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H: Permutations and symmetric groups
Suppose that a permutation $f$ is the product of disjoint cycles $f_1,f_2,\dots, f_m$. Show that $o(f)$ is the least common multiple of $o(f_1), o(f_2),\dots, o(f_m)$.
Really lost with the question..
Say if there are 3 disjoint cycles, then $f=f_1*f_2*f_3$ say $f=(a,b)(c,d)(e,f)$ would that mean that $o(f)=\text{lcm}(o(f_1),o(f_2),o(f_3))$? How do you go about proving that?
I also have a second part that wants to know the order of a product of $k$-cycles;
$(1,6,4,9)(2,7,11)(3,5,8)(10,12)$? I got 12 because $\text{lcm}(4,3,3,2)$ is that true?
AI: Disjoint cycles commute, so, if $f_1,\dots f_m$ are disjoint, then $$(f_1f_2\dots f_m)^k=f_1^kf_2^k\dots f_m^k.$$ Since the cycles are disjoint, this product is $1$ if and only if all $f_i^k$ are $1$, so $k$ has to be a multiple of all of $o(f_1),\dots o(f_m)$. Then, the rank would be the smallest of those natural numbers, so it has to be their least common multiple.
In your first example, the rank is $2$, and the least common multiple of $2,2$ and $2$ is $2$. In your second example, $12$ is indeed the correct answer. |
H: Is there a closed form for...
I was working on an analysis question, and was wondering if there's a closed form for $\sum_{i=0}^{log(n)}(1/2^i)log(i)$? Unless I have made a mistake, I am trying to show that $n\sum_{i=0}^{log(n)}(1/2^i)log(i) \in \Theta(n)$.
Thanks!
AI: You are interested in the sum
$$S(n) = \sum_{i=1}^n \dfrac{\log(i)}{2^i}$$
Note that we have $\log(i) < i$ and hence we have
$$S(n) < \sum_{i=2}^n \dfrac{i}{2^i} = \dfrac32 - \dfrac{n+2}{2^n} < \dfrac32$$ |
H: What is the proof to show that $3^n$ divides $(3n)!$?
Working on a number theory question and I am quite stuck. How would you prove that $(3n)!/3^n \in \mathbb{N}$ with $\mathbb{N}$ being the natural numbers?
AI: HINT: How many multiples of $3$ are there in the set $\{1,2,\ldots,3n\}$? |
H: What does multiplying an inverse of a quotient ring by a ring mean?
I am attempting to prove that a ring $R$ is the intersection of all $R_p$, where $R_p=S^{-1}R$ for S=R\P and the intersection is over all prime ideals P of R. The trouble right now is that I don't quite understand the statement. I understand that S is a quotient ring, but what does it mean to take $S^{-1}$? And should the product $R_p=S^{-1}R$ be interpreted as a direct product with ordered pairs, or differently?
AI: No, $R\color{Red}{/}P$ is a quotient ring. $S=R\color{Red}{\setminus}P=\{x|x\in R\wedge x\not\in P\}$ is the complement of $P$ in $R$.
If $P$ is a prime ideal of $R$ then $S$ is multiplicative: if $s_1,s_2\in S$ then $s_1s_2\in S$. This is important, and you should prove it yourself as a warm-up exercise.
And if $S$ is a multiplicative set, then $S^{-1}R$ is the localization of $R$ with respect to $S$. In the case that $R$ is a domain, it has a fraction field $F$, and $S^{-1}R$ is the smallest subring of $F$ containing $R$ and all inverses of elements in $S$. For example if $S=\{s,s^2,s^3\cdots,\}$ then $S^{-1}R=R[s^{-1}]$. In this situation the localizations $R_P=(R\setminus P)^{-1}R$ all lie inside $F={\rm Frac}(R)$ and so you can take their intersection. I am not familiar with what the intersection means if $R$ is not a domain. |
H: Metric Space- open sets
$\qquad\mathit{(i)}\,$ We know that $\sin:\Bbb R\to\Bbb R$ is continuous. Show that, if $\,U=\Bbb R$, then $U$ is open, but $\sin U$ is not.
$\qquad\mathit{(ii)}\,$ We define a function $f:\Bbb R\to\Bbb R$ as follows. If $x\in\Bbb R$, set $\langle x\rangle=x-[x]$ and write $$\langle x\rangle=.x_1x_2x_3\ldots$$ as a decimal, choosing the terminating form in case of ambiguity. If $x_{2n+1}=0$ for all sufficiently large $n$, let $N$ be the least integer such that $x_{2n+1}=0$ for all $n\ge N$, and set $$f(x)=(-1)^N\sum_{j=1}^\infty x_{2N+2j}10^{N-j}.$$ We set $f(x)=0$ otherwise.
$\qquad$ Show that if $U$ is a non-empty open set, $f(U)=\Bbb R$ and so $f(U)$ is open. Show that $f$ is not continuous.
Note: consider $\Bbb R$ with the ordinary Euclidean metric.
I've done the first part, it's the second part where I'm badly stuck, I tried showing $f(U)=R$ but I seem to get reals that $f(x)$ can't get.
Any help will be appreciated.
AI: For any $x \in \mathbb{R}$, $-1 \leq \sin(x) \leq 1$, so the image of $\sin$ over the reals is exactly $[-1,1]$ (since equality is made at many points, like $\frac{\pi}{2}$ and $-\frac{\pi}{2}$). The closed interval is not open because the points $-1$ and $+1$ are not interior points. |
H: Is $n \choose k$ defined when $k < 0$? What about $n < k$?
I know that ${n \choose 0} = 1$, and this makes sense to me based on my understanding of combinatorics.
But what about ${n \choose -1}$? My instinct is that this is undefined, since it is equivalent to $\frac{n!}{k!(n - k)!}$, which when $k = -1$ equals $\frac{n!}{(-1)!(n + 1)!}$, and factorials are only defined for non-negative integers.
Similarly, ${5 \choose 6} = \frac{5!}{6!(5 - 6)!} = \frac{5!}{6!(-1)!}$ should also be undefined. However, intuitively it seems to me that "5 choose 6" could also equal $0$, since there are no combinations of 6 objects to be made from a set of only 5 objects.
So, is $n \choose k$ defined when $k < 0$? What about $n < k$? And in either case, what's the reasoning?
AI: It is customary to define $\binom{n}k=0$ for $0\le n<k$. If you restrict $n$ and $k$ to the non-negative integers and think of $\binom{n}k$ as the number of $k$-element subsets of an $n$-element set, then $\binom{n}k$ should certainly be $0$ when $n<k$. If instead you use the more general algebraic definition of the binomial coefficient,
$$\binom{x}k=\frac{x^{\underline k}}{k!}=\frac{x(x-1)(x-2)\ldots(x-k+1)}{k!}$$
for real $x$ and non-negative integer $k$, you automatically get $\binom{n}k=0$ when $n$ is a non-negative integer less than $k$. (Here $x^{\underline k}$ is a falling factorial.)
Not everyone defines $\binom{n}k$ for negative integers $k$, but when it is defined for them, it’s defined to be $0$. This ensures that the Pascal triangle identity,
$$\binom{n}k=\binom{n-1}{k-1}+\binom{n-1}k\;,$$
holds even for $k=0$ (and for integers $k<0$ as well, of course); this is convenient in some induction proofs, for instance. |
H: Find $\lim_{(x,y) \to (0,0)}\frac{\sin(x - y)}{\sqrt{x} + \sqrt{y}}$
I'm trying to find the $\lim_{(x,y) \to (0,0)}$ $\dfrac{\sin(x - y)}{\sqrt x + \sqrt y}$ and I know the answer is $0$ (used WolframAlpha for that), but I need to understand why.
AI: Note that for $x,y>0$, we have
$$\dfrac{\sin(x-y)}{\sqrt{x} + \sqrt{y}} = \dfrac{\sin(x-y)}{x-y} \cdot (\sqrt{x} - \sqrt{y})$$
Can you now conclude what you want? |
H: Real Analysis: Continuity of a Composition Function
Suppose $f$ and $g$ are functions such that $g$ is continuous at $a$, and $f$ is continuous at $g(a)$. Show the composition $f(g(x))$ is continuous at $a$.
My idea: Can I go straight from definition and take $\delta=\min\{\delta_1,\delta_2\}$, where $\delta_1$ is used for the continuity of $g$ at $a$ and $\delta_2$ is used for f being continuous at $g(a)$. In my proof I just treat $g(a)$ as a point when referring to the composition. So it goes like this:
Proof: Given $\epsilon>0$, take $\delta=\min\{\delta_1,\delta_2\}$. Then $0<|x-g(a)|<\delta$ which implies $|f(g(x))-f(g(a))|<\epsilon$.
AI: Since $f$ is continuous at $g(a)$, our definition of continuity tells us that for all $\varepsilon > 0$ there is some $\delta_1$ such that $$|g(x) - g(a)| < \delta_1\implies|f(g(x))-f(g(a))|<\varepsilon.$$
Also, since $g$ is continuous at $a$, there is some $\delta$ such that $$|x-a|<\delta \implies |g(x)-g(a)|<\delta_1.$$
I've taken $\varepsilon =\delta_1$ here. Now this tells us that for all $\varepsilon > 0$ there is some $\delta > 0$ (and a $\delta_1 > 0$) such that $$|x-a| < \delta\implies|g(x)-g(a)|<\delta_1\implies|f(g(x)) - f(g(a))|<\varepsilon,$$
which is what we wanted to show. |
H: Find the triple integral of $f(x,y,z)=(x^8+y^6+z^4)$sin$(z^3)+2y-3$.
I am trying to find the triple integral of $f(x,y,z)=(x^8+y^6+z^4)$sin$(z^3)+2y-3$ over the unit ball. I am completely stuck. I have tried converting to cylindrical and spherical coordinates and neither seems to help. Does anyone have any ideas?
AI: Since the first half of the function is proportional to $\sin(z^3)$ which is odd, and since for every choice of $x,y$ the interval of integration over $z$ is symmetric, the integral over the first part is equal to $0$. The same logic holds for $2y$. So now you just need to calculate:
$$\int_B -3 \ \ dx dy dz = -3V = -4\pi$$ |
H: Solving logarithms with different bases?
How would I go about getting an exact value for a question like: $\log_8 4$
I know that $8^{2/3} = 4$ but how would I get that from the question?
AI: The logarithm can be rewritten as
$$\log_8(4) = x \iff 8^x = 4$$
Now note that both $8$ and $4$ are powers of $2$ to get
$$(2^3)^x = 2^2$$
or alternatively,
$$2^{3x} = 2^2$$
So $3x = 2$. |
H: Finding $H'(1)$
Given $H(x)=F(x)G(x)$ Find $H'(1)$
Suppose:
$F(1)=2$
$F'(1)=3$
$G(1)=5$
$G'(1)=-2$
Then using the product rule I assumed $H(1)=11$ because:
$H(1)=((2)(-2))+((3)(5))=11$ using the product rule. However I am unsure as to how to derrive $H'(1)$ from $H(1)$ Would anyone mind giving me some insight on this?
AI: $$H'(x)=F(x)G'(x)+G(x)F'(x),$$ and so
$$H'(1)=F(1)G'(1)+G(1)F'(1)=2 \cdot (-2)+5 \cdot 3=11.$$
There is not much else to say here. |
H: Discarding lower cardinality subset doesn't change infinite cardinality?
This is a basic question about set theory.
I am of the belief that if $A$ is a set of some infinite cardinality and $B$ is a subset with lower cardinality, $A\setminus B$ has the same cardinality as $A$.
This is true, right? How is it proven?
Thanks in advance.
(Thoughts: the statement is equivalent to saying if $A$ is infinite and $B$ is a subset that doesn't biject onto $A$, then its complement does biject, right? This is actually kind of an unbelievable statement to me! How could the failure of $B$ to biject be enough to know that its complement bijects??)
AI: It’s true if you’re talking about well-ordered cardinals (assuming, of course, that $\lambda\ge\omega$). Let $\kappa$ and $\lambda$ be cardinals with $\kappa<\lambda$. Clearly $$f:\lambda\setminus\kappa\to\lambda:\xi\mapsto\xi$$ is an injection. The map
$$g:\lambda\to\lambda\setminus\kappa:\xi\mapsto\kappa+\xi\;,$$
where the addition is ordinal addition, is also an injection, so $|\lambda\setminus\kappa|=\lambda$ follows from the Schröder-Bernstein theorem.
In the absence of the axiom of choice you can have an amorphous set set $X$, an infinite set that is not the union of two disjoint infinite sets. Let $X_k=X\times\{k\}$ for $i\in\{0,1\}$, and let $Y=X_0\cup X_1$. Then $|X_0|<|Y|$, since there is no injection of $Y$ into $X$, but $$|Y\setminus X_0|=|X_1|=|X_0|\ne |Y|\;.$$
Added: In fact, as Asaf reminds me, the result is equivalent to the axiom of choice. Suppose that $X$ is a set that cannot be well-ordered, and let $\kappa$ be the Hartogs number of $X$: $\kappa$ is the smallest well-ordered cardinal that cannot be injected into $X$, and its existence is provable in $\mathsf{ZF}$. Let $Y=X\sqcup\kappa$. Then $|X|<|Y|$, but $Y\setminus X=\kappa$, and $|Y|>\kappa$, since there is no injection $Y\to\kappa$: if there were one, $X$ would be well-orderable. |
H: Structure of $(\mathbb{Z}/p^k\mathbb{Z})^\times$ other than via $p$-adics?
I think I heard somewhere long ago that $\mathbb{Z}/p^k\mathbb{Z}$'s unit group is cyclic if $p$ is odd and $C_2\times C_{2^{k-2}}$ if $p=2$. I remember trying to prove it and finding it surprisingly difficult.
This morning I think I realized you can prove this using the $p$-adic exponential and log maps. Proof sketch below. In a way, this result seems like the $p$-adics would be its natural home, because it's about $\mathbb{Z}$ mod $p^k$ for arbitrary $k$. However, it also seems like exactly the sort of completely classical result that Gauss should have a proof of somewhere. So, my question:
How is this usually proven? And, is there a classical, Gauss-typed or at least 19th-century-typed proof?
Proof sketch for $p$-adic proof: if $p$ is odd, then $\exp_p$ is a continuous group isomorphism from $(p\mathbb{Z}_p)^+$ to $(1+p\mathbb{Z}_p)^\times$. Since $p\mathbb{Z}\subset p\mathbb{Z}_p$ is cyclic and dense, this means that the $1$-units of $\mathbb{Z}_p$ have a dense cyclic subgroup. Reducing mod $p^k$ means that the $1$-units of $\mathbb{Z}/p^k\mathbb{Z}$ are cyclic, since the group is discrete so a dense subset is everything. Now I just have to show that some generating $1$-unit is a $(p-1)$th power in $\mathbb{Z}/p^k\mathbb{Z}$, and this I can do. If $p=2$, then I don't have all the details worked out, but $\log_p$ is $2$-to-$1$ from $1+2\mathbb{Z}_2$ to $2\mathbb{Z}_2$; and I think the preimage of $2\mathbb{Z}$ is dense, whereupon since it is $\{\pm 1\} \times$ cyclic, generated by a lift of $2$, the same argument works (without the shenanigans about the $1$-units, since every unit in $\mathbb{Z}/2^k\mathbb{Z}$ is a $1$-unit).
AI: This is due to Gauss. See Wikipedia and the references there. |
H: A non-UFD where we have different lengths of irreducible factorizations?
A unique factorization domain (UFD) is defined to be an integral domain where each element other than units and zero has a factorization into irreducibles, and if we have two such factorizations then they are of the same length (the sum of the powers of the irreducible factors) and the irreducibles involved are associated to each other.
My knowledge about algebra is extremely limited, but the only non-UFD integral domain I have seen is $\mathbb{Z}[\sqrt{-5}]$, where one shows $9=3\cdot 3=(2+\sqrt{-5})(2-\sqrt{-5})$ gives two different factorizations of $9$ into irreducibles with non-associated factors.
