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H: When can a mathematician have $3$ friends?
At a mathematics convention there are $n$ mathematicians. Each one has $3$ friends (friendship is symmetric). For what values of $n$ is this possible?
AI: For all even $n$ except $2$.
$n$ cannot be odd because if there are $k$ friendship-pairs, we must have $2k=3n$ (by counting person-pair incidences in two ways).
Also, clearly $n=2$ is impossible.
If $n>2$ (or $n=0$) is even, arrange $n$ mathematicians in a circle and assume that everyone is friend with their left and their right neighbour and additionally with the one diametrically opposite. |
H: If $(a_{2n+1})$ and $(a_{2n})$ converge to $a$ then $(a_n)$ converges to $a$.
If $(a_{2n+1})$ and $(a_{2n})$ converge to $a$ then $(a_n)$ converges to $a$.
So far I realize that if $(a_{2n+1})$ and $(a_{2n})$ converge then for each $\epsilon>0$, there exists $N$ such that for all $n>N$, $|a_{2n+1}|, |a_{2n}| < \epsilon$. As eventually even and odd values of $n$ converge (which is to say that eventually for all natural numbers $n>N$) $a_n$ converges.
I just don't know how to prove it. Perhaps I could show that the set of all odd and even numbers is the set of Natural numbers but I'm struggling to prove it properly.
P.S. Sorry for the simplicity of the question, I'm not used to proving such things.
AI: By assumption, for $\epsilon>0$ there exists $N_o$ such that $n>N_o$ implies $|a_{2n+1}-a|<\epsilon$ and there exists $N_e$ such that $n>N_e$ implies $|a_{2n}-a|<\epsilon$.
I claim that for $N:=\max\{2N_o+1,2N_e\}$ we have $|a_n-a|<\epsilon$ for all $n>N$.
Indeed, assume $n>N$. Then either $n$ is even or $n$ is odd. In the first case $n=2m$ for some $m$ and $m=\frac n2 > \frac N2\ge \frac{2N_e}2=N_e$, hence $|a_n-a|=|a_{2m}-a|<\epsilon$. In the other case $n=2m+1$ for some $m$ and $m=\frac{n-1}2>\frac{N-1}2\ge\frac{(2N_o+1)-1}2=N_o$, hence $|a_n-a|=|a_{2m+1}-a|<\epsilon$. |
H: Describing a simple set
If I were asked to describe set A if:
$(∀x ∈ A)(∃n ∈ N) (2n = x)$
Would I be correct in saying that: for all values of x in A, there exists a natural number that is half of x.
Is there a way I can improve this?
AI: HINT: Every element of $A$ is an insert term here natural number. |
H: Show that $d(x,y)=\min \{1,|x-y|\}$ is a metric on $\mathbb R$
Is the following just a matter of showing the 3 properties that make up a metric??
Define d on $\Bbb R\times\Bbb R$ by $d(x,y)=\min \{1,|x-y|\}$. Show that $d$ is a metric on $\Bbb R$
$d(x,y)=0$ if $x=y$
$d(x,y)=d(y,x)$ for every $x,y \in X$
$d(x,y)\le d(x,z)+d(z,y)$ for every $x,y,z \in X$
AI: That's correct (though your first condition should read "if and only if $x=y$"). This is what is known sometimes as the standard bounded metric induced by the metric $\rho(x,y)=|x-y|$. More generally, if $\langle X,\rho\rangle$ is any metric space, then the function $$d(x,y)=\min\{1,\rho(x,y)\}$$ is again a metric on $X$, called the standard bounded metric induced by $\rho$, and induces the same metric topology.
To prove that it is in fact a metric, the first two properties are relatively straightforward. For the triangle inequality, note that if $\rho(x,y)\ge1$ or $\rho(y,z)\ge 1,$ then $$d(x,y)+d(y,z)\ge1\ge d(x,z).$$ Otherwise, we have $$d(x,y)+d(y,z)=\rho(x,y)+\rho(y,z)\ge\rho(x,z)\ge d(x,z)$$ by definition of $d,$ since $\rho$ is a metric. |
H: How do I solve this recurrence equation using substitution?
f(1)=1, and f(n) = f(n-1)+2(n-1)
Using substitution, here are the first few steps:
f(n-1) = f((n-1)-1) + 2((n-1)-1)
f(n-1-1) = f((n-1-1)-1-1) + 2((n-1-1)-1-1)
And then eventually I see that f(n+(-1)*2^j) = f(n+(-1)*2^(j+1)) + 2n + 2(-1)*2^(j+1), where j is an increasing integer >=1
What do I do now? It looks like 2(-1)*2^(j+1) will diverge to negative infinity...
