the four faces, a $180^{\circ}$ rotation for each of the three opposite pairs of edges, and the trivial element) and $12$ reflections (one for each diagonal line on a face), corresponding to even and odd permutations of the four vertices (respectively). We can say $T\cong A_4$ and $T_h\cong S_4$. Moreover, every symmetry of the tetrahedron induces an automorphism of the graph, and $T_h\to\mathrm{Aut}(K_4)$ is an isomorphism.
If one connects the midpoints of the edges of the tetrahedron according to which edges share an endpoint, one gets the line graph $L(K_4)$ represented by an inscribed octahedron. The octahedron's faces may be partitioned into two alternating subsets, four corresponding to the tetrahedron's vertices and four corresponding to the tetrahedron's.
Note that the dual tetrahedron flips around the original's vertices and faces. That is, if one connects the midpoints of the faces of the tetrahedron according to which faces share an edge one gets a dual tetrahedron. If one rescales the dual tetrahedron until it's the same size as the original, their overlap will be the inscribed octahedron and their convex hull a cube (every cube has a dual pair of inscribed tetrahedra). The octahedron and the cube share the same full symmetry group $O_h$ in which $T_h$ is an index $2$ subgroup. If one applies one of the reflections in $O_h\setminus T_h$ it swaps the two tetrahedra but preserves the octahedron, corresponding to automorphisms of $L(K_4)$ that does not come from an automorphism of $K_4$.
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H: Combinatorial identity: $\sum\limits_{k=0}^{i\land j}\binom ik(-1)^k\binom{i+j-k}i=1$
Let $i,j\in\mathbb Z_{\ge0}$ be nonnegative integers. How can we prove
$$\sum_{k=0}^{i\land j}\binom ik(-1)^k\binom{i+j-k}i=1?$$
(Here, $i\land j=\min(i,j)=\min\{i,j\}=\min(\{i,j\})$ is the minimum of $i$ and $j.$ This problem comes from my study of stationary distributions of birth-death chains.)
By the identity
$$\binom ik\binom{i+j-k}i=\frac{(i+j-k)!}{k!(i-k)!(j-k)!}=\binom{i+j-k}{k,i-k,j-k},$$
we have
$$\sum_{k=0}^{i\land j}\binom ik(-1)^k\binom{i+j-k}i=\sum_{k=0}^{i\land j}(-1)^k\binom{i+j-k}{k,i-k,j-k}.$$
I was thinking of using the trinomial theorem, but I don't see how -- the form of the sum seems a bit different.
AI: Suppose that you want to count the $i$-element subsets of $[i]=\{1,2,\ldots,i\}$. Of course there’s only one of them, but we can also count them by the following roundabout procedure. We first expand the set from which we’re drawing the $i$-element subset to $[i+j]=\{1,\ldots,i+j\}$. Now for each $\ell\in[i]$ let $A_\ell$ be the family of $i$-element subsets of $[i+j]$ that do not contain $\ell$; $\bigcup_{\ell=1}^iA_\ell$ is the family of $i$-elements subsets of $[i+j]$ that are not subsets of $[i]$. By the inclusion-exclusion principle we have
$$\begin{align*}
\left|\bigcup_{\ell=1}^iA_\ell\right|&=\sum_{\varnothing\ne I\subseteq[i]}(-1)^{|I|+1}\left|\bigcap_{\ell\in I}A_\ell\right|\\
&=\sum_{k=1}^i\binom{i}k(-1)^{k+1}\binom{i+j-k}i\;,
\end{align*}$$
since each non-empty $I\subseteq[i]$ has cardinality in $[i]$, for each $k\in[i]$ there are $\binom{i}k$ subsets of $[i]$ of cardinality $k$, and if $|I|=k$,
$$\left|\bigcap_{\ell\in I}A_\ell\right|=\binom{i+j-k}i\;.$$
There are $\binom{i+j}i$ $i$-element subsets of $[i+j]$ altogether, so after we throw out the ones not contained in $[i]$, we have left
$$\begin{align*}
\binom{i+j}i&-\sum_{k=1}^i\binom{i}k(-1)^{k+1}\binom{i+j-k}i\\
&=\binom{i+j}i+\sum_{k=1}^i\binom{i}k(-1)^k\binom{i+j-k}i\\
&=\sum_{k\ge 0}\binom{i}k(-1)^k\binom{i+j-k}i\;,
\end{align*}$$
and we already know that this is $1$.
Note that there is no need to specify an upper limit on the summation: $\binom{i}k=0$ when $k>i$, and $\binom{i+j-k}i=0$ when $k>j$, so all terms with $k>i\land j$ are $0$ anyway.
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H: Interpretation of $\mathbb P(A|X=x)$ in two ways
Let $X:(\Omega,\mathscr A) \to (\mathbb R,\mathscr{B})$ be a random variable between two measurable spaces (the latter being the Borel measurable space over $\mathbb R$). Let $x\in \mathbb R$. Let $\mathbb P$ be a probability on $(\Omega,\mathscr A)$. Assuming $\mathbb P(X=x)\ne 0$ ,I have two interpretations for $\mathbb P(A|X=x)$:
(1) Naive definition: Simply use the elementary definition of the conditional probability:
$$\mathbb P(A|X=x)=\frac{\mathbb P(A\cap[X=x])}{\mathbb P([X=x])}.$$
(2) Official definition of such conditional probability:
$$\mathbb P(A|X=x)=\mathbb E(\mathbb 1_A|X=x)=\varphi(x),$$
where the conditional probability $\mathbb E(\mathbb 1_A|X=x)=\varphi(x)$ is defined via the "factorization lemma", which states that there exists a measurable $\varphi:(\mathbb R,\mathscr{B})\to (\mathbb R,\mathscr{B})$ such that $\varphi(X)=\mathbb E(\mathbb 1_A|X):=\mathbb E(\mathbb 1_A|\sigma(X))$. ($\mathbb E(\mathbb 1_A|\sigma(X))$ is the conditional probability in the usual sense).
The definition in (2) can be found for example in Probability Theory: A Comprehensive Course, by Achim Klenke, page 180-181.
Do the conditional probabilities in (1) and (2) agree (at least almost surely)? If they only agree under some further assumptions, please let me know.
AI: Given that the set $G := \{\omega: X(\omega) = x\}$ has positive probability, by conditional probability definition, $\varphi(X)$ satisfies
$$P(A \cap G) = \int_G \varphi(X(\omega))dP = \int_{\{x\}}\varphi(y)\mu(dy) = \varphi(x)\mu{(\{x\})} = \varphi(x)P[X = x],$$
where $\mu$ is the induced probability measure on $(\mathbb{R}, \mathscr{B})$.
In above, the second equality follows the change-of-variable formula. The third equality follows the definition of integration.
Hence your conjecture is correct, as it should be.
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H: If $f'(c)=0$ and $f''(c)\gt0$, then $f$ has a local minimum at $c$
Question: Let $f$ is differentiable on $I$. For $c\in I$, if $f'(c)=0$ and $\exists f''(c)\gt0$, then show that $f$ has a local minimum at $c$.
As you know, this is regarded as a fundamental theorem and is useful when graphing the function. But, I had some trouble because I only learned in the case that it holds if $f''$ is continuous near c.
When $f''$ is continuous near $c$, then there exists $\delta$ such that $\forall x\in(c-\delta, c+\delta)\implies f''(x)\gt0$, or $f'$ is increasing. So, $\forall x\in (c-\delta, c), f'(x)<f'(c)=0$ and $\forall x\in (c, c+\delta), f'(x)>f'(c)=0$, which in turn we conclude that $f$ has a local minimum at $c$.
But, what can we imply if there's no such condition? More basically, does the statement hold even though the condition($f''$ is continuous near $c$) is absent? Thanks a lot.
AI: We don't need to assume that $f''$ is continuous, essentially because it's already defined as a limit. Specifically, we know that
$$
f''(c)=\lim_{x \to c} \frac{f'(x)-f'(c)}{x-c}=\lim_{x \to c}\frac{f'(x)}{x-c}
$$
Let $\varepsilon = \frac{1}{2}f''(c)$. Then there is some $\delta$ such that, whenever $0 < |x-c| < \delta$, we have
$$\left|f''(c)-\dfrac{f'(x)}{x-c}\right|<\varepsilon
$$
It follows that
$$
\frac{f'(x)}{x-c} > f''(c)-\varepsilon = \frac{1}{2}f''(c)
$$
and so $\dfrac{f'(x)}{x-c}$ is positive when $0 < |x-c| < \delta$. That is, $f'(x)$ is negative when $x \in (c-\delta,c)$ and positive when $x \in (c, c+\delta)$, which as you've noted is what we need.
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H: If $\lim (f(x) + 1/f(x)) = 2 $ prove that $\lim_{x \to 0} f(x) =1 $
Let $f:(-a,a) \setminus \{ 0 \} \to (0 , \infty) $ and assume $\lim_{x
\to 0} \left( f(x) + \dfrac{1}{f(x) } \right) = 2$. Prove using the
definition of limit that $\lim_{x \to 0} f(x) = 1$
Attempt:
Let $L = \lim_{x \to 0} f(x) $. Let $\epsilon > 0$ be given. If we can find some $\delta > 0$ with $|x| < \delta $ such that $|f(x) - 1 | < \epsilon $ then we will be done.
We know that since $f(x) > 0$, then $\lim 1/f(x) $ is defined. In fact, applying the limit to hypothesis, we end up with
$$ L+ \dfrac{1}{L} = 2 $$
and certainly $L=1$ as desired. I am having difficulties making this proof formal in $\delta-\epsilon$ language. Can someone assist me?
AI: First, note that:
$$a + \frac{1}{a} - 2 = \frac{a^2 - 2a + 1}{a} = \frac{(a - 1)^2}{a}.$$
So, roughly speaking, given we can make $\left|f(x) + \frac{1}{f(x)} - 2\right|$ as small as we like, we should be able to make $\frac{(f(x) - 1)^2}{|f(x)|}$ as small as we like. There are two ways for this to happen: either $f(x)$ is exceptionally close to $1$ or $|f(x)|$ is extremely, unreasonably large. However, in the latter case, $|f(x)|$ being very large means that $(f(x) - 1)^2$ is even larger, which will prevent the whole fraction from being small, so there really was only one option.
Let's formalise this. First, we need to eliminate the possibility of $|f(x)|$ becoming overly large, so let's try to show that $f(x)$ is bounded on some neighbourhood of $x = 0$.
I'm going to pick, fairly arbitrarily, $\varepsilon = 1$, and apply it to the limit definition of $f(x) + 1/f(x) \to 2$. Then, there exists some $\delta_0 > 0$ such that
\begin{align*}
0 < |x| < \delta_0 &\implies \left|f(x) + \frac{1}{f(x)} - 2\right| = \frac{(f(x) - 1)^2}{|f(x)|} < 1 \\
&\implies f(x)^2 - 2f(x) + 1 < |f(x)| \\
&\implies f(x)^2 + (\pm 1 - 2)f(x) + 1 < 0,
\end{align*}
where the $\pm$ depends on the sign of $f(x)$. In either case, we have a convex parabola, and the solutions to the above inequalities will be bounded. That is, there exists some $M$ such that $|f(x)| \le M$ for all $x \in (-\delta_0, \delta_0) \setminus \{0\}$. Since $f(x) \neq 0$ for at least some $x$ near $0$, we know $M > 0$.
Now we prove the limit. Suppose $\varepsilon > 0$. We have, for $0 < |x| < \delta_0$,
\begin{align*}
|f(x) - 1| < \varepsilon &\impliedby (f(x) - 1)^2 < \varepsilon^2 \\
&\impliedby \frac{(f(x) - 1)^2}{M} < \frac{\varepsilon^2}{M} \\
&\impliedby \frac{(f(x) - 1)^2}{|f(x)|} < \frac{\varepsilon^2}{M} \\
&\impliedby \left|f(x) + \frac{1}{f(x)} - 2\right| < \frac{\varepsilon^2}{M}.
\end{align*}
Using the definition of the known limit, there is some $\delta_1$ such that
$$0 < |x| < \delta_1 \implies \left|f(x) + \frac{1}{f(x)} - 2\right| < \frac{\varepsilon^2}{M},$$
and hence
$$0 < |x| < \min\{\delta_0, \delta_1\} \implies |f(x) - 1| < \varepsilon,$$
completing the proof.
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H: Confusion on a lemma
So the lemma I had trouble understanding was this:
For all real numbers r, $−|r| ≤ r ≤ |r|$
The solution was something like this(used the definition of absolute values + division into cases)
Suppose r is any real number. We divide into cases according to
whether r ≥ 0 or r < 0.
Case 1 (r ≥ 0):
$|r| = r$. (By definition of absolute value)
$−|r| < r$.(since $r$ is positive and $−|r|$ is negative)
Thus it is true that $−|r| ≤ r ≤ |r|$.
Qn: But based on what it said above, shouldn't it be more of $−|r| < r$ and $r=|r|$ not $−|r| ≤ r ≤ |r|$?
Case 2 (r < 0):
$|r|=−r$. (By definition of absolute value)
Multiplying both sides by −1 gives that $−|r| = r$.
$r < |r|($since $r$ is negative and $|r|$ is positive)
Thus it is also true in this case that $−|r| ≤ r ≤ |r|$.
Qn: But based on what it said above, shouldn't it be more of $r < |r| $ and $-r=|r|$ not $−|r| ≤ r ≤ |r|$ ?
Hence, in either case, $−|r| ≤ r ≤ |r|$
Is the answer to my question in anyway linked to abusing inequalities?
Eg: Something like $1≤2$, although correct, it is imprecise. (sorry if this is totally off)
AI: Case $(1)$:
We have $$-|r|< r =|r|$$
but $-|r| < r$ would imply that $-|r|\le r$ and $r=|r|$ would imply that $r \le |r|$.
Hence we can write $-|r| \le r \le |r|$.
Note that $\le $ means less than or equal to.
Similarly for case $(2)$.
Explicit numerical example:
If $r=2$, then we have $-2 < 2 = 2$, we also have the inequalities $-2 \le 2 \le 2$.
If $r=-2$, then we have $-2 = -2 < 2$, which implies that $-2 \le -2 \le 2$.
Notice that for $-|r| \le r \le |r|$, equality is attained on at least one side. If both equalities are attained, then we have $r=0$.
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H: Poisson Distribution and Conditional Probability
I'm stuck with a problem for my Statistics class. The problem says that there is a hospital where patients arrive at a constant rate of 2 patients per hour, and there is a doctor that works 12 hours, from 6 am to 6 pm. If the doctor has treated 6 patients by 8 am, what is the probability that he treats 9 patients by 10 am?
So, I was planning to solve this problem using Poisson distribution and conditional probability. I identified two events: A, where after 2 hours the doctor treats 6 patients, and B, where after 4 hours the doctor treats 9 patients.
For A, m=lambda*hours=2*2hrs=4, and P(X=6)= (e^-4 * 4^6)/6! = 0.104
For B, m=lambda*hours=2*4hrs=8, and P(X=9)= (e^-8 * 8^9)/9! = 0.124
So, given that events are independent in Poisson distributions, I try to calculate P(B|A) as P(B)/P(A) but I get a number > 1.
Also, I tried working the answer considering the time interval from 8 am to 10 pm, where the number of patients increased by (9-6)=3
In that scenario, m=lambda*hours=2*2hrs=4, and P(X=3)= (e^-4 * 4^3)/3! = 0.53 , but I think it must be wrong too, since it does not consider the event A.
Could someone help me with this?
Thanks!
AI: The number of patients treated is cumulative, so the outcome of interest here is that the doctor has treated three additional patients between $8$ and $10$ a.m.
One of the properties of the Poisson process is that, at any given point in time, the occurrence of future events is independent of any past events. So all you have to do is compute the probability that the number of arrivals $X(t)$ in the next $t = 2$ hours equals $3$, when the rate is $\lambda = 2$ arrivals per hour. This is given by $$\Pr[X(2) = 3] = e^{-\lambda t} \frac{(\lambda t)^3}{3!} = \frac{32}{3e^{4}} \approx 0.195367.$$
But suppose we don't do it this way, and try to model the problem more along the lines of what you were thinking. Then let $A$ be the event that exactly $6$ arrivals occurred by time $t = 2$ hours, and let $B$ be the event that exactly $9$ arrivals have occurred by time $t = 4$ hours. Then $$\Pr[B \mid A] = \frac{\Pr[B \cap A]}{\Pr[A]}.$$ The denominator is easy; with $\lambda t = 2(2) = 4$, we have $$\Pr[A] = e^{-4} \frac{4^6}{6!} \approx 0.104196.$$ This is the same as what you computed. Where we differ is the numerator; in particular, $$\Pr[B \cap A] \ne \Pr[B].$$ The reason is because the event $B \cap A$ excludes outcomes in which, say, only $5$ arrivals occurred in the first two hours but $4$ occurred in the next two hours; whereas $\Pr[B]$ considers whether a total of $9$ arrivals occurred within the first four hours irrespective of their distribution within the first and second halves of the four-hour interval. That's why your probability exceeds $1$: your numerator becomes too large because $\Pr[B]$ includes events you should be excluding.
This leads us to conclude that the intuitive way to compute $\Pr[B \cap A]$ is to reason that $\Pr[B \cap A] = \Pr[B \mid A]\Pr[A]$ and $\Pr[B \mid A] = \Pr[X(2)=3]$, which is of course circular because what we were after from the beginning was $\Pr[B \mid A]$. So is there some other way? Well, yes, but it's not simple. The idea is to consider the distribution of the $9$ arrivals in the first four hours according to whether they happen in the first two or last two hour intervals. In other words, we think of how many ways they can be distributed between the two intervals, and count only the one where $6$ happen in the first half and $3$ in the second.
Because the intervals have equal length, given that an arrival occurred in the first four hours, the probability that it occurs in the first half equals the probability it occurs in the second. So the probability that exactly $Y = 6$ events occurred in the first half is a binomial probability with $n = 9$ and $p = 1/2$: $$\Pr[Y = 6] = \binom{9}{6} (1/2)^6 (1 - 1/2)^3 = \frac{21}{128} \approx 0.164063.$$ Then we have to also multiply this by the Poisson probability of seeing $9$ events in the first place, i.e. $$Pr[B \cap A] = \Pr[Y = 6]\Pr[B] \approx 0.164063(0.124077) \approx 0.0203564.$$ And when we finish the calculation, you find $$\Pr[B \mid A] = \frac{0.0203564}{0.104196} \approx 0.195367,$$ which matches our original calculation at the beginning.
What the equation $\Pr[B \cap A] = \Pr[Y = 6]\Pr[B]$ suggests is that in fact, $$\Pr[Y = 6] = \frac{\Pr[B \cap A]}{\Pr[B]} = \Pr[A \mid B].$$ This is the conditional probability that, if $9$ events were observed in the first $4$ hours, that $6$ of them happened in the first two hours. And this is indeed binomially distributed, but we did not really formally prove this, resorting only to intuition.
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H: Stirling's formula to bound infinite series
I'm trying to show the following bound, $$\sum_{n=k}^{\infty} \frac{e^{-n\alpha}{(n\alpha)}^{n-1}}{n!} \leq \frac{1}{\alpha}\sum_{n=k}^{\infty} \frac{e^{-kI}}{\sqrt{2\pi n^3}} e^{-{(n-k)I}}$$
where $I=\alpha-1-\log(\alpha)$. I understand that I'm supposed to use Stirling's Formula but I'm lost as to how to do this and am not getting anywhere close. Any suggestion would be great.
AI: Hint: For starting One may give simple bounds valid for all positive integers $n$, rather than only for large $n$ using this inequality:
$\sqrt{2\pi n}n^{n}e^{-n}\leq n!\leq e n^{n+1/2}e^{-n}$, you can use the LHS of that inequality which is :$\sqrt{2\pi n}n^{n}e^{-n}\leq n!$ implies :
$$\frac{1}{n!}\leq \frac{1}{\sqrt{2\pi n}n^{n}e^{-n}}\leq \frac{1}{\sqrt{2\pi n^3}e^{-n}}$$ , Mutltiplying both sides by your needed factors and do summation you will come up to your bounds for $n^n$ in your dominator wich it would be still valid for $n^3$ in dominator
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H: Convert linear distance to steering angle
I need to calculate the angle of the front steering wheel using a collapsible piston(linear sensor). 'x' is used to represent the length in inches of the movable part of the sensor and is the independent variable.
θ represents the steering angle. The angle is 0 when the wheels are perfectly straight and increases as the wheels turn to the left and decreases as the wheels turn to the right.
This is the image where the θ is zero:
diagram 1
This is when it's steering all the way to the left:
diagram 2
and this is when it's steering all the way to the right:
diagram 3
I'm trying to solve it using the parametric equations of the circle; Relating the piston with the equations using the euclidean distance. As it turns out, I'm having a hard time solving for theta.