However in this case the two factorizations are of the same length.
So just curious, can we find integral domains where some elements have factorizations into irreducibles with different lengths?
Thanks!
AI: One standard example is the ring $\mathbb Q[X^2,X^3]$. This is a Noetherian domain and hence atomic (that is, every non-unit has a factorization into irreducibles). The elements $X^2$ and $X^3$ are irreducible and so $X^6$ has two factorizations of different length: $X^2 X^2 X^2 = X^6 = X^3 X^3$. |
H: Question About Lipschitz Maps and Measure Zero
I'm having difficulty with a problem involving measure.
Suppose $F : B^k \rightarrow B^n$ is a Lipschitz map from the unit ball in $\mathbb{R}^k$ to the unit ball in $\mathbb{R}^n$. If $k \lt n$, show that the image of $F$ has measure zero in $B^n$.
Starting from the fact that $F$ is a Lipschitz map, I figured out that for all $\epsilon \gt 0, \forall F(x), F(y) \in B^n$, if $F(x), F(y) \in B_\epsilon$ in $B^n$, then $x, y \in B_{\epsilon/C}$ in $B^k$ where $C$ is the Lipschitz constant.
From here I'm thinking that one can cover the $k$ unit ball with some $n$-balls and then transfer those $n$-balls to $B^n$ where they cover image $(F)$. Then reducing epsilon reduces the volume of the $n$-balls faster than their number increases so the volume goes to zero.
I would appreciate any help or tips. Thank you.
AI: Considering $B^k$ as a subset of $B^n$, you can extend $F$ to a map $G:B^n\to B^n$ such that $G(x_1,\ldots,x_n)=F(x_1,\ldots,x_k)$. Note that $G$ is Lipschitz with the same Lipschitz constant as $F$. The measure of $B^k$ in $B^n$ is $0$, so given $\varepsilon>0$ there is a sequence of balls $\{B(\mathbf{x}_i,r_i)\}_i$ in $B^n$ such that $B^k\subseteq \cup_i B(\mathbf{x}_i,r_i)$ and $\sum_i m(B(\mathbf{x}_i,r_i))<\varepsilon$.
Let $C>0$ be the Lipschitz constant of $F$, hence of $G$. Then $G(B(\mathbf{x}_i,r_i))$ has diameter at most $2Cr_i$, and therefore it fits inside a ball of radius $2Cr_i$. This implies that $m(G(B(\mathbf{x}_i,r_i))\leq (2C)^nm(B(\mathbf{x}_i,r_i)$. Since
$$F(B^k)=G(B^k)\subseteq G(\cup_iB(\mathbf{x}_i,r_i))=\cup_i G(B(\mathbf{x}_i,r_i)),$$
this implies that $m(F(B^k))<(2C)^n\varepsilon$. Since $C$ and $n$ are constants and $\varepsilon$ was arbitrary, this implies that $m(F(B^k))=0$.
Regarding your remarks, note that $d(F(x),F(y))<\varepsilon$ does not imply that $d(x,y)<\dfrac{\varepsilon}{C}$. (The converse holds.) For example, consider a constant map. |
H: Find the probability that the largest number shown by any throw is r
So I have a pretty hard dice question on a past exam that has thrown me a bit:
A dice is rolled a random number $n$ of times. Let $A_i$ be the event that $n=i$ and suppose that $P(A_i)=1/2^i$
a.) How would i find the probability that the largest number shown is $r$?
b.) How to find the probability that $n=2$, given that the sum of the scores is $5$ and the first throw showed $2$.
So for part a.) I have that $P(A_1)=1/2$, $P(A_2)=1/4$, $P(A_3)=1/8$, $P(A_4)=1/16$ $P(A_5)=1/32$, $P(A_6)=1/64$
Is this correct or have I made a huge error? I cannot find any similar style questions about throwing a die a random number $n$ times and am thus struggling to work out how i would solve a question like this.
Many thanks for your help
AI: An outline of a solution of the first problem has been given by Ross Millikan. We consider the second problem.
Let $T$ be the event $n=2$. Let $S$ be the event the sum was $5$ and the first throw was $2$. We want $\Pr(T|S)$. We have
$$\Pr(T|S)=\frac{\Pr(T\cap S)}{\Pr(S)}.$$
Now we want to calculate the two probabilities on the right.
For $\Pr(T\cap S)$, note that for this to happen, we must have $n=2$ (probability $\frac{1}{4}$). Given that $n=2$, the probability that the first toss is $2$ and the sum is $5$ is $\frac{1}{6^2}$. Thus $\Pr(T\cap S)=\frac{1}{144}$.
Next we find $\Pr(S)$. The event $S$ can happen in several ways, with (i) $n=2$; (ii) $n=3$; (iii) $n=4$.
We have already calculated the probability of (i). For (ii), we must have $3$ tosses (probability $\frac{1}{8}$), then a $2$ (probability $\frac{1}{6}$), then $1$ and $2$ or $2$ and $1$ (probability $\frac{2}{36}$), for a total of $\frac{1}{8}\cdot \frac{1}{6}\cdot\frac{2}{36}$.
For (iii), we must have $4$ tosses, and the pattern $2,1,1,1$. The probability is $\frac{1}{16}\cdot \left(\frac{1}{6}\right)^4$.
To calculate $\Pr(S)$, add together the probabilities (i), (ii), (iii). Now you have all the ingredients for calculating $\Pr(T|S)$. |
H: Prove that $\phi(n) \geq \sqrt{n}/2$
So I'm trying to prove the following two inequalities: $$\frac{\sqrt{n}}{2} \leq \phi(n) \leq n.$$ The upper bound we get from simply noting that $\phi(n) = n \prod_{p | n}\left( 1 - \frac{1}{p}\right)$ and the fact that $(1 - \frac{1}{p}) \leq 1$. But how can we get the lower bound? I tried using the same expression for $\phi(n)$ but it seems to not really give me the inequality. Can you help?
AI: We have
$$
\frac{\phi(n)^2}{n}=
\prod_{p|n \ \text{prime}} \frac{(p^{a_p-1}(p-1))^2}{p^{a_p}} =
\prod_{p|n \ \text{prime}} p^{a_p-2} (p-1)^2 \geq
\prod_{p|n \ \text{prime}} \frac{(p-1)^2}{p}
$$
Now for $p\geq 3$ we have $p^2-3p+1=1+p(p-3)\geq 0$ so
$p^2-2p+1 \geq p$ and hence $\frac{(p-1)^2}{p} \geq 1$.
So
$$
\frac{\phi(n)^2}{n} \geq \prod_{p|n \ p=2} \frac{(p-1)^2}{p}
$$
When $2$ does not divide $n$, this gives a lower bound of
$1$. When $2$ divides $n$, this gives a lower bound of $\frac{1}{2}$.
In any case, we always have $\frac{\phi(n)^2}{n} \geq \frac{1}{2}$. We deduce
the stronger inequality
$$
\phi(n) \geq \sqrt{\frac{n}{2}}
$$ |
H: Prove that $\lim_{n\to\infty}\frac1{n}\int_0^{n}xf(x)dx=0$.
Let $f$ be a continuous, nonnegative, real-valued function and $$\int_0^{\infty}f(x)dx<\infty.$$ Prove that $$\lim_{n\to\infty}\frac1{n}\int_0^{n}xf(x)dx=0.$$
A start: If $\lim\limits_{n\to\infty}\int_0^{n}xf(x)dx$ is finite, then it's obvious. Otherwise, perform L'Hopital's rule, we get $\lim nf(n)$, which we want to show is $0$.
AI: Let $u_n=\frac1{n}\int_0^{n}xf(x)dx \geq 0$. Let $\varepsilon > 0$.
We need to show that $u_n \leq \varepsilon$ for large enough $n$.
By hypotesis, there is a $a$ such that
$\int_a^{\infty}f(x)dx \leq \frac{\varepsilon}{2}$.
Let $n$ be an integer such that
$n\geq {\sf max}(a,\frac{2\int_0^a xf(x)dx}{\varepsilon})$. Then
$$
u_n=\int_0^a \frac{xf(x)}{n}dx
+ \int_a^n \frac{xf(x)}{n}dx \leq
\frac{\varepsilon}{2}+ \int_a^n \frac{nf(x)}{n}dx
\leq \frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon
$$
and we are done. |
H: Finding Binet's formula using generating functions
$\newcommand{\fib}{\operatorname{fib}}$
I am trying to solve the Fibonacci recurrence using generating functions, but I've run into a bit of a snag. Here's what I've done so far:
$$\begin{align}
&\fib(0) = 0\\
&\fib(1) = 2\\
&\fib(n) = \fib(n-1) + \fib(n-2)
\end{align}
$$
$$\begin{align}
A(x) &= \sum_{i=0}^\infty a_ix^i=\sum_{i=1}^{\infty}a_ix^i\\
&=x+\sum_{i=2}^\infty a_ix^i\\
&=x+\sum_{i=2}^\infty (a_{i-1} + a_{i-2})x^i\\
&=x+x\sum_{i=1}^\infty a_ix^i + x^2\sum_{i=0}^{\infty}a_ix^i\\
\\
A(x)&=x+xA(x)+x^2A(x)\\
\\
&=\frac{-x}{x^2+x-1}
\end{align}
$$
I then need to rearrange $A(x)$ back into the form $A(x)=\sum_{i=0}^\infty f(i)x^i$ to obtain $f(i)$, which should be Binet's closed-form equation for the $i$'th Fibonnacci number, but I don't know any techniques to do so. Is there a method for rearranging arbitrary rationals into infinite sums?
AI: The general technique is to use partial fractions. You may not have had much experience decomposing things into slightly messy partial fractions, so I’ll go into a bit of detail.
I would write it $$A(x)=\frac{x}{1-x-x^2}\;.$$ Now factor the denominator: $1-x-x^2=(1-\alpha x)(1-\beta x)$; equating coefficients, we see that $\alpha\beta=-1$ and $\alpha+\beta=1$, so $\beta=1-\alpha$, and $\alpha(1-\alpha)=-1$, i.e., $\alpha^2-\alpha-1=0$. The quadratic formula then gives us
$$\alpha=\frac{1\pm\sqrt5}2\;,$$
so we can set $\alpha=\frac12(1+\sqrt5)$ and $\beta=\frac12(1-\sqrt5)$. Now a partial fraction decomposition of $A(x)$ is possible:
$$A(x)=\frac{x}{1-x-x^2}=\frac{A}{1-\alpha x}+\frac{B}{1-\beta x}\;,$$
so $A(1-\beta x)+B(1-\alpha x)=x$. Thus, $A+B=0$, and $-\beta A-\alpha B=1$, and you can solve for $A$ and $B$. Finally, you can easily convert each of these rational functions to an infinite series:
$$\frac{A}{1-\alpha x}=A\sum_{n\ge 0}(\alpha x)^n=A\sum_{n\ge 0}\alpha^nx^n\;,$$
for instance. |
H: Find $a_{2012}-3a_{2010}/3 a_{2011}$ where the sequence $a_n$ is determined by roots of a quadratic equation
If $\alpha$ and $\beta$ are the roots of $x^2-9x-3=0$, $a_n=\alpha^n-\beta^n$ and $b_n=\alpha^n+\beta^n$, then find the value of $\dfrac{a_{2012}-3a_{2010}}{3 a_{2011}}$ and $\dfrac{b_{2012}-3b_{2010}}{3b_{2011}}$.
Please help me solve this problem.
I tried finding relation between a1, a2, a3 and b1, b2, b3. But I could go no further.
AI: We have$$a_{2012}-3a_{2010}=(\alpha^{2012}-\beta^{2012})-3(\alpha^{2010}-\beta^{2010})=$$
$$=\alpha^{2010}(\alpha^2-3)-\beta^{2010}(\beta^2-3)$$
But $\alpha^2-3=9\alpha$ and $\beta^2-3=9\beta$, since each of them satisfy the quadratic equation. Hence we may simplify as $$\alpha^{2010}(9\alpha) - \beta^{2010}(9\beta)=9(\alpha^{2011}-\beta^{2011})=9a_{2011}$$
Dividing by $3a_{2011}$ gives just 3. The other problem is similar. |
H: $\log(n)$ is what power of $n$?
Sorry about asking such an elementary question, but I have been wondering about this exact definition for a while. What power of $n$ is $\log(n)$. I know that it is $n^\epsilon$ for a very small $\epsilon$, but what value is $\epsilon$ exactly?
AI: No: If $\epsilon$ is any positive number, then $n^{\epsilon}$ grows faster than $\log{n}$. This can be made precise in the statement
$$\lim_{n \to \infty} \frac{\log{n}}{n^{\epsilon}} = 0$$
for all $\epsilon > 0$. To prove this, just note that by L'Hospital's rule,
$$\lim_{n \to \infty} \frac{\log{n}}{n^{\epsilon}} = \lim_{n \to \infty} \frac{\frac{1}{n}}{\epsilon n^{\epsilon - 1}} = \frac{1}{\epsilon} \lim_{n \to \infty} \frac{1}{n^{\epsilon}} = 0$$ |
H: How to prove some statements about divisibility and the $\gcd$ function
Struggling with some number theory homework. Could use a helping hand. The two statements are as follows
$\gcd(c, ab) \mid \gcd(c,a)\gcd(c,b)$
If $c \mid ab$ and $\gcd(a,b)=1$, then $c=\gcd(a,c)\gcd(b,c)$
I'm supposed to figure out a proof for each, but cannot even seem to get them started.
AI: HINT
Use the fundamental theorem of arithmetic (FTA). These hints will guide you to a solution to 2.
Write out $a$ and $b$ with the FTA. Do they share any primes?
In general, what is $\gcd(x,y)$ in terms of the factorizations of $x$ and $y$?
Put these together to conclude the answer to 2. |
H: Expected Value and Summation
The question is suppose $X$ ~ $Geometric(\frac{1}{2})$ and I have computed $P(X > 1) = \frac{1}{2}$ and $Var(X) = 2$ where $E[X] = 2$ and $E[X^2] = 6$. How can I verify that $E[X] = \sum_{n=0}^{\infty}P(X > n)$. I know that this formula holds in general but which theorem relates all of these for this case?
AI: We have $\Pr(X\gt n)=\left(\frac{1}{2}\right)^n$. Add up, $0$ to $\infty$.
Why is $\Pr(X\gt n)=\left(\frac{1}{2}\right)^n$? We have $X\gt n$ if and only if we start with $n$ consecutive "failures." |
H: Why rows should be proportional for independence?
Suppose that $X$ and $Y$ have discrete distributions; that $X$ can take the values $1,2,\dots,r$; that $Y$ can take the values $1,2,\dots,s$; and that $$Pr(X=i \text{ and } Y=j)=p_{ij}\text{ for }i=1,2,\dots,r \text{ and }j=1,2,\dots,s.$$Then for $1,2,\dots,r$, let $$Pr(X=i)=\sum_{j=1}^sp_{ij}=p_{i+}.$$Also, for $1,2,\dots,s$, let $$Pr(Y=j)=\sum_{i=1}^rp_{ij}=p_{+j}.$$Therefore, $X$ and $Y$ will be independent if and only if the following relationship is satisfied for all values of $i$ and $j$: $$p_{ij}=p_{i+}p_{+j}.$$It should be noted that $X$ and $Y$ will be independent if and only if for matrix $$P=\begin{pmatrix}
p_{11} & p_{12} & p_{13} & \cdots & p_{1s} \\
p_{21} & p_{22} & p_{23} & \cdots & p_{2s} \\
\vdots & \vdots& \vdots & \ddots & \vdots \\
p_{r1} & p_{r2} & p_{r3} & \cdots & p_{rs}
\end{pmatrix},$$ rows are proportional to one another, or equivalently, columns are proportional to one another.
I can't understand why rows(and columns) should be proportional to one another to assure independence.
AI: It's quite easy to understand if you look at how the matrix $P$ is formed in the light of the information $p_{ij} = p_i p_j$.