The answer is f(n) = (n-1)n + 1 (using wolfram) but i have no idea what they did..
AI: This is how it expands:
$$f(n) = f(n - 1) + 2(n - 1)$$
$$f(n) = f(n - 2) + 2(n - 2) + 2(n - 1)$$
And you can see that we’ll get to $f(1) = 1$ eventually, and that will be when $n - a = 1$. So what this is actually saying is:
$$f(n) = 1 + 2(n - (n - 2)) + … + 2(n - 1)$$
i.e.
$$f(n) = 1 + 2(1) + … + 2(n - 1)$$
So one plus twice the sum of $1$ to $n - 1$ inclusive, which is one plus twice the $n-1$th triangle number. The nth triangle number is $\dfrac{n^2 + n}2$, so the whole thing is:
$$1 + 2\left(\dfrac{(n-1)^2+(n-1)}{2}\right)$$
$$1 + (n - 1)^2 + n - 1$$
$$(n - 1)^2 + n$$
$$n^2 - 2n + 1 + n$$
$$n^2 + 1 - n$$
$$(n - 1)n + 1$$
… which is indeed what Wolfram got you! |
H: What finite fields are quadratically closed?
A field is quadratically closed if each of its elements is a square.
The field $\mathbb{F}_2$ with two elements is obviously quadratically closed.
However, testing some more finite fields with this property, I didn't find any more. Hence my question is:
Which finite fields $\mathbb{F}_{p^n}$ are quadratically closed and
why?
AI: Consider the squaring map from the multiplicative group of a finite field $F$ to itself. The kernel is $\{\pm1 \}$, i.e., it is trivial if and only if the characteristic of $F$ is $2$. Since this map is surjective if and only if it is injective, every element of $F$ is a square if and only if the characteristic of $F$ is $2$. |
H: Two differentiable functions
Prove or disprove the following statement: Let $f,g:[a,b]\to\mathbb{R}$ be differentiable with $g'\neq 0$, then there exists $c\in (a,b)$ s.t.
$$\frac{f(a)-f(c)}{g(c)-g(b)}=\frac{f'(c)}{g'(c)}.$$
I think this statement is true, it looks very similar to Cauchy's mean value theorem and I believe it might be used in the proof. But I cannot go further at this point. Any idea?
AI: Define $h(x)$ as follows:
$$
h(x)=(f(a)-f(x))(g(x)-g(b))
$$
clearly $h(a)=h(b)=0$ and therefore there is $c\in(a,b)$ such that $h'(c)=0$, hence:
$$
h'(c)=(f(a)-f(c))g'(c)-f'(c)(g(c)-g(b))\implies \frac{f'(c)}{g'(c)}=\frac{f(a)-f(c)}{g(c)-g(b)}
$$ |
H: Prove that it's a subspace of $\mathbb{R}^3$
I have the following definition of $V_a$:
$$V_a := \{(x, y, z)^T \in \mathbb{R}^3 : y = 3x - az\}, \quad \text{for $a \in \mathbb{R}$}.$$
My first problem: I don't understand this definition. Which role does $a$ play in this formula? How do the vectors look like? Why there's only $y$ given, what about $x$ and $z$?
Second problem (which is no wonder taking into account the first one): How can I find a basis of $V_a$ for all $a \in \mathbb{R}$
And one more: How can I prove that $V_a$ is a subspace of $\mathbb{R}^3$?
Please, help me to understand and to solve these problems.
AI: For every $a \in \mathbb{R}$ you take, you get a different $V_a$. For example,
$$V_0 = \{ (x,y,z)^T \in \mathbb{R}^3 \colon x,y,z \in \mathbb{R}, \ y = 3x \} = \{ (x,3x,z)^T \colon x,z \in \mathbb{R} \}.$$
For a general $a$, you have
$$V_a = \{ (x,y,z)^T \in \mathbb{R}^3 \colon x,y,z \in \mathbb{R}, \ y = 3x-az \} = \{ (x,3x-az,z)^T \colon x,z \in \mathbb{R} \}.$$
Note that
$$(x,3x-az,z)^T = x(1,3,0)^T + z(0,-a,1)^T$$
for all $x,z \in \mathbb{R}$. So, the set $\{(1,3,0)^T, (0,-a,1)^T\}$ spans $V_a$ for every $a \in \mathbb{R}$. To prove that it's basis, you need to show that these two vectors are linearly independent.