In the following two images you can see my derivations:
image 1
image 2
I'm stuck here:
$$b\cdot cos(\theta) - a\cdot sin(\theta) = \frac{a^2 + b^2 - d^2 + r^2}{2\cdot r}$$
Any help would me appreciate it!
AI: Just taking your calculation further:
$$b\cdot cos(\theta) - a\cdot sin(\theta) = \frac{a^2 + b^2 - d^2 + r^2}{2\cdot r} =Q $$
$$\dfrac {b\cdot cos(\theta) - a\cdot sin(\theta)}{\sqrt{a^2+b^2}} = \dfrac{Q}{\sqrt{a^2+b^2}}= R$$
$$ \sin \beta \ cos\theta - \cos \beta \sin \beta = R$$
$$ \sin (\beta-\theta)=R$$
$$ \theta =\beta-\sin^{-1}R$$
where
$$ \beta = \tan^{-1}\frac{b}{a}$$
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H: Maximize $\sum\limits_{k =1}^n x_k (1 - x_k)^2$
Given problem for maximizing
\begin{align}
&\sum_{k =1}^n x_k (1 - x_k)^2\rightarrow \max\\
&\sum_{k =1}^n x_k = 1,\\
&x_k \ge 0, \; \forall k \in 1:n.
\end{align}
My attempt: first of all i tried AM-GM, or we can just say, that $x_k (1 - x_k)^2 \le x_kx_k^2 =x_k^3$, but now we got just sum of cubes. Can we say, that $\sum\limits_{k =1}^n x_k^3 \le \sum\limits_{k =1}^n x_k,$ because $x_i \le 1$ and get our maximum - 1?
It looks very easy, but i suppose i'm wrong
AI: For $x_1=x_2=...=x_n=\frac{1}{n}$ we get a value $\frac{(n-1)^2}{n^2}.$
We'll prove that it's a maximal value.
Indeed,
\begin{align}
\frac{(n-1)^2}{n^2}-\sum_{k=1}^nx_k(1-x_k)^2&=\sum_{k=1}^n\left(\frac{(n-1)^2}{n^3}-x_k(1-x_k)^2\right)\\
&=\frac{1}{n^3}\sum_{k=1}^n(1-nx_k)(n^2x_k^2-n(2n-1)x_k+(n-1)^2)
\end{align}
\begin{align}
&=\frac{1}{n^3}\sum_{k=1}^n\left((1-nx_k)(n^2x_k^2-n(2n-1)x_k+(n-1)^2)-(1-nx_k)(n^2-4n+3)\right)\\
&=\frac{1}{n^3}\sum_{k=1}^n(1-nx_k)^2(2n-2-nx_k)\geq0\text{ for any } n\geq2.
\end{align}
For $n=1$ we have $x_1=1$ and $$x_1(1-x_1)^2=0=\frac{(n-1)^2}{n^2},$$
which says that $\frac{(n-1)^2}{n^2}$ is the answer.
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H: Calculating the number of poker hands that have exactly 4 different denominations
I'm been working through some examples of questions mentioned in the title and came across this one that I wasn't entirely if I was approaching this correctly.
My initial approach to solving this question was to calculate the probability of drawing such a hand and then multiply this by the total number of hands possible.
To get the probability of such a hand I did this
$$\frac{52}{52} * \frac{48}{51} * \frac{44}{50} * \frac{40}{49} * \frac{12}{48} = 0.05634253701$$
Then to get the total number of possible hands I simply did
$$\binom{52}{5} = 2598960$$
And finally, the possible hands by the probability
$$2598960 * 0.05634253701 = 146432$$
Did I approach this question correctly or is the a simpler way I should be doing a question like this?
AI: Choose the 4 ranks (denominations), which 1 of these ranks appears twice, which 2 suits appear for that rank, and which 1 suit appears for each of the other 3 ranks:
$$\binom{13}{4}\binom{4}{1}\binom{4}{2}\binom{4}{1}^3=1,098,240$$
Also known as one pair. An alternative approach is to choose the rank for the pair, the 2 suits for the pair, the other 3 ranks, and the suit for each of the other 3 ranks:
$$\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3=1,098,240$$
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H: A function that satisfies Cauchy-Riemann but is not holomorphic
I'm attempting Chapter 1, Exercise 12 in Stein & Shakarchi's Complex Analysis, which is as follows:
Consider the function defined by $$f(x+iy) = \sqrt{|x||y|}$$ whenever $x, y \in \mathbb{R}$. Show that $f$ satisfies the Cauchy-Riemann equations at the origin, yet $f$ is not holomorphic at $0$.
I think that I have solved it, but since I don't have much experience with complex analysis, I'm not sure if my argument is valid/correct:
If, for $x, y \in \mathbb{R}$, we write $$f(x+iy) = \sqrt{|x||y|} = u(x,y)$$ and $v(x,y) = 0$, since $f$ is a real-valued function, then, for $h \in \mathbb{R}$, $$\frac{\partial u}{\partial x} = \lim_{h\rightarrow 0} \frac{u(x+h,y)-u(x,y)}{h} = \lim_{h\rightarrow 0}\frac{0}{h} = 0 = \frac{\partial v}{\partial y}$$ and similarly we find that $\partial u/\partial y = 0 = -\partial v/\partial x$, so this function satisfies the Cauchy-Riemann equations.
Now, for $h = h_1 + ih_2 \in \mathbb{C}$, $$\lim_{h\rightarrow 0} \frac{f(z+h)-f(z)}{h} = \lim_{h\rightarrow 0} \frac{\sqrt{|h_1||h_2|}}{h_1+ih_2}$$ at the origin.
Suppose that $h_1 = ab = h_2$ for real numbers $a,b > 0$. Then $$\lim_{h\rightarrow 0} \frac{f(z+h)-f(z)}{h} = \lim_{ab\rightarrow 0} \frac{ab}{ab + iab} = \frac{1}{1+i}.$$ But if instead $h_1 = -ab = -h_2$, then $$\lim_{h\rightarrow 0} \frac{f(z+h)-f(z)}{h} = \lim_{ab\rightarrow 0} \frac{ab}{-ab + iab} = \frac{1}{-1+i},$$ so the limit does not exist and hence $f$ is not holomorphic at the origin.
I would greatly appreciate if somebody could check the above for correctness. (In particular, can I just assume that the $ab$ factorization exists? And is it sufficient to show that the limit is not the same by approaching from different directions?)
AI: I don't get why you used a product $ab$. It suffices to just consider the two paths where $(h_1, h_2) = (t,t)$ for $t> 0$ and $(h_1, h_2) = (t,-t)$ for $t>0$. Then, show that the limit as $t \to 0^+$ along these two directions yields different answers. So, basically, the idea of your computation is right, but the presentation seems weird.
Anyway, here's a more general overview of why your function fails to be holomorphic.
Let $U \subset \Bbb{C}$ be open and $f:U \to \Bbb{C}$ be a map. Then, $f$ is holomorphic at a point $\alpha \in U$ if and only if
When you consider $\Bbb{C} = \Bbb{R}^2$, the map $f: U \subset \Bbb{R}^2 \to \Bbb{R}^2$ is $\Bbb{R}$-differentiable in the standard real-multi-variable sense, and
$f$ satisfies the Cauchy-Riemann equations (so that the real derivative $Df_{\alpha}$ is actually a $\Bbb{C}$-linear map)
But the function $f$ you have, is $f(x,y) = (\sqrt{|xy|}, 0)$, and this is not even real-differentiable as a map from an open subset of $\Bbb{R}^2$ into $\Bbb{R}^2$. Why? Notice that $f$ is positively homogeneous, which means for every $t\geq 0$, we have $f(tx,tx) = t f(x,y)$. By this answer, if $f$ was $\Bbb{R}$-differentiable at the origin, it would have to be an $\Bbb{R}$-linear map, which it clearly isn't.
By the way, you should really take note of the statement proven in the linked answer, because it's often a quick way to prove that several functions defined on $\Bbb{R}^2$ are not $\Bbb{R}$-differentiable at the origin (for example, things like $g(x,y) = \dfrac{x^3}{x^2 + y^2}$ if $(x,y) \neq (0,0)$, and $g(0,0) = 0$ else, or other functions like this are positively homogeneous, but non-linear, hence not $\Bbb{R}$-differentiable). Once you know that theorem, the non-differentiability of such functions becomes almost obvious by inspection.
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H: Convergent Improper integral whose integrand tends to a non zero finite limit as x tends to infinity.
Let $f:\mathbb{R} \to \mathbb{R}$ be a continuous function such that $\int\limits_0^\infty f(x)dx$ exists. If $f(x)\ge 0 \,\forall x\, \in \mathbb{R}$, then prove or disprove that $\lim\limits_{x\to \infty}f(x)$ exists and is zero.
If $f(x)$ is any function,then taking $f(x)=\sin(x^2),$
I am able to conclude that the result is false because $\lim\limits_{x\to \infty}\sin(x^2)$ does not exist even though $\int\limits_0^\infty \sin(x^2)dx$ is convergent.
But in the case of non negative functions,I am neither able to prove the result nor do I get a counter example.Thank you in advance for your help.
AI: Define a function $ f $ as the following graph shows :
It is possible to define $ f $ explicitly, but that's not a big deal.
Its graph is formed of triangles centered at integers, each one centered at $ n\in\mathbb{N} $, has a base $ B_{n}=\frac{1}{n^{2}} $, and an altitude $ h=2 $, which means has an area $ \mathcal{A}_{n}=\frac{B_{n}\times h}{2}=\frac{1}{n^{2}} \cdot $
Then $$ \int_{0}^{+\infty}{f\left(x\right)\mathrm{d}x}=\sum_{n=1}^{+\infty}{\mathcal{A}_{n}}=\sum_{n=1}^{+\infty}{\frac{1}{n^{2}}}=\frac{\pi^{2}}{6} $$
$ \int_{0}^{+\infty}{f\left(x\right)\mathrm{d}x} $ converges, $ f\geq 0 $, but $ \lim\limits_{x\to +\infty}{f\left(x\right)}\neq 0 $ since $ \left(\forall n\in\mathbb{N}^{*}\right),\ f\left(n\right)=2 \cdot $
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H: Showing images and inverse images don't invert each other
While studying analysis I got the following:
If $f : Z → Z$ is the map $f(x) = x^2$, then $f^{−1}({0, 1, 4}) = \{−2,−1, 0, 1, 2\}$.
Note that $f$ does not have to be invertible in order for $f^{−1}(U)$
to make sense. Also note that images and inverse images do not quite
invert each other, for instance we have $f^{−1}(f({−1, 0, 1, 2})) \neq \{−1, 0, 1, 2\}$ (why?)
Even though it's not a formal question, the author put "why?" so we think about it, I did but I want to try and answer it formally as a general case (if this is doable). I know one counter example is enough to show that a statement isn't true but I would like both sides of things if possible.
For instance in this particular function this happens because $+x$ or $-x$ give the same result for $f(x)=x^2$ but is there a general way to state this for any function?
AI: In general, $f^{-1}(f(S)) = S$ holds iff $f$ is injective (and, of course, your example $x \mapsto x^2$ is not one of those).
Indeed, if $f: X \to Y$ is an injective function and $S \subset X$, the inclusion $S \subset f^{-1}(f(S))$ is obvious (because $f(s) \in f(S)$ for any $s \in S$ by definition). For the other inclusion, notice that any $s \in f^{-1}(f(S))$ satisfies $f(s) \in f(S)$, so that $f(s) = f(x)$ for some $x \in S$, and since $f$ is injective this means $s = x \in S$ as desired.
Conversely, if for every $S \subset X$ the inclusion $f^{-1}(f(S)) = S$ holds, then $f(a) = f(b)$ for $a, b \in X$ implies that $f(b) \in f(\{a \})$, therefore $b \in f^{-1}(f(\{a\})) = \{a\}$, whence $a = b$, showing as desired that $f$ is injective.
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H: Using implicit differentiation to find equation of tangent at arbitrary (a,b)
For the following equation $\sqrt x + \sqrt y = 2$
(1) Find equation of tangent at point (a, b) on curve
Using implicit differentiation: $$y' = - \frac{√y}{√x}$$
Equation at (a, b) is: $$y - b = - \frac{\sqrt b}{\sqrt a}(x - a)$$
$$y = - x\frac{\sqrt b}{\sqrt a} + a\frac{\sqrt b}{\sqrt a} + b$$
(2) Find points where the tangent intersects at x and y axes and show the sum of these is always 4
Let y-intercept be (0, $y_0$)
and x-intercept be ($x_0$, 0)
Slope = $\frac{y_0}{x_0}$
Equation of line through (a, b) is: $$y = x\frac{y_0}{x_0} –
a\frac{y_0}{x_0} + b$$
Equating it to the equation of tangent: $$- x\frac{\sqrt b}{\sqrt a} + a\frac {\sqrt b}{\sqrt a} + b = x\frac{y_0}{x_0} – a\frac{y_0}{x_0} + b$$
$$\frac{\sqrt b}{\sqrt a} = - \frac {y_0}{x_0}$$
Not sure what to do after this
AI: From $y-b=-\sqrt{\frac{b}{a}}(x-a)$ for $y=0$ we obtain $x_0=a+\sqrt{ab}$
and for $x=0$ we get $y_0=b+\sqrt{ab}.$
Id est, $$x_0+y_0=a+b+2\sqrt{ab}=(\sqrt a+\sqrt b)^2=4.$$
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H: Independent random variables - Finding $P(X = Y )$ and $P(X ≤ Y )$
Let $X$ and $Y$ be independent random variables in a probability space $(\Omega, P)$ with the following distributions:
Calculate:
(a) $P(X = Y ),$
(b) $ P(X ≤ Y ).$
Can you please check my solutions? I have the following:
(a) $P(X = Y )= 0.20 \cdot 0.15 + 0.36 \cdot 0.26 + 0.26 \cdot 0.37 + 0.18 \cdot0.22 = 0.2584.$
(b) $ P(X ≤ Y )= P( X = Y) + P (x < y) = 0.2584 + 0.20 \cdot (0.26 + 0.36 + 0.22) + 0.36 \cdot (0.37 + 0.22) + 0.20 \cdot 0.22 =0.688. $
Thank you for your time.
AI: Almost. There appear to be some typos. Recheck your figures.
(a) $P(X = Y )= 0.20 \cdot 0.15 + 0.36 \cdot 0.26 + 0.26 \cdot 0.37 + 0.18 \cdot0.22 = 0.2584.$
(a) $P(X = Y )= 0.20 \cdot 0.15 + 0.36 \cdot 0.26 + 0.26 \cdot 0.37 + 0.18 \cdot0.22 = 0.25\mathbf 94.$
Have not
(b) $ P(X ≤ Y )= P( X = Y) + P (x < y) = 0.2584 + 0.20 \cdot (0.26 + 0.36 + 0.22) + 0.36 \cdot (0.37 + 0.22) + 0.20 \cdot 0.22 =0.688. $
$ P(X ≤ Y )= P( X = Y) + P (x < y) = 0.25\mathbf 94 + 0.20 \cdot (0.26 + 0.3\mathbf 7 + 0.22) + 0.36 \cdot (0.37 + 0.22) + 0.2\mathbf 6 \cdot 0.22 =0.6\mathbf{99}$
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H: Hint as to how to formalize that the curvature of a curve inside the unit circle is bounded below by the curvature of the circle
I am trying to self teach differential geometry and to that effect I am trying to do the Homework in the MIT open course.
The specific question I am struggling with is:
Let $c$ be a regular curve such that $|c(s)| ≤ 1$ for all $s$. Suppose
that there is a point $t$ where $|c(t)| = 1$. Prove that the curvature at that
point satisfies $|κ(t)| ≥ 1$
This is what I have so far:
The curve being regular implies that it can be arc length parametrized, which means that it's curvature is just the norm of the second derivative, or in 2D $|x'y'' - x''y'|$.
The curve magnitude being upper bounded by 1 means the curve is fully contained within the unit disk.
The curve being one at $t$ means that point is ON the unit circle. Thus my intuition says this curve can only "bend inwards" at a rate equal or higher than the circle, otherwise it will be an epsilon outside of the circle at a point infinitesimally close to $t$. In other words it seems to me that a if a curve with a point on the unit circle has a smaller curvature than that of the unit circle (i.e 1) the curve will "pop out" of it for some infinitesimal amount. But I am not sure if A) this is correct B) how to formalize it.
I am looking mostly for a hint and advice, not for the full solution. Thank you lots in advance.
AI: Let $t_0$ be the point satisfying $\| \alpha(t_0)\| = 1$. Then, at $t_0$, the function $$t \mapsto \|\alpha(t) \|^2$$
assumes a maximum. Thus, its first order derivative at $t_0$ vanishes, that is:
$$2 \langle \alpha(t_0), \alpha'(t_0)\rangle = 0$$
And by the second derivative test, its second order derivative at $t_0$ is at most $0$, that is:
$$2 \langle \alpha'(t_0), \alpha'(t_0)\rangle + 2 \langle \alpha''(t_0), \alpha'(t_0) \rangle \leq 0$$
Assuming without loss of generality that $\alpha$ has unit speed, we get:
$$1 \leq \langle \alpha''(t_0), \alpha'(t_0) \rangle $$
Taking the norm of both sides and using the Cauchy-Schwarz inequality, we have:
$$1 \leq | \langle \alpha''(t_0), \alpha'(t_0) \rangle | \leq \|\alpha''(t_0)\| \cdot \|\alpha'(t_0)\| = |\kappa(t_0)|$$
as desired.
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H: Why is the stereographic projection bijective?
I know this might be a very basic question, but I am just not able to wrap my head around it.
Why is the map
$$ S:\mathbb{S}^n-\{e_{n+1}\}\rightarrow \mathbb{R}^n \quad \textrm{such that } \bar{x}\mapsto (\frac{x_1}{1-x_{n+1}},...,\frac{x_n}{1-x_{n+1}})$$
a bijective function? I tried to understand it geometrically as well, but I just could not see how it was formulated in that manner.
Here $\bar{x}=(x_1,...,x_n)$ and $e_{n+1}=(0,...0,1)$.
Any help will be much appreciated!
AI: It's certainly not obvious (except, like many things in mathematics, in hindsight). One can show directly that this map is both injective and surjective. It's easier perhaps to simply write down the inverse function $S^{-1}\colon \Bbb R^n \to \Bbb S^n - \{e_{n+1}\}$:
$$
S^{-1}\big( (y_1,\dots,y_n) \big) = \frac1{1+y_1^2+\cdots+y_n^2}\big( 2y_1,\dots,2y_n,y_1^2+\cdots+y_n^2-1 \big).
$$
(Of course one has to verify that the domain/codomain of this function are correct and that both $S\circ S^{-1}$ and $S^{-1}\circ S$ are the identity maps.)
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H: Meaning of "extension field is a vector space"
For example it is clear that the case $\mathbb{Q}\big(\sqrt 2\big) = \{ a+b\sqrt2 \vert a,b \in \mathbb{Q}\}$
(Here the $a,b$ is a scalar)
Then if we more generalize this thought, we can get a well known fact that:
(*) Given the field extension L / K, the larger field L is a K-vector space. (Source : https://en.wikipedia.org/wiki/Field_extension)
Hence, I've understood the statement, (*) means $\forall l(\in L) = \sum k_i l_i $ for some $k_i \in K$, $l_i \in L$.
(i.e. Considering the $L$ as a vector space, all the element of the $L $ can be expressed as a linear combination with scalars $k_i(\in K)$ ).
Does it still hold the $L$ like not a finite extension or not a algebraic extension etc like a $\mathbb{Q}(\pi)$ ? In my guess $\mathbb{Q}(\pi)$ is a vector space but not expressed as a finite linear combination.
So, is my thought right? I want to check my idea that I've known.
Any comment or advice would be appreciated. Thanks.
AI: While your "understanding" statement is technically true, I would say that your interpretation misses most of the useful properties of thinking of $L$ as a $K$-vector space. It makes more sense to think of it in terms of the definition of a vector space.
How I would interpret it: $L$ satisfies all of the axioms of a vector space over $K$. I.e. you can consider the elements of $L$ to be "vectors", with addition and scalar multiplication (multiplication by elements of $K$) satisfying all of your familiar vector space properties. This lets you define the dimension of $L$ over $K$, if the dimension is finite you can define a basis, and you can represent $K$-linear transformations with matrices (over $K$).
Under this interpretation, $\mathbb{Q}(\pi)$ is an infinite dimensional vector space over $\mathbb{Q}$.
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H: Multinomial Coefficient of $x^{1397}$ in expansion of $(x^3+x^4+x^5+...)^6$
I have the following problem:
Find the Coefficient of $x^{1397}$ in expansion of $(x^3+x^4+x^5+...)^6$
I know how to solve these kind of questions using Multinomial Theorem but since the polynomial in this one is infinite I’m lost!
Thanks in advance.