Suppose the probabilities for values of $X$ and $Y$ are arranged into two column matrices:
$$
P_X = [p_x(1)~~p_x(2)~~p_x(3)~~...~~p_x(r)]^T \\
P_Y = [p_y(1)~~p_y(2)~~p_y(3)~~...~~p_y(s)]^T
$$
Then the matrix $P$ is obtained by:
$$
P = P_X \cdot P_Y^T
$$
Here, $P_X$ is a column vector and $P_Y^T$ is a row vector. This means that $\forall i, j \in \{1, 2, 3, ...r\}$, the $i$th and $j$th rows of $P$ are precisely in the proportion $p_x(i):p_x(j)$. Similarly, $\forall i, j \in\{1, 2, 3, ..., s\}$, the $i$th and $j$th columns of $P$ are in the proportion $p_y(i):p_y(j)$.
Hope this throws some light on the issue. |
H: Fixed field of automorphism $t\mapsto t+1$ of $k(t)$
I'm working on the following problem:
Determine the fixed field of the automorphism $t\mapsto t+1$ of $k(t)$. (Ex 7, section 14.1, Abstract Algebra by Dummit & Foote).
Here is my attempt of a solution:
Let be $\sigma$ the automorphism determined by $t\mapsto t+1$ and fixing $k$. An arbitrary element of $\alpha(t)\in k(t)$ has the form $\alpha(t)=\dfrac{p(t)}{q(t)}$ with $p,q\in k\left[ t\right]$. We can take $p,q$ such that $(p,q)=1$ and $q$ monic. If $\alpha$ is a fixed element by $\sigma$, then we have $\alpha(t)=\alpha(t+1)$, so that $\dfrac{p(t)}{q(t)}=\dfrac{p(t+1)}{q(t+1)}$ and finally $p(t)q(t+1)=p(t+1)q(t)$. As $(p,q)=1$, $q(t)$ must divide $q(t+1)$. By the same reason, $q(t)$ must divide $q(t+1)$, and as we took $q$ monic, we conclude that $q(t+1)=q(t)$, ie, $\sigma(q(t))=q(t)$. Now we note that, as $\sigma$ fixes both $\alpha(t)$ and $q(t)$, it must also fix $p(t)$.
Now I'm getting stuck. If the characteristic of the field $k$ is $0$, I can say both $p$ and $q$ have an infinite number of roots (if they're not constant polynomials), so both have to be constant. But if the characteristic is non-zero, I can't conclude in the same way. What can I say about the polynomials?
Thanks in advance!
AI: While you can't come to a contradiction when $\operatorname{char}k=p$, you can still conclude quite a bit using the same reasoning.
Suppose that $p(t)$ has a root $\alpha\in\overline{k}$. Then you've shown that it must have roots $\alpha+1, \alpha+2,\ldots$. So you can conclude that $p(t)$ is divisible by the product of $(x-\alpha+n)$ for $n=0,1,\ldots p-1$. Can you compute this product explicitly, and conclude that $p(t)$ must have a very specific form? |
H: Question about distinctness of coset given $H \leq K \leq G$
This question is regarding a detail I am unsure about in a problem in the Dummit and Foote abstract algebra textbook that I am currently working on. The problem itself is that if $H \leq K \leq G$ I need to show $|G:H| = |G:K|\cdot|K:H|$. Within this problem I am not sure about the uniqueness of how the cosets are counted in the case of both sides being finite.
If $H \leq K \leq G$ as defined and I let $x_i$ for $i = 1, \dots, |G:K|$ be the coset representatives of $K$ in $G$ and $y_j$ for $j=1,\dots,|K:H|$ be the coset representatives of $H$ in $K$, why is it that each pair of cosets, $x_iy_j$ is a unique coset representative of $H$ in $G$?
Suppose $x_{i_0}y_{j_0}$ and $x_{i_1}y_{j_1}$ are two such cosets representatives where $x_{i_0}y_{j_0} = x_{i_1}y_{j_1}$. What makes it necesarry for $i_0=i_1$ and $j_0=j_1$?
AI: Hint: argue $~~x_{i_0}y_{i_0}H=x_{i_1}y_{i_1}H~~~\Rightarrow~~~ x_{i_0}K=x_{i_1}K~~~\Rightarrow~~~ x_{i_0}=x_{i_1}~~~\Rightarrow~~\cdots$
In the language of set partitions, this should be "obvious." Consider a map with "nations" and within each nation are "territories." One can speak of all the territories in a single nation or all the territories in the world. Let $A$ and $B$ be two territories. If $A=B$ in the context of the whole map, then they both inhabit the same nation, and within that nation they are both the same territory.
This metaphor is applicable to $H\le K\le G$, where $G$ is the whole map, $G/K$ is the set of nations and $G/H$ is the entire map's set of territories. The systems of representatives basically amount to labels associated with each nation or territory. Or, in direct algebraic language:
$$G=\bigsqcup_i x_iK=\bigsqcup_i x_i\left(\bigsqcup_j y_jH\right)=\bigsqcup_i\bigsqcup_jx_i(y_jH)=\bigsqcup_{i,j}x_iy_jH.$$
Check that bijections (e.g. left multiplications) preserve disjoint unions so the above is valid. |
H: Complex Analysis Problem
Find the image under the map f(z) = $ e^{iz} $ of the square
S = {z $\in$ $C$ | 0 $\leq$ Re(z) $ \leq $ 1, 1 $\leq$ Im(z) $\leq$ 2}
and label the corresponding portions of the boundary.
How would I go about doing this?
AI: The paths are relatively straightforward ; the bottom of the square is given by:
$z(t)=t ; 0\leq t\leq1 $ Then the image of the bottom of the square is $e^{it}=cost+isint$.
Now, parametrize the other sides of the rectangle and do the same process. |
H: Problem on Lagrange multipliers
This problem has two parts:
$a)$ Let $k>0$, find the minimum of the function $f(x,y)=x+y$ over the set S=$\{(x,y) \in \mathbb R^2_{> 0}:xy=k\}$. $b)$ Prove that for every $(x,y) \in \mathbb R^2_{> 0}$ the inequality $\frac{x+y} {2}\geq \sqrt{xy}$ holds.$$$$
I want to find the minimum of $f$ restricted to the level set $g(x,y)=k$. The function $f$ and $g$ are differentiable and $\nabla g(x,y)=(y,x)\neq0$ for every $(x,y) \in \mathbb R^2_{> 0}$, then I can apply lagrange multipliers to find the critical points of f restricted to that region. By the equation $\nabla f(x,y)=\lambda\nabla g(xy)$ I get that $(1,1)=\lambda (y,x)$, which means $x=y$. By the conditions of the problem, it is required that the points of the form $(x,x)$ live in the level set. So, the only critical point is $(\sqrt{k},\sqrt{k})$. $$$$My question is, how can I prove that at $(\sqrt{k},\sqrt{k})$, $f|_S$ attains a minimum?$$$$For part $b)$, let's assume I could prove that $f(\sqrt{k},\sqrt{k})$ is a minimum of $f|_S$.Then, for all $(x,y) \in S$ it immediately follows that $x+y\ge \sqrt{k}+\sqrt{k}=2\sqrt{k}$, so dividing both sides by $2$, I get $\frac{x+y} {2}\ge\sqrt{k}=\sqrt{xy}$. I don't see why this inequality would also hold for $(x,y) \in \mathbb R^2_{>0}\setminus S$
AI: For (b), suppose that $x,y\in\Bbb R_{>0}$, and let $k=xy$; from (a) you know that $x+y\ge 2\sqrt{k}$ and hence that $\frac{x+y}2\ge\sqrt{xy}$.
For (a) you don’t really need anything as fancy as Lagrange multipliers. You have $y=\frac{k}x$, so $x+y=x+\frac{k}x$; its derivative with respect to $x$ is $1-\frac{k}{x^2}$, which is increasing on $x>0$ and equal to $0$ at $x=\sqrt{k}$, so you have a minimum. |
H: extending a limit to an integral case
For a finite set of postitive real numbers $\{g_i\}$, it's clear that
$$
\lim_{\lambda->\infty} \frac{1}{\lambda} \ln\left(\sum_i e^{\lambda g_i} \right) = \max \{g_i\},
$$
since the argument of the logarithm becomes dominated by the largest of the $e^{\lambda g_i}$ terms. In the case where there are $n$ $g_i$'s that all share the same maximum value, the limit becomes $\frac{1}{\lambda}(\ln e^{\lambda g_\text{max}} + \ln n)$, and the $\ln n$ term vanishes.
Intuitively, this should generalise to the case where the sum becomes an integral, yielding
$$
\lim_{\lambda->\infty} \frac{1}{\lambda} \ln\left(\int_{0}^{\infty} e^{\lambda f(x)}dx \right) = \max_x f(x),
$$
where $f(x) \ge 0$ for all $x\ge 0$.
However, I don't know how to show this formally, and I'm uncertain what properties $f(x)$ must have in order for this limit to hold. For example, does it still work if the second derivative vanishes at the maximum point?
If anyone has any insight into how to prove this formula, and how to determine the class of functions for which it holds, it would be much appreciated.
AI: Actually the result makes no sense for integrals on $[0,\infty)$, thanks to @AntonioVargas for having spotted that. Rather, the following holds:
Let $B$ denote a Borel set of finite Lebesgue measure $m$, and $f$ a measurable function defined on $B$ and bounded from above by $S=\sup\{f(x)\mid x\in B\}$. For every positive $\lambda$, consider
$$
L(\lambda)=\frac{1}{\lambda} \ln\left(\int_B\mathrm e^{\lambda f(x)}\mathrm dx \right).
$$
Then $f(x)\leqslant S$ for every $x$ in $B$ hence
$$
L(\lambda)\leqslant\frac{1}{\lambda} \ln\left(\mathrm e^{\lambda S}m \right) =S+\frac{1}{\lambda}\log m,
$$
In particular,
$$
\limsup_{\lambda\to+\infty}L(\lambda)\leqslant S.
$$
In the other direction, assume that:
$(\dagger)$ For every $\varepsilon\gt0$, the set $\{x\in B\mid f(x)\geqslant S-\varepsilon\}$ has positive measure.
Then, calling $B_\varepsilon$ this set and $m_\varepsilon$ its measure, one sees that
$$
\int_{B}\mathrm e^{\lambda f(x)}\mathrm dx\geqslant\int_{B_\varepsilon}\mathrm e^{\lambda f(x)}\mathrm dx\geqslant\int_{B_\varepsilon}\mathrm e^{\lambda (S-\varepsilon)}\mathrm dx\geqslant m_\varepsilon\mathrm e^{\lambda(S-\varepsilon)},
$$
hence
$$
L(\lambda)\geqslant S-\varepsilon+\frac1{\lambda}\ln m_\varepsilon.
$$
In particular,
$$
\liminf_{\lambda\to+\infty}L(\lambda)\geqslant S-\varepsilon.
$$
This proves that $\lim\limits_{\lambda\to+\infty}L(\lambda)=S$.
Finally, note that if $f$ is continuous, then $(\dagger)$ holds. |
H: Poincaré inequality for $W_0^{1,\infty}$
In the book A first course in Sobolev spaces by Leoni, the following Poincaré inequality for $W_0^{1,p}(\Omega)$ is stated:
Suppose $\Omega\subset \mathbb{R}^n$ has finite width (lies between two parallel hyperplanes) and $p\in [1,\infty)$. Then for all $u\in W^{1,p}_0(\Omega)$,
$$\|u\|_{L^p} \leq C \|\nabla u\|_{L^p}$$
My question is: Does the Poincaré inequality above still hold for $p=\infty$?
If yes, how to prove it? And if no, what is the counterexample?
Thank you.
AI: By the fundamental theorem of calculus we have that, if $u\in C_c^\infty(\mathbb{R}^n)$ and $x_0$ is such that $u(x_0)=0$ then
$$
u(x)=\int_0^1 \nabla u(tx+(1-t)x_0)\cdot (x-x_0)dt
$$
from which we get
$$
|u(x)|\leq \| \nabla u \|_\infty |x-x_0|
$$
Now if $\Omega$ has finite width and $u\in C_c^\infty(\Omega)$, then for every $x\in \Omega$ there is an $x_0 \notin \Omega$ such that $|x-x_0| \leq D$ (where $D$ is the distance between the two parallel hyperplanes bounding $\Omega$). We conclude that
$$
\| u\|_{\infty} \leq D \| \nabla u\|_{\infty}.
$$ |
H: What is the Probability that the Lowest Card out of 4 Cards is X?
If I have four different value cards, what is the probability that the lowest card (ace, lowest -> king, highest) is some value X?
Here is what I have so far:
I know that the lowest value card cannot be a king, queen, or jack, as there must exist a value that is 3 above. And I can guess that the probability of an ace will be the highest and a 10 will be the lowest. Can someone please give me some hints on how I should approach this?
AI: HINT: Since you’re looking at sets of $4$ cards of different denominations, you might as well just look at sets of denominations: in effect you’re choosing $4$ cards from one suit and asking for the probability that the lowest denomination chosen is $x$. There are $\binom{13}4$ possible sets of $4$ denominations. How many have an $x$ as lowest denomination? You must choose the other $3$ from how big a set of values? |
H: are all dynamical systems described by differential equations?
we defined in lecture a dynamical System as a one-parameter family of maps $\phi^t:M\rightarrow M$ such that $\phi^{t+s}=\phi^t\circ\phi^s$ and $\phi^0=Id$, where $M$ is some (smooth) manifold and $s,t\in (a,b)\subset\mathbb{R}$.
Of course, if we consider some vector field $V:M\rightarrow TM$, then the flow of that vector field around some point $x_0\in M$ is a (local) dynamical system.
Now I'm wondering if all dynamical systems can be described that way. Can we find for all dynamical systems $\phi^t$ a vector field $V$, s.t. $\phi^t$ is the flow of $V$? Maybe you know some argument.
Regards
AI: Assuming differentiability, define a vector field $V(t)$ by the differential of $\phi^t$, i.e. $V(t)=D\phi^t$. The ($t$-dependent) flow of $V(t)$ is what you're looking for. |
H: Boolean Algebra A + AB = A
Hi I have a question about the following algebra rule
A + AB = A
My textbook explains this as follows A + AB = A This rule can be proved as such:
Step 1:
Dustributive Law:
A + AB = A*1 = A(1+B) Huh...? Where do they get the one(1) from?
Step 2:
1 + B = 1 {Question:1 + B = B right? so how is this posible(1 + B = 1)?}
Step 3:
A + 1 = A
Thus A + AB = A
If anyone can clarify this for me it would be greatly appreciated
AI: Well, note that $A$ and $AB$ have a common factor--namely $A$--and factoring it out, we have $$A+AB=A(1+B)=A(1)=A$$ |
H: Is $(\mathbb R,\tau)$ compact?
Let us consider $\tau=\{G\subset \mathbb R: \mathbb R\setminus G$ is compact set in ($\mathbb R,\tau_u)\}$, where $\tau_u$ denotes the usual topology on $\mathbb R$. Then $\tau$ is a topology coarser than $\tau_u$. Since $(\mathbb R, \tau_u)$ is connected, so $(\mathbb R, \tau)$ is connected. I tried to prove or disprove that $(\mathbb R, \tau_u)$ is compact. But couldn't get any clue. Any hint will be of great help.
AI: Suppose that $\mathscr{U}$ is a $\tau$-open cover of $\Bbb R$. Pick any $U_0\in\mathscr{U}$, and let $K=\Bbb R\setminus U_0$; $K$ is $\tau_u$-compact, and $\mathscr{U}\setminus\{U_0\}$ is a $\tau$-open cover of $K$. And $\tau\subseteq\tau_u$, so $\mathscr{U}\setminus\{U_0\}$ is also a $\tau_u$-open cover of $K$. |
H: Question related to the spectrum of a bounded operator
If $A$ is a bounded linear operator on a Banach space $X$ and $\lambda\in \sigma(A)$, is it true that for all $\epsilon>0$, there is $ x\in X$ and $||x||=1$ such that
$$ ||(A-\lambda I) x|| <\epsilon \ ?$$
If not, is it true if we consider $X$ a Hilbert space and $A$ self adjoint? Thanks.