I'll let you take over now. Feel free to ask if you get stuck. |
H: If $\gcd(f(x), g(x)) = 1$, then $\gcd(h(x)f(x), g(x)) = \gcd(h(x), g(x))$
This is not homework, but I would just like a hint. The question asks
Let $f(x), g(x), h(x) \in F[x]$ (where $F$ is a field), and $\gcd(f(x), g(x)) = 1$. Show that $\gcd(f(x)h(x), g(x)) = \gcd(h(x), g(x))$.
Say $d(x) = \gcd(f(x)h(x), g(x))$ and $t(x) = \gcd(h(x), g(x))$. Then $t(x) \mid h(x)$ and $t(x) \mid g(x)$, so $t(x) \mid f(x)h(x)$ and $t(x) \mid g(x)$, which implies $t(x) \mid d(x)$. If I can show that $d(x) \mid t(x)$ then I'd be done, but I am having trouble with this part. Could someone give me just a hint please?
AI: Since $\gcd(f(x), g(x)) = 1$, there exists polynomials $u(x), v(x) \in F[x]$ such that
$$ 1 = f(x)u(x) + g(x)v(x), $$
which implies
$$ h(x) = f(x)h(x)u(x) + g(x)h(x)v(x). $$
Since $d(x) \mid h(x)f(x)$ and $d(x) \mid g(x)$, we have
$$ h(x) = d(x)m(x)u(x) + d(x)n(x)h(x)v(x) $$
for some $n(x), m(x) \in F[x]$. Now $d(x) \mid h(x)$ and $d(x) \mid g(x)$, which implies $d(x) \mid t(x)$. Coupled with the result $t(x) \mid d(x)$, as well as the fact that $d(x)$ and $t(x)$ are monic, this implies $d(x) = t(x)$. |
H: For what values of a and b does the following limit equal 0?
I understand I need to make the sum of the individual limits equal 0 - but I'm a little lost. I computed the limit of the first term to be -4/3 via L'Hospitals Rule but Wolframalpha contradicts me (http://www.wolframalpha.com/input/?i=limit+x-%3E+0+%28sin%282x%29%2F%28x%5E3%29%29).
'a' obviously should be left for last - and $b/($x^2) is some constant - so I get 0? Assuming 'b' is positive, or anything other than 0, I get that term is 0.
Thus, -4/3 + 0 + a = 0
Simple algebraic manipulating would lead me to believe a = 4/3 and then I'd plug that back in to find 'b'.
Questions
Why does Wolfram state what it does?
Is this solution correct? (I don't have the answer for the problem.)
AI: Hint: you're correct that $a$ can be taken care of when you have found
$$
\lim_{x\to0}\frac{\sin(2x)+bx}{x^3}
$$
that you can compute by applying l'Hôpital's theorem. But you have to compute derivatives right.
For instance, just to show your computations were wrong,
$$
\lim_{x\to0}\frac{\sin(2x)}{x^3}=
\lim_{x\to0}\frac{2\cos(2x)}{3x^2}=\infty
$$
You can't go on with l'Hôpital here, because it's not a $0/0$ form any more. |
H: Prove that $1^2 + 3^2 + 5^2+\cdots+(2n-1)^2 = (4n^3-n)/3$ for all $n \in \mathbb{N}$
Prove that $1^2 + 3^2 + 5^2+\cdots+(2n-1)^2 = (4n^3-n)/3$ for all $n \in \mathbb{N}$.
How can I solve this with induction? I've been working through a couple examples and for this one I can't relate the base case to the induction hypothesis.
I realize the base case is $n = 1$, which I check by putting $n = 1$ directly into $(4n^3-n)/3$ and $(2n-1)^2$, which proves the base case.
Then I tried to compose the last two terms of the sequence by: $\cdots + (2k-3)^2 + (2k-1)^2$. Am I approaching this correctly? How would I do this? A solution would be helpful as I've tried many other things such as expanding, making an inequality, etc.
Thanks!