AI: This is the number of solutions to the equation $Z_1 + \dots + Z_6 = 1397$ where $Z_1$, $\dots$, $Z_6$ are positive integers $\geq 3$.
Subtracting three from each of the $Z_i$'s, each solution corresponds to a solution of $$X_1 + \dots + X_6 = 1397 - (6 \cdot 3) = 1379$$
where each $X_i$ is a non-negative integer.
In general, the number of solutions to $X_1 + \dots + X_k = n$ for integer $X_i \geq 0$ is $\binom{n + k - 1}{k-1}$. So in your particular case, $k = 6$ and $n = 1379$; the answer is $\binom{1384}{5} = 42010498234776$.
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H: Logic and set theory books
I have studied real analysis, linear algebra, and how number sets are constructed from N, but now, i want to learn the foundations of math and some more advanced set theory (cardinals, ordinals), because my brain is full of questions such as "what is a property?, how does logic work in math?, What is ZFC?" and things like that. Can you recommend any books that could help me with that?
AI: Axiomatic Set Theory by Patrick Suppes is an easy intro to the basics. Last year it was available as a free PDF. Maybe still is.
Lectures In Set Theory. Various authors. Edited by Morley. I found the essay on the definition of L (Godel's constructible class) to be the easiest and clearest intro to L that I've seen.
Introduction To Set Theory And Modern Analysis by Simmons.
Set Theory: An Introduction To Independence Proofs by K. Kunen. A thorough axiomatic development from the bottom up.
You will need to learn about Godel's incompleteness theorems. These are like the axiomatic foundation of the properties of the reals, in that you study it once and then take it for granted. Do NOT read Godel, Escher,Bach: An Eternal Golden Braid by Hofstader. For a long time Godel's Proof by Nagy & Newman was the only "popular" exposition in English.
Something on Model Theory and on Logic. Sorry I can't name a book.
Stories About Sets by V'Lenkin (Vilenkin). Good fun.
50 years ago Dover Publications (formerly Dover Press) was an excellent source of cheap re-prints of math & science books. It still is.
BTW. You will meet the Schroeder-Bernstein, Cantor-Bernstein, and Cantor-Schroeder-Bernstein theorems. These are all the same theorem. The Simmons book has a nice presentation of the short proof. There is also a long proof, which I saw & ran away from.
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H: If we have $f : x→ y$ with $S \subset x$ what can we say about $f^{-1}f(S)$ and $S$
Here is the full question:
Let $f : X → Y$ be a function from one set $X$ to another set $Y$, let $S$ be a subset of $X$, and let $U$ be a subset of $Y$. What, in general, can one say about $f^{−1}(f(S))$ and $S$? What about $f(f^{−1}(U))$ and $U$?
My reasoning:
-$f^{−1}(f(S))=S$ if and only if $f(S)$ is a one-to-one function since no two different inputs will map to one output, so we will always get our original input when we reverse the process.
-$f(f^{−1}(U))=U$ if and only if $f(U)$ is a bijective function, this way we can guarantee and there is an input to "go back to" when we reverse the process AND we will go back to the correct input (since it's unique).
Why I posted this: I feel like I am missing something in the question, I am new to analysis and always pretty unsure about my reasoning especially that there is no solutions available for this, I want to make sure about building on the correct blocks of understanding. If someone can help me point out my mistakes if there is any or suggest improvements about the way of writing proofs (I guess I use too much English) I would be grateful.
AI: Since $S$ and $U$ are fixed, your statements are actually false. That is, the following are true:
$f^{-1}(f(S)) = S$ for all $S \subseteq X$ if and only if $f$ is injective.
$f(f^{-1}(U)) = U$ for all $U \subseteq Y$ if and only if $f$ is surjective.
However, since $S$ and $U$ are fixed in the problem statement, we could have $f^{-1}(f(S)) = S$ without $f$ being injective, and we could have $f(f^{-1}(U))=U$ without $f$ being surjective.
Instead, this type of question, usually, is asking you to verify which of the following are necessarily true:
$f^{-1}(f(S)) \subseteq S$
$f^{-1}(f(S)) \supseteq S$
$f(f^{-1}(U)) \subseteq U$
$f(f^{-1}(U)) \supseteq U$
As a final note, for a fixed $S$ and $U$ we still have the direction of
$f^{-1}(f(S)) = S$ if $f$ is injective.
$f(f^{-1}(U)) = U$ if $f$ is surjective.
If you want to additionally prove those, I'm sure your instructor wouldn't mind, as the question is certainly open-ended.
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H: Prove that $f(x) = \frac{x}{2} + c$ if $|f(x+y)-f(x-y)-y|\le y^2\ \forall \ x,y \in \mathbb{R}$
Problem Statement: Let $f:\mathbb{R} \to \mathbb{R}$ be a function such that $\forall \ x,y \in \mathbb{R}$,
$$|f(x+y)-f(x-y)-y|\le y^2$$
Prove that $f(x) = \frac{x}{2} + c$ for some $c\in \mathbb{R}$
My try:
$$|f(x+y)-f(x-y)-y|\le y^2 \\ \implies \frac{y-y^2}{2y}\le \frac{f(x+y)-f(x-y)}{2y} \le \frac{y+y^2}{2y}\ \ \text{when}\ y \ne 0$$
Now, taking the limit as $y \to 0$ in the above inequality and using the Squeeze theorem we have
$$\lim_{y\to 0}{\frac{f(x+y)-f(x-y)}{2y}} = \frac{1}{2}\ \ \forall x \in \mathbb{R}\ \ \ (*)$$
Now, if it is given that $f$ is differentiable on $\mathbb{R}$, then we have $f'(x)=\frac{1}{2}\ \forall \ x \in \mathbb{R} $ and hence $f(x) = \frac{x}{2} + c$ for some $c \in \mathbb{R}$
But since the problem does not explicitly state that $f$ is differentiable on $\mathbb{R}$, I am not sure on how to proceed after step $(*)$.
Thanks for any answers!!
AI: Put $x = a+h/2$, $y = h/2$ to have $|f(a+h) - f(a) - h/2|< h^2/4$, or $$\left| \frac{f(a+h)-f(a)}{h} - \frac12\right| < \frac{h}{4}$$. What can you say about the diffentiablity?
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H: Proof of limits of sequences tending to infinity
How can we prove this :
Prove that if $\quad\lim_{n\to+\infty} y_n = \lim_{n\to+\infty} z_n=+\infty$
then $\quad\lim_{n\to+\infty} v_n = \lim_{x\to+\infty} w_n=+\infty$ .
With $$w_n=\frac{n}{\sum_{i=1}^n\frac{1}{z_i}},\quad v_n = \frac{y_1+y_2+....y_n}{n} $$.Can we solve this with "epsilon delta" proof? Or can we solve this with another way? I don't find anything to get started
AI: $y_n \to \infty \implies$ For every $\epsilon \gt 0 \;\;\exists N$ such that for all $n\gt N$, $|y_n|\gt \epsilon$ For sufficiently large $n$, We have, $v_n=
\frac{\sum_{i=0}^{i=n}y_i}{n}=\frac{\sum_{i=0}^{i=N}y_i+\sum_{i=N+1}^{i=n}y_i}{n}\gt \frac{\sum_{i=0}^{i=N}y_i}{n}+\frac{\epsilon (n-N)}{n}$
As $n\to \infty$, we have, $\lim v_n=\lim \frac{\sum_{i=0}^{i=n}y_i}{n}\ge \epsilon $
Since, $\epsilon \gt 0$ is arbitrary, $\lim v_n$ tends to infinity.
Hint for $w_n$: You just need to show that if $(1/z_n)\to 0$, then $(1/w_n)\to 0$. Proceed as above to prove this.
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H: Find $\sup\left\{ \frac{m}{|m| + n }\right\} $ where $m \in \Bbb{Z}$ and $n \in \Bbb{N}$
Im trying to do $\sup A $ and $\inf A$ if $A= \left\{ \frac{m}{|m| + n } : m \in \Bbb{Z} , n \in \Bbb{N}\right\} $ rigoriously.
my try:
Let $f(m,n) = \dfrac{m}{|m|+n} $. Notice that $f(1,1) = \dfrac{1}{2}$ and $f(-1,1) = \dfrac{-1}{2}$
If $m \geq 0$, then $f(m,n) = \dfrac{m}{m+n} < \dfrac{m}{m} = 1$
Next, if $m<0$, then $f(m,n) = \dfrac{m}{n-m} > \dfrac{m}{n} $
Claim: $\sup A = 1 $ and $\inf A = - \infty ?$
Proof: To prove $\sup A = 1$ we show no upper bound $( \dfrac{m}{|m|+n} \leq u )$ is less than $1$
if $1 > u$ then there is some integer $N>1$ such that $N(1-u) > 1 $ or equivalently
$$ 1-u > \dfrac{1}{N} \iff 1 - \dfrac{1}{N} > u \iff \dfrac{N-1}{N} > u \iff \dfrac{ N-1}{N-1 + 1 } > u$$
But with $m=N-1$ and $n=1$ we have found an element of $A$ that is greater than upper bound $u$. Thus $\boxed{ \sup A = 1 }$.
Now, how am I still unsure about the infimum? Any suggestion? Am I on the right track?
AI: Infimum:
Note : The Denominator is $>0.$
Let $m <0$.
Then
$\inf ( \dfrac {m}{|m|+n} )= -\sup (\dfrac {-m}{|m| +n} )$.
Find $\sup ( \dfrac{|m|}{|m|+n}).$
$\dfrac{|m|}{|m|+n} <1$, i.e. $1$ is an upper bound.
Fix $n$:
$\lim_{|m| \rightarrow \infty}\dfrac{|m|}{|m|+n}=1.$
Hence $\sup_{|m|>0} (\dfrac{|m|}{|m|+n})=1.$
Can you finish?
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H: Convergence weak * (star) in a metric space $X$
Let $X$ be a normed space and $\{x^*_n\} \subseteq X^*$,$\{y_n\}\subseteq X$
a) if $\{x^*_n\} \rightharpoonup x ^*$ then $\{x^*_n\}$ is strongly bounded and $\|x^*\| \le \liminf ||x^*_n||$?
b) if $\{x^*_n\} \rightharpoonup x ^*$ and $\|y_n-y\|_X \to 0$ then $(y_n,x^*_n) \to (y,x^*)$?
My attempt:
a) $\{x^*_n\} \rightharpoonup x ^*\iff(x,x^*_n) \to (x,x^*) \quad \forall x \in X$
$\forall x \in X$ $|(x,x^*_n)|\le\|x^*_n||\,||x\|$
because $\{x^*_n\} \rightharpoonup x ^*$ then
$|(x,x^*)|=\lim_{n\to +\infty} |(x,x^*_n)|\le\liminf(\|x^*_n\|\,\|x\|)=(\liminf \|x^*_n\|)\|x\|$
so I have $\|x^*\|=\sup_{\|x^*\|=1}|(x,x^*)| \le \liminf \|x^*_n\|$?
and for b)?
AI: Here,I shall denote linear functionals by $f$ and vectors by $v$.First of all, if $f_n$ converges in the weak$^{*}$ sense to $f$ then $f_n$ is uniformly bounded, as stated in the first part.Now say, $v_n$ converges to $v$ in norm.Then $|f_n(v_n) - f(v)|= |f_n(v_n)-f_n(v)+f_n(v)-f(v)| \leq ||f_n||.||v_n - v|| + |f_n(v) - f(v)| \leq M||v_n - v|| + |f_n(v)-f(v)|$ for some $M>0$.I could bound it above like this due to the uniform boundedness of $f_n$.Now the first term tends to zero due to norm convergence of $v_n$ and the second one tends to zero due to weak$^{*}$ convergence of $f_n$.
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H: are there rules that cant be rigorously explained?
this is sort of a shower thought question that came to mind, and I would prefer if it wasn't taken violently seriously, but are their rules that cant be rigorously explained. like for example the limit of a constant times a function is equal to the constant times the limit of a function. Intuitively it makes sense and can be explained. but how about rigorously???
AI: Pretty much any "rule" in maths is either a definition, an axiom, or can be rigorously proved.
Let's look at your example, $\lim_{x\to a}cf(x)=c\lim_{x\to a}f(x)$. To prove it rigorously, we may use the $\epsilon-\delta$ definition for limits:
If for any $\epsilon \gt 0$, there is a $\delta \gt 0$ such that $|f(x)-L|\lt \epsilon$ whenever $|x-a|\lt \delta$, then we say $\lim_{x\to a}f(x)=L$.
Proof:
Suppose $\lim_{x\to a}f(x)=L.$ In this case $ |cf(x)-cL|<|c|\epsilon$ whenever $|x-a|\lt \delta$. By the arbitrariness of $\epsilon$, the "rule" is now proven.
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H: Express a Probability through $\Phi(t)$
Using Poisson i.i.d. random variables and having $Z$ as standard Gaussian r.v. $~N(0,1)$,
I am struggling to express $P(|Z|≤t)$ in terms of $Φ(r)=P(Z≤r)$ for $t>0$.
I understand that the absolute value of Z means non-negativity and therefore the range is bound by 0 and t, but this doesn't seem to help.
My guess was $P(|Z|≤t) = 2*\Phi(t)$, but this is incorrect.
I assume the solution is simple, but I cannot get there. Any input appreciated.
AI: Note, first of all, that this question doesn't have anything to do with Poisson variables.
Anyway, for $t>0$:
$$
P(|Z|<t)=P(Z<t,Z>-t)=P(Z<t)-P(Z<-t)=\Phi(t)-\Phi(-t)
$$
If you want to get even more clever about it, you might note that
$$P(Z<-t)=P(-Z>t)=1-P(-Z<t)=1-\Phi(t),$$
since $-Z$ is also standard Gaussian.
Hence, it is also true that
$$
P(|Z|<t)=2\Phi(t)-1
$$
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H: Every irreducible is product of irreducibles
I'm studying some commutative algebra and learning some of the bases. While taking a look at the proof of Theorem 1, proposition 2, i've stumbled upon the following: "Suppose $d$ is not a product of irreducible elements. Hence, $d$ isn't irreducible". I'm trying to understand this statement, and it seems pretty trivial, but my head is just not working anymore. If anyone could help, I'd appreciate.
Theorem 1: given $D$ an integral domain.
If $D$ is and UFD, and $\pi \in D$ is an element such that $\pi \neq 0$ and $\pi \not\in D^{\times}$, then $\pi$ is prime if, and only if, $\pi$ is irreducible;
Suppose that
(i) Every ideal of $D$ is finitely generated (i.e. $D$ is Noetherian);
(ii) Every irreducible element is a prime element.
Then, $D$ is an UFD.
AI: Important: In this contexts, products of irreducibles can consist of one factor only!
Thus if $a$ is irreducible, then $a$ is an irreducible product.
Do you see why the statement is trivial then?
Compare with the integers: Every non-unit integer is the product of primes. Here every prime $p$ is a trivial product. The statement would no longer be true if we did not allow such products (with one factor) and that's definitely not something we want.
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H: How to show that a (metric) space having a countable dense subset is a topological property?
I have to show that a (metric) space having a countable dense subset is a topological property.
Given that A property P of a space is said to be a topological property if home-omorphic spaces share the same properties.
I think if i can show that two seperable spaces are homeomorphic then i can say that this is a topological property. I need help to understand this and how to show that?
AI: You must show this property is preserved by homeomorphisms, i.e. if $X$ has this property and $f: X\to Y$ is a homeomorphism, then $Y$ has this property.
Given a countable dense subset $D$ of $X$, what can you tell about $f(D)$?
By the way, it is not true that two separable spaces are homeomorphic. A finite space and a countable infinite space are both trivially separable but there cannot be a bijection between these two.
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H: Find the solutions to the $w''-z^2w=3z^2-z^4$ as Taylor series where $w(0)=0$ and $w'(0)=1$
We need to find the solutions of the
$w''-z^2w=3z^2-z^4$
where
$w(0)=0;w'(0)=1$
I wrote down the series that we can use to find the answer ($w$ as Taylor series):
$w=\sum_{n=0}^\infty C_nz^n$
$w'=\sum_{n=0}^\infty nC_nz^{n-1}$
$w''=\sum_{n=0}^\infty n(n-1)C_nz^{n-2}$
It is easy to find $C_0$ and $C_1$:
$w(0)=C_0=0$
$w'(0)=C_1=1$
I found this problem in Isaac Aramanovich "Collection of problems on the theory of functions of a complex variable", problem #3.112
AI: Now insert into the equation and compare the coefficients of equal power
$$
z^n:~~~~(n+2)(n+1)c_{n+2}-c_{n-2}=3\delta_{n,2}-\delta_{n,4}
$$
with $c_n=0$ for $n<0$. This then allows you to compute the coefficients step-by-step.
This gives equations
$$
2c_2=0\\
6c_3=0\\
12c_4-c_0=3\\
20c_5-c_1=0\\
30c_6-c_2=-1\\
42c_7-c_3=0\\
...
$$
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H: Square root with rational exponent
It might seem very stupid question.
If $x^2=9$ then to solve for $x$ we take both principal $n$-th root of $9$, i.e. $3$ and the negative $n$-th root of $9$, i.e. $-3$. This is right until I found about rational exponents.
If I try to solve the same equation using rational exponents, then $x^2=9$ will be $(x^2)^{1/2} =9^{1/2}$ which leads to $x= 3^{2\cdot (1/2)}$. Here solution will only be $x=3$ the principal $2$-nd root only.
I know the $x$ should equal $\pm3$. What am I doing wrong solving equation using rational exponents?
AI: The problem is that $(x^2)^{1/2} = \lvert x\rvert$, not $x$. So $\lvert x \rvert = 3$ and therefore $x = \pm 3$.
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H: MLE for an undirected network degree distribution
I have an empirical undirected network. I assume, that a degree distribution is $ F(k) = 1 - e^{1 - \frac{k}{m}} $. and would like to estimate $m$.
The only method I'm aware of for such task is MLE.
If I write a likelihood function $ L = \prod_{i=1}^{N}{p(x_i)} $ (where $ p(k) = \frac{1}{m}e^{1 - \frac{k}{m}} $ and ${x_i}$ is a degree of node $i$), then I can estimate $ m = \bar{x} $, but as I see the result is twice as much as it should be.
I believe this is because I should not write the likelihood function as I did, because degrees of two connected vertices are not independent, so I should take this into account, but I don't quite undestand how can I do this.
Could you confirm or reject this, and if I am right, please could you help me to build the correct likelihood function?
AI: The problem here is that the support of the variable is not $(0;\infty)$ but it is $k \in [m;\infty)$, I suppose (you didn't say where the variable is defined).
So your density is this (I prefer to indicate $x$ the variable instead of $k$)
$f(x,m)=\frac{1}{m}e^{1-\frac{x}{m}}\mathbb{1}_{[m;\infty)}(x)$
and, I suppose, $m >0$
Now, as you can see, the support depends on the parameter. Are you able to continue by yourself?
EDIT: anyway...here is the solution
The likelyhood is the following
$L(m)=\frac{1}{m^n}e^{n-\frac{\sum_{i}x_i}{m}}\mathbb{1}_{(0;x_{(1)}]}(m)$
so the MLE estimator is
$\hat{m}=min[x_{(1)};\bar{x}]=x_{(1)}$
where $x_{(1)}=min(x)$
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H: Connectedness of a union of convex sets
Let $p_1,...,p_k$ be points in $\mathbb{R}^n$, let $S_1,...,S_m$ be subsets of $\{p_1,...,p_k\}$, and let $V_i$ be the convex hull of $S_i$. Is it true that \begin{equation}
V:=\bigcup_{i=1}^m V_i
\end{equation} is path-connected if and only if any two $p_i,p_j\in V$ can be joined in $V$ by a concatenation of line segments whose endpoints belong to $\{p_1,...,p_k\}$?
In other words, we can define a graph $G$ whose vertices are the points $p_i$ appearing in at least one $S_j$, and put an edge between $p_{i_1},p_{i_2}$ if and only if $\{p_{i_1},p_{i_2}\}\subseteq S_j$ for some $j$. Then my question is, is $V$ path-connected if and only if $G$ is a connected graph?
I feel like it should be easy to argue that yes but I have a hard time deducing something combinatoric from something topological.
AI: This implication: "$\Leftarrow$" is obviously true.
But the other implication is false. Consider 4 points forming a square in $\mathbb{R}^2$:
$$p_1=(0,0),\ p_2=(0,1),\ p_3=(1,0),\ p_4=(1,1)$$
and let $S_1=\{p_1,p_4\}$ and $S_2=\{p_2,p_3\}$. Then $V$ is a union of two diagonals which is path connected. Note that $V_1\cap V_2=\{(1/2,1/2)\}$ which is not a vertex. And thus clearly $p_1$ and $p_2$ cannot be joined by concatenating line segments ending at any $(p_i,p_j)$. In this case such lines coincide with $V_1,V_2$.