AI: The shift operator $Se_n = e_{n+1}$, where $\{e_n\}$ is an orthonormal basis on a separable infinite-dimensional Hilbert space is an isometry, despite not being invertible. As for your second question, the residual spectrum of a normal operator is empty (Exercise 79 of Halmos's A Hilbert Space Problem Book). This would imply that every element of the spectrum is either in the point spectrum (an eigenvalue) or in the approximate point spectrum (the condition you want).
Edit: Citation found. |
H: Proof or disproof that $\frac{\pi^{8}}{3150}=\sum\limits_{k=0}^{\infty} \int\limits_{0}^{\infty}\frac{t^{3} e^{-4(k + 1)t}}{1-e^{-4(k + 1)t}}\,dt$
I need a proof or disproof of- $$\frac{\pi^{8}}{3150}=\sum_{k=0}^{\infty} \int_{0}^{\infty}\frac{t^{3} e^{-4(k + 1)t}}{1-e^{-4(k + 1)t}}\,dt$$
AI: For every $\alpha\gt0$,
$$
\int^{\infty}_{0}\frac{t^3\mathrm e^{-\alpha t}}{1-\mathrm e^{-\alpha t}}\mathrm dt=\frac1{\alpha^4}\int^{\infty}_{0}\frac{t^3 \mathrm e^{-t}}{1-\mathrm e^{-t}}\mathrm dt.
$$
Furthermore,
$$
\int^{\infty}_{0}\frac{t^3 \mathrm e^{-t}}{1-\mathrm e^{-t}}\mathrm dt=\sum_{n\geqslant1}\int^{\infty}_{0}t^3 \mathrm e^{-nt}\mathrm dt=\sum_{n\geqslant1}\frac1{n^4}\int^{\infty}_{0}t^3 \mathrm e^{-t}\mathrm dt.
$$
Since
$$
\int^{\infty}_{0}t^3 \mathrm e^{-t}\mathrm dt=\Gamma(4)=6,
$$
all this implies
$$
\sum^{\infty}_{k=0} \int^{\infty}_{0}\frac{t^3 \mathrm e^{-4(k+1)t}}{1-\mathrm e^{-4(k+1)t}}\mathrm dt=\frac1{4^4}\cdot\left(\sum_{n\geqslant1}\frac1{n^4}\right)^2\cdot6.
$$ |
H: Symbol for "if any"
I am looking for a symbol if any for the following equation in my algorithm
This is to find closed pattern where $p_i$ is longer than $p$ and $p$ is a sub-pattern of $p_i$ and $support(p) = support(p_i)$
so can I say
if $\forall \; p \subset p_i$ and $support(p) = support(p_i)$ or
if $ \left | \{ p_i | p \subset p_i, support(p) = support(p_i) \} \right | > 0$
Or any other better way to represent?
AI: The sentence
$\exists \pi\, P(\pi)$
translates to "there is a $\pi$ such that $\pi$ has property $P$." In your example we could take $P(\pi)$ to say $\text{length}(\pi) = \text{length}(p) + 1 \wedge \ldots.$
The sentence
If $\exists \pi\, P(\pi)$, then $Q$
translates to "if there is a $\pi$ such that $\pi$ has property $P$, then $Q$ is true." (You didn't say what the conclusion $Q$ would be in your case.) You could also write it more colloquially as "if some $\pi$ has property $P$, then $Q$ is true."
This sounds like what you want, unless to conclude $Q$ you need the stronger hypothesis that all $\pi$ have property $P$.
By the way, I would avoid using the word "any" in mathematical writing. It is usually clearer to use "some" or "all" depending on what you mean. |
H: how do I solve this seperable equation with so many terms?
Solve given differential equation by separation of variables
$$\frac{dy}{dx}=\frac{xy+3x-y-3}{xy-2x+4y-8}$$
I started by multiplying each side by the denominator to get
$$(xy-2x+4y-8) dy = (xy+3x-y-3) dx$$
Now that there is $xy$ on each side of the equation do they cancel off or no because they are being multiplied by $dy$ or $dx$. How do I proceed? I'm getting a little discourage because this is the third separable differential equation I have come across and they all have taken a long time and involved lots of writing and are error prone. Is that right, in general differential equations are long and tedious (e.g. involve two integratoins and more)?
AI: $$\frac{dy}{dx}=\frac{xy+3x-y-3}{xy-2x+4y-8}=\frac{y(x-1)+3(x-1)}{x(y-2)+4(y-2)}=$$
$$=\frac{(x-1)(y+3)}{(y-2)(x+4)}\iff\frac{y-2}{y+3}dy=\frac{x-1}{x+4}dx$$ |
H: Let $\sum_{n=1}^\infty \frac{a_n}{3^n}.$ Determine (numerically or not) the limit of the infinite series by choosing $a_n=0$ or $2$ randomly.
The problem I have is essentially in the title.
What I'm trying to do in Matlab is to have a set which has two elements $0$ and $2$ and to choose either $0$ or $2$ randomly and plug it into the series at some $n$.
To just generate either $0$ or $2$ I've used the following code:
r=randi([1 2],10,1)
if r==1
r=r-1
end
This was how I figured I would get either $0$ or $2$, yet I still output a $1$ as well. Is my approach even correct using the randi function or is there a more specific function to use for my purpose?
Also, if there's a way to do this without actually having to run any calculations that would be great too.
Thanks for any help or feedback!
AI: So
r = randi([1,2], 10, 1)
generates a $10 \times 1$-vector of random entries, if you now compare $r$ to $1$ by r==1 this will never be true, as $r$ is a vector, not the scalar $1$, what you can do is find all 1s in $r$ and replace them by zeros doing
r(find(r==1)) = 0;
or apply a map to $r$ witch fixes 2 and maps 1 to 0 e. g.
r = 2*r - 2; |
H: characterization of weakly convergent to zero sequences on $l^p$ for $1\le p < \infty$
Let $1\le p< \infty$. Show that a sequence $t_k = ({t_{kj}})_{j=1}^{\infty}\in l^p$ converges weakly to 0 iff $||t_k||_p$ is bounded and $\lim_k t_{kj}=0$. I proved that if $t_k$ converges weakly to 0 then we conclude that. I want to prove the reciprocal.
Let's assume that $1<p<\infty$ If I assume that $(t_k)$ it's weakly cauchy I can prove that it's weakly convergento to 0, but I don't know how to prove that. Under that assumption I used the reflexivity of the space $l^p$ ($l^1$ is not reflexive)
If $p=1$ since $l^1$ is not reflexive my arguments are not valid here. I don't really now if the result it's also true here. Please help me!
AI: You can find a proof of more general result here.
Appply that theorem to the case $p\in(1,+\infty)$ with $S=\{f_j\in (\ell_p)^*:j\in\mathbb{N}\}$, where
$$
f_j:\ell_p\to\mathbb{K}: t\mapsto t_j
$$
For $p=1$ see this answer, where it was proved that weak convergence is equivalent to strong convergence. It is remains to note that every strongly convergent sequence is bounded and pointwise convergent. |
H: Show that if $10$ divides into $n^2$ evenly then $10$ divides into $n$ evenly
I'm not sure how to show that if $10$ divides into $n^2$ evenly, then $10$ divides into $n$ evenly.
AI: This answer is an answer to the original form of this question "show that if $20$ divides into $n^2$ evenly, then $20$ divides into $n$ evenly". The second half of this answer is automatically an answer to the new form of this question "show that if $10$ divides into $n^2$ evenly, then $10$ divides into $n$ evenly".
It's not true - consider $n = 10$ - and the reason it fails is that $20$ is divisible by $p^k$ for some prime $p$ and some $k > 1$ (namely by $4 = 2^2$).
In general, if $m = p_1 \dots p_l$ is a product of distinct primes, then $m|n^2 \implies m|n$. This follows, for instance, by looking at the prime number decomposition of $n$. Alternatively, look at a single prime factor $p_i$ of $m$. This divides $n^2$ and because it's prime it also divides $n$. Hence $m$ also divides $n$. |
H: Prove the following identity for Fibonacci numbers
Prove this:
for any positive integer $a,b,c$,
$F_{a+b+c+3}=F_{a+2}(F_{b+2}F_{c+1}+F_{b+1}F_c)+F_{a+1}(F_{b+1}F_{c+1}+F_bF_c)$
Is there any way other than induction to prove this?
AI: Consider matrix formula from here. Let $M=\pmatrix{1 & 1\\ 1 & 0}$. Then $M^{a+1}M^{b+1}M^{c+1}=M^{a+b+c+3}$. However, we can express all of these matrices in terms of the Fibonacci numbers. Compare both sides. |
H: If $A \in Mat_{n \times n} (\mathbb{Q}) $ is a matrix, where $n$ is odd show that $A^2 \neq 2I.$
I don't quite see the use of the fact that $n$ is odd.
Anyway, I give a counterexample: Take a matrix $A$ where $a_{ii}$ are the non-zero rational numbers, and all other entries are zero. Clearly, $A^2 \neq 2I.$ If it was, $a_{ii}$ will be irrational. Contradiction.
If this enough? Or should I prove the statement. If I should prove, please help me!
AI: If $A^2 = 2I$, then the eigenvalues of $A$ lie in the set $\{\pm \sqrt 2 \}$; since $A$ has rational entries, $\det(A)$ is rational; but $-\det(A)$ is the product of these eigenvalues. An odd number of factors of the form $\pm \sqrt 2$ will always be of the form $\pm 2^k \sqrt 2$ for some non-negative integer $k$, not rational. Thus $A^2 = 2I$ may be ruled out. QED.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!! |
H: orthonormality and transpose of vectors
Let $ \{u_1, u_2, \ldots , u_k \} $ be $k$ vectors in $\mathbb{R}^n$. Show that $\{u_1, u_2, \ldots, u_k\}$ is an orthonormal family if and only if the $n \times k$ matrix $U = \begin{bmatrix}
u_1,u_2, \ldots, u_k
\end{bmatrix}$ satisfies the equation $U^{\intercal} U = I_k$, where $I_k$ denotes the $k\times k$ identity matrix.
Can anyone help me prove the statement above?
AI: Note that $[U^T U]_{ij} = u_i^T u_j $.
Hence $U^T U = I$ iff $u_i^T u_j = \delta_{ij} $ iff $u_k$ are orthonormal. |
H: product of two numbers ending in 6 also ends with 6
We have to prove that
"the product of two numbers ending in 6 also ends with 6" mathematically.
I have no clue how to start. I don't want you to prove it for me! but some hints would be very helpful as I'm totally stuck on how to start this prove.
AI: HINT: If $a$ ends in $6$, then $a=10m+6$ for some integer $m$. If $b$ also ends in $6$, then $b=10n+6$ for some integer $n$. Now what’s $ab$? |
H: Groups with no abelian centralizer
Suppose $G$ is a finite group with no abelian centralizers. Is it true that $G$ must be a 2-group?
Thanks for any help.
AI: No, this is not necessarily the case. If $G$ is a group such that there are no abelian centralizers in $G$, then $G \times H$ also has this property for any group $H$. |
H: Does there exist an injection from $P(S)$ to $u(S)$
Let $S$ be an uncountable set , let $u(S)$ denote the set of all uncountable subsets of $S$ and let $P(S)$ denote , as usual , the powerset i.e. the set of all subsets of $S$, then does there exist an injection $f:P(S)→u(S)$ ?
AI: Yes. $|S\times\{0,1\}|=|S|$, so replace $u(S)$ by $u(S\times\{0,1\})$. If $A\subseteq S$ is countable, let $\varphi(A)=(S\times\{0\})\cup(A\times\{1\})$; otherwise let $\varphi(A)=A\times\{1\}$. (I am assuming the axiom of choice here.) |
H: Minimization of Sum of Squares Error Function
Given that $y(x,{\bf w}) = w_0 + w_1x + w_2x^2 + \ldots + w_mx^m = \sum_{j=0}^{m} w_jx^j$ and there exists an error function defined as $E({\bf w})=\frac{1}{2} \sum_{n=1}^{N} \{y(x_n, w)-t_n\}^2$ (where $t_n$ represents the target value). I'm having trouble making sense of a passage in my textbook. (Note: ${\bf w}$ represents a vector of the polynomial's coefficients.) I've listed the passage below:
We can solve the curve fitting problem by choosing the value of ${\bf w}$ for which $E({\bf w})$ is as small as possible. Because the error function is a quadratic function of the coefficients ${\bf w}$, its derivatives with respect to the coefficients will be linear in the elements of ${\bf w}$, and so the minimization of the error function has a unique solution, denoted by ${\bf w^*}$, which can be found in closed form.
How do we know that the minimal solution exists and is unique? What guarantees this? Any help understanding this would be appreciated.
AI: Let's, as the cited passage suggest, look at the derivative of $E$. First we note that $y$ is linear in $w$, as
$$ y(x,w+\mu w') = \sum_i (w_i +\mu w_i')x^i = \sum_i w_ix^i +\mu \sum_i w'_ix^i= y(x,w) + \mu y(x,w') $$
Now we have for $w, h \in \mathbb R^{m+1}$ that
\begin{align*}
E(w+ h) &= \frac 12\sum_{n=1}^N \bigl(y(x_n, w) + y(x_n, h) - t_n\bigr)^2\\
&= E(w) + \sum_{n=1}^N \bigl(y(x_n, w) - t_n)y(x_n, h) + \sum_{n=1}^N y(x_n, h)^2\\
&= E(w) + \sum_{n=1}^N \bigl(y(x_n, w) - t_n)\bigr)y(x_n, h) + o(h)
\end{align*}
so $E'(w)h = \sum_{n=1}^N \bigl(y(x_n, w) - t_n)\bigr)y(x_n, h)$. The second derivative is
$$ E''(w)[h,k] = \sum_{n=1}^N y(x_n, k)y(x_n, h) $$
Now for $h \in \mathbb R^{m} -\{0\}$
$$ E''(w)[h, h] = \sum_{n=1}^N y(x_n, h)^2 $$
and this is positive if $N \ge m+1$ and all $x_i$ are different (as a polynomial of degree $m$ cannot have $N \ge m+1$ zeros). So $E''(w)$ is positive definite for every $w$, as $E''$ is constant, hence every zeros of $E'$ is a minimum for $w$. Now lets look at $E'$, we have $E'(w) = 0$, if $E'(w)e_i = 0$ for each $i$ ($e_i$ denoting the $i$th standard basis vector), it holds
$$ E'(w)e_i = \sum_{n=1}^N \bigl(y(x_n, w) - t_n\bigr) x_n^i $$
That is we want $w$ to be such that $y(x,w)- t$ is orthogonal to $(x_1^i, \ldots, x_N^i)$ for all $i$. Projection of $t$ onto the subspace generated by this vectors, gives us an unique point, as $w \mapsto y(x,w)$ is injective, $w$ is unique. |
H: Problem about pointwise and uniformly convergence
We define $f_n$ define the functions with $n \in N$, as follows:
$$f(n) = \begin{cases}
\sin(x) &\mbox{if } 2\pi n \leq x \leq 2\pi(n+1) \\
0 & \mbox{if other way} \end{cases}
$$
The functions $f_n$ are uniformly bounded and equicontinuous and converge
pointwise, but not uniformly, to $0$.
I try dividing the inequalities between $ n $ but is not useful, please help me.
AI: It is easy to see that $f_n(x) \in [0,1]$ for all $n,x$, hence they are uniformly bounded.
$f_n$ is Lipschitz continuous with rank $1$ for all $n$, hence equicontinuous.
Finally, $f_n(2\pi n + \frac{\pi}{2}) = 1$ for all $n$. Hence the $f_n$ cannot converge to $0$ uniformly;. |
H: number of compositions of [n] that each contain a largest part
I am trying to generalize for any [n] the number of compositions that each contain a largest part. [1] has only one composition with a unique largest part, [2] has 1, [3] has 3 compositions with a unique largest part, [4] has 6 compositions with a unique largest part, [5] has 12, [6] has 23, and so on... I want to determine if there is a generalization of how many compositions have a unique largest part relative to the total number of compositions 2^(n-1).