AI: The induction step is just a matter of assuming that
$$1^2+3^2+5^2+\ldots+(2k-1)^2=\frac{4k^3-k}3$$
and proving that
$$1^2+3^2+5^2+\ldots+(2k-1)^2+\big(2(k+1)-1\big)^2=\frac{4(k+1)^3-(k+1)}3\;,$$
i.e., that
$$1^2+3^2+5^2+\ldots+(2k-1)^2+(2k+1)^2=\frac{4(k+1)^3-(k+1)}3\;.\tag{1}$$
By the induction hypothesis the lefthand side of $(1)$ is equal to
$$\frac{4k^3-k}3+(2k+1)^2\;,\tag{2}$$
so you need only do the algebra necessary to show that $(2)$ is equal to the righthand side of $(1)$. |
H: What Does $\cong$ (Congruence?) Mean in Linear Algebra
I've tried to look it up, but I don't know what it means to say that $V^* \cong V$. Also, what do I say, "$V$ dual is congruent to $V$"?
AI: In functional analysis there is a concept of the (topological) dual space $V^*$ of a normed vector space which is the set of continuous linear functionals $L:V\to \mathbb{R}$ (or $\mathbb{C}$, but in this case the isomorphism will end up being anti-linear). There is a natural way to equip $V^*$ with a linear structure so that it becomes a normed vector space too.
In very special circumstances, we may find that $V$ as a normed vector space is isometrically isomorphic to its dual $V^*$. In particular, if $V$ is a Hilbert space, then we get this for free. When this happens we write $V^* = V$ abusively, or if the author is more careful $V^* \cong V$.
It would be pronounced "$V$ star is isometrically isomorphic to $V$", or "$V$ is self-dual." |
H: Proving Cauchy Given Given Sequence Terms Arbitrarily Close
I need to show that $\{x_{n}\}$ is Cauchy given that there exists $0<C<1$ s.t. $|x_{n+1}-x_{n}|\leq C|x_{n}-x_{n-1}|$. Intuitively, that statement clearly implies $\{x_{n}\}$ is Cauchy, since it implies the sequence terms become arbitrarily close. But how to make it precise?
Couldn't it also be said from the given information that the sequence is either monotone increasing or monotone decreasing and bounded? Then it would converge, which means it is Cauchy.
Thanks for any assistance!
AI: Expanding robjohn's hint,
if
$|x_{n+1}-x_{n}|\leq C|x_{n}-x_{n-1}|
$,
then
$|x_{n+2}-x_{n+1}|
\leq C|x_{n+1}-x_{n}|
\leq C(C|x_{n}-x_{n-1}|)
\leq C^2|x_{n}-x_{n-1}|
$.
Continuing this,
what can you show about
$|x_{n+3}-x_{n+2}|
$?
$|x_{n+4}-x_{n+3}|
$?
$|x_{n+k}-x_{n+k-1}|
$?
Why is it essential
that
$0 < C < 1$? |
H: More series help, this time a telescopic sum.
$$\sum_{n=2}^\infty \frac{2}{n^2 - 1}.$$
I Tried setting it up as a telescoping sum as $$\frac{2}{n} - \frac{2}{n-1}.$$ but now i'm sure that cannot be correct. Mayhaps I need to complete the square? or pull out an N? I'm sorry for asking so many questions. I just really need to get a grasp on these sort of things...
AI: We know
$$
\frac{2}{n^2-1} = \frac{2}{(n-1)(n+1)} = \frac{A}{n-1} + \frac{B}{n+1}
$$
for some $A,B$. Solving for $A,B$ we find $A=1,B=-1$. Thus we want to calculate
$$
\sum_{n=2}^\infty (\frac{1}{n-1} - \frac{1}{n+1})
$$
which telescopes. All the terms cancel except for $\frac{1}{2-1}$ and $\frac{1}{3-1}$ so the sum is $\frac{3}{2}$. |
H: Alternative way of expressing the limit of a function
When proving that the limit
$$ \lim_{h\to 0} \frac {f(x+h) - f(x-h)}{2h} = f'(x) $$
as $h \to 0$, am I correct in assuming that the limit
$$ \lim_{h\to 0}\frac {f(x)-f(x-h)}{h} $$
as $h \to 0$ is equal to $f'(x)$? If so, why?
Fixed: Put the wrong fraction in the original question. Reversed the numerator.