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H: Radius of convergence for n^n
Given is a power series. We need to find the radius of convergence for this series. The series given is:
$$ \sum_{n=0}^{\infty}n^n(x-1)^n$$
To find the radius of convergence, I have first tried to substitute $y = x - 1$, since this was explained the the course notes. Then I took the limit of the absolute value of $\frac{c_k}{c_{k+1}}$ where $n \xrightarrow[]{} \infty$, where $c_k$ is $n^n$. Then I rewrote the sum as:
$$ \sum_{n=0}^{\infty}n^n(y)^n $$
Thuss the limit if I'm not mistaken is:
$$ \lim_{n \xrightarrow[]{} \infty} \mid{\frac{n^n}{(n + 1)^{n + 1}}}\mid $$
This is where I'm quite stuck. I have build up this limit, but then the course says we must use the following where
$$ \lim_{n \xrightarrow[]{} \infty} \left( \frac{n + 1}{n} \right)^n = e$$
Can anyone explain to me how this is related to each other?
Thanks in advance.
AI: The Root Test on this series can help illuminate the radius of convergence. As long as the limit described by this test is greater than 1, the series diverges. Thus we have:
$$\lim_{n\to\infty} |n^ny^n|^{\frac{1}{n}} > 1$$
Simplifying we get:
$$\lim_{n\to\infty} ny$$
If y is not zero then this limit goes off to infinity for any fixed y (implying the limit is greater than 1 unless y is 0). Thus, the original series only converges if $x = 1$ .
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H: Probability - Bayes' Theorem
99% of the restaurants practice good hygiene. Each time you eat in a clean restaurant, there is a $1\%$ chance that you will get sick, independent of your previous visits. Each time you eat in a restaurant that does not practice good hygiene, on the other hand, there is a $50\%$ chance that you will get sick, independent of your previous visits.
Question:
You eat at a random restaurant and get sick. You go to the same restaurant for a second time, and you get sick again. What is the probability of the restaurant practicing good hygiene?
When it is conditional independent, I am assuming the probability is the same no matter it is the 1st, 2nd, or 3rd visit.
By Bayes' Theorem, $P(A\mid B)=\frac{0.99\cdot0.01}{(0.99\cdot0.01)+(0.01\cdot0.50)}\approx0.6644$.
Do I need to consider if I was sick after the first visit?
AI: Assume there are $100$ restaurants, one of which doesn't practice good hygiene, and you go to each restaurant $10,000$ pairs of times. You will get sick twice in $2599$ of those pairs. Of those, $99$ times you'll have picked a restaurant with good hygiene. So the probability that you picked a restaurant that practices good hygiene is $\frac{99}{2599}$, or a little under $4\%$.
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H: What is the integral of the fractional part of a variable -- related to integration by parts?
Let $\{x\}$ denote the fractional part of a variable, i.e. $\{x\}=x-\lfloor x\rfloor$. Would the integral of $\{x\}-\frac{1}{2}$ from $0$ to $1$ evaluate to $1$? That is
$$
\int_{0}^1 \{x\} -\frac{1}{2} = 1
$$
Am asking this because I want to apply it to an arbitrary integrable function $f(x)$, where I want to derive its integral using the 'long-cut' of integration by parts.
Particularly, does the following hold for any number $b$?
$$
\int_b^{b+1}f(x) = \Big(\{x\}-\frac{1}{2}\Big)f(x)|^{b+1}_b - \int_b^{b+1} \Big(\{x\}-\frac{1}{2}\Big)f'(x)
$$
i.e. I take the integral of $f(x)$ from $b$ to $b+1$, and set $u=\{x\}-\frac{1}{2}$, $du=1$, and $v=f(x)$, $dv=f'(x)$, instead of the usual substitution of $x$ for $u$....
AI: Note that for $x \in [0, 1)$, we have $\{x\} = x$ and so, we have
$$\int_0^1\left(\{x\}-\dfrac12\right)dx = \int_0^1 \left(x-\dfrac12\right)dx = 0.$$
(The value at $1$ does not matter.)
In fact, the function $\{x\}-\dfrac12$ is a periodic function with period $1$ such that the integral over a full period is $0$. So you have
$$\int_b^{b+1}\left(\{x\}-\dfrac12\right)dx = 0$$
for any $b \in \Bbb R$. (Not just $b \in \Bbb Z$.)
I'm not quite sure what you are doing in the later part with by-parts. However, I'll point this out-
There exists no $F:\Bbb R\to\Bbb R$ such that $F'(x) = \{x\} - \dfrac12$.
This is because $\{x\} - \dfrac12$ does not satisfy the intermediate value property and so, by Darboux's Theorem, it cannot have an anti-derivative.
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H: If $\vert x \vert > 1$ then $x > 1$ or $x < –1$
Prove that if $\vert x \vert > 1$ then $x > 1$ or $x < –1$ for all $x\in\mathbb{R}$.
I can't wrap my head around as to how it is provable, I could just put some values but that wouldn't be a concrete generalized proof
AI: Since$$|x|=\begin{cases}x, &\text{ if } x\geq0\\-x, &\text{ if } x<0\end{cases},$$
so $|x|>1$ implies $x >1$ or $-x>1,$ i.e. $x>1$ or $x<-1.$
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H: I need to find the spectrum of an operator.
I need to prove that the spectrum of operator A in $L_{2}[0,1]$ is [-1;1] where $Ax(t) = \sin(\frac{1}{t})x(t)$ if $ t > 0$ and$ Ax(0) = 0$
Elementary - norm of $A$ is less or equal to $1$. Hence, all $\lambda$ from spectrum is such that $|\lambda| \leq$ 1.
$g(t)$ := $\sin(\frac{1}{t})$ if $t > 0$ and $0$ if $4t = 0$.
If there exists $y(t)$ such that $(A -\lambda I)x(t) = y(t)$ then if $y_{0} = 1$ almost everywhere then $x(t) = \frac{1}{g(t) - \lambda}$ $\notin L_{2}[0,1]$ if $\lambda \in [-1;1]$.
So, this means that $1 \notin$ Im(A-$\lambda I$) which means that Im(A - $\lambda I$) $\neq$ $L_{2}[0,1]$ and $\lambda$ is in spectrum.
But i can't seem to proof the fact that i used:
$x(t) = \frac{1}{g(t) - \lambda}$ $\notin L_{2}[0,1]$ if $\lambda \in [-1;1]$ And that's where i need help. Why is $x(t) = \frac{1}{g(t) - \lambda} \notin L_{2}[0,1]$ where $\lambda $ in $[-1;1]$
AI: Hint: First make the substitution $u=\frac 1 t$ in $\int |x(t)|^{2}dt$. Write $\lambda$ as $\sin y$. Use the fact that $\sin u -\sin y=2\sin (\frac {u-y} 2) \cos (\frac {u+y} 2)$. In an interval around $y$ the function $\frac 1 {\sin (\frac {u-y} 2)}$ behaves like $\frac 2 {u-y}$ and the square of this is not integrable around $y$.
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H: Distribution of $\Big(Y_1+Y_2\Big)^2$ and $\Big(Y_1-Y_2\Big)^2$ where $Y_i \sim N(0,1)$
Does anyone know what is the distribution of $(Y_1+Y_2)^2$ and $(Y_1-Y_2)^2$ where $Y_i \sim N(0,1)$ are independent variables? I have tried to go through the joint pdf, but when trying to change variables I have multiples cases because it is squared.
AI: $(Y_1+Y_2)\sim N(0;2)$
$\frac{Y_1+Y_2}{\sqrt{2}}\sim N(0;1)$
$(\frac{Y_1+Y_2}{\sqrt{2}})^2\sim \chi_{(1)}^2=Gamma(\frac{1}{2};\frac{1}{2})$
$(Y_1+Y_2)^2 \sim 2Gamma(\frac{1}{2};\frac{1}{2})=Gamma(\frac{1}{2};\frac{1}{4})$
Show you efforts for the second case
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Here are the detailed calculations
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Let's $Z=\frac{Y_1+Y_2}{\sqrt{2}}$
applying properties of Gaussian distribution $Z \sim N(0;1)$
Deriving the Law of $W=Z^2$ via Fundamental Transformation Theorem you get
$f_W(w)=\frac{1}{\sqrt{2\pi}}e^{-\frac{y}{2}}\frac{1}{2\sqrt{y}}+\frac{1}{\sqrt{2\pi}}e^{-\frac{y}{2}}\frac{1}{2\sqrt{y}}$
Now re-organizing $f(w)$ we have
$f_W(w)=\frac{(\frac{1}{2})^{\frac{1}{2}}}{\sqrt{\pi}}y^{-\frac{1}{2}}e^{-\frac{y}{2}}=\frac{(\frac{1}{2})^{\frac{1}{2}}}{\Gamma(\frac{1}{2})}y^{\frac{1}{2}-1}e^{-\frac{y}{2}}\sim Gamma(\frac{1}{2};\frac{1}{2})$
(also known as $\chi_{(1)}^2$)
Multiplying this density by 2 you get again a Gamma, a $Gamma(\frac{1}{2};\frac{1}{4})$
This result can be easily derived again via Fundamental Transformation Theorem
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H: The set of real numbers whose product is rational is Borel in $\mathbb{R}^2$
Let $A=\{(x,y)|xy\in \mathbb{Q}\}$ be a set in $\mathbb{R}^2$. Show that A is a Borel set, and find its Lebesgue measure $m_2(A)$.
This is an exercise of the chapter of Fubini and Tonelli's theorem, so I wonder if we consider $f(x,y)=xy$, and the integral $\int_A f(x,y)dm_2$, which may be somewhat helpful. But I can't go on from there as I don't know how to link this to the Borel quality of the set.
Thanks for any suggestions.
AI: Split A into $$A=\cup_{r\in\mathbb{Q}}\{(x,y)|xy=r\}$$ and we can see that $A$ is a countable collection of zero-measure closed sets. Closed sets are Borel sets so we can conclude that $A$ is a Borel set and $m_2(A)=0$.
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H: Dice rolled 6 times. What is the probability to roll 2 twice, 4 twice, and 6 twice? Also, what is the probability to roll 2 thrice, and roll 4 thrice?
Here's what I currently have, however im not sure its correct.
a) $\mathbb{P}(2 \text{ twice}, 4 \text{ twice}, 6 \text{ twice}) = \{a, a, b, b, c ,c\}$ where $a$, $b$, $c$ are numbers between $1$ and $6$.
Say $a = 2$, $b = 4$, and $c = 6$. The order of the rolls don't matter and so we can order $6$ elements in $6$ fac ways. Therefore the probability $= \frac{6!}{6^6} = \frac{720}{46656}$
b) The same result, simply changing the set above with ${a, a, a, b, b, b}$ with $a = 2$ and $b = 4$.
Is this correct?
AI: a) Taking $6!$ as numerator you are overcounting. In stead you should take: $$\binom6{2,2,2}=\frac{6!}{2!2!2!}=90$$
b) Here as numerator you should take: $$\binom63=\frac{6!}{3!3!}=20$$
To get some insight try to count how many sixtuples there are containing $2$ and $4$ both three times.
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H: Proving $\sqrt 1+\sqrt 2+..+\sqrt n \approx\frac{2}{3} n^{\frac{3}{2}}$ asymptotically
Let $n$ be a positive intiger, prove this asymptotic formula for large $n$
$$\sqrt1+\sqrt2+\cdots +\sqrt n=\frac{2}{3}n^{\frac{3}{2}}+\text{lower order}$$
using a Riemann sum.
AI: We can obtain your approximation as follows:
$$\sum\limits_{k=1}^nk^\frac12=n^\frac32\sum\limits_{k=0}^n\left(\frac kn\right)^\frac12\frac1n=n^{\frac32}g\left(n\right)$$
where $g\left(n\right)=\sum\limits_{k=0}^n\left(\frac kn\right)^\frac12\frac1n$. Now for large $n$, we have
$$\lim\limits_{n\to\infty}g\left(n\right)=\lim\limits_{n\to\infty}\sum_{k=0}^n\left(\frac kn\right)^\frac12\frac1n=\int_0^1x^\frac12\mathrm dx=\frac23$$
So for large $n$, we have $g\left(n\right)=\frac23+\left(\text{things vanish at large }n\right)$. Thus for large $n$,
$$\sum_{k=1}^nk^\frac12=\left(\frac23+\text{things that vanish at large }n\right)n^\frac32\\
=\frac23n^\frac32+\text{lower order}$$
|
H: Textbooks in which determinant is defined as an alternating multilinear map
I'm interested in this abstract definition of determinant, i.e. determinant is defined as an alternating multilinear map. Could you please suggest me some Linear Algebra textbooks that define determinant in such way.
Thank you so much for your references!
AI: Linear Algebra by Hoffman and Kunze does determinant theory in the way you mentioned. Further it treats matrices not just over fields but over arbitrary commutative rings with unity.
|
H: What number's factorial is $i$?
I am trying to find the solution to the equation- $$\Gamma(z)=i$$
I have tried doing it the following way- LHS is-
$$\displaystyle \int_{0}^{\infty}t^ze^{-t}\ dt$$
Taking $z=a+ib$, we get-
$$\displaystyle \int_{0}^{\infty}t^{a+ib}e^{-t}\ dt$$
or
$$\displaystyle \int_{0}^{\infty}t^{a}t^{ib}e^{-t}\ dt$$
Using Euler's formula- $e^{i\theta}=\text{cos}\ \theta +\ i\ \text{sin}\ \theta$, we have-
$$\displaystyle \int_{0}^{\infty}t^{a}(\text{cos}\ (b\ \text{ln}t) +\ i\ \text{sin}\ (b\ \text{ln}t)) e^{-t}\ dt$$
After this point, I am not being able to solve this integral. I have tried graphing it on desmos, but it doesn't seem like this question has any solution. How can I approach further in this?
AI: To find the zero of function
$$f(z)=\Gamma(z)-i$$ I used Newton method with (why not ?) $z_0=1+i$. The iterates are
$$\left(
\begin{array}{cc}
n & z_n\\
0 & 1.00000+1.00000\,i \\
1 & 2.56016+2.59498\,i\\
2 & 2.24930+1.16218\,i \\
3 & 2.72561+2.04341 \,i \\
4 & 2.78276+1.60761\,i \\
5 & 2.84247+1.68952\,i \\
6 & 2.84550+1.68427 \,i \\
7 & 2.84550+1.68429 \,i
\end{array}
\right)$$
Using @Robjohn's approximation, we should obtain
$$z\sim e^{1+W(t)}+\frac 12\qquad \text{where} \qquad t=\frac 1 e \log \left(\frac{i}{\sqrt{2 \pi }}\right)$$ which, numerically, is $2.84071 +1.69496\, i$
Using it as $z_0$, Newton iterates are
$$\left(
\begin{array}{cc}
n & z_n \\
0 & 2.84071+1.69496\, i \\
1 & 2.84547+1.68422\, i \\
2 & 2.84550+1.68429\, i
\end{array}
\right)$$
Take care : this is only one of the many possible solutions (as usual when dealing with complex numbers).
Warning : In the previous edit of this answer, I had a terrible numerical mistake using @robjohn's superb approximation.
|
H: Radius of convergence and uniformly convergence
I think the radius of convergence for $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)x^n$, $x\in \mathbb R$ is:
$r^{-1}$=$\lim_{n\to \infty}$$|\frac{a_{n+1}}{a_n}$|=1 so we get that $r$=1.
But how can I show it formally?
After that, I have to show that $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)x^n$ is uniformly convergent in the interval $[-r,r]$. Can I maybe use Weierstrass majoranttest? What can I use as majorant serie?
AI: Since$$\lim_{n\to\infty}\frac{\frac1n-\sin\left(\frac1n\right)}{\frac1{n^3}}=\frac16$$and since the radius of convergence of the power series $\sum_{n=1}^\infty\frac{z^n}{n^3}$ is $1$, it follows from the comparaison test that the radius of convergence of your series is $1$ too.
By the same argument, together with the help of the Weierstrass $M$ test, it follows that the series converges uniformly on $\overline{D(0,1)}$. In fact, since the sequence$$\left(\frac{\frac1n-\sin\left(\frac1n\right)}{\frac1{n^3}}\right)_{n\in\Bbb N}$$converges, it is bounded. Take $M\in(0,\infty)$ such that$$(\forall n\in\Bbb N):\left|\frac{\frac1n-\sin\left(\frac1n\right)}{\frac1{n^3}}\right|<M.$$Then you have, if $|z|\leqslant1$,$$\left|\left(\frac1n-\sin\left(\frac1n\right)\right)z^n\right|\leqslant\frac M{n^3}|z^n|\leqslant\frac M{n^3}$$and the series $\sum_{n=1}^\infty\frac M{n^3}$ converges.
|
H: Why is $\Big\{\frac{1}{n^2}\Big\}_{n \in \mathbb{N}} \bigcup \hspace{0.2cm} \{-1,0 \}$ compact?
We have a set on Real numbers:
$$\Big\{\frac{1}{n^2}\Big\}_{n \in \mathbb{N}} \bigcup \hspace{0.2cm}\{-1,0 \}$$
I can't seem to understand why this is a compact set as if we look at the interval I think that there are tiny spaces in between the points of this interval that aren't in it. Isn't it so that a compact set must contain it's points and be somehow continuous?
Does that mean even let's say for example: $A = \{{1,2,3}\}$ that only contains this points is compact? Am I missing something here?
AI: First note that points of $\Bbb R$ are compact. Therefore the subset $\{-1\}$ is compact. Now $\lim_{n \to \infty} \frac{1}{n^2} = 0$, therfore $0$ is a limit point of the set $S = \{\frac{1}{n^2}: n \in \Bbb N\}$. Now if $x$ is a limit point of $S$, then there is a subsequence of $\frac{1}{n^2}$ which converges to $x$. But this forces $x = 0$. Therefore $0$ is the only limit point of $S$. Therefore $S\cup \{0$} is closed. Since it is trivially bounded, $S\cup \{0\}$ is compact(Heine Borel Theorem). Therefore $S\cup \{0\}\cup\{-1\}$ is compact.
|
H: $A$ and $B$ are two subnormal $p$-subgroups of $G$, how to show that $\langle A,B\rangle$ is a $p$-subgroup of $G$?
$A$ and $B$ are two subnormal $p$-subgroups of $G$, how to show that $\langle A,B\rangle$ is a $p$-subgroup of $G$?
It is not true in general if $A$ and $B$ are not subnormal. For example, $A:=\langle (12)\rangle$ and $B:=\langle (13)\rangle$ are two $2$-subgroups of $G:=S_3$, but $\langle A,B\rangle=G$ is not a $2$-group.
I know that since $A$ and $B$ are subnormal in $G$, $A$ and $B$ are contained in $O_p(G)$. That perhaps helps, but I don’t know what to do with it.
Thank you!
AI: $A$ and $B$ must lie in $O_p(G)$. We can actually generalize this to $\pi$ being a set of primes. What we will use is the fact that if $X$ is a characteristic subgroup of $Y$, and $Y \unlhd Z$ then $X \unlhd Z$, Here $X$ is characteristic in $Y$ if every automorphism of $Y$ maps $X$ onto $X$ and we write $X$ char $Y$. Now recall that $O_\pi(G)$ is the unique largest normal $\pi$-subgroup of $G$ (or, differently put, it is the intersection of all Hall $\pi$-subgroups). One can easily show that $O_\pi(G)$ is characteristic in $G$.
Let $S$ be a subnormal $\pi$-subgroup of $G$, say $S=H_0 \lhd H_1 \lhd \cdots \lhd H_r=G$. Since $S$ is normal, $S \subseteq O_\pi(H_1)$. Observe that $O_\pi(H_1) \text{ char } H_1 \lhd H_2$, so $O_\pi(H_1) \lhd H_2$ and this yields $O_\pi(H_1) \subseteq O_\pi(H_2)$. In its turns, $O_\pi(H_2) \text{ char } H_2 \lhd H_3$, so $O_\pi(H_2) \lhd H_3$ and this yields $O_\pi(H_2) \subseteq O_\pi(H_3)$. Now work up your way till $H_r=G$ is reached and we conclude $S \subseteq O_\pi(H_1) \subseteq O_\pi(H_2) \subseteq \cdots \subseteq O_\pi(G)$.
As a corollary we can also conclude that the subgroup generated by two subnormal $\pi$-subgroups of $G$ is again a $\pi$-subgroup.
|
H: Does the convergence in probability imply the following limit is $1?$
Let $X_n \in \mathbb{R}$ be a sequence of non-constant random variable with continuous PDF converging in probability to $c,$ but not necessarily convergence almost surely, i.e.
$$\lim\limits_{n \to \infty}P[|X_n - c| \le \epsilon]=1 \forall \epsilon > 0.$$
What can we then say about:
$$\lim\limits_{n \to \infty}P[|X_n - c| \le \frac{1}{n}]?$$
In general, choose a sequence $s_n \to 0$ as $n \to \infty.$ Then what can we then say about:
$$\lim\limits_{n \to \infty}P[|X_n - c| \le s_n]?$$
Is it always $1?$ If not, can we have a counter example?