AI: This is the sequence OEIS A$097979$ offset one place; the correct value for $[6]$ is $23$. A very ugly generating function is given, but no recurrence and no closed form.
The $23$ compositions for $6$ are:
$$\begin{array}{l|l}
6&1\text{ permutation}\\
5+1&2\text{ permutations}\\
4+2&2\text{ permutations}\\
4+1+1&3\text{ permutations}\\
3+2+1&6\text{ permutations}\\
3+1+1+1&4\text{ permutations}\\
2+1+1+1+1&5\text{ permutations}\\ \hline
\text{Total}:&23\text{ permutations}
\end{array}$$ |
H: Let $\sum_{n=1}^\infty \frac{a_n}{3^n}.$ Determine (numerically or not) the limit of the infinite series by choosing $a_n=0$ or $2$ randomly.
This is a follow up question to Let $\sum_{n=1}^\infty \frac{a_n}{3^n}.$ Determine (numerically or not) the limit of the infinite series by choosing $a_n=0$ or $2$ randomly.
From that question I received help in choosing randomly from the set $\{0,2\}$. Now I'm attempting to implement the code with the sum and I'm familiar with the symsum command, but it's giving an error.
My Matlab code is:
r = randi([1 2],1000,1)
r(find(r==1))=0
syms n
u = symsum(r(n)/(2^(n-1)),n,1,1000)
ans
What's weird is if I just run,
u = symsum(1/(2^(n-1)),n,1,1000)
I get a huge number, but if I do
u = symsum(1/(2^(n-1)),n,1,inf)
it outputs the right number. Is there something I'm missing about the symsum function?
I figure what I wrote above should make sense since the indexing on the vector I make with the randi function is the same indexing on the sum. Am I even using the right function to begin with?
Thanks for any help or feedback!
AI: Look at the error message Matlab gives you, we have
Indexing input must be numeric, logical or ':'.
So: No indexing with symbolic variables. If you want to evaluate it for your $r_n$, just do
sum(r .* 2.^(-(0:length(r)-1)'));
That does the trick. Regarding your second question, what Matlab tells you is:
$$ \sum_{k=0}^{999} \frac 1{2^k} = \frac{
107150860718626732094842504906000181056140481170553360744375\\ \ \ 038837035105112493612249319837881569585812759467291755314682\\\ \ \ 518714528569231404359845775746985748039345677748242309854210\\ \ \ \ \ 746050623711418779541821530464749835819412673987675591655439\\ \ \ \ \ \ 46077062914571196477686542167660429831652624386837205668069375}{535754303593133660474212524530000905280702405852766803721875\\ \ \ 194185175525562468061246599189407847929063797336458776573412\\ \ \ \ 593572642846157021799228878734928740196728388741211549271053\\ \ \ \ \ 730253118557093897709107652323749179097063369938377958277197\\ \ \ \ \ \ 3038531457285598238843271083830214915826312193418602834034688}
$$
That's not a big number, just look for the / in the answer. |
H: Restrict a metric, gives same topology as subspace topology from larger space $X$
Let $A\subseteq X$. If $d$ is a metric for the topology of $X$, show that $d\restriction_{A\times A}$ is a metric for the subspace topology on $A$.
I've shown that $d'=d\restriction_{A\times A}$ is a metric on $A$.
I am letting $\tau_B$ denote the subspace topology on $A$ that is induced by the metric topology on $X$ and $\tau_{B'}$ denote the topology on $A$ induced by the metric $d'$. My goal is to show $\tau_B=\tau_{B'}$.
Let $B=\{A \cap B_\varepsilon^d(x) \mid x \in X, \varepsilon>0\}$, a basis for $\tau_B$ and $B'=\{B_\varepsilon^{d'}(a) \mid a \in A\}$, a basis for $\tau_{B'}$. I've shown $B' \subseteq B$ and so $\tau_{B'} \subseteq \tau_B$.
For showing $\tau_B \subseteq \tau_{B'}$ I've picked some $a \in A$ and $A\cap B_\varepsilon^d(x)$, a basis element of $\tau_B$ such that $a \in A\cap B_\varepsilon^d(x)$. My goal is to find some basis element $C$ of $\tau_{B'}$ such that $a \in C \subseteq A\cap B_\varepsilon^d(x).$ If $x \in A$, then
$$A\cap B_\varepsilon^d(x)=B_\varepsilon^{d'}(x)$$
and so I have my needed basis element $C$. The part where I have been having trouble is if $x \notin A$. In this case, what I have tried so far is to see if I can find some $\delta >0$ such that
$$a \in B_\delta^{d'}(a) \subseteq A\cap B_\varepsilon^d(x)$$
but in searching for a delta and trying to show via set containment that the $\delta$-ball is contained in the $\varepsilon$-ball intersected with $A$ has been difficult. I've tried using triangle inequality but I am running into a problem of not knowing how to show that if $y \in B_\delta^{d'}(a)$, meaning $d'(a,y)$ that $y \in A\cap B_\varepsilon^d(x)$, mainly that $d(x,y) < \varepsilon$. It seems with the different $\delta$'s I have tried, I can use the triangle inequality to show something like $d(x,y) < \frac{3}{2}\varepsilon$ but not quite $\varepsilon$.
AI: Suppose that $a\in A\cap B_\epsilon^d(x)$ for some $x\in X$. You know that $\{B_r^{d'}(a):r>0\}$ is a local base at $a$ in $A$, so there must be some $\delta>0$ such that $B_\delta^{d'}(a)\subseteq B_\epsilon^d(x)$, and hence $B_\delta^{d'}(a)$ is a $d'$-nbhd of $a$ contained in $A\cap B_\epsilon^d(x)$. Since you can do this for each $a\in A\cap B_\epsilon^d(x)$, $A\cap B_\epsilon^d(x)\in\tau_{B'}$. |
H: When is the product of two points on a circle is still on the circle?
Suppose we have a pair $(x,y)\in\Bbb{R}^2$ such that $x^2+y^2=1$. I want to find a pair $(a,b)$ such that $a^2+b^2=1$ and $(xa)^2+(yb)^2=1$. What is $a$ and $b$ in terms of $x$ and $y$?
AI: Given $x^2+y^2=1$ and $a^2+b^2=1$, multiplying, we have $(xa)^2+(yb)^2+(xb)^2+(ya)^2=1$
Now if we also have $(xa)^2+(yb)^2=1$, we must have $(xb)^2+(ya)^2=0$. So what could satisfy this as well as be on the unit circle?
Addition:
For $(xb)^2+(ya)^2=0$, we must have both $xb, ya = 0$. However one among $(a^2, b^2)$ must be $1$, and similarly for $(x^2, y^2)$. Hence we have solutions
$(a, b) = (\pm1, 0)$ when $(x, y) = (\pm1, 0)$ and
$(a, b) = (0, \pm1)$ when $(x, y) = (0, \pm1)$ |
H: Some question about even function
Given that $f$ is an even function, $f''(x)>0$. Which of the following are true?
I. $f(0)<f(1)$
II. $f(4)-f(3)<f(6)-f(5)$
III. $f(-2)<\frac{f(-3)+f(-1)}{2}$
I understand I is true since $f$ is increasing for $x>0$, and II is also true since $f'(x)>0$ for $x>0$. How about III? Any hint for it? Thanks!
AI: HINT: $\frac{f(-3)+f(-1)}2$ is the $y$-coordinate of the midpoint of the line segment whose endpoints are $\langle-3,f(-3)\rangle$ and $\langle-1,f(-1)\rangle$; the $x$-coordinate of that midpoint is clearly $-2$. And the curve is concave up, so ... ? |
H: Is this notation correct?
I am writing a paper and I have an expression something like this:
$$\begin{equation}
\notag
x = \text{large_expression} + \begin{cases}
y & \text{if } a<b \\
z & \text{if } a=b \\
\end{cases}
\end{equation}$$
Will this notation be understood just as well as this more common, undoubtly correct way?
$$ x = \begin{cases}
\text{large_expression} + y & \text{if } a<b \\
\text{large_expression} + z & \text{if } a=b \\
\end{cases}$$
Note that splitting up $\text{large_expression}$ and $y$/$z$ will hurt the understanding a lot, as well as break the flow of the paper.
AI: Rather than indicator expressions I would suggest the equivalent but somewhat more readable (once you are used to them) Iverson brackets
$$
x = \text{large_expression} + [a<b]y + [a=b]z\quad.
$$
But the solution you suggested is also perfectly understandable an unambiguous, so I don't see much wrong in using it. After all the normal use of the case distinction braces is just after "$=$", which is also in the middle of a syntactic construct; I don't see a fundamental difference when using it after "$+$". |
H: Help with a question regarding basis and dimension?
I am presented with:
Let $a_1$ and $a_2$ be linearly independent vectors in $R^3$ and let x be a vector in $R^2$.
a) Describe geometrically Span($a_1, a_2$)
b) If A=($a_1,a_2$) and b=Ax, then what is the dimension of Span($a_1, a_2$, b)? Explain.
For a), since there are only two vectors $a_1$ and $a_2$, wouldn't the Span($a_1, a_2$) only be a plane in $R^3$?
For b), since b can be represented as b=Ax, wouldn't the dimension of Span($a_1, a_2$, b) only be 2 because b is linearly dependent?
These are really only guesses at the questions. I'm having a terrible time understanding span, spaces, basis, and dimension (as well as most of linear algebra in general). Can anyone either confirm or reject my answers and perhaps provide a little more information?
AI: Ad a) Yes, you are right, it is just a plane in $\mathbb R^3$. Maybe you want to be a bit more precise and say which plane it is? It will be a plane containing the origin, for example. And it will be the unique plane that contains the vectors $a_1$ and $a_2$. Convince yourself that any point in this plane can be written as a combination of the $a_i$ and that they are the only ones!
Ad b) Again correct. If $x=(x_1,x_2)$, then $Ax$ is nothing else but $Ax=x_1a_1+x_2a_2$ (this is purely algebraic). Thus $b$ already lives in the plane described in a).
As a remark: don't despair. It is completely normal to be confused by these things when you encouter them for the first time. Just keep asking yourself similar questions and it will become very intuitive quickly! |
H: Hartshorne's proof that $\mathcal{O}_{\operatorname{Spec} A}(D(f)) \cong A_f$, Prop II.2.2(b)
Hartshorne, "Algebraic Geometry," Proposition II.2.2(b) on page 71 reads (roughly):
$\mathcal{O}_{\operatorname{Spec} A}(D(f)) \cong A_f$
The relevant section of the proof reads (after some simplification):
We define $\psi : A_f \to \mathcal{O}_{\operatorname{Spec} A}(D(f))$ by sending $a/f^n$ to the section which assigns to each $\mathfrak{p}$ the image of $a/f^n$ in $A_\mathfrak{p}$. First we show that $\psi$ is injective. If $\psi(a/f^n) = 0$, then for every $\mathfrak{p} \in D(f)$ we have $a/f^n = 0$ in $A_\mathfrak{p}$, so by definition there is some $h \not \in \mathfrak{p}$ such that $h a = 0$ in $A$. Let $\mathfrak{a}$ be the annihilator of $a$ in $A$. Then $h \in \mathfrak{a}$ and $h \not \in \mathfrak{p}$, so $\mathfrak{a} \not \subseteq \mathfrak{p}$. We conclude that $V(\mathfrak{a}) \cap D(f) = \emptyset$. Therefore $f \in \sqrt{\mathfrak{a}}$, so $f^\ell \in \mathfrak{a}$ for some $\ell$, so $f^\ell a = 0$. Since $f$ is a unit in $A_f$, this says that $a = 0$ in $A_f$. The hard part is to show that $\psi$ is surjective...
I follow the argument fine, but it strikes me as particularly ingenious, despite Hartshorne's implicit assertion that it's the "easy part" of the argument. (To be fair, the proof of surjectivity takes a whole page.) In particular, the fact that we're checking a commutative algebra result to apply to a problem in algebraic geometry by applying algebraic geometry to the commutative algebra problem kind of blows my mind at the moment.
Is this really just a very clever argument, or is there some perspective from which it's straightforward? Alternatively, is there some other way of seeing this result?
AI: If $S\subset A$ is a multiplicative system, and $x\in S^{-1}A$ is such that $x=0$ in every $A_P$ with $P\cap S=\emptyset$, then $x=0$ since it is $0$ in all localizations of $S^{-1}A$. |
H: Subset of $\mathbb{Z}$ not a subgroup of $(\mathbb{Z},+)$
I have this question for homework and I can't figure it out.
Find a subset of $\mathbb{Z}$ that is not a subgroup of $(\mathbb{Z},+)$ but adheres to closure.
In other words, a subset whose elements when added are elements of the subset, but it is not a subgroup. I've been stumped on this. All $n\mathbb{Z}, n\in{\mathbb{N}}$ are subgroups, the odd numbers don't work since they sum to even numbers. I tried utilizing some cosets of $n\mathbb{Z},$ but can't see them there either. ANy hints or help is appreciated.
AI: The naturals $\;\Bbb N\;$
The positive multiples of $\;n\;,\;\;n\in\Bbb N\;$
The negative multiples of $\;n\;,\;\;n\in\Bbb N\;$ ... |
H: Graph Cut Problem
I have a problem as such:
$2n$ players, each of whom has an odd number of friends, are
distributed into two teams. A player is happy if more of his friends
are on the other side than on his own side, and a distribution is
stable if everyone is happy. The cut number of a distribution is the
number of friends belonging to the opposite teams.
Give an example of two different stable distributions of the same set of players that have different cut numbers.
I don't know how to attack this. I've tried drawing graphs and trying to construct two such matchings for low values of $n$, but no success so far. I need a hint.
AI: I think I've got an example. Take $12$ vertices; call them them $a_i,b_i,c_i,\;d_i,1\le i\le3$. Draw $24$ edges as follows: $9$ edges $a_ib_j$, $9$ edges $c_id_j$, $3$ edges $a_id_i$, $3$ edges $b_ic_i$. This is a $4$-regular bipartite graph. Put the $a$'s and $c$'s on one team, the $b$'s and $d$'s on the other; everyone is happy, and the cut number is $24$. Next, put the $a$'s and $d$'s on one team, the $b$'s and $c$'s on the other; again everyone is happy, but the cut number is only $18$. Right?
P.S. That example was bigger than it needed to be. Here is an example with $8$ vertices ("players") and $12$ edges ("friendships"): the cube, with vertices $000$, $001$, $010$, $011$, $100$, $101$, $110$, $111$.
Partition the vertices into the Even Team $\{000,011,101,111\}$ and the Odd Team $\{001,010,100,111\}$; everyone is happy; each player sees all three friends on the other team; all $12$ edges are "cut".
Partition the vertices into the Blue Team $\{000,001,110,111\}$ and the Red Team $\{010,011,100,101,\}$; each player sees one friend on his own team and two friends on the other team; everyone is still happy, but only $8$ edges are "cut." |
H: What use are Fermat’s Little Theorem and Wilson’s theorems in number theory?
Do these theorems have any real life applications? We cannot use them to find primes as both are pretty inefficient for large numbers.
AI: Fermat’s Little Theorem plays a key role in modern life in the proof of correctness of the RSA algorithm for public-key cryptography. |
H: Find the radius and centre of the circle $x^2 -6x +y^2 -2y -6=0$
Find the radius and centre of the circle $x^2 -6x +y^2 -2y -6=0$
Can someone please help me with this question? I'm quite lost with what I have to do.
AI: HINT Try to complete the square $x^2-6x$ to get it into the form of $(x-a)^2-c$, do the same for $y$ and collect the numbers on the RHS. |
H: Do monotone operators have positive Frechet derivatives?
If a scalar function $f\colon \mathbb R \to \mathbb R$ is monotone and differentiable, then $f'\geq 0$.