AI: By definition, we have $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}h.$$ Now, let's make the substitution $k=-h,$ and observe that $h\to 0$ precisely as $k\to 0,$ so that $$f'(x)=\lim_{k\to 0}\frac{f(x-k)-f(x)}{-k}=\lim_{k\to 0}\frac{f(x)-f(x-k)}{k}.$$ Look familiar? |
H: Why don't I need to use chain rule to find the derivative of arccos (x^2)?
I thought that I would need to use chain rule because of x^2 but apparently this is not the case, according to the work-through on calcchat.com.
It shows the derivative as:
[-1 / sqrt(1-x^4)] * 2x
Thanks in advance.
AI: That was the chain rule! The $2x$ is the "derivative of the inside." |
H: Series question, telescoping.
Okay i'm starting to get a handle on this. my new question is $$\sum_{n=1}^\infty \frac{3}{n(n + 3)}.$$
I know i have to use Partial Fraction Decomposition, and what i came up with is
$$\sum_{n=1}^\infty (\frac{1}{n} - \frac{2}{n+3}).$$
Am I on the correct track? It doesn't seem to be working out correctly.
AI: Almost, but
$$
\frac3{n(n+3)}=\frac1{n}-\frac1{n+3}
$$ |
H: Differentiating $f(x)^{g(x)}$
Is there any general rule for what the derivative of $f(x)^{g(x)}$ (where $f(x),g(x)$ are differentiable functions) is in terms of $f(x),g(x),f'(x),g'(x)$.
In other words is there something analogous to product,chain and quotient rules for such expressions?
AI: Indeed there is, note that
$$
f(x)^{g(x)} = e^{g(x)\log f(x)}
$$
so differentiating gives
$$
\frac{d}{dx}(\cdot) =f(x)^{g(x)}\frac{d}{dx}(g(x)\log f(x)) = f(x)^{g(x)}(g'(x)\log f(x) + g(x) \frac{f'(x)}{f(x)})
$$
Make sure to be careful about this derivative existing though. |
H: Showing that intersections are not defined
I'm checking to see why intersections are not defined when looking at the class $A$ defined by:
$$ A = ON \cup [ON]^2\;,$$
where $ON$ is the class of ordinals and $[ON]^2$ the class of unordered pairs of distinct ordinals. Intersections are defined in $A$ if for any $x , y \in A$, we can find a $z \in A$ for which $\forall u \in A[u \in z \leftrightarrow u \in x \wedge u \in y].$
Here's what I've been thinking so far:
I have to find some $x, y \in A$ with the intersection of $x$ and $y$ not being in $A$. If I'm taking $x$ and $y$ in $A$, then they have to either be ordinals, or ordered pairs of ordinals. Since the intersection of two ordinals is an ordinal, I can't choose both $x$ and $y$ to be ordinals. I'm guessing I have to let $x$ be an ordinal and $y$ be an ordered pair.
This is where I'm getting stuck. I've tried several choices for $x$ and $y$, with $x$ being an ordinal, and $y$ being an ordered pair of ordinals, but I'm not getting the result I want.
Can anyone help? Thanks in advance!
AI: HINT: What is $2\cap\{1,2\}$? I’ve left the answer spoiler-protected.
$2\cap\{1,2\}=\{0,1\}\cap\{1,2\}=\{1\}\notin A$. |
H: Convex Function Inequality
Let $f: I \to \mathbb{R}$ where $I$ is an interval. We say that $f$ is convex if for every $a,b\in I$ and every $\lambda : 0<\lambda < 1.\\$
Prove that for any $x\in (a,b)$ $f(\lambda b + (1-\lambda)a) \leq \lambda f(b) + (1-\lambda )f(a) \implies f(x) \leq \frac{(x-a)}{(b-a)}f(b) + \frac{(b-x)}{(b-a)}f(a)\\$
I can see how this works clearly geometrically but I am unsure where to start. It is clear that $\frac{x-a}{b-a}$ and $\frac{b-x}{b-a} \leq 1$. And by the triangle inequality that $f(x)\leq f(a) + f(b)$ but I don't know where to even begin. Can anyone help?