AI: Take $P(X_n = 1) = 1- P(X_n = \frac{2}{n}) = p_n$ with $(p_n)\subset (0,1)$ your favorite sequence satisfying $\lim_{n\to \infty} p_n = 0$. Then $X_n \to 0$ in probability but $P(|X_n|\leq \frac{1}{n}) = 0$.
Edit: if you want the $X_n$ to have a density, let $X \sim \mathcal{U}([0,1])$ be uniformly distributed and set $X_n = \frac{2}{n}X$. Then $X_n \to 0$ in probability (and a.s.) but
$$P\left(|X_n|\leq \frac{1}{n}\right) = P\left(X\leq \frac{1}{2}\right) = \frac{1}{2}.$$
|
H: Flat extension of local rings with a specified extension of residue field
Let $(R, \mathfrak m_R, k)$ be a Noetherian local ring and $K$ be a field containing $k$.
Then is it true that there is a Noetherian local ring $(S, \mathfrak m_S)$ and a flat ring homomorphism $f: R\to S$ such that $f(\mathfrak m_R)S=\mathfrak m_S$ and $S/\mathfrak m_S\cong K$ ?
AI: Yes. You can see a proof in Bourbaki, Algèbre Commutative, chapitre IX, Appendice, n.2, Corollaire du Théorème 1.
|
H: $f(x) = ax^3 + bx^2 + cx + d,$ with $a > 0. $ If $f$ is strictly increasing, then the function $g(x) = f′ (x) −f′′(x) + f′′′(x)$ is
QUESTION: Consider the function $f(x) = ax^3 + bx^2 + cx + d,$ where $a, b, c$ and $d $ are real numbers with $a > 0. $ If $f$ is strictly increasing, then the function $g(x) = f′ (x) −f′′(x) + f′′′(x)$ is
$(i)$ zero for some real $x$.
$(ii)$ positive for all real $x$.
$(iii)$ negative for all real $x$.
$(iv)$ strictly increasing.
MY ANSWER: This is an easy question. I just want to know where I went wrong. This is what I did-
$$f'(x)=3ax^2+2bx+c$$$$f''(x)=6ax+2b$$ and, $$f'''(x)=6a$$ therefore, $$g(x)=3ax^2+2bx+c-6ax-2b+6a$$ which is nothing but, $$g(x)=3ax^2+(2b-6a)x+(6a+c-2b)$$ therefore, it represents an upward opening parabola $(\because a>0)$. So we can see that the function is not strictly increasing and not always negative. Now whether or not the function attains zero, depends on the discriminant of the quadratic equation. But it comes out to be $$\sqrt{(2b-6a)^2-4(6a+c-2b)3a}$$ and this becomes too complicated.. also we have another piece of information that $f'(x)=3ax^2+2bx+c>0$. But I couldn't use it anywhere..
I am quite sure there are smarter ways to tackle this one.
Any help will be much appreciated. Thank you so much.
AI: Since $f'(x) = 3ax^2 + 2bx + c \geq 0$, the discriminant of $f'$ should be negative:
$$4b^2 - 4\cdot 3a \cdot c = 4b^2 - 12ac \leq 0$$
Now the discriminant of $g$ equals
$$(2b-6a)^2 - 4\cdot 3a \cdot (6a+c-2b) = 4b^2 - 24ab + 36a^2 - 72a^2 - 12ac + 24ab = 4b^2 - 12ac - 36a^2$$
but since $4b^2 - 12ac \leq 0$ and $-36a^2 < 0$, this discriminant is striclty negative.
This implies that $g(x) > 0$ for all $x$, since $g$ can't have any zeroes.
|
H: Why is set $ \bigcup\limits_{n = 1}^{\infty} \{(\frac{k}{n},\frac{1}{n}):k=0,1,\dots,n\} $ closed.
We have a set:
$$ \bigcup_{n = 1}^{\infty} \left\{\left(\frac{k}{n},\frac{1}{n}\right):k=0,1,\dots,n\right\} $$
I don't understand the notation, does it mean that when $n=1$ then $k=0$ and both raise equivalently or can we fix one number and change the other?
Also, it says that the set isn't closed. Why not if it's made out of isolated points, shouldn't it be closed?
AI: I am interpreting $(a,b)$ as a point with coordinates $a$ and $b$ in $\mathbb R^{2}$.
$(\frac n n,\frac 1n)$ belongs to this set for every $n$ and $(\frac n n,\frac 1n) \to (1,0)$. Since $(1,0)$ does not belong to the set it follows that the set is not closed. [The second coordinate of every point in the set is strictly positive].
|
H: The "o" notation and limit
I just want some explanation about this notation ,it's very new to me ,and I've been also been given this exercise :
"Is this true or false : $x=o(\sqrt x)$ , $x\to 0$"
And it is true in the hint of the exercice ,and here is the hint : $\lim_{x\to 0} $=$\frac{x}{\sqrt x}$=$0$
How do we get this expression $\frac{x}{\sqrt x}$
And what is the real meaning of this "o" notation?
AI: The $o(\ )$ notation : $f(x)=o(g(x))$ means $f(x)$ is negligible compared to $g(x)$.
Notice that the context is often implicit, I should say when "$x\to 0$" or "$x\to +\infty$" but that is generally stated in the question.
From now on, I will assume that we are studying in a neighbourhood of $0$.
Thus $f(x)=o(1)$ simply means $\lim\limits_{x\to 0} f(x)=0$.
This notation is multiplicative, i.e. $f(x)=o(1)\iff x^kf(x)=o(x^k)\iff f(x)g(x)=o(g(x))$
In particular (disregarding zeroes of $g$) then $f(x)=o(g(x))\iff \dfrac{f(x)}{g(x)}=o(1)\to 0$
You can also see $f(x)=o(1)$ has $\forall \epsilon>0,\ |f(x)|<\epsilon$ in the epsilon-delta proofs, so you can see that the multiplicative property comes from $|f(x)g(x)|<\epsilon |g(x)|$
Regarding addition though it is "absorbant".
For instance if you have $f(x)=1+x^2+o(x)$, you can see that $x^2\ll x$ when $x$ is small, so we can simply ignore this term. and write $f(x)=1+o(x)$.
Similary when you add two quantities $f(x)=1+x+o(x^2)$ and $g(x)=3-7x+5x^2+x^3+o(x^3)$ the terms that are too small will be discarded:
$f(x)+g(x)=4-6x+5x^2+o(x^2)$
|
H: Circular track problem (LCM)
A question in my text book:-
A circular field has a circumference of $360 \;km$. Three cyclists start together and can cycle $48$, $60$ and $72$ km a day, round the field. When will they meet again?
Solution: We first find out the time taken by each cyclist in covering the distance.
Number of days $1^{st}$ cyclist took to cover $360 \;km = 360/48 = 7.5$ days.
Number of days taken by $2^{nd}$ cyclist to cover same distance $= 360/60 = 6$ days.
Number of days taken by $3^{rd}$ cyclist to cover this distance =$ 360/72 = 5$ days.
Now, $LCM$ of $7.5, 6$ and $5 = 30$ days
My question is: Why is the distance covered taken as $360 \;km$? Does this involve the pre-assumption that they are going to meet at the starting point only.
Another question, can we solve this question by taking LCM in terms of distances?
Thanks.
AI: You are right. It is not necessarily the case that the first time they meet happens at the starting location. The solution method you quote seems to make that assumption so will not always give the correct answer.
Imagine you are the slowest rider. The other two riders ride $12$ and $24$ km more than you each day. They will meet you again if they travel a whole number of laps more than you. The second rider laps you every $360/12=30$ days, and the third rider laps you every $360/24=15$ days. They first time they both lap you at the same time occurs after $\text{LCM}(30,15)=30$ days.
It just so happens that in those $30$ days you have travelled a whole number of laps ($4$ laps actually) so that it is the starting point where you all meet, but that would not have been the case with different speeds, e.g. $40, 52, 64$.
Note that every time the second rider laps you, so does the third, so the LCM does not do much in the calculation above, but if the speeds were different, it could have taken a lot longer before their laps coincided.
|
H: $1^{-1}+2^{-1}+\dots+\Big(\frac{p-1}{2}\Big)^{-1} \equiv -\frac{2^p - 2}{p} \mod p$ for an odd prime $p.$
I've reduced a problem down to proving this identity. Unfortunately, I don't know where to even start. There has to be some way of expanding the RHS or combining terms on the LHS, but I don't see it. Any hints?
AI: David W. Boyd writes:
There is a relationship between values of certain $H_n$ and Fermat's quotient $q_a = {(a^{p-1}-1)}/p \bmod p$. For example, a result from Eisenstein from 1850 [Dickson 1952, p. 41] states that $H_{(p-1)/2} \equiv -2q_2 \mod p$; this is easily seen from the binomial expansion of ${(1 + 1)}^p$.
This is your hint, and if that's not enough, the reference where to find a proof.
[Dickson 1952]: L. E. Dickson, History of the Theory of Numbers.
[Boyd 1994]: David W. Boyd, A $p$-adic Study of the Partial Sums of the Harmonic Series.
|
H: Showing the collection of subsets is the neighborhood system of a given topology
Suppose that we have a topological space $(X,\tau)$, a subset $A\subseteq X$ is a neighborhood of $x \in X$ if there exists an open subset $U_x$, containing $x$ such that $U_x \subseteq A$. Also, suppose that the collection of all neighborhoods of $x \in X$ in $(X,\tau)$ is denoted by $\mathcal{N}_x$.
Each $\mathcal{N}_x$ has the following interesting properties, which I managed to prove.
$\forall A \in \mathcal{N}_x$, $x \in A$
$\forall A,B \in \mathcal{N}_x$, $A \cap B \in \mathcal{N}_x$
$\forall A \subseteq X$ and $\forall B \in \mathcal{N}_x$, if $B \subseteq A$, then $A \in \mathcal{N}_x$
$\forall A \in \mathcal{N}_x$, there exists $N \in \mathcal{N}_x$ such that, $\forall y \in N$, $A \in \mathcal{N}_y$
Let us denote the collection of subsets of $X$ satisfying the four above-mentioned properties by $\mathcal{U}_x$.
If we define $\tau_{\mathcal{N}} = \{G \subseteq X \:{:}\: \forall x \in G, G \in \mathcal{U}_x \}$, then it can easily be shown that $\tau_{\mathcal{N}}$ is a topology on $X$. Moreover, $\tau_{\mathcal{N}} = \{G \subseteq X \:{:}\: \forall x \in G, \exists U_x \in \mathcal{U_x}, U_x \subseteq G \}$ is also an equivalent expression.
My struggle starts from proving that $\mathcal{U}_x = \mathcal{N}_x$ where $\mathcal{N}_x$ is the collection of all neighborhoods of $x \in X$ in $(X,\tau_{\mathcal{N}})$. One way is easy because each $\mathcal{N}_x$ satisfies the four conditions of $\mathcal{U}_x$.
For the other way, suppose $A \in \mathcal{U}_x$, we need to show that $A \in \mathcal{N}_x$. Firstly, there exists $N \in \mathcal{U}_x$ such that, $\forall y \in N$, $A \in \mathcal{U}_y$, which implies that $N \subseteq A$. Since $x \in N$, all we need to show is that $N \in \tau_{\mathcal{N}}$, which means that I need to show that either
$\forall z \in N$, $N \in \mathcal{U}_z$, or
$\forall z \in N$, $\exists U_z \in \mathcal{U_z}$ such that $U_z \subseteq N$
I spent hours but could not make any progress to show that $N \in \tau_{\mathcal{N}}$. Any help will be greatly appreciated.
AI: This is a subtle matter and I already devoted a pretty comprehensive answer to it here, where you can also double check your proof that $\tau$ is a topology.
The basic idea is for $A \subseteq X$ to define
$$A^\ast = \{y \in X: A \in \mathcal{N}_y\}$$
(a neighbourhood version of the interior) and note that the $N\in \mathcal{N}_x$ promised by 4. to exist for $A$ is a subset of $A^\ast$, so $A^\ast \in \mathcal{N}_x$ too. Then show that $A^\ast \in \tau$ and so $A \in \mathcal{U}_x$.
To see that $A^\ast \in \tau$: let $z \in A^\ast$ be arbitrary.
By definition, $A \in \mathcal{N}(z)$.
Apply axiom 4. to $A$ to get an $N \in \mathcal{N}(z)$ such that $\forall y \in N: A \in \mathcal{N}(y)$, so by definition $N \subseteq A^\ast$ and so $A^\ast \in \mathcal{N}(z)$ by the enlargement property. Hence $\forall z \in N^\ast: A^\ast \in \mathcal{N}(z)$ which means that $A^\ast$ is open, as claimed.
|
H: How does a transitive extension differ from a transitive closure?
Quoting an example from C.L Liu's Discrete Mathematics: Let R be a binary relation on A. The transitive extension of R (let's denote it as $R_1$) is a binary relation on A such that $R_1$ contains R. Doesn't that make $R_1$ the transitive closure to R or is a transitive extension different from a closure?
Also, in the following two matrices, how is the second one the transitive extension to the first one?
$$
\begin{matrix}
\verb|bmatrix| & \begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix} \\[15pt]
\end{matrix}
$$
$$
\begin{matrix}
\verb|bmatrix| & \begin{bmatrix}
0 & 1 & 1 & 0 \\
0 & 1 & 1 & 1 \\
0 & 1 & 1 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix} \\[15pt]
\end{matrix}
$$
AI: Yes, transitive extension and transitive closure are different notions. If you start from a relation $R$ on some set, then the transitive extension $R_1$ of $R$ is obtained by adding to $R$ a pair $(a,c)$ whenever $(a,b)$ and $(b,c)$ are in $R$ for some $b$. Note that the transitive extension is in general not-transitive, while transitive closures are. For example, consider the two matrices of your question, and let $(i,j)$ denote the place corresponding to row $i$ and column $j$. Let $R$ be the relation represented by the first matrix, $R_1$ the relation represented by the second one. Then we say that $i$ is related to $j$ under $R$ if there is $1$ in place $(i,j)$ of the matrix representing $R$, we say they are not related if there is $0$. The same for $R_1$. In the first matrix, you have $1$ at place $(1,2)$, and $1$ in place $(2,3)$, meaning that $1$ is related to $2$ and $2$ to $3$. In the matrix, you have $0$ at place $(1,3)$, meaning that $R$ is not transitive, but you have $1$ in matrix $R_1$, which represents the transitive extension of $R$. Still $R_1$ is not transitive, because for example, using the same notation, $(1,2)$ and $(2, 4)$ are in $R_1$, but $(1,4)$ is not.
To obtain a transitive relation starting from a general relation $R$, you have to build its transitive closure, which is the least relation containing $R$ which is transitive, or equivalently, the set intersection of all transitive relations containing $R$.
You can relate the two notions by observing that, if you denote by $R_1$ the transitive extension of $R$, by $R_2$ the transitive extension of $R_1,\ldots,$ by $R_{i+1}$ the transitive extension of $R_i\ldots$, then the transitive closure of $R$ is the set union of all $R_i$'s.
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H: Is this weighted average smaller than the corresponding arithmetic average?
Is it true that the following weighted-average is smaller than the respective arithmetic average
$$
\sum_{n=1}^{N}\left(\frac{b_{n}}{\sum_{m=1}^{N}b_{m}}\right)\frac{a_{n}}{b_{n}}\overset{?}{\le}\frac{1}{N}\sum_{n=1}^{N}\frac{a_{n}}{b_{n}}
$$
for all $a_n \in [0,1]$, $b_n \in [1, D]$, and $D \ge 2$?
Without the $a_n$'s it would simply be true by Jensen's inequality and the convexity of $x\mapsto 1/x$ (for positive $x$'s), but with the $a_n$ in the mix it is not so clear and it seems one cannot directly apply Chebyshev's sum inequality either.
Edit: it is easy to see that one can upper bound the numerator in the parenthesis by $D$ and lower bound the corresponding denominator by $N$, obtaining the desired upper bound with an extra $D$ term. I was wondering however if one can do better in general.
AI: Testing the claim for $N=2$, suppose $b_2 > b_1$.
\begin{align*}
\text{Then}\;\,
&
\left(\frac{b_1}{b_1+b_2}\right)\left(\frac{a_1}{b_1}\right)
+
\left(\frac{b_2}{b_1+b_2}\right)\left(\frac{a_2}{b_2}\right)
\le
\frac{1}{2}\left(\frac{a_1}{b_1}+\frac{a_2}{b_2}\right)\\[4pt]
\iff\;&
\frac{a_1+a_2}{b_1+b_2}\le\frac{a_1b_2+a_2b_1}{2b_1b_2}\\[4pt]
\iff\;&
(b_1+b_2)(a_1b_2+a_2b_1)-(2b_1b_2)(a_1+a_2)\ge 0\\[4pt]
\iff\;&
(b_2-b_1)(a_1b_2-a_2b_1)\ge 0\\[4pt]
\iff\;&a_1b_2-a_2b_1\ge 0\\[4pt]
\iff\;&a_1b_2\ge a_2b_1\\[4pt]
\iff\;&\frac{a_1}{a_2}\ge \frac{b_1}{b_2}\\[4pt]
\end{align*}
which need not be true.
|
H: Decide whether this is correct, using the method of resolution. If not, provide a counter example.
I am new to logic and wanted to decide whether the following is correct using the method of resolution: |= p → ¬ (p → (p ∧ (p ∨ q)))
My attempt to this I answered that the conclusion is incorrect, though the premises hold true. Thus, I begin with the negation of the conclusion and solve for validity.
¬ (p → ¬ (p→(p ∧(p ∨ q)))) ↔
¬ p ∧ ¬ (¬(p→(p ∧(p ∨ q)))) ↔
¬ p ∧ (p→( p→(p ∧(p ∨ q)))) ↔
¬ p ∧ (p→(p ∨ p)) ↔
¬ p ∧ (p→p) ↔ ¬ p ∨ p
Therefore, I deduced two clauses from the negated conclusion, ← p and p ←
Resolution gives the clauses in one step. ← p and p ←
AI: Your error is already in the first step. You simplified $\neg(a \rightarrow b)$ to $\neg a \wedge \neg b$ but it should be $\neg(\neg a \vee b) = a \wedge \neg b$.
$$\neg (p \rightarrow \neg (p\rightarrow(p \wedge (p \vee q))))$$
$$p \wedge \neg\neg(p\rightarrow(p \wedge (p \vee q))))$$
$$p \wedge (p\rightarrow(p \wedge (p \vee q))))$$
$$p \wedge (p \vee q)$$
$$p$$
|
H: Holomorphic function, existence of a sequence
Suppose $f$ is holomorphic on $D$. Prove that there exists a sequence $\lbrace z_n\rbrace$ such that $|z_n|\rightarrow 1$ and $\lbrace f(z_n)\rbrace$ is bounded.
Now, what I did is
$f$ is analytic on $D$ and $|f(z)|<1$. Let $z_0,z_1,z_2,...$ be zeros on $D$ then each $|z_i|<1$ on $D$ and let $z_n$ be the zeros of $f$ and I am stuck. I know that I should put the zeros in a sequence form, any ideas?
AI: If $f$ has infinitely many zeros, you are done (because $f$ has only finitely many roots in any compact subset of $D$).
Else, you can can divide $f$ by a polynomial $g$ so that $f/g$ has no roots on $D$, and you can check that the problem is equivalent for $f$ and $f/g$.
Now, assume for the sake of contradiction that the statement doesn’t hold for $f/g$, it follows that $h=(f/g)^{-1}$ is holomorphic on $D$, without any roots, and such that $h$ goes to zero on the border of $D$. As a consequence, by the Cauchy formula, $h(0)=0$, a contradiction.
|
H: Closely minimize error bound $\frac{4 \pi \exp(\cosh(a))}{\exp(a N) -1}$
I'm trying to minimize an error bound $$\frac{4 \pi \exp(\cosh(a))}{\exp(a N) -1},$$ where $N$ is the step size for the trapezoidal rule and $-a < Im < a, a > 0$ is a strip bound which may be adjusted for different $N$. Now the paper with this example states that one can show with calculus that the bound is minimized with a value of $a$ close to $a = \log(2N)$.
I tried taking the derivative and setting it $0$; however this didn't yield any proper result.
How did they manage to find that $a = \log(2N)$ closely minimizes this bound?
AI: Of no interest at all ! Just to show how good is the approximation.
@Gary explained why the approximation is valid.