Monotonicity is generalized for an operator $A\colon V \to V^*$, where $V$ is a Banach spaces with its dual space $V^*$, via:
The operator $A\colon V \to V^*$ is called monotone if $$ \langle A(u) - A(v), u-v \rangle \geq 0,$$ for all $u$, $v \in V$.
So my question is: Does monotonicity of $A$ imply, that the Frechet derivative of $A$ is positive, i.e. $\langle A'(w)v,v\rangle \geq 0$ for any $w \in V$ and for all $v\in V$.
Any idea or reference is appreciated.
AI: If the operator $A$ is Fréchet differentiable, then its derivative is positive (semidefinite).
For suppose to the contrary there were a point $w$ and a $v\in V$ with
$$c := \langle A'(w)v,v\rangle < 0.$$
Then for small real $t \neq 0$, we have
$$t^{-2}\langle A(w + tv) - A(w), tv\rangle = t^{-2} \langle A'(w)(tv) + o(t), tv\rangle = c + \frac{t\cdot o(t)}{t^2} \to c < 0,$$
contradicting the monotonicity of $A$. |
H: Generating $A_4$ with two distinct $3$-cycles $x$ and $y$ in $S_4$
Given two distinct $3$-cycles $x$ and $y$ in $S_4$ such that $x\neq y^{-1}$, I would like to show that $A_4$ is generated by any such pair of $x$ and $y$.
The only times where $3$-cycles in $S_4$ can be the inverse of each other is when they can be written in the form $(a\;b\;c),(a\;c\;b)$. Excluding this case, the possible pairs of cycles should only be of the form
$$(a\;b\;c),(a\;b\;d)\qquad\text{or}\qquad(a\;b\;c),(a\;d\;b).$$
Suppose the two $3$-cycles are of the form $(a\;b\;c),(a\;b\;d)$, then I can say the element
$$(a\;b\;c)(a\;b\;d)=(c\;a)(d\;b)$$
is in the subgroup generated by $(a\;b\;c),(a\;b\;d)$.
From here I am not sure how I should go about showing that the generated set must be $a_4$. Should I keep discovering every possible element in the generated set display that the sets are equivalent? Or is there a more clever way than brute forcing it like this for every valid pair of 3-cycles?
AI: Given a pair $x,y$ with property $x \neq y^{-1}$ you have distinct elements $e,x,y,x^2,y^2$. It is sufficient to show that $xy$ and $x^2y^2$ are two new elements distinct from $e,x,y,x^2,y^2$. And we obtain 7 elements in the respective subgroup of $A_4$. So, this subgroup is $A_4$ (by Lagrange's theorem). |
H: Order of operations: Matrix product and hadamard product
Is there any convention of the order of operations in a term with both ordinary matrix multiplication and hadamard (elementwise) multiplication?
Obviously,
$$ A ( b \circ c ) \ne (A b) \circ c $$
But how is $ A b \circ c $ (without parentheses) conventionally interpreted?
AI: There is probably no convention, and if there is it is not sufficiently well known to be unambiguous. Authors using such a mixture would either have to declare their convention or (better still) use brackets for both cases. |
H: How to solve the inequality $\frac {5x+1}{4x-1}\geq1$
Please help me solve the following inequality.
\begin{eqnarray}
\\\frac {5x+1}{4x-1}\geq1\\
\end{eqnarray}
I have tried the following method but it is wrong. Why?
\begin{eqnarray}
\\\frac {5x+1}{4x-1}&\geq&1\\
\\5x+1&\geq& 4x-1\\
\\x &\geq& -2
\end{eqnarray}
Thank you for your attention
AI: As Carlos Eugenio Thompson Pinzón has commented, we need $\displaystyle4x-1\ne0$
Method $1:$
If $\displaystyle4x-1\ne0, (4x-1)^2>0$
Multiplying either sides of $\displaystyle\frac{5x+1}{4x-1}\ge1$ by $(4x-1)^2$
we get $\displaystyle(5x+1)(4x-1)\ge(4x-1)^2$
$\displaystyle\implies (4x-1)\{5x+1-(4x-1)\}\ge0 $
$\displaystyle\implies \left(x-\frac14\right)\{x-(-2)\}\ge0$
We know if $(x-a)(x-b)\ge0$ where $a<b$ either $x\le a$ or $x\ge b$
Method $2:$
As we know $a\ge b\implies \begin{cases} ac\ge bc &\mbox{if } c> 0 \\
ac\le bc &\mbox{if } c< 0 \end{cases}$
If $\displaystyle 4x-1>0\iff x>\frac14, \frac{5x+1}{4x-1}\ge1\implies 5x+1\ge 4x-1\iff x\ge-2$
$\displaystyle\implies x>\frac14$ is one of the solutions
If $4x-1<0\iff x<\frac14, \frac{5x+1}{4x-1}\ge1\implies 5x+1\le 4x-1\iff x\le-2$
$\displaystyle\implies x\le -2$ is the other solution |
H: Proof that a subset of R has same cardinality as R
I am sitting with a task where I have to prove the following:
Claim:
Every subset of $\mathbb{R}$, that contains an interval $I$ with $a < b$, has the same cardinality as $\mathbb{R}$.
So I think that I should prove that there exist a bijection from $I$ to $\mathbb{R}$? I'm kinda lost, and don't know how to start.
There is a lemma 3 in the book saying:
Let $a,b \in \mathbb{R}$ with $-\infty\neq a < b \neq \infty$. There exist a $f\colon ]-1; 1[ \to ]a; b[$ that is bijektiv. The intervals $]-1; 1[$ and $]a; b[$ has same cardinality.
Another lemma 4 says:
$f\colon ]0; 1[ \to ]1; \infty[$, $x\to \frac{1}{x}$, is bijective. The intervals $f\colon ]0; 1[ \to ]1; \infty[$ have same cardinality.
(There is an image added to lemma4)
The above interval $]-1; 1[$ has same cardinality as $\mathbb{R}$, and the $f$ is bijective
Any help is highly appreciated
AI: Hints:
First prove that any two non-empty open intervals have the same cardinality.
Second, pass now to use the nice interval $\;(-\pi/2\,,\,\pi/2)\;$ and a rather nice, simple trigonometric function to show equipotency with $\;\Bbb R\;$ |
H: Prove that this is a legitimate PMF
I know that these are two properties of PMF.
Non-negativity
Sum over the support equals 1
However I can't show that this PMF's sum over the support equals 1.
AI: Using the Taylor series for $e^{\lambda}$ you have
$$
\sum_{k=0}^{\infty} p_X(k) =
\sum_{k=0}^{\infty} e^{-\lambda} \frac{{\lambda}^k}{k!}
= e^{-\lambda} \sum_{k=0}^{\infty} \frac{{\lambda}^k}{k!}
= e^{-\lambda} e^{\lambda} = 1 $$ |
H: A point P moves so that AP and BP are perpendicular. Find the equation of the locus of P
A point p(x,y) moves so that AP and BP are perpendicular, given A=(3,2) and B =(-4,1). Find the equation of the locus of P.
Can someone please advise me on what to do for this question. Just need a direction. Thanks.
AI: Direction: The locus is the circle with diameter $AB$. |
H: non-transitive, antisymmetric and reflexive binary relation on $\mathbb Z$
Does anybody know about a reflexive, antisymmetric, but not transitive relation on $\mathbb Z$? I really cant figure any out and I am having doubts that something like that exists.
AI: Let $R \subseteq \mathbb Z^2$ denote the relation to find, if we want $R$ to be reflexive, we need $$ \Delta_{\mathbb Z} = \{(z,z)\mid z \in \mathbb Z\} \subseteq R $$
$R$ is allready antisymmetric, we just must pay attention not destroying this property. Now for the not-transtivity: Add two points, say $(0,1), (1,2) \in R$, but $(0,2) \not \in R$. As $R := \Delta_{\mathbb Z}\cup\{(0,1), (1,2)\}$ is antisymmetric, we are done.
Addendum. If you want to have a formula, this can be done, our $R$ is given by
$$ x\mathbin R y \iff (x-y)^2\cdot\bigl(x^2+(y-1)^2\bigr)\cdot \bigl((x-1)^2+(y-2)^2\bigr) = 0. $$ |
H: Unknown terms of the proportion
please help me solving this problem. The question is, find the unknown terms of the proportion $$\frac 23 = \frac x{12} = \frac y{15}.$$
AI: $$\dfrac 23 = \cfrac x{12}$$ $$\dfrac 23 = \dfrac y{15}$$
Cross multiplying gives us the following equalities:
$$2\cdot 12 = 3x$$ $$2 \cdot 15 = 3y$$
Solve for $x$ and $y$. |
H: Upper bound on a sum of complex numbers
Let $A=\{z_1, z_2, z_3,\ldots \} $ be a set of complex numbers with $|z_j|\ge 1$ such that the number of elements of $A$ with modulus $<r$, denoted $N_A(r)$, satisfies
$$
N_A(r) \le C_0r^N
$$
for some positive integer $N$ where $C_0$ is some constant. For any $\delta >0$ I wish to prove an inequality of the form
$$
\sum _{\substack{z_j\in A \\ |z_j| <r}}\frac{1}{|z_j|^{N+\delta }} \le \eta r,
$$
where $0<\eta <1$.
Is this even true and in that case could anyone share a hint as how to show it?
AI: I think that this is true. Since this problem is approximation so I will not comment exact number. And we use $C$. In probelm, we don't need to use complex number.
Fix $N$. Then $r$-ball contains $Cr^N$ integer points in $ {\bf R}^N$.
So given quantity $\sum \frac{1}{|z_j|^{N+\delta }}$ is approximately $$ C\int_{B_r(0)}
\frac{1}{|x|^{N+\delta}} t^{N-1} dt d\theta = C\int_0^r \frac{1}{t^{1+\delta}}\ dt = Cr^{-\delta}\leq \eta r$$ |
H: Brownian Motion hitting random point
I got a problem that seems to be quite standard and easy, but I have lots of problems with it. I do already know that $T_a:=\inf\{t\geq 0: B_t=a\}$ is a stopping time for any $a\in\mathbb{R}$ fixed, $B_t$ being a standard BM.
But what about for instance $T_{B^{o}_1}$, where $B_t^{o}$ is an BM independent from $B_t$? Clearly $\{T_{B_1}\leq t, B^{o}_1=x\}=\{T_{-x}\leq t\}\in\mathcal{F}_t$. But the union over all $x$ is uncountable and I'm not sure if it could work if I only look at rational $x$.
Any suggestions? Thanks!
AI: Assuming that $B_t^o$ is adapted to $F_t$ and $t > 1$ then the classical $\mathbb{Q}$ trick works because of the continuity of brownian paths:
\begin{align*}
\lbrace T_{B_1^o} \leq t \rbrace
&= \lbrace \max_{s \in [0,t]}{B_s \geq B_1^o} \rbrace \cap \lbrace B_1^o \geq 0 \rbrace \cup \lbrace \min_{s \in [0,t]} B_s \leq B_1^o \rbrace \cap \lbrace B_1^o \leq 0 \rbrace
\\&= \cup_{q \in \mathbb{Q}^+} \lbrace{\max_{s \in [O,t]} B_s \geq q\rbrace} \cap \lbrace q \geq B^o_1 \geq 0 \rbrace \cup \cup_{q \in \mathbb{Q}^-} \lbrace{\min_{s \in [O,t]} B_s \leq q\rbrace} \cap \lbrace q \leq B^o_1 \leq 0 \rbrace
\end{align*}
The "max" and "min" sets are in $F_t$ because $B$ is $F_t$-adapted. $B_1^o$ is adapted also so the sets $\lbrace q \geq B_1^o \geq 0 \rbrace$ and $\lbrace q \leq B_1^o \leq 0 \rbrace$ are in $F_1 \subseteq F_t$ because $t > 1$.
In hindsight $\lbrace \max_{s \in [O,t]} B_s \geq B_1^o \rbrace$ is $F_t$ measurable since the max and $B_1^o$ are. This property is shown by the $\mathbb{Q}$-trick early in a course of measure theory. |
H: Plot implicit equation
I'm working with a frequency-response curve of a nonlinear oscillator and came across the following equation (Kovacic & Brennan 2011, p. 179):
$$
A^2 = \frac{f^2}{4 \xi^2 \omega^2 + (\omega^2 - 1 - \frac{3}{4}A^2)^2 }.
$$
Question: what is the strategy for plotting of $A$ vs. $\omega$?
I am only interested in real roots with $A > 0$. The parameters $\xi, f$ and $\omega$ are fixed and known. Furthermore $0 <\xi \ll 1$, and $0 < f, \omega$.
AI: Set $$f_\omega(A) = A^2 - \frac{f^2}{4 \xi^2 \omega^2 + (\omega^2 - 1 - \frac{3}{4}A^2)^2 }.$$
Now look for zeros of this function. One way to do this is to use Newton's method (wiki). If there are multiple zeros for fixed $\omega$ you would have to choose different starting points $A_0$ for every $\omega$, e.g. $A_0=0,0.1,0.2,\ldots$ and repeat Newton's method for all of those. But note that then the thing you plot is not a function. |
H: Isometries of the plane
Let $m$ be an orientation-reversing isometry. Prove algebraically the $m^2$ is a translation.
What I attempted: We know that $m$ is an orientation-reversing isometry i.e. it is either a reflection or a glide-reflection.
I started by consider the case when $m$ is a reflection.
The matrix representing this transformation is given by $\begin{pmatrix} \cos \phi & \sin \phi\\ \sin \phi & -\cos \phi \end{pmatrix}$
I calculated the square of the above matrix and obtained the identity matrix, which means that $m^2$ is a translation of $v=\begin{pmatrix}0 \\ 0 \end{pmatrix}$
I am not sure if this is correct..
With the glide-reflection, I am not sure how to proceed, since I don't know how to express neither the endomorphism "glide-reflection" in terms of the endomorphism rotation, reflection, translation and neither the matrix representing this transformation.
AI: Hint: Consider an orientation-reversing isometry as $m:x \mapsto Ax+b$, where $A$ is a reflaction matrix (so, $A^2=I$), $b$ is a translation vector. And try to find $m^2$. |
H: Proving ${\sim p}\mid{\sim q}$ implies ${\sim}(p \mathbin\& q)$ using Fitch
I am struggling with proving something in Fitch. How can I prove from the premise ~p | ~q , that ~(p & q). Any ideas on how I should proceed?
AI: I'll have my first stab at an answer here (I've been won over from Stack overflow- first answer!).
I believe on math.stackexchange it's not tradition to completely give the game away if it's a homework question?
notation: noticed you're using single bar for vel, I'll use $\vee$ and "|" to distinguish Fitch hypothesis marks until I can work out math stackexchange formatting.
So,
Hint: your last step will be $¬(p \wedge q)$, of course, and obviously your first
1 $(¬p \vee ¬q)$
Then we prove by two classical reductio.
First, on the conclusion
2 | $p \wedge q$
Which gives us
3 | $p$
Then, remember the $\vee$-elim rules (you're working from (1) now): so we assume:
4 | | $¬p$
5 | | $\bot$(by 3)
Well, I'll leave you to fill in the details, but we can do the same for the q side of 2 and 1.
This means we can $\vee$ eliminate to
9) | $\bot$
Which gives us the discharge by reductio of (2)
10) $¬(p \wedge q)$
Hope this helps you fill in the minimal gaps.
If you wanted to cheat you could have a look at de morgan's laws as you're going one way here, or ask whether you can prove this in all logics.
Apologies for my first attempt at formatting, I'll try to edit as I go |
H: Opposite function definition
How to define an opposite of a function. If for example I have the function F(x) = y how can use it to define the function f(y) = x
AI: What you are seeking is called the inverse function $f^{-1}(x)$ of a bijective function $f(x)$. A function's inverse exists if and only if the function is bijective: if and only if it is both one-to-one and onto.