AI: If $$\lambda= \frac{x-a}{b-a},$$ then $$1-\lambda= \frac{b-a}{b-a}-\frac{x-a}{b-a}=\frac{b-x}{b-a}.$$
Mathematical statements are usually Extremely general... the above inequality holds for $\textit{all}$ $0<\lambda<1,$ in particular even for $\lambda =\frac{x-a}{b-a},$ or $\lambda=\frac{b-x}{b-a}.$ |
H: definition of limsup of a function
I already have some idea on what $\limsup_{x\rightarrow\infty} f(x)=\infty$ means but I would like to hear other ideas from the math stack exchange community. Maybe some intuition would be helpful.
AI: If you are talking about $\limsup,$ I assume you have a good bit of familiarity with $\sup.$
As an example: the supremum of a function on the interval $[0,\infty)$ is just its maximum, if the maximum exists, and $\infty$ or some real number if it does not exist (imagine a function going to infinity, or oscillating but getting slightly bigger with each oscillation, never reaching a maximum nor going to infinity). There are tons of other ways such a function can behave, but the supremum will always exist.
However, the supremum doesn't tell you anything about how $f(x)$ behaves as $x\to\infty.$ This is the job of $\limsup.$ $\limsup$ tries to ignore what happens on any finite interval $[0,a]$ and tells you the supremum once you look further and further away. In other words, it is the limit of the supremums as we look further away: $$\limsup_{x\to\infty} f(x) = \lim_{y\to\infty} \sup_{x\geq y} f(x).$$ |
H: $E$ an $n$-dimensional vector space. Find all endomorphisms $f$ of $E$ which satisfy $f\circ f = \operatorname{Id}_E$.
Let $E$ be a vector space of dimension $n$. Find all endomorphisms $f$ of $E$ which satisfy $f\circ f = \operatorname{Id}_E$.
Is trivial that $f = \operatorname{Id}_E$ is a solution, but I don't know how to see others solutions.
AI: Choose an ordered basis $\mathcal{B}$ for $E$. With respect to this basis, for every endomorphism $f$, there is a matrix $A_f \in \operatorname{M}(n, \mathbb{F})$ such that $[f(e)]_{\mathcal{B}} = A_f[e]_{\mathcal{B}}$, where $[\, \cdot\, ]_{\mathcal{B}}$ means the component vector with respect to $\mathcal{B}$. As $f\circ f = \operatorname{id}_E$, we must have $A_f^2 = I$. Such a matrix is an involutory matrix. In addition, every such matrix defines such an endomorphism, so the number of endomorphisms $f$ that satisfy $f\circ f = \operatorname{id}_E$ is equal to the number of involutive matrices. This number depends on the field $\mathbb{F}$. |
H: Determine all group homomorphisms from $\mathbb{Z}_{17}^{\times}$ into $\mathbb{Z}_7^{\times}$.
Determine all the group homomorphisms from $\mathbb{Z}_{17}^{\times}$ into $\mathbb{Z}_7^{\times}$.
I noticed that the group is cyclic and found that the generator of both groups is $\langle 3 \rangle$ but how can I find all the group homomorphism?
AI: HINT: $\Bbb Z_{17}^\times$ is a group of order $16$, and $\Bbb Z_7^\times$ is a group of order $6$. If $\varphi:\Bbb Z_{17}^\times\to\Bbb Z_7^\times$ is a homomorphism, the order of $\ker\varphi$ must divide $16$, and the order of $\varphi[\Bbb Z_{17}^\times]$ must divide $6$. Moreover, $\Bbb Z_{17}^\times/\ker\varphi\cong\varphi[\Bbb Z_7^\times]$, so the product of these two orders is ... what? |
H: Real Analysis question in Second Fundamental Theory of Calc
Define
$$F(x)=\int_1^x \frac{1}{2\sqrt{t}-1} dt \quad \text{for all $x\ge 1$}.$$
Prove that if c>0, then there is a unique solution to the equation
$$F(x)=c, \quad x>1.$$
Attempt at a solution: I am not sure how to "prove" this. What I have so far is that $\int_1^x \frac{1}{2\sqrt{t}-1} dt= \sqrt{x}+\frac{1}{2} ln(2\sqrt{x}-1)-1$. If F(x)=c, and x>1, then F(x) will always be positive so clearly there will be a unique solution to F(x)=c...there must be a trick I am not seeing.
AI: Note that
$$F'(x) = {1\over 2\sqrt{x}+1}$$
for $x > 1$.
The function $F$ is strictly increasing, and $F(0) = 0$. Note that as $x\to\infty$, $F(x) \uparrow