You could even have something closer using a Taylor expansion of the derivative
$$f'=\sinh (a)-\frac{n \,e^{a n}}{e^{a n}-1}$$ around $a=\log(2n)$. Limited to first order
$$f'=\left(-\frac{2^n n^{n+1}}{2^n n^n-1}+n-\frac{1}{4 n}\right)+\left(\frac{2^n
n^{n+2}}{\left(2^n n^n-1\right)^2}+n+\frac{1}{4 n}\right) (a-\log (2
n))+O\left((a-\log (2 n))^2\right)$$ and solving for $a$ gives
$$a=\log(2n)+\frac{1}{1+\frac{4 \left(2^n \left(2^n n^n+n-3\right) n^n+2\right) n^2}{\left(2^n
n^n-1\right) \left(2^n n^n+4 n^2-1\right)}}$$ For $n=1$, the correction term is $\frac{5}{13}$ dropping to $\frac{465}{4337}$ for $n=2$ and to $\frac{53965}{1733653}$ for $n=3$. For $n=10$, it is $0.0025$.
|
H: Which matrices can or cannot be obtained by matrix multiplication?
I was reading this post, Can you transpose a matrix using matrix multiplication?, and I thought it was interesting that we can't get a "transpose matrix" $B$ such that $BA = A^T$, at least not for all $A$. My questions are:
Given a matrix $A$, what kinds of matrices can or cannot be obtained by left multiplication $BA$? What about right multiplication $AC$? What if we allow both left and right multiplication, like $BAC$?
Does it depend heavily on the matrix $A$, or are there certain types of matrices that can always be obtained or can never be obtained?
I suppose what I am asking is what sorts of linear maps we can get by left composition and right composition, and why certain maps cannot be achieved.
I would also appreciate if someone could let me know what area of math this kind of question belongs to or is related to (e.g. some subfield of abstract algebra?), and how it might be important.
Edit: as pointed out in the comments, there are some familiar examples: For $A = 0$, left or right multiplication by anything still gives $0$. For $A = I$, it is the identity, so any matrix is obtainable. For invertible matrices, we can get the identity by left or right multiplication by its inverse; for non-invertible matrices, this is impossible.
AI: You are entering into the land of ring theory.
The $n\times n$ matrices over a ring $R$, denoted $M_n(R)$ forms a ring. The set of matrices $\{BA\mid B\in M_N(R)\}$ is what is called a left ideal of $M_n(R)$. In fact, one would call it a principal left ideal.
Looking at what $BA$ means, you can say that the possible results will be every matrix whose rows lie in the row space of $A$.
The set $\{BAC\mid B,C\in M_n(R)\}$ is not very special in itself. If you take the collection of finite sums of elements of this form, you do get something special: a (two-sided) ideal.
|
H: Winning strategy for player $0$ in the game $G_3(\mathfrak{A_1},\mathfrak{A_2})$ $\mathfrak{A_i}=(\mathbb{N},\{(n,n+i),n\in \mathbb{N}\})$
This is an Ehrenfeucht–Fraïssé game
Is it possible that player $0$ or the attacker/competitor can win against the defender/player $1$/duplicator in $3$ rounds ?
I have thought about it but I could not find a winning strategy:
my attempt: (let A be the first set and B the second set)
0 chooses $0$ in first set 1 chooses some $a$ in B if $a\geq2$ then 0 wins by choosing $a-2$ in B in round 2. Player 1 cannot make a move and lost.
Hence if 1 chooses an $a<2$ then we have either $0$ or $1$ in B.
What can 0 do in round 2 and round 3 to win the game? Is it possible?
The competitor can win in 2 rounds therefore the attacker needs at least 3 rounds to win the game.
AI: If the attacker plays $k$ in $A$, the defender can play $2k$ in $B$ to mimic the structure, so at some point the attacker needs to play in $B$ to win, and it seems plausible that it’s advantageous to play in $B$ right away.
Indeed, there is a winning strategy for the attacker that starts by playing $3$ in $B$. If the defender plays $a\ge2$ in $A$, the attacker then plays $a-1$ and $a-2$ in $A$ to win. On the other hand, if the defender plays $0$ in $A$, the attacker then plays $1$ in $B$ to win. So the defender has to play $1$ in $A$. But then the attacker plays $0$ in $B$. If the defender responds with $a\not\in\{0,2\}$ in $A$, the attacker then plays $a-1$ to win; if the defender responds with $0$ in $A$, the attacker then plays $2$ in $B$ to win; and if the defender responds with $2$ in $A$, the attacker then plays $5$ in $B$ to win.
|
H: Test if a point on a hexagonal lattice falls on a specified superlattice?
Based on previous answers (1, 2, 3) integers $i, j$ produce a hexagonal lattice using
$$x = i + j/2$$
$$y = j \sqrt{3} / 2.$$
From a point $k, l$ I can make a superlattice from integers $I, J$ using
$$i_{sup} = I k + J (-l)$$
$$j_{sup} = I l + J (k+l)$$
and
$$x_{sup} = i_{sup} + j_{sup}/2$$
$$y_{sup} = j_{sup} \sqrt{3} / 2.$$
If I have a point $m, n$ on the lattice, is there a simple test I can apply to find out if it is on the superlattice as well?
Example for $k, l = 3, 1$
Python script
AI: This one's all linear algebra all the time: your two lattices can be considered as transformations of a square lattice, taking integer $(i,j)$ through a linear transformation.
The two lattices coincide at a particular point if the $(i,j)$ for a particular lattice point, passed through the transform of that lattice and then the inverse transform of the other lattice, give integer results.
So in this particular case:
the blue lattice is the transformation of the unit square lattice through
$$A = \begin{pmatrix}
1 & \frac{1}{2} \\
0 & \frac{\sqrt{3}}{2} \\
\end{pmatrix}
$$
and the red lattice is the transformation of the unit square lattice through
$$B = \begin{pmatrix}
\frac{7}{2} & -1 \\
\frac{\sqrt{3}}{2} & 2\sqrt{3} \\
\end{pmatrix}
$$
To get the test matrix, we find $$B^{-1}A = \frac{\begin{pmatrix}
4 & 3 \\
-1 & 3 \\
\end{pmatrix}}{15}$$
Now, to discover if a particular point from the blue lattice is also on the red lattice, we simply apply this new matrix to the $(i,j)$ coordinates, and see if the result is in the integers: $(3,1)$ gives $(1,0)$ so is on the red lattice; $(4, -7)$ gives $(\frac{1}{3}, -\frac{5}{3})$ which is not integer so it is not on the red lattice.
|
H: How to write a matrix from $SU(2)$ in terms of one angle and one complex number $z$ , where $z$ is from sphere $S^{2}$
For given a matrix from $SU(2)$ , how can represent it in terms of two parameters: one angle and one complex number $z$ from the sphere $S^{2}$ ?
Does this have any links with : $\mathrm{SU}(2)$ axis and angle representation ?
Can someone help me do this?
Thank you in advance!
AI: A matrix in $\mathrm{SU}(2)$ is of the form
$$
\left(
\begin{matrix}
a & -\overline{b}\\
b & \overline{a}
\end{matrix}
\right), \qquad \lvert a\rvert^2 + \lvert b\rvert^2 = 1, \quad a,b \in \mathbb C.
$$
Write $a=x+yi, b=z+wi$. Note that the equation $\lvert a\rvert^2 + \lvert b\rvert^2 = 1$ says that $$x^2+y^2+z^2+w^2=1,$$ which is the equation of a three-sphere $\mathbb S^3$ in $\mathbb R^4$. Identify $\mathbb{R}^4$ with the space of quaternions $\mathbb H$, so that $(x,y,z,w)$ corresponds to $x+yi+zj+wij \in \mathbb H$ (this is only a correspondence of vector spaces). Now, since $i,j,ij$ span an $\mathbb R^3$ in $\mathbb H$, define $u$ as a unit vector in their span which is proportional to $yi+zj+wij$, so that $u \in \mathbb{S}^2 \subset \mathbb R^3$ (there are two choices for this vector). You can write the quaternion $x+yi+zj+wij$ as $$\cos \vartheta 1+ \sin \vartheta u, \qquad \text{ for some } \vartheta \in \mathbb S^1.$$ Obviously $x^2 = \cos^2 \vartheta$ and $y^2+z^2+w^2=\sin^2 \vartheta$. Therefore, $(\vartheta,u) \in \mathbb S^1 \times \mathbb S^2$ is one of the possible pairs you are looking for.
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H: Prove $\left|\frac{a_1 + ... + a_n}{b_1 + ... + b_n} - c \right| \le \max\limits_{k \in 1:n}\left|\frac{a_k}{b_k} - c\right|$
Given two sets of numbers - ${a_1, ..., a_n}$ and ${b_1, ..., b_n},b_i \ge 0 \; \forall i \in 1:n$ and some constant $c$.
I'm trying to prove that
$$\left|\frac{a_1 + ... + a_n}{b_1 + ... + b_n} - c \right| \le \max_{k \in 1:n}\left|\frac{a_k}{b_k} - c\right|.$$
Will it be right to say we need to prove, that we need to find $\max\frac{a_k}{b_k}$ for $c\le 0$ and $\min\frac{a_k}{b_k}$ for $c\ge 0$?
Or can we say just we need to prove
$$\left|\frac{a_1 + ... + a_n}{b_1 + ... + b_n} \right| \le \max_{k \in 1:n}\left|\frac{a_k}{b_k}\right|?$$
And how can i proceed?
AI: EDIT: By using @David C. Ullrich idea, the proof can be greatly simplified (credit goes to his deleted post):
Let $M=\max_{k \in 1:n}\left|\frac{a_k}{b_k}-c\right|$ it follows that:
$|a_i-cb_i|\le Mb_i$ for all $i=1,2,\ldots,n$
$|a_1+\ldots+a_n -c(b_1+\ldots+b_n)|\le|a_1-cb_1|+\ldots+|a_n-cb_n|\le M(b_1+\ldots+b_n)$
And one gets the desired result by dividing both sides by $b_1+\ldots+b_n$
INITIAL ANSWER:
To prove the last inequality,drop first the absolute value, as you deal with positive numbers. Then, without loss of generality, reorder the indices such that $\frac{a_1}{b_1}\le\frac{a_2}{b_2}\le\ldots\le\frac{a_n}{b_n}$ and proceed by induction.
The second step is to observe that you can drop the requirement $a_i\ge 0$, as we always have $\frac{|a_1+\ldots+a_n|}{b_1+\ldots+b_n}\le\frac{|a_1|+\ldots+|a_n|}{b_1+\ldots+b_n}\le\max_{k \in 1:n}\frac{|a_k|}{b_k}$
As the last step, you may apply the last inequality for $a_1\leftarrow a_1-cb_1, \ldots a_n\leftarrow a_n-cb_n$ to get your desired result.
|
H: Finding $a$ such that $a^2x^2+3x-5\frac{1}{a}=0$ has exactly one solution
For what value of $a$ would the following function have exactly one solution?
$$a^2x^2+3x-5\frac{1}{a}=0$$
I know that it needs to become
$$\frac{3}{2}x^2+3x+\frac{3}{2}=0$$
but how can one find value of parameter $a$ for this to happen ?
AI: For exactly one solution, we need $D=0$ i.e. $$3^2 -4\cdot a^2\cdot\left(-\frac 5a\right)=0$$
|
H: Show that $f(A ∩ B) ⊆ f(A) ∩ f(B)$, can the relation be improved to equality?
I have the following question:
Let $A,B$ be two subsets of a set $X$, and let $f : X → Y$ be a function. Show that $f(A ∩ B) ⊆ f(A) ∩ f(B)$. Is it true that the $⊆$ relation can be improved to $=$?
While I know that in these type of questions I should start by going back to the formal definition of everything stated and work from there I am having trouble stating these formally.
Usually a function can be written as ${\displaystyle \{(x,f(x)):x\in X\}}$ but how do I alter this to show that $x \subseteq A ∩ B$
Does the following work: ${\displaystyle \{(x,f(x)):x\in (A ∩ B) \subseteq X\}}$
In case it does, here is attempt:
${\displaystyle \{(x,f(x)):x\in (A ∩ B) \subseteq X\}}$
is equivalent to: ${\displaystyle \{(x,f(x)):x\in A \subseteq X \wedge x\in B \subseteq X\}}$
which in turn is equivalent to: ${\displaystyle \{(x,f(x)):x\in A \subseteq X\}} \wedge {\displaystyle \{(x,f(x)):x\in B \subseteq X\}}$
which is equivalent to $f(A) ∩ f(B)$
The biggest flaw in my work is that I am working as if I am showing equality, can someone guide me or give hints?
AI: The first half of proving the containment is addressed by @devianceee's comment. Listed here (plz give his/her comment upvote if this was useful to you: Prove $f(S \cap T) \subseteq f(S) \cap f(T)$ )
The second half of the question asks about improving the containment to an equality.
I don't believe the containment can be improved to an equality.
Consider $f(x) = x^2$ On the set $A = \lbrace -1, 2 \rbrace$ and $B=\lbrace 1, 2 \rbrace$
We have that $A \cap B = \lbrace 2 \rbrace$
So $f(A \cap B) = \lbrace 4 \rbrace$
Now $f(A) = \lbrace 1,4 \rbrace$ and similarly $f(B) = \lbrace 1,4 \rbrace$
so $f(A) \cap f(B) =\lbrace 1,4 \rbrace$
Clearly $\lbrace 1,4 \rbrace \ne \lbrace 4 \rbrace$ so we do NOT have that
$ f(A) \cap f(B) = f(A \cap B)$
In general.
Interesting questions to ask:
What conditions on $f$ and the sets $A,B$ need to be added so that equality again becomes possible? How can we play around with those conditions. This can lead to an interesting rabbit hole I believe.
|
H: Marginal distributions for a standard bivariate Cauchy distribution
Consider the following:
Let $X$, $Y$ be two jointly continuous random variables with joint PDF
$$f\left(x,y\right)=\frac{c}{2\pi}\frac{1}{\left(c^{2}+x^{2}+y^{2}\right)^{\frac{3}{2}}}\quad\text{(Standard Bivariate Cauchy Distribution)}.$$
Find the marginal PDF's of X and Y.
I know, I checked with Mathematica, the marginal PDF's of $X$ and $Y$ to be
$$f_{X}(x)=2\int_{0}^{\infty}\frac{c}{2\pi}\frac{1}{\left(c^{2}+x^{2}+y^{2}\right)^{\frac{3}{2}}}dy=\frac{c}{\pi(c^{2}+x^{2})}$$
ad
$$f_{Y}(y)=2\int_{0}^{\infty}\frac{c}{2\pi}\frac{1}{\left(c^{2}+x^{2}+y^{2}\right)^{\frac{3}{2}}}dx=\frac{c}{\pi(c^{2}+y^{2})}$$
but how? I mean I can't even begin to imagine how one computes those integrals
AI: Okay, I won't calculate exactly those integrals, but something which would be sufficient.
You need to calculate integral of type ($a>0$)$$\int_0^\infty \frac{dx}{(a+x^2)^{\frac{3}{2}}}$$
Substitute $t=\frac{x}{\sqrt{a}}$ you get $$\frac{\sqrt{a}}{a^{\frac{3}{2}}} \int_0^\infty \frac{1}{(1+t^2)^{\frac{3}{2}}}dt$$
So we'll just calculate $$ \int_0^\infty \frac{1}{(1+t^2)^{\frac{3}{2}}} dt$$
And when one see $t^2+1$, one should think about $t=\tan(u)$ substitution, getting:
$$ \int_0^\frac{\pi}{2} \frac{1+\tan(u)^2}{(\tan(u)^2+1)^{\frac{3}{2}}} du = \int_0^\frac{\pi}{2} \cos(u)du = 1$$
By that our first integral is equal to $\frac{1}{a}$
In your case it gives $$\frac{c}{\pi}\int_0^\infty \frac{1}{(c^2+z^2 + t^2)^{\frac{3}{2}}}dt = \frac{c}{\pi(c^2+z^2)}$$ where either $(z,t) = (x,y)$ or $(z,t)=(y,x)$
|
H: Is $\sum\limits_{n=0}^{\infty} \frac{n}{n+1} (-1)^{n}$ convergent?
I will like to know whether $\sum\limits_{n=0}^{\infty} \frac{n}{n+1} x^{n}, \: x \in \mathbb{R}$ is convergent for $x=\pm1$ or not.
I found the radius of convergence to be $r=1$. So I have to check each ($x=1$ and $x=-1$) case separately. $x=1$ is easy. $x=-1$ is not so easy. I have used the limit test, M-test, root test, Leibniz test, comparison tes. But none if them work. I do not think I have other tests in my arsenal.
Any help is welcome.
AI: You've already noted @Tom explained that, if $\sum_{n\ge0}a_n$ converges, the triangle inequality implies $\lim_{n\to\infty}a_n=0$. Whatever name that test has, it proves your series diverges. Indeed$$a_n:=\frac{(-1)^nn}{n+1}\implies\lim_{n\to\infty}|a_n|=1\implies\lim_{n\to\infty}a_n\ne0.$$In fact $\lim_{n\to\infty}a_n$ doesn't exist, because the subsequences for even (odd) $n$ have respective limits $1$ ($-1$), which differ. (That test appears to have a name too.)
|
H: Is there a real valued positive function such that it and its square integrate to $1$
Does there exist a function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that $f > 0$ and
$$
\int_{-\infty}^\infty f(x) dx = \int_{-\infty}^\infty f(x)^2 dx = 1.
$$
I suspect the answer is yes. I have looked at taking $f$ to be the PDF of a normal distibution $\mathcal N(0, \sigma)$ to guarantee that $f > 0$ and integrates to give $1$. I'm thinking about using IVT to find a value of $\sigma$ that works. However, in order to do this, I'm not entirely sure how to integrate $e^{x^4}$.
AI: It is possible. Take
$$f(x) = 2 e^{-4|x|}$$
$\int_{\Bbb{R}} f \:dx = \int_{\Bbb{R}} f^2 \:dx = 1$ as desired.
|
H: Variation of a functional
I have to find the variation of the following functional:
There are two conditions a > 0 and b > 0.
The question is "find the differential equation with respect to x(t), so that the functional is minimized".
AI: Set $y(\tau) = x(\tau) + \varepsilon \eta(\tau)$ where $\eta(\tau)$ is an arbitrary but twice differentiable function. Then
$$
S[y(\tau)]=\int_0^t a \left(\dot{x}(\tau)+\varepsilon \dot{\eta}(\tau)\right)^2 + b\left(\ddot{x}(\tau)+\varepsilon \ddot{\eta}(\tau)\right)^2 + c(x(\tau)+\varepsilon \eta(\tau))^4 d\tau.
$$
Take the derivative wrt to $\varepsilon$ and let $\varepsilon \rightarrow 0$, yielding
$$
\nabla_\eta S[x(\tau)] = \int_0^t 2a\dot{x}(\tau)\dot{\eta}(\tau) + 2 b \ddot{x}(\tau)\ddot{\eta}(\tau) + 4 c x(\tau)^3\eta(\tau) d\tau.
$$
This is the directional/Gateaux derivative of $S[]$ wrt $\eta$ evaluated at $x$.
Are you optimizing the functional? The idea is that if you are at $x$ and take a step $\varepsilon$ in the direction $\eta$ in the appropriate space of functions, you are evaluating the value of moving from $x$ towards $y$. Setting $\nabla_\eta S[x^*]=0$ means that there are no directions away from $x^*$ that locally improve $S$. Do you have initial conditions, or other constraints?
|
H: Do there exist sudoku's with clues that only contain $1$'s and $2$'s
I read this article saying it was mathematically proven that if a sudoku has a unique solution, then it has at least $17$ initial hints.
In which case, is it possible to have a sudoku with $17$ initial hints that consist of nine $1$'s and eight $2$'s (of course the $1$'s and $2$'s are chosen arbitrarily, I could have said $5$'s and $3$'s)?
This was just a small thought I had. Apologies if the answer is obvious in any case. Is there any examples, or, if not, reasons as to why we can't have this.
AI: A sudoku with a unique solution has to have at least eight distinct numbers amongst
its initial clues. If it were missing two, say $8$ and $9$, then if we had
a valid solution, then swapping all $8$s and $9$s therein would give a second
valid solution,
|
H: Discrete measure and Lebesgue measurability
According to wikipedia
https://en.wikipedia.org/wiki/Discrete_measure
a driscrite measure is defined in the following way:
Let's consider a real line $\mathbb{R}$. For some (possibly finite) sequences $s_{1}, s_{2}, \dots$ and $a_{1}, a_{2}, \dots$, s.t. $a_{i}>0$ and $\sum_{i}a_{i} = 1$, let
$$
\begin{equation}
\delta_{s_i}(X)=\begin{cases}
1, & \text{if $s_{i}\in X$}\\
0, & \text{if $s_{i}\not\in X$}
\end{cases}
\end{equation}
$$
for any Lebesgue measurable set $X$. Then
$$
\mu = \sum_{i}a_{i}\delta_{s_i}
$$
is the discrete measure on $\mathbb{R}$.
The question: why do we need Lebesgue measurability?
AI: You don't. Such a measure makes sense on any $\sigma$-algebra.
The authors of that page probably have in mind a context where one is studying some larger class of measures on the Lebesgue $\sigma$-algebra (or perhaps the Borel $\sigma$-algebra), and where one wants to say something about which measures in that class are discrete.