The process of finding the inverse depends on the bijective function. We can take something rather simple, for example $$f(x) = y = x^3$$
We first solve to express it as a function of $y$ ($f(y)): \sqrt[\large 3]{y} = \sqrt[\large 3]{x^3} = x$. Now we have $x = \sqrt[\large 3] y$. Then the inverse of the original function $f(x)$ is given by exchanging the positions of "x" and "y". $$f^{-1}(x) = \sqrt[\large 3]{x}$$
One can always check one's answer by evaluating the composite function to confirm that $$f^{-1}\Big(f(x)\Big) = f\left(f^{-1}(x)\right) = x$$ |
H: Rolling a die, probability problem
So this is a problem I'm stuck on,
You roll a fair 4-sided dice and with probability 1/3 you get to roll once more, and with probability 2/3 you have to stop. Assume that you get as many coins as the sum of the rolls. What is the probability you will win an even number of coins?
Can somebody help?
PS: You can get as many extra rolls as possible!
AI: $\frac{1}{3}\times\frac{2}{3}+\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3}\times\frac{2}{3}+\cdots=\frac{2}{9}\sum_{n=0}^{\infty}9^{-n}=\frac{2}{9}\times\frac{1}{1-\frac{1}{9}}=\frac{2}{9}\times\frac{9}{8}=\frac{1}{4}$
$P(RS \; or\; RRRS \; or\; ......)=P(RS)+P(RRRS)+...$
This because the events are disjoint. Secondly they are independent so that for instance: $P\left(RRRS\right)=P\left(R\right)\times P\left(R\right)\times P\left(R\right)\times P\left(S\right)$ |
H: Equivalence of zero divisor in commutative ring
Let $x$ be a nonzero element in a commutative ring, then $\exists y\neq0(xy=0)$ ($x$ is a zero divisor) iff $\exists y\neq 0(x^2y=0)$. $(\rightarrow)$ part is pretty trivial. How to prove the other way?
AI: Hint: $x^2y=x(xy)$. Consider the cases $xy = 0$ and $xy \ne 0$. |
H: How to calculate Frenet-Serret equations
How to calculate Frenet-Serret equations of the helix
$$\gamma : \Bbb R \to \ \Bbb R^3$$
$$\gamma (s) =\left(\cos \left(\frac{s}{\sqrt 2}\right), \sin \left(\frac{s}{\sqrt 2}\right), \left(\frac{s}{\sqrt 2}\right)\right)$$
I know the following info about Frenet-Serret equations:
$$\frac{\mathrm{d}}{\mathrm{d}s} \begin{bmatrix} t \\ n \\ b \end{bmatrix} = \begin{bmatrix} 0 & \kappa & 0 \\ - \kappa & 0 & \tau \\ 0 & -\tau & 0 \end{bmatrix}\begin{bmatrix} t \\ n \\ b \end{bmatrix}$$
AI: If $|\gamma'(s) |=1$ then let $t= \gamma'(s)$ which is tangent.
And curvature is $\kappa \leq 0$ : $\gamma''(s)=\kappa n(s)$ where $n$ is normal and $|n|=1$.
Define binormal $b$ : $b=t\times n$.
Define torsion $\tau$ : $$(a)\ b'(s)=\tau n$$
Now we will prove (a) : $b' =t'\times n + t\times n' = \kappa n\times n + t\times n' = t\times n' = t\times n'$
Since $n\perp n'$ so $ t\times n'$ is parallel to $n$.
Note $n =b\times t$ so that $n'=b'\times t + b\times t' =\tau n\times t +\kappa b\times n =-\tau b - \kappa t $ |
H: Clarification of L'hospital's rule
I have a question regarding L'hospital's rule.
Why can I apply L'hospital's rule to $$\lim_{x\to 0}\frac{\sin 2x}{ x}$$ and not to $$\lim_{x\to 0} \frac{\sin x}{x}~~?$$
AI: The reason you cannot use L'Hopital on the $\sin(x)/x$ limit has nothing to do with calculus, and more with logic, and the problem is subtle.
To use L'Hopital you need to know the derivative of $\sin(x)$. What is that derivative? You'd say $\cos(x)$ on reflex, and then you'd miss the problem. See, to calculate the derivative of $\sin(x)$ in the first place, you need to calculate
$$
\lim_{h \to 0}\frac{\sin(x + h) - \sin(x)}{h}
$$
We use the formula for $\sin(u + v)$ and simplify the fraction:
$$
\frac{\sin(x + h) - \sin(x)}{h} = \frac{\sin(x)\cos(h) + \sin(h)\cos(x) - \sin(x)}{h}\\\\
= \frac{\cos(h) - 1}{h}\sin(x) + \frac{ \sin(h)}{h}\cos(x)
$$
Thus to justify the use of L'Hopital, you need to know the limit of the two fractions as $h \to 0$, one of which is the limit we tried to use L'Hopital on in the first place.
If someone says that you cannot use it on $\sin(x)/x$, but you can use it on $\sin(2x)/x$, then that is because the former is in some way needed to justify L'Hopital in the first place, while when solving the latter you implicitly assume somehow that the derivative of $\sin(x)$ is already known. It all comes down to trying to read the mind of whoever poses the problem and try to see how heavy machinery they allow you to use. |
H: Prove that $\lfloor0.999\dots\rfloor= ?$ $0$ or $1$?
I think $\lfloor0.999\dots\rfloor= 1$, as $0.999\dots=1$,but I have doubt, as $\lfloor0.9\rfloor=0$,$\lfloor0.99\rfloor=0$,$\lfloor0.9999999\rfloor=0$, etc.
AI: Your first assertion is correct. The other observation just says that the function $x\mapsto\lfloor x\rfloor$ is not continuous. |
H: Uniform convergence of $\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{n!}$ on $\mathbb{R}$?
For every $r>0,$ the series $f(x)=\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{n!}$ converges uniformly on $[-r,r]$. May I know how to prove/disprove that $\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{n!}$ converges uniformly on $\mathbb{R}$ ?
Thank you.
AI: Hint: The limiting function is $f(x)=e^{-x^2}$ which is bounded, while each partial sum is a polynomial - which is unbounded. |
H: How to find the $f^{-1}(x)$ of $f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}}$
It is a question from a quiz.
The following is the whole question.
Let
\begin{eqnarray}
\\f(x)=x^{3}-12x+\frac{48}{x}-\frac{64}{x^{3}} , \space x\in (-\infty ,0),
\end{eqnarray}
find $f^{-1}(x)$. Hint : $f(x)$ can be written in the form, like$(A+B)^{3}$.
The first thing I think is $(A+B)^{3}=A^3+3A^2B+3AB^2+B^3$, then try to make it become the the form of $A^3+3A^3B+3AB^3+B^3$. However, it it so difficult to obtain this form.
I need help.
Update :
Now I have $\left(x - \frac 4x\right)^3$ but how to find the $f^{-1}(x)$ of $f(x)=\left(x - \frac 4x\right)^3$?
Thank you for your attention
AI: Hint: Try matching first and last terms: $A^3 = x^3$ and $B^3=-\frac{64}{x^3}$ and check if it fits the other terms. |
H: Limit induction proof
I need help to verify the following
Prove that if does not equal 0
lim of x approaches a: x^-n = a^-n
I know to prove lim of x approaches a: x^n = a^n requires induction so I believe that this problem requires the same.
lim of x approaches a: [f(x)]^-n ... = k^-n
inductive step
lim of x approaches a: [f(x)]^-n-1 * lim of x approaches a: [f(x)] =
k^-n-1*k = k
Does this prove the question?
AI: If the statement is known to you with positive exponents (as you say), then you can prove the statement for negative ones by using that the function $x\mapsto\dfrac1x$ is continuous. |
H: Prove that $ x_{k}=2^{k} \cdot \sin \frac{\pi}{2^{k}}$ equals $ x_{1}'=2, x'_{2}=2 \sqrt{2}, x_{k+1}=x_{k} \sqrt{\frac{2x_{k}}{x_{k}+x_{k-1}}}$
Prove that $ x_{k}=2^{k} \cdot \sin \frac{\pi}{2^{k}}$ equals $ x_{1}'=2, x'_{2}=2 \sqrt{2}, x_{k+1}=x_{k} \sqrt{\frac{2x_{k}}{x_{k}+x_{k-1}}}$.
This is what I've managed:
$x_{k+1}=x_{k} \sqrt{\frac{2x_{k}}{x_{k}+x_{k-1}}}= 2^{k} \cdot \sin \frac{\pi}{2^{k}} \sqrt{\frac{2^{k+1} \cdot \sin \frac{\pi}{2^{k}}}{2^{k} \cdot \sin \frac{\pi}{2^{k}}+2^{k-1} \cdot \sin \frac{\pi}{2^{k}}}}=2^{k} \cdot \sin \frac{\pi}{2^{k}} \sqrt{\frac{2}{1+\cos \frac{\pi}{2^{k}}}}$
And I don't see how to proceed....
AI: You can use that
$$\cos (2x) = \cos^2 x - \sin^2 x,$$
and hence
$$1 - \cos \frac{\pi}{2^k} = 1 - \cos^2 \frac{\pi}{2^{k+1}} + \sin^2 \frac{\pi}{2^{k+1}} = 2\sin^2 \frac{\pi}{2^{k+1}}.$$
Then write
$$\sqrt{\frac{2}{1+\cos \frac{\pi}{2^k}}} = \sqrt{\frac{2(1 - \cos \frac{\pi}{2^k})}{1 - \cos^2 \frac{\pi}{2^{k}}}} = \sqrt{\frac{4\sin^2 \frac{\pi}{2^{k+1}}}{\sin^2 \frac{\pi}{2^k}}} = \frac{2\sin \frac{\pi}{2^{k+1}}}{\sin \frac{\pi}{2^k}},$$
since all the sines involved are non-negative. |
H: Can we replace the condition $d(E_1,E_2)>0$ with $E_1\cap{E_2}=\emptyset$ in an property of outer measure?
Can we replace the condition $d(E_1,E_2)>0$ with $E_1\cap{E_2}=\emptyset$ in "If $E=E_1\cup{E_2}$,and $d(E_1,E_2)>0$,then $m_\star(E)=m_\star(E_1)+m_\star(E_2)$."? Obviously,$E_1\cap{E_2}=\emptyset$ can not imply $d(E_1,E_2)>0$.I think we can't replace the condition,but I can't find a counter-example. Can you give me a counter-example?
AI: The counterexample is (necessarily) a nonmeasurable set. Try working with Vitali sets. |
H: Laurent series of $\frac{1-\cos(z)}{z^2}$
How do I calculate the Laurent series of $\frac{1-\cos(z)}{z^2}$? (I know the general formula as is shown here )
AI: If you don't want to use the general formula, but you know the series for $\cos z$, then you can use them.
I understand that $f(z)=\frac{1-\cos z}{z^2}$ is not defined at $z=0$, so the function should be competed as
$$f(0)=\lim_{z\to0}\frac{1-\cos z}{z^2}\quad\text{if exists}$$
which makes it difficult for a direct approach. In this case the limit exists and is $1/2$.
update (to show the general formula to get the Laurent series)
Let's assume that in a neighborhood of $z=c$, then \begin{align}f(z)&=a_0+a_1(z-c)+a_2(z-c)^2+\ldots\\&=\sum_{k=0}^\infty a_k(z-c)^k\end{align}
If we derivate $n$ times we have:\begin{align}f'(z)&=\frac{d}{dz}\sum_{k=0}^\infty a_k(z-c)^k\\&=\sum_{k=0}^\infty ka_k(z-c)^{k-1}\\&=\sum_{k=1}^\infty ka_k(z-x)^{k-1}\\&=\sum_{k=0}^\infty(k+1)a_{k+1}(z-c)^k\end{align}
Evaluated at $z=c$ then $f'(c)=a_1$.
Repeat the process $n$ times: \begin{align}f''(z)&=\sum_{k=0}^\infty(k+1)(k+2)a_{k+2}(z-c)^k\\f'''(z)&=\sum_{k=0}^\infty(k+1)(k+2)(k+3)a_{k+k}(z-c)^k\\f^{(n)}(z)&=\sum_{k=0}^\infty\frac{(k+n)!}{k!}a_{k+n}(z-c)^k\end{align}
Evaluate at $z=c$ and $f''(c)=2!a_2,f'''(c)=3!a_3\ldots f^{(n)}(c)=n!a_n$.
So you can get $a_n=\frac1{n!}f^{(n)}(z)|_{z=c}$. |
H: Meaning of "relative likelihood".
I am quoting this from DeGroot's "Probability and Statistics".
Let $X$ and $Y$ be random variables. The joint probability density function(p.d.f.) $f$ defines a surface over the $xy$-plane for which the height
$f(x, y)$ at each point $(x, y)$ represents the relative likelihood of that point. For
instance, if it is known that $Y = y_0$, then the point $(x, y)$ must lie on the line $y = y_0$ in
the $xy$-plane, and the relative likelihood of any point $(x, y_0)$ on this line is $f(x, y_0)$.
I don't know the meaning of "relative likelihood".Please explain.
AI: I think the quote is using the term "likelihood" rather loosely. Likelihood is a separate concept from probability. What the joint density function gives at each point is a....density (hence the name). I don't know why the text said relative likelihood, which is $L_{relative}(a,b)=\frac{L(a)}{L(b)}$, where the likelihood function is evaluated given a set of data with the parameters becomming the variables: i.e., L(a|X) = P(X|a), where X is a set of data or outcomes and a is a parameter of the distribution.
The quote above appears to be talking about the conditonal probability of x given $y_{0}$, which is $p_{X|Y}(x|y_{0}) = p_{XY}(y_{0},x)/\int_x p_{XY}(y_{0},x) dx$, in that case the joint density evaluated at $y_{0}$ is acting in a similar manner to the likelihood function in Bayesian inference, but the two are rather different concepts.
Hope that helps. |
H: how to determine which cell in a grid a point belongs to
I have a square area which is divided into an N X N grid.I need to insert a point (x,y) into this area.I tried to find out if there is a relation between the value of N and x,y coordinates sothat I can say this particular point belongs in the cell (0,3) or some other.
I tried plotting on a graph paper, assuming origin at left bottom corner,and tried to find which cell the point P(5.5,4.5) belongs to. When N=2 , it seems that the point P should be in cell(1,1) .If N=6, the point P belongs in cell(5,4) of grid.
I could not make out a relation between the values of N,coordinates of point and grid cell indices.Can someone please point me in the right direction?
AI: It depends on the size of the square. If the square $S$ consists of the points having $0 \le x,y < a$ (that is $S = [0,a)^2$), then the cell $C(i,j)$ (with $0 \le i,j < N$) has
$$ C(i,j) = \left\{(x,y) \biggm| i\frac aN \le x < (i+1)\frac aN, \quad j \frac aN \le y < (j+1) \frac aN \right\} $$
That gives that
$$ i = \def\fl#1{\left\lfloor#1\right\rfloor} \fl{\frac{xN}a},\; j = \fl{\frac{yN}a} $$
gives the cell of the point $(x,y) \in S$. |
H: Hints for a complex limit: Prove if $\lim_{z \to \infty} f(z)/z = 0$ then $f(z)$ is constant.
(To clarify, I would just like a hint. Please do not give me the answer to this problem. )
The solution to the following problem has really evaded me here:
Problem:
Assume that $f$ is entire and that $\lim_{z \to \infty} f(z)/z = 0.$ Prove that $f(z)$ is constant.
My Thoughts and Work So Far: We know that proving that $f'(z) = 0$ or that for some fixed $c \in \mathbb{C}$, $f(z) = c$ for all $z\in \mathbb{C}$. My first approach was to use Liouville's Theorem; If I could could show that $f$ is bounded then I am done. Since $\lim_{z \to \infty} f(z)/z = 0$, for all $\varepsilon > 0$ there exists a $N \in \mathbb{C}$ so large that if $z \geq N$ then $|f(z)/z| \leq \varepsilon$. Thus, if $C_R$ is the circle of radius $R$ centered at the point $z$, then, as long as z is large enough by Cauchy's Inequality
$$
|f'(z)| \leq \frac{1}{2\pi i} \oint_{C_R} \frac{f(\zeta)}{(\zeta - z)^2}\ d\zeta \leq \bigg | \frac{1}{2\pi i} \bigg | \oint_{C_R} \bigg | \frac{f(\zeta)}{(\zeta - z)^2} \bigg | d \zeta \leq \frac{1}{2\pi} \frac{|\zeta|\varepsilon 2\pi R}{R^2} = \frac{|\zeta|\varepsilon}{R}.
$$
Now taking the limit as $R \to \infty$ (which to me says, "let our circle around our point z dilate to an infinite radius so that it covers all of $\mathbb{C}$)
$$|f'(z)| \leq \lim_{R \to \infty} \frac{|\zeta|\varepsilon}{R} = 0.$$
Thus $f'(z) = 0$ and $z$ was arbitrary, so $f$ must be constant.