Note that they discuss a more general setting further down - they want to call a measure $\mu$ discrete with respect to $\nu$ only when it is of the form you describe, and all the singletons $\{s_i\}$ are $\nu$-null. So this part requires saying something about the $\sigma$-algebra $\Sigma$ on which $\nu$ is defined. The measure $\mu$ will still make sense and behave well for sets $X$ which are not $\Sigma$-measurable, but in this context we are just not interested in such sets.
|
H: Prove that $U(E_{\lambda})=E_{\lambda}$ and $U(K_{\lambda})=K_{\lambda}$.
Let T be a linear map on a finite-dimensional vector space V , and let $\lambda$ be an eigenvalue of T with corresponding eigenspace and generalized eigenspace $E_{\lambda}$ and $K_{\lambda}$. Let U be an invertible operator on V that communutes with T(i.e. TU=UT) Prove that $U(E_{\lambda})=E_{\lambda}$ and $U(K_{\lambda})=K_{\lambda}$.
Theorem:Let T be a linear map on a finite-dimensional vector space V such that the characteristic polynomial of T splits. suppose that $\lambda$ is an eigenvalue of T with multiplicity m. Then $dim(K_{\lambda}) \leq m$ and $K_{\lambda}=N((T-\lambda I)^m)$
Theorem: Let T be a linear map on a finite-dimensional vector space V, and let $\lambda$ be an eigenvalue of T, then $K_{\lambda}$ is a T-invariant subspace of V containing $E_{\lambda}$ (the eigenspace of T corresponding to $\lambda$).
Since U is an inverse linear oeprator on V and TU=UT=I and since U is linear, then we have $U(E_{\lambda})=E_{\lambda}$ and $U(K_{\lambda})=K_{\lambda}$.
I am also thinking about the fact that if $v \in ker T$, then $U(v) \in Ker T$ and that U commutes with $(T-\lambda I)^k$ for all $k \geq 0$. I am not sure how to finish this proof.
AI: The relation $UT=TU=I$ is false, since you don't know whether $U$ is the inverse of $T$.
If $v\in E_\lambda$, then $Tv=\lambda v$ and
$$
T(U(v)) = U(T(v)) = \lambda U(v)
$$
so $U(v)\in E_\lambda$. This is enough to say $U(E_\lambda)\subseteq E_\lambda$ and since $U$ is invertible,
it must necessarily hold $U(E_\lambda)=E_\lambda$.
There exists an index $m$ for which
$K_\lambda = N((T-\lambda I)^m)$. Notice that
$$
UT^k = T^kU \quad\forall k\implies U(T-\lambda I)^m = (T-\lambda I)^mU
$$
so if $v\in K_\lambda$, then
$$
0 = U(T-\lambda I)^m(v) = (T-\lambda I)^m(U(v))
$$
and $U(v)\in N((T-\lambda I)^m)=K_\lambda$, so
$U(K_\lambda)\subseteq K_\lambda$ and since $U$ is invertible,
it must necessarily hold $U(K_\lambda)=K_\lambda$.
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H: complex non algebraic manifold local ring of holomorphic functions is noetherian?
Consider $X$ a complex manifold. Denote $x\in X$ a point and $O_x$ as the local holomorphic function ring at $x$.
Assume $X$ is not algebraic.
$\textbf{Q1:}$ Is $O_x$ Noetherian? If it is Noetherian, then my guess is that it can be locally defined by equations and it follows from Chow's theorem that $X$ is algebraic as well. Is there something wrong here?
$\textbf{Q2:}$ Is $O$ ring of holomorphic functions over non-compact $X$ Noetherian?
AI: For arbitrary complex manifold, locally, $\mathcal O_{X,x}=\mathcal O_{\mathbb C^n,0}$ is noetherian.
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H: Confusion with mathematical objects (sets and multisets in particular)
The two sets $\{1,2,3\}$ and $\{1,1,2,3\}$ are equal, but the two multisets $<1,2,3>$ and $<1,1,2,3>$ are not equal.
Multisets are generalizations of sets that allow repeated elements. A natural mapping exists from sets to multisets: we can write $\{1,2,3\} = <1,2,3>$ since these two objects have identical properties in all respects. Likewise, $\{1,1,2,3\} = <1,1,2,3>$
So we have $$<1,1,2,3> =\{1,1,2,3\} = \{1,2,3\} = <1,2,3>$$
But we noted that $<1,1,2,3> \neq <1,2,3>$, so we have a contradiction
What am I misunderstanding?
AI: There is a natural mapping from sets to multisets, but not all multisets are images of sets under this mapping. The mapping takes $\{1,2,3\}$ to the multiset $\{\!\{1,2,3\}\!\}$ in which each of the three elements occurs once, but it also takes the set $\{1,1,2,3\}$ to $\{\!\{1,2,3\}\!\}$, because the set $\{1,1,2,3\}$ also contains each of $1$, $2$, and $3$ exactly once: it is equal to the set $\{1,2,3\}$. The multiset $\{\!\{1,1,2,3\}\!\}$ is not the image of any ordinary set, because no ordinary set has two distinguishable copies of a single element.
A set is completely determined by its members, so if $A$ is a set, then (for instance) the number $1$ either is or is not a member of $A$: it can’t be a member twice or three times. Multisets are specifically designed to overcome this limitation: in a multiset we can have multiple instances of a single element, but when we do, we have a multiset that is not the natural counterpart of an ordinary set.
Thus, your example should read that the set $\{1,1,2,3\}=\{1,2,3\}$ corresponds to (not is equal to) the multiset $\{\!\{1,2,3\}\!\}$.
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H: What does the notation of $2^{\mathbb{N}}$ mean?
I've learned over the course of the last years that some mapping $\lambda$ denoted :
$$ \lambda : \mathbb{N} \rightarrow 2^{\mathbb{N}}.$$
essentially means that for every natural number, you assign (or map) some set of natural numbers to it. At multiple times, I've wondered if there's some meaning behind this notation (more precisely the $2^{\mathbb{N}}$ part) that I'm not understanding, as it does not seem as straightforward as say :
$$ f : n \rightarrow 2n$$
(also just noted as $f(n) = 2n$), where you clearly map any $n$ to $2n$. Is this $2^{\mathbb{N}}$ just some notation that one has to get used to, or does it make sense somehow to note it this way? If it does make sense, why the choice of exponentiation, and why the $2$?
AI: $2^{\mathbb N}$ denotes the power set of $\mathbb N$, i.e. the set of all subsets of $\mathbb N$. This notation is used because, for a finite set $A$, the size of the power set $2^A$ of $A$ is $2^{|A|}$.
However, $\lambda:n\to 2^{\mathbb N}$ still seems like weird notation to me: the $n$ is an element of $\mathbb N$ (presumably) but $2^{\mathbb N}$ is a set. Should it maybe be $\lambda:\mathbb N\to 2^{\mathbb N}$?
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H: How many ways to pick pairs from a group of six?
How many ways can we group six people into pairs?
I expect the answer to be $6!/(2!)^3$ but the textbook gives the solution as 6!/(2!)33!
I'm confused, as the way I got my answer was to do
(6 pick 2) * (4 pick 2) * (2 pick 2)
(6!)/(2!4!) * (4!/(2!2!))
AI: There are $6!$ distinct ways to place $6$ items into six slots. Consider the first two slots as the first pair, the second two slots as the second pair, and the last two slots as the last pair. Within each pair there are $2!$ equivalent orders (e.g., $35$ is the same as $53$). This is true for each of the three pairs, so we must divide by $(2!)^3$.
But you can also re-order the three pairs in $3!$ ways. (That is, ${\bf 35}12{\bf 64}$ is the same as ${\bf 64}12{\bf 35}$, and so forth.) Thus we must also divide by $3!$.
Thus the final number is:
$$\frac{6!}{2! 2! 2! 3!} = \frac{6!}{(2!)^3 3!}$$
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H: For angles $A$ and $B$ in a triangle, is $\cos\frac B2-\cos \frac A2=\cos B-\cos A$ enough to conclude that $A=B$?
Brief enquiry:
$$\cos\frac B2-\cos \frac A2=\cos B-\cos A$$
Optionally
$$\sqrt\frac{1+\cos B}{2}-\cos B=\sqrt\frac{1+\cos A}{2}-\cos A$$
Is above equality sufficient to prove that it implies $A=B$?
Detailed explanation and motivation for this question:
Consider a triangle with bisectors of equal length:
By definition:
$\lvert AE\rvert = \lvert BD\rvert = D\\\frac A2+\frac A2 = A,\space\space\space\frac B2 + \frac B2 = B$
By cosine law:
$$\lvert AE\rvert^2=y^2+a^2-2ya\cos B\\\lvert BD\rvert^2=x^2+a^2-2xa\cos A\\x^2=\lvert BD\rvert^2+a^2-2\lvert BD\rvert a\cos \frac B2\\y^2=\lvert AE\rvert^2+a^2-2\lvert AE\rvert a\cos \frac A2$$
By sine law:
$$\frac{x}{D}=\frac{\sin \frac B2}{\sin A};\space\space\space \frac{y}{D}=\frac{\sin \frac A2}{\sin B}$$
$$\bigl[D=\lvert AE\rvert = \lvert BD\rvert\bigr]$$
Since bisectors are equal:
$$x^2-2xa\cos A=y^2-2ya\cos B\implies \lvert BD\rvert \cos \frac B2 +x\cos A = \lvert AE\rvert \cos \frac A2 +y\cos B $$
$$D (\cos\frac B2-\cos \frac A2)=y\cos B-x\cos A $$
Dividing by D and substituting y and x we obtain:
$$\cos\frac B2-\cos \frac A2=\frac{\sin \frac A2\cos B}{\sin B}-\frac{\sin \frac B2\cos A}{\sin A}$$
Consider triangles $\Delta$ABE and $\Delta$BAD
Area of triangle $\Delta$ABE:
$$A = \frac{aD}{2}\sin\frac A2 = \frac{aD}{2}\sin B \implies \sin\frac A2 = \sin B$$
Similarly for triangle $\Delta$BAD
$$A=\frac{aD}{2}\sin\frac B2 = \frac{aD}{2}\sin A\implies \sin\frac B2 = \sin A$$
Therefore:
$$\cos\frac B2-\cos \frac A2=\cos B-\cos A$$
Optionally
$$\sqrt\frac{1+\cos B}{2}-\cos B=\sqrt\frac{1+\cos A}{2}-\cos A$$
Is above equality sufficient to prove that it implies A = B?
AI: After more thinking, I realized that the answer is in fact "Yes".
As before, let $a=\cos\frac{A}{2},b=\cos\frac{B}{2}, 0<a,b<1$
The identity becomes:
$2a^2-a=2b^2-b$
which can be rewritten as:
$(a-b)\left(a+b-\frac{1}{2}\right)=0$
So either $a=b$, which gives $A=B$, as $\cos$ is $1:1$ on $\left[0,\frac{\pi}{2}\right]$
Or $a+b=\frac{1}{2}$
But the latter is impossible, because $A+B<\pi$, so at least one of the angles $A$ and $B$ has to be acute, which gives:
$a=\cos\frac{A}{2}>\cos\frac{\pi}{4}>\frac{1}{2}$
The whole point is that we look at the interior bisectors only. But if we consider the exterior bisectors as well, it is possible to find $A\ne B$ satisfying $\cos\frac{A}{2}+\cos\frac{B}{2}=\frac{1}{2}$
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H: Verifying a Topological Property
Let (X,T) be any Topological Space. Verify that Intersection of any finite number of members of T is a member of T.
I tried to prove using that intersection of any two sets of T belongs to T. So the result of the intersection can Intersect with any other subset of T, the result of which will also belong to T. This will continue for any finite number of sets of T. Hence, verified. Please Help
Definition of Topological Space:$$1. X \in T\ and\ \emptyset \in T$$$$2.Union\ of \ any(finite\ or\ Infinite)\ members\ of\ T\ belongs\ to\ T$$$$3.Intersection\ of\ any\ two\ members\ of\ T\ belongs\ to\ T$$
AI: The base case for induction is the intersection of $n=2$ members of $T$ is also in $T$. This is in your definition
Inductive hypothesis is: Given $n \geq 2$ members of $T$, call them $A_1 , ... , A_n$, whose intersection $\cap_{k=1}^{n} A_k$ is also a member of $T$. Then given another member $A_{n+1}$ of $T$, $\cap_{k=1}^{n+1} A_k$ is a member of $T$
Now to prove it. By assumption we have the set $B = \cap_{k=1}^{n} A_k$ is a member of $T$. Now by definition of a topology, intersections of $2$ members are a member. Namely,
$$B \cap A_{n+1}$$
is a member of $T$. Notice that $B \cap A_{n+1} = (\cap_{k=1}^{n} A_k) \cap A_{n+1} = \cap_{k=1}^{n+1} A_k $ as desired
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H: Negative-base logarithm, where's the issue here
Here we go again, basics of the basics. Faced with the following question.
Definitions
Logarithm base a of x is by definition a number such as:
$$a^{\log_a x} = x$$
i.e. that answers the question "what power do I have to raise a in order to get x". I've read this answer about why logarithms can't be negative-based but that leaves out the case of "well-behaved" negative numbers.
$$(-2)^{3} = -8$$
so it is natural to expect that using the definition above and setting a = -2 and x = -8 we can define a logarithm in this case:
$$\log_{-2} (-8) = 3$$
which breaks the rule that requires the base of the logarithm to be positive and not 1, but sort of makes sense.
Question
Following the rule of the base change, we can now do the following:
$$3 = \log_{-2} (-8) = {\log_2 (-8)\over \log_2 (-2)}$$
and now, that definitely makes no sense because on the left side we have something that's well-defined and on the right side we have something that's not defined at all, because there is no power that 2 can be raised to to yield -2 or -8. So my question is - where is the mistake? Which definition went wrong?
AI: Definitions aren't right or wrong; claims about their consequences are. In this case, you've chosen $a,\,b,\,c$ so that $\log_ab=\frac{\log_ca}{\log_cb}$ breaks down; indeed, so will the proof of it.
The tricky thing about negative-base logarithms is that, unless you're prepared for them to be complex-valued, the set of values they can take isn't continuous. For example, what's $\log_{-2}3$? Well unfortunately, no real $x$ solves $(-2)^x=3$; in fact, $x,\,(-2)^{x}$ can only both be real if $x$ is a rational number which, in its lowest terms, has an odd denominator.
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H: Cauchy sequence is bounded? (Do we need any element in a sequence to be finite?)
This question is related to the question Is every cauchy sequence bounded?
The sequence $\{a_n\}$ used in that question
$$a_n=\frac{1}{n-1}$$
has the first element $a_1\rightarrow\infty$. As shown in the answer of that question, $a_1=\infty$ is not well-defined, so we can not say it is an unbounded Cauchy sequence.
However, in an unbounded sequence $$a_n=n,$$ we also have $$\lim_{n\rightarrow\infty} a_n\rightarrow\infty.$$ Why $\infty$ works here?
AI: $\lim_{n \to +\infty} a_n = +\infty$ means that for each $M \in \mathbb{R}$, there exists a $n_0 \in \mathbb{N}$ such that for all $n \geq n_0$
$$a_n > M.$$
This means that the terms of the sequence get arbitrarily large.
Note that in your first sequence you somehow claim that $\infty$ is an element of that sequence, but it is a concept to denote unboundedness, it can not be an element of a sequence.
edit: proof that $\lim_{n\to \infty}a_n = + \infty$ with $a_n = n \in \mathbb{N}$ :
Choose $M \in \mathbb{R}$ arbitrarily, then there exists a natural number $n_0$ with $n_0 > M$ (fundamental property of $\mathbb{R}$. Let $n \geq n_0$, then
$$a_n = n \geq n_0 > M.$$
This proves that $\lim_{n\to \infty}a_n = + \infty$.
However, it can be 'the limit of a sequence', as the definition above states.
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H: Represent $\sum_{k=1}^{n} k^{2}$ in terms of binomial coefficient.
Came across a probability problem that is sort of challenging for a beginner in a sense that I may have not seen or came across a lot of binomial identities.
What I am looking for is to see if there is any way to represent $\sum_{k=1}^{n} k^{2}$ in terms of binomial coefficient.
AI: Yes. Notice the following:
$$k^2=k\cdot (k-1)+k=2\cdot \frac{k(k-1)}{2}+\binom{k}{1}=2\binom{k}{2}+\binom{k}{1}.$$ Doing this we have
$$\sum _{k=1}^n\left (2\binom{k}{2}+\binom{k}{1}\right )=2\binom{n+1}{3}+\binom{n+1}{2},$$ using the Hockey-Stick identity.
Notice that this is part of a greater picture in which one can go from polynomials like $x^k$ to polynomials like $\binom{x}{k}$ as a change of basis.
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H: Evaluate: $\int_0^1 \frac{e^x(1+x) \sin^2(x e^x)}{\sin^2(x e^x)+ \sin^2(e-x e^x)} \,dx $
Evaluate: $\int_0^1 \frac{e^x(1+x)\sin^2(x e^x)}{\sin^2(x e^x)+ \sin^2(e-x e^x)} \,dx $
My assumption put $t= x e^x $ then $\int_0^e \frac{\sin^2(t)}{\sin^2(t)+ \sin^2(e-t)} \,dt$ How can I proceed from here? Thanks in advance.
AI: The key with the integral in terms of t is to exploit symmetry by letting $u=e-t$:
$$\int_0^e \frac{\sin^2{\left(e-u\right)}}{\sin^2{\left(e-u\right)}+\sin^2{\left(u\right)}} \; du$$
But, $u$ is a dummy variable so you can just switch $u$ to $t$, and add this integral to the integral before the $u$ substitution:
$$2I=\int_0^e \frac{\sin^2{\left(e-t\right)}+\sin^2{\left(t\right)}}{\sin^2{\left(e-t\right)}+\sin^2{\left(t\right)}} \; dt= \int_0^e \; dt=\boxed{\frac{e}{2}}$$
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H: How to prove if $(A \cup B) - (A \cap B) = A$ is true, then $B = \emptyset$?
I have figured out that this is true, but I'm not quite able to prove it.
I have tried direct proof, contrapositive and contradiction.
For contradiction, I assumed that B is not the empty set.
Then, I wrote that the difference between A and the empty set is the only operation that equals back A. (For the premise) $(A-\emptyset = A)$
Then $A\cap B$ only equals empty set when B is the empty set. Thus B has to be the empty set, therefore contradicting and proving that B is the empty set.
Does this make sense?
I also tried direct proof, where I simplified the premise to $(\overline A \cap B) \cup (A\cap \overline B)$ through set identities, but this doesn't seem to help me determine that $ B = \emptyset$
Any help is appreciated, thanks!
AI: Proof. By contradiction. Suppose $B\ne \varnothing$. Then there exists some element $x\in B$. Now, we have two possible cases:
Case 1. $x\in A$. In this case, $x\in A\cup B$ and $x\in A\cap B$, then $x\notin (A\cup B)-(A\cap B)=A$, a contradiction to $x\in A$.
Case 2. $x\notin A$. In this case, $ x\in A\cup B$ and $x\notin A\cap B$, then $x\in (A\cup B)- (A\cap B)=A$, a contradiction to $x\notin A$.
Since we have a contradiction in both cases, we conclude $B=\varnothing.$
Also, $\varnothing$ is not the only set such that $A-B=A$. Any disjoint set (that is $A\cap B=\emptyset$ will check $A-B=A$.
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H: How do I finish deriving this property of normal subgroups?
Socratica has a fantastic video on normal subgroups and quotient groups, but there’s one part of which I can’t convince myself.
Let $G$ be a commutative group under juxtaposition, let $N$ be a normal subgroup of $G$, and let the quotient group $G/N$ be equipped with the operation $\cdot$ for clarity. The members of $G/N$ are of the form $gN$, where $g\in G$, and the quotient group operation is defined by $$xN\cdot yN = zN \iff \forall g_1\in xN\ \ \forall g_2\in yN\ :\ g_1g_2\in zN$$
We know that the coset containing $g_1g_2=g_1g_2e$ is $g_1g_2N$ since $e\in N$. Therefore $\{g_1g_2N : g_1\in xN\wedge g_2\in yN\}\subseteq xN\cdot yN$. However, the video claims $xyN = xN \cdot yN$.
How does one go from $\{g_1g_2N\}\subseteq xN\cdot yN$ to $xyN = xN \cdot yN$? I tried to compare $G/N$ to $\Bbb Z/5\Bbb Z$, but that didn’t make this step any clearer.
AI: Let $G$ be a group and $N$ a normal subgroup of $G$. $N$ being normal by definition means $gng^{-1} \in N$ for all $g \in G$ and for all $n \in N$.
This is equivalent to the statement $gNg^{-1} \subset N$ for all $g \in G$. And this is further equivalent to the statement $gNg^{-1} = N$ i.e. $gN = Ng$ for all $g \in G$.
Using this together with the fact that $N = NN$ (since $N$ is a subgroup), we have
$$(xN)(yN) = x(Ny)N = x(yN)N = xyNN = xyN.$$
A more direct way to see this is the following: Let $xN, yN \in G/N$ where $x, y \in G$. Then
$$ \begin{align} xN \cdot yN &= \{ xnyn : n \in N \}
\\ & = \{ xy(y^{-1}ny)n: n \in N\}
\\ & = \{ xy(y^{-1}n(y^{-1})^{-1})n: n \in N\}
\\ &= \{ xyn'n: n,n' \in N\}
\\ &= \{xy n'': n'' \in N \}
\\ &= xyN,
\end{align}$$
where in the second equality we used the fact that $e = yy^{-1}$ and in the fourth equality is where we used the normal property of $N$.
Addendum: Note that here, we don't even need $G$ to be commutative! Since it was given the fact that $G$ is a commutative group, we immediately have
$$xN \cdot yN = \{xnyn: n \in N \} = \{xynn: n \in N \} = xyN.$$
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H: A variation in the construction of the tensor product of modules
Let $A$ be a ring, $E$ a right $A$-module and $F$ a left $A$-module. Consider the free $\mathbf{Z}$-module $\mathbf{Z}^{(E\times F)}$ which comes with the injective canonical mapping $\phi:E\times F\rightarrow\mathbf{Z}^{(E\times F)},\,(x,y)\mapsto e_{x,y}$, where $e_{x,y}:=(\delta_{(x,y),z})_{z\in E\times F}$ for $(x,y)\in E\times F$.
Bourbaki defines the tensor product of $E$ and $F$ as the quotient $\mathbf{Z}$-module $(\mathbf{Z}^{(E\times F)})/C$, where $C$ is the submodule of $\mathbf{Z}^{(E\times F)}$ generated by the elements of the form $(e_{x_1+x_2,y}-e_{x_1,y}-e_{x_2,y})$, $(e_{x,y_1+y_2}-e_{x,y_1}-e_{x,y_2})$ and $e_{x\lambda,y}-e_{x,\lambda y}$ for $x,x_1,x_2\in E$ and $y,y_1,y_2\in F$ and $\lambda\in L$.
Elsewhere, I have seen the element of the form $ne_{x,y}-e_{xn,y}$, with $x\in E$, $y\in F$ and $n\in\mathbf{Z}$, added to the list above. Is this necessary? Why does Bourbaki leave it out?
AI: It's indeed not necessary, as these elements will already be in $C$ even for Bourbaki's definition.
Specifically, for $n\ge 1$, use induction to see it (let $x_1=nx$ and $x_2=x$ in the induction step).
For $n\le 0$, use the rule $e_{x\lambda, \, y} - e_{x,\, \lambda y} \in C$ with $\lambda=0$ and $\lambda=-1$.
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H: To what extent does Goedel's 2nd incompleteness theorem extend?
In chapter 8 of Shoenfield's matheamtical logic[1967], He proves that The formula of P which states that P is consistent is not a theorem of P, where P stands for Peano Arithmetic. And then He says, without a proof, that this result can be extended to more general theories.
I don't really know what 'more general theories' exactly mean.
'more general theories' mean recursive consistent extensions of P?
Thank you.
AI: Certainly it applies to any (consistent, computably axiomatizable) extension of $\mathsf{PA}$ (that's the standard modern notation for first-order Peano arithmetic). It also holds for many vastly weaker systems of arithmetic, as well as theories which are not about arithmetic (directly, anyways) at all, like $\mathsf{ZFC}$.
On the other hand, unlike the first incompleteness theorem (whose extent is discussed in section $4$ of this paper of Beklemishev), there are reasonably-strong systems to which it doesn't apply. These have been studied by Willard; briefly speaking, they are systems which do not prove that multiplication is always defined, which kills the argument of the second incompleteness theorem due to a "size blowup" issue. See this answer of mine for more about this.
The short version, though, is: the second incompleteness theorem applies to any "natural" theory which can "do basic arithmetic."
OK, but what does that actually mean? (The following is unfortunately pretty technical, but I've tried to keep it still readable.)
Here's one fairly general - indeed, the most general I'm aware of - way to tackle the second incompleteness theorem:
Suppose $T$ is any (consistent, computably axiomatizable) theory which interprets Robinson arithmetic via some formula tuple $\Phi$. Then "$\Phi$'s version of" the consistency statement of $T$ is not $T$-provable.
This takes some unpacking. First of all, Robinson arithmetic is just a very weak theory of arithmetic; if you like, you can replace that with $\mathsf{PA}$ for simplicity for now. The real issue is around interpretations. The idea here is initially to handle things like $\mathsf{ZFC}$, which are not directly about the natural numbers but within which we can implement the natural numbers. The definition is fairly technical, but basically an interpretation of one theory $\mathsf{A}$ into another theory $\mathsf{B}$ is a tuple $\Psi$ of formulas in the language of $\mathsf{A}$ such that $\mathsf{A}$ proves "The structure described by $\Psi$ satisfies $\mathsf{B}$." For example:
The usual construction of the rational numbers as equivalence classes of ordered pairs of integers with nonzero second coordinate - with $(a,b)$ intuitively standing for $a\over b$ - amounts to an interpretation of the theory of $(\mathbb{Q};+,\cdot)$ in the theory of $(\mathbb{Z};+,\cdot)$.
Writing down the formulas defining the finite ordinals and ordinal addition and multiplication yields an interpretation of $\mathsf{PA}$ into $\mathsf{ZFC}$. (Note that actually $\mathsf{ZFC}$ proves a lot more than $\mathsf{PA}$ for this interpretation, but it doesn't matter that we've "overshot" $\mathsf{PA}$; extra strength is fine.)
So far so good - "interprets $\mathsf{Q}$" is also the standard strength condition for the first incompleteness theorem. But now we come to the new bit, which is basically how we express "$T$ is consistent" in the language of $T$. We already know how to do this in the context of arithmetic, but what if our theory in question is not about arithmetic at all, at least on the face of things? For example, $\mathsf{ZFC}$ falls into this category. Or, maybe we're looking at some complicated theory of geometry (much stronger than simple Euclidean geometry of course) which can in fact talk about arithmetic in some roundabout way; how should consistency be expressed there?
The idea is to use the interpretation we already have as a kind of "translation" mechanism. If we let $Con(T)$ be the usual formulation of "$T$ is consistent" in the language of arithmetic, the interpretation $\Phi$ whips up a corresponding sentence $\mathfrak{Con}(T)$ in the language of $T$ itself such that we can run Godel's argument "through" $T$'s version of arithmetic and "back again" into $T$, with the result that $T$ does not prove $\mathfrak{Con}(T)$.
Perhaps more cleanly, the statement of the second incompleteness theorem above can also be phrased as:
Suppose $T$ is any consistent, computably axiomatizable theory. Let $Con(T)$ be the usual statement in arithmetic asserting the consistency of $T$. Then $T$ does not interpret $\mathsf{Q}+Con(T)$.
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H: Confusion about A1, IMO 2002
The following is question A1 from the 2002 IMO:
$S$ is the set of all $(h,k)$ with $h,k$ non-negative integers such that $h+k<n$. Each element of $S$ is colored red or blue such that if $(h,k)$ is red, and $h'\leq h,k'\leq k$, then $(h',k')$ is also colored red. A Type 1 subset of $S$ has $n$ blue elements with different first member, and a Type 2 subset of $S$ has $n$ red elements with different second member. Show that there are the same number of Type 1 and Type 2 subsets.
Is there any way in which the elements are all not just one color? For example, let $(a,b)$ be colored red and $(c,d)$ be colored blue. Clearly, $0\leq a,c$ and $0\leq b,d$. Hence, $(0,0)$ should be colored both red and blue, which is impossible.
If my explanation is true, then both Type 1 and Type 2 sets cannot exist at the same time, making the whole question confusing
AI: They never said anything about a pair being blue forcing anything else to be blue. Only a red pair forces all "smaller" pairs it to be red.
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H: How long to catch up to a stream started 1 hour ago at 1.5x speed?
I opened a stream that started an hour ago. Not wanting to miss anything, I started from the beginning and set it to 1.5x speed. How long will it take for me to catch up?
I know that it will take 40 minutes to watch the hour that I missed ($ \frac {60}{1.5}=40$) but during that time, the stream has generated another 40 minutes that I need to watch. This tells me I need to do some calculus, but it's been a decade since I took that course. Can someone help me come up with an equation?
AI: To make this easier to understand, I will solve this as if you are an object travelling at $1.5m/s$, and the stream is an object which started one hour before you travelling at $1m/s$ in the same direction. So you are travelling at $1.5\times$ the speed of the stream.
We know that disance ($d$), speed ($s$), and time ($t$), are related as follows:
$$\Delta d=\Delta s\Delta t$$
You are trying to find time, so rearranging for $t$ gives
$$\Delta t=\frac{\Delta d}{\Delta s}$$
The initial "distance" between you and the stream is $3600$ metres, based on a speed of $1m/s$ for one hour. So
$$\Delta d=3600m$$
The difference between your speed and the stream's speed is $1.5m/s-1m/s=0.5m/s$. So
$$\Delta s=0.5m/s$$
Now solving for $\Delta t$:
$$\Delta t=\frac{3600m}{0.5m/s}=7200s$$
So, it will take you $7200s$, or $2$ hours, to catch up with the stream.
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H: Show $f:\mathbb{Z}\to\mathbb{Z}[i]/(3+2i)$ via $c\mapsto c+(3+2i)$ is surjective
Write $x=a+bi+(3+2i)\in\mathbb{Z}[i]/(3+2i)$. I want to find a $c\in\mathbb{Z}$ such that $f(c)=x$. I think we need to use the fact that there exist $m,n\in\mathbb{Z}$ so that $2m+3n=1$, but I am not sure how. I'd like to do something similar to the following trick for when the codomain is $\mathbb{Z}[i]/(3-i)$: Let $z=a+bi+(3-i)$. Note $a+bi=a+3b-3b+bi=(a+3b)-b(3-i)$, and thus $f(a+3b)=a+bi+(3-i)=z$. How can I adapt this example to when the codomain is a quotient of $\mathbb{Z}[i]$ by $(3+2i)$ instead of $(3-i)$?
I think it is a matter of rewriting $a+bi$ in a clever way, like adding and subtracting the same quantity to obtain a nice form. I have played around with it for an hour and keep getting stuck. Any hints?
AI: Here's a way to follow your nose to find a suitable preimage. Using your notation, fixing a given $x=a+bi\in\mathbb{Z}[i]$, you want to find $c\in\mathbb{Z}$ such that $x-c\in (3+2i)$. Write $x-c=(a-c)+bi=d+bi$.
Now $d+bi\in (3+2i)$ if $\frac{d+bi}{3+2i}\in\mathbb{Z}[i]$. Computation shows
$$
\frac{d+bi}{3+2i}=\frac{d+bi}{3+2i}\cdot\frac{3-2i}{3-2i}=\frac{(3d+2b)+(3b-2d)i}{13}\in\mathbb{Q}[i].
$$
Since $c$ is yet to be determined, you have freedom to adjust $d$ as you'd like, and you need $13\mid 3d+2b$ and $13\mid 3b-2d$ for the above to be in $\mathbb{Z}[i]$. If $13$ does divide both these numerators, it divides their sum, which is $d+5b$, so a natural guess for such $d$ is $d=13-5b$. Unless I've made a bad algebra error, if you set $d=13-5b$, a quick check shows both ratios are integers. This translates to saying $a-c=13-5b$, or $c=a+5b-13$. This $c$ is then a suitable preimage for $x+(3+2i)$.
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H: Is this switch of order of summation legal?
$$\sum_{k=1}^{\infty} \frac{1}{k^3} \sum_{m=0}^{\infty} \frac{(-1)^m}{k^{3m}} \to \sum_{m=0}^{\infty} (-1)^m \sum_{k=1}^{\infty} \frac{1}{k^3k^{3m}} \to \sum_{m=0}^{\infty} (-1)^m \sum_{k=1}^{\infty} \frac{1}{k^{3(m+1)}} \to \sum_{m=0}^{\infty} (-1)^m \sum_{k=2}^{\infty} \frac{1}{k^{3m}} \to (A) \sum_{m=0}^{\infty} (-1)^m (\zeta(3m) - 1)$$
I’m trying to evaluate
$$(B) \sum_{k=1}^{\infty} \frac{1}{k^3 + 1}$$
So I compared A and B on wolfram alpha and they’re off by one. Is it the weird switching of order of summation I did? I’d like to know.
AI: Your manipulative mistake is in the third step:
$$\sum_{m=0}^\infty(-1)^m\sum_{k=1}^\infty\frac1{k^{3(m+1)}}\ne\sum_{m=0}^\infty(-1)^m\sum_{k=2}^\infty\frac1{k^{3m}}\;.$$
The shift from $m+1$ to $m$ in the exponent requires a change in the limits on $m$, not those on $k$. In fact
$$\sum_{m=0}^\infty(-1)^m\sum_{k=1}^\infty\frac1{k^{3(m+1)}}=\sum_{m=1}^\infty(-1)^{m-1}\sum_{k=1}^\infty\frac1{k^{3m}}\;.$$
You also don’t want to allow $k=1$ in your first double summation, as the inner series diverges then.
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H: Finding residues of $\frac{z}{e^z+e^{-z}}$.
I am having trouble finding the residues of
$$\frac{z}{e^z+e^{-z}}.$$
I have found the poles (at $z= ±i(\frac{\pi}{2} +n\pi)$) but I cannot find a way to expand the function into a Laurent series to find the residues.
AI: You don't need such a Laurent expansion, since\begin{align}\require{cancel}\operatorname{res}\left(i\left(\frac\pi2+n\pi\right),\frac z{e^z+e^{-z}}\right)&=\frac{i\left(\frac\pi2+n\pi\right)}{e^z-e^{-z}|_{z=i\left(\frac\pi2+n\pi\right)}}\text{ (because }(e^z+e^{-z})'=e^z-e^{-z}\text)\\&=\frac{\cancel i\left(\frac\pi2+n\pi\right)}{(-1)^n2\cancel i}\\&=(-1)^n\left(\frac\pi4+\frac\pi2n\right).\end{align}
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H: Is $f(a_1 + p\mathbb{Z},a_2 + p^2 \mathbb{Z}, \ldots) = (0,pa_1 + p^2 \mathbb{Z},\ldots)$ surjective?
Let $p$ be prime and the $\mathbb{Z}$-module $M=\prod_{n=1}^\infty\mathbb{Z}_{p^n}$ where $\mathbb{Z}_{p^n}$ is $\mathbb{Z}$ modulo $p^n$. Define a map $f : M\to M$ by $f(a_1 + p\mathbb{Z},a_2 + p^2 \mathbb{Z},a_3 + p^3 \mathbb{Z},\ldots) = (0,pa_1 + p^2 \mathbb{Z},pa_2 + p^3 \mathbb{Z},\ldots)$
The question: Is the map $f$
(1) surjective,
(2) injective?
Answer: The image of $f, f(M)=Mp$. This implies that $f$ is a multiplication map by $p$ and so is injective. However I am failing to verify whether it is surjective or not.
AI: The map $f$ is not surjective, because any element in the image of $f$ has zero first component.
Checking injectivity requires a little bit of work. $f$ is a group homomorphism, so let's check that the kernel of $f$ is zero. Suppose $f(a_1 + p\mathbb{Z},a_2 + p^2 \mathbb{Z},a_3 + p^3 \mathbb{Z},\ldots) = (0,pa_1 + p^2 \mathbb{Z},pa_2 + p^3 \mathbb{Z},\ldots)$ is the zero vector. So $pa_1\equiv 0\pmod{p^2}$. Dividing by $p$ we see that $a_1$ is a multiple of $p$. Similarly any $pa_i\equiv 0\pmod{p^{i+1}}$ and $a_i$ is a multiple of $p^i$. So the initial vector $(a_1, a_2,...)$ is the zero element of $M$.
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H: Polynomials that form $1+xy+x^2 y^2$
Show that there is no polynomials $a(x), b(x) \in R[x]$ and $c(y), d(y) \in R[y]$ such that $1+xy +x^2 y^2 = a(x) c(y) + b(x) d(y) $
AI: Expanding on @MikeDaas's comment we have$$\begin{align}1&=a(x)c(0)+b(x)d(0),\\1+x+x^2&=a(x)c(1)+b(x)d(1),\\1-x+x^2&=a(x)c(-1)+b(x)d(-1)\\\implies x&=\frac{c(1)-c(-1)}{2}a(x)+\frac{d(1)-d(-1)}{2}b(x),\\x^2&=\frac{c(1)-2c(0)+c(-1)}{2}a(x)+\frac{d(1)-2d(0)+d(-1)}{2}b(x).\end{align}$$Since $1,\,x,\,x^2$ are all linear combinations of $a,\,b$, these two functions span a $3$-dimensional space, a contradiction.
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H: Finding the asymptotic of an integral
I came across the following exercise on asymptotic behavior of integrals:
$$I(a) = \int_0^\infty\frac{\cos x}{x^a} \, dx, \text{ where } a\to0^+.$$
I have tried integrating by parts or replacing $\cos x$ with the first summands of its Taylor series, but I end up with something equal to infinity (independent of $a$ and that is not what we want).
I just thought about the substitution $x \leftrightarrow\frac{1}{1+x^2}$ and I get the following result:
$$I(a) = \frac{\cos(1/2)}{2^{2-a}} + \int_0^1\frac{2x^3 \cos\left(\frac{1}{1+x^2}\right)}{(1+x^2)^{3-a}} \, dx - \int_0^1 \frac{2x^3\sin\left(\frac{1}{1+x^2}\right)}{(1+x^2)^{2-2a}}\,dx$$
If I could prove that the last summands are $o(I(a))$ then it would be OK, but I think, in general my strategy won't work.
Apparently this is supposed to be an easy exercise, but I can't come up with anything right now.
AI: Integration by parts is a good idea (as it usually is when one factor of the integrand is oscillating): when $0<a<1$,
$$
I(a) = \int_0^\infty \frac{\cos x}{x^a} \,dx = \frac{\sin x}{x^a}\bigg|_0^\infty - \int_0^\infty \frac{-a\sin x}{x^{1+a}} \,dx = a \int_0^\infty \frac{\sin x}{x^{1+a}} \,dx.
$$
This is enough to prove convergence, but not yet enough to get an upper bound that goes to $0$ with $a$. However, let's integrate by parts again!—being careful to choose an antiderivative that still allows us to control functions' behavior at the lower endpoint $0$.
\begin{align*}
I(a) = a \int_0^\infty \frac{\sin x}{x^{1+a}} \,dx &= a\frac{1-\cos x}{x^{1+a}}\bigg|_0^\infty - a \int_0^\infty -(1+a)\frac{1-\cos x}{x^{2+a}}\,dx \\
&= a(1+a) \int_0^\infty \frac{1-\cos x}{x^{2+a}}\,dx.
\end{align*}
This new integrand is nonnegative (showing $I(a)\ge0$) and
\begin{align*}
I(a) &= a(1+a) \int_0^\infty \frac{1-\cos x}{x^{2+a}}\,dx \\
&= a(1+a) \int_0^1 \frac{1-\cos x}{x^{2+a}}\,dx + a(1+a) \int_1^\infty \frac{1-\cos x}{x^{2+a}}\,dx \\
&\le a(1+a) \int_0^1 \frac{x^2/2}{x^{2+a}}\,dx + a(1+a) \int_1^\infty \frac{1}{x^{2+a}}\,dx \\
&= \frac{a(1+a)}{2(1-a)} + 2a = a \cdot\frac{5-3a}{2(1-a)};
\end{align*}
since that last factor is continuous at $a=0$,
these inequalities are enough to show that $\lim_{a\to0+} I(a) = 0$.
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H: Find 5 zero divisors of $\mathbb{Z}_6 \times \mathbb{Z}$
I want to find 5 zero divisors of $\mathbb{Z}_6 \times \mathbb{Z}$.
So far, I have found the zero divisors of $\mathbb{Z}_6$ as follows:
First, note that the non-zero elements of $\mathbb{Z}_6$ are $\{1, 2, 3, 4, 5\}$. Now, check for the various possibilities. Note that $2 \times 3 = 3 \times 2 \equiv 3 \times 4 = 4 \times 3 \equiv 0$ in $\mathbb{Z}_6$. This gives that the zero divisors of $\mathbb{Z}_6$ are $\{2, 3, 4\}$. This is where am getting stuck. I can't adopt a similar process to find the zero divisors of $\mathbb{Z}$, of course. How should I proceed?
AI: Note that $(a,0)$ with $a \neq 0$ can be a zero divisor as well. Because with $b \neq 0$, we can have $(a,0) \times (0,b)=(0,0)$ (assuming componentwise multiplication)
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