Why I Think Im Wrong: I say $z$ was arbitrary, but really it is "any $z \geq N$" which really isn't all that arbitrary.
This is where I am stuck. Am I right, wrong, close, or totally lost? Any hints would be great.
Edit: I am very sorry but I accidentally posted this before I was done typing the problem.
AI: Hint: Use the Cauchy Integral Formula for the second derivative and show that $f'' \equiv 0$. So $f$ is a polynomial of degree at most __, and .... |
H: smallest algebra generated by ring
Let $\Omega\neq \emptyset$ and $\mathcal E$ be a ring in $\Omega$ and $\mathcal R_0 := \mathcal E\cup \{E^c:E\in\mathcal E\}$. Show that $\mathcal R_0$ is the smallest algebra in $\Omega$ containing the ring $\mathcal E$, i.e. $\mathcal R_0=\mathcal R_0(\mathcal E)$.
I have been able to show that $\mathcal R_0$ is indeed an algebra, but I'm not able to show that it really is the smallest one containing $\mathcal E$.
The inclusion $\mathcal R_0(\mathcal E)\subset \mathcal R_0$ is trivial since $\mathcal E\subset\mathcal R_0$ by definition and $\mathcal R_0(\mathcal E)=\bigcap_{\mathcal E\subset\mathcal A, \mathcal A\mathrm{\ algebra}}\mathcal A$.
My attempt to prove the inclusion $\mathcal R_0\subset\mathcal R_0(\mathcal E)$ failed, though:
Let $A\in\mathcal R_0$, then we either have $A\in\mathcal E$ or $A\in\{E^c|E\in\mathcal E\}$. If $A\in\mathcal E$ it is clear that $A\in\mathcal R_0(\mathcal E)$. What I am not able to show is that if $A\in\{E^c|E\in\mathcal E\}$ the inclusion $A\in\mathcal R_0(\mathcal E)$ holds.
Does anyone know a way to show this last part? Or a different way to prove the statement? Thanks.
AI: Hint: If $A \in \mathcal R_0$, then, as $\mathcal R_0$ is an algebra, we have $A^c \in \mathcal R_0$. |
H: What is the asymptote for the positions of the largest Stirling numbers of the second kind?
The infinite lower triangular array of Stirling numbers of the second kind starts:
$$\begin{array}{llllllll}
1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
1 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
1 & 3 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} \\
1 & 7 & 6 & 1 & \text{} & \text{} & \text{} & \text{} \\
1 & 15 & 25 & 10 & 1 & \text{} & \text{} & \text{} \\
1 & 31 & 90 & 65 & 15 & 1 & \text{} & \text{} \\
1 & 63 & 301 & 350 & 140 & 21 & 1 & \text{} \\
1 & 127 & 966 & 1701 & 1050 & 266 & 28 & 1
\end{array}$$
https://oeis.org/A008277
Leaving only the largest elements of the table, we have:
$$\begin{array}{llllllll}
1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
1 & 0 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
0 & 3 & 0 & \text{} & \text{} & \text{} & \text{} & \text{} \\
0 & 7 & 0 & 0 & \text{} & \text{} & \text{} & \text{} \\
0 & 0 & 25 & 0 & 0 & \text{} & \text{} & \text{} \\
0 & 0 & 90 & 0 & 0 & 0 & \text{} & \text{} \\
0 & 0 & 0 & 350 & 0 & 0 & 0 & \text{} \\
0 & 0 & 0 & 1701 & 0 & 0 & 0 & 0
\end{array}$$
The column index of those elements is a sequence starting:
$$a(n) = 1, 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, \
9, 10, 10, 10, 10, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 14, 14, \
14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, \
18, 19, 19,...$$
http://oeis.org/A024417
For the first few terms this agrees with the nearest integer of:
$$\frac{n}{W(n)}-1$$ but starts to differ at $n=54$ and $n=58$.
The plot up to $n = 160$:
Where the sequence $a(n)$ is the blue dots and the asymptotic is the red curve.
So is: $$a(n) \sim \frac{n}{W(n)}-1$$ where $W(n)$ is the LambertW function, a good asymptotic?
Mathematica code for the plot:
Clear[a, n, t, m, aa, bb, x]
a[n_] := (m = Max[t = Table[StirlingS2[n, k], {k, 1, n}]];
Position[t, m][[1, 1]]);
aa = Table[a[n], {n, 1, 160}]
(*From Jean-François Alcover, Nov 15 2011*)
bb = N[Table[n/ProductLog[n] - 1, {n, 1, 160}]]
Show[ListPlot[aa], ListLinePlot[bb, PlotStyle -> Red]]
AI: Wikipedia has a different bound, namely $\frac{n}{\log n}$, and gives a paper that proves this bound. |
H: Chebysev inequality proof
Let $(X,\Omega,\mu)$ be a measure space and let $f$ be an extended real valued measurable function defined on $X$. Proof that
$$
\mu\left(\{x\in X : |f(x)|\geq t\}\right)\leq \frac{1}{t}\int_X |f|~d\mu
$$
for any t>0.
Then, conlude that the measure of the set $$\{x\in X : |f(x)|\geq t\}$$
is finite for every t
AI: Define $A$ as $A=\{x\in X : |f(x)|\geq t\}$. Then we have
$$
\mu\left(\{x\in X : |f(x)|\geq t\}\right)=\int_{A} ~d\mu =\int_{X} \chi_A ~d\mu
$$
where $\chi_A(x)=\mathbf 1(x\in A)$. Then we have:
$$
\mu\left(\{x\in X : |f(x)|\geq t\}\right)=\int_{X} \chi_A ~d\mu \\
= \int_{X} \frac{t}{t}\chi_A ~d\mu \leq \int_{X} \frac{|f|}{t}\chi_A ~d\mu \\
\leq \int_{X} \frac{|f|}{t}~d\mu \\
$$
Another inequality is as follows:
$$
\mu\left(\{x\in X : |f(x)|\geq t\}\right)=\int_{X} \chi_A ~d\mu \\
= \int_{X} \frac{t^2}{t^2}\chi_A ~d\mu \leq \int_{X} \frac{f^2}{t^2}\chi_A ~d\mu \\
\leq \int_{X} \frac{f^2}{t^2}~d\mu \\
$$ |
H: on two dimensional graded vector spaces
I read the following statement:
Consider a graded vector space $V$ with basis $\{a, b\}$ such that $a \in V^2$
and $b \in V^5$.
Does this mean that $V=\bigoplus_{i\geq 0}V^i$ such that all $V^i$ are $0$ except $V^2$ and $V^5$ hence we can simply write $V=V^2\oplus V^5$ and if yes why don't we say directly that $V$ is a two dimensional vector space that can be written as the direct sum of two vector spaces $V=A\oplus B$ where $A$ is a one dimensioanl vector space with basis $\{a\}$ and
$B$ is a one dimensional vector space with basis $\{b\}$.
AI: Graded vector spaces need to retain their grading information. It might be that $V=V^1\oplus V^2$ is isomorphic as a vector space to $V'=V'^1\oplus V'^3$ but they are not isomorphic as graded vector spaces because an isomorphism would have to respect the degrees of elements in the grading. If were to just say $V$ is two-dimensional and isomorphic to the direct sum of $A$ and $B$, as you mention, then you have lost the graded structure. |
H: Prove linear operator is a reflection
Prove that a linear operator on $\mathbb{R}^2$ is a reflection if and only if its eigenvalues are $1$ and $-1$ and the eigenvectors with these eigenvalues are orthogonal.
$\Rightarrow$: Let $r: \mathbb{R}^2\ \rightarrow \mathbb{R}^2$ such that $\forall x= \begin{pmatrix} x_1 \\ x_2\end{pmatrix}, r(x)= \begin{pmatrix} \cos \phi & \sin \phi \\ \sin \phi & - \cos \phi \end{pmatrix} \begin{pmatrix} x_1 \\ x_2\end{pmatrix}$
Let $\lambda \in \mathbb{R}$.
$\det\begin{pmatrix} \cos \phi- \lambda & \sin \phi \\ \sin \phi & - \cos \phi- \lambda \end{pmatrix}=0 $
and after calculation we find that $\lambda= ±1$
But now I am having some trouble.
Let $\lambda=1$
$\begin{pmatrix} \cos \phi- 1 & \sin \phi \\ \sin \phi & - \cos \phi- 1\end{pmatrix} \begin{pmatrix} x_1 \\ x_2\end{pmatrix}=0$
$\Leftrightarrow \begin{cases} (\cos \phi-1)x_1+\sin \phi x_2=0 \\\sin \phi x_1-(\cos\phi+1)x_2=0\end{cases}$
I don't seem to see what $x= \begin{pmatrix} x_1 \\ x_2\end{pmatrix}$ works.
Another question that I have is the following: if I find the two eigenvectors associated with each eigenvalue, which matrix do I have to verify that it is orthogonal. Since the eigenvectors found in each case are of order $2 \times 1$, they can't be orthogonal. So I suppose that the matrix which I have to verify is the one whose first column is constituted of the eigenvector associated with $\lambda=1$ and the second column with $\lambda=-1$.
I haven't tried the converse yet, so please don't give me a hint on that.
AI: If you find non-$0$ vectors $x,y\in\Bbb R^2$ such that $r(x)=x$ and $r(y)=-y,$ then to verify that they are orthogonal, you need only take their dot product.
As for actually finding them, proceed by cases. If $\sin\phi=0,$ then you shouldn't have any trouble, so suppose not. Solving your first equation for $x_2$ gives $$x_2=\frac{1-\cos\phi}{\sin\phi}x_1,$$ whence substitution into your second equation yields $$x_1\sin\phi-\frac{1-\cos^2\phi}{\sin\phi}x_1=0\\x_1\sin\phi-\frac{\sin^2\phi}{\sin\phi}x_1=0\\0=0.$$ Uninformative, at first glance. What that means, though, is that you have a free variable (in fact, you have to, since the eigenspace associated to $1$ is one-dimensional). Thus, you can take any $x_1$ you like, then put $x_2=\frac{1-\cos\phi}{\sin\phi}x_1,$ and you're set!
Addendum: Personally, I'd let $x_1=\sin\phi,$ for simplicity, for then $x_2=1-\cos\phi.$ In fact, $x=\begin{pmatrix}\sin\phi\\1-\cos\phi\end{pmatrix}$ works even in the case that $\sin\phi=0,$ so that's a nice general solution that lets you out of having to proceed by cases! |
H: Find an Inverse function
I need to find the inverse of those functions:
$x \mapsto \sin e^{x}$
$x \mapsto e^{\sin x}$
I know that the way is to solve the equation $y = f(x)$ for $x$, and I did it with functions like ($x \mapsto x^2$, $2x+3$) but I can't do it with those too.
Thanks for help!
AI: You are looking for a function that when composed yields the identity, so $(f\circ f^{-1})(x)=(f^{-1}\circ f)=x$. In this case we have composed function or $f\circ g$, what is the inverse? It is $g^{-1}\circ f^{-1}$.
For problem 1 $f(x)=\sin(x)$, $g(x)=e^x$, now what are the inverses of these functions?
For $\sin(x)$ we have $\arcsin(x)$ and for the exponential function the inverse is the natural logarithm ln($x$).
Composing the inverses we then get $(g^{-1}\circ f^{-1})(x)$=ln($\arcsin(x)$).
Double check to convince yourself that this is correct and then attempt problem 2 using the same logic. |
H: Calculate the 1 in a value
I am getting different values from a computer program that I designed. I then want to formulate an algorithm to calculate the 1th value below it (and I'm not really sure the terminology for this so bear with me).
Examples if I have:
43535 the result = 10000,
76765 = 10000,
5674 = 1000,
344 = 100,
56 = 10,
8 = 1,
0.435 = 0.1, etc.
If someone could educate me a little that would be most appreciated!
AI: These numbers are called powers of ten.
To decide which one, calculate M=floor(log(N)/log(10)).
Lastly, if you want, calculate 10^M. |
H: A common eigenvector of $A^2$ and $A^{-1}$
Show that the eigenvector of $A^2$ is the same as the one of $A^{-1}$, where
$$A = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}.$$
All that I can find is that
$$A^2 = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}, \quad A^{-1} = \begin{bmatrix} 1/2 & -1/4 \\ 0 & 1/2 \end{bmatrix}.$$
The eigenvalue of $A^2$ is $4$, and of $A^{-1}$ is $1/2$.
I don't know how to do the rest.
*Edit
According to my lecturer's note,
$$(A-\lambda I)X = 0 \rightarrow \begin{bmatrix} a-\lambda & b \\ c & d-\lambda \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
If I use $A^2$ 's eigenvalue which is $\lambda = 4$.
By using the note provided,
$$\begin{bmatrix} 4-\lambda & 4 \\ 0 & 4-\lambda \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix} 4-4 & 4 \\ 0 & 4-4 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix} 0 & 4 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$
Then I'm stuck here given what I know and what I've done is correct...
AI: Hint: The actual form of $A$ doesn't even matter. All you need is that $A$ is invertible. Then you have
$v$ is eigenvector of $A^{-1}\Leftrightarrow$ $v$ is eigenvector of $A\Leftrightarrow$ $v$ is eigenvector of $A^{2}$
One of the four directions: assume $$Av=\lambda v$$ then $$\lambda^{-1}v=\lambda^{-1}A^{-1}Av=\lambda^{-1}A^{-1}\lambda v=A^{-1}v$$
Note that $A$ being invertible implies that $\lambda\neq 0$. |
H: Define domain $X,$ predicate $A(X)$ and $B(X)$
I'm having trouble creating a domain $X$ and the predicates $A(X)$ and $B(X)$ to for this set of sentences to be evaluated to be true or false.
$(T)\quad \forall x \in X, (A(x) \rightarrow B(x))$
$(F)\quad \exists x \in X (A(x) \land B(x))$
I've tried letting X be the set of all natural numbers, A(x): "x is even", B(x): "x is odd" as it worked in past questions. However, I'm stumped on this one. I also want to know if I can use a more generic domain and predicate rather than being limited to numbers? If so, how? Any help is appreciated!
AI: Define $A(x):$ "x is $\dfrac 13$".
Define $B(x):$ "x is an integer."
And let the domain be the set of integers. Then the first sentence is true, but the second sentence is false. The first sentence is true vacuously, because $\frac 13 \notin \mathbb Z$, and an implication is true whenever the antecedent is false. The second is false, since $\lnot \exists x \in \mathbb Z$ such that $x = \frac 13$, and a conjunction is true if and only if both conjuncts are true. |
H: Prove that every integer ending in 3 or 7 has a prime factor that also ends in 3 or 7
Prove that every integer ending in 3 or 7 has a prime factor that also ends in 3 or 7.
I have that such an integer n has n=3 or 7(mod 10) but don't know where to go from there.
Then show that there are infinitely many prime numbers n with n=3 or 7 (mod 10)
AI: HINT:
First of all, any integer$(N)$ ending in $3$ or $7$ will not be divisible by $2,5$(why?)
So, the factors of $N$ must be $\equiv1,3,7,9\pmod {10}$
Observe that $1\cdot1 \equiv1, 9\cdot9\equiv1, 1\cdot9\equiv9 \pmod{10}$
So, the product of the primes(or their powers) $\equiv1,9\pmod{10}$ will be $\equiv1$ or $9\pmod{10}$ |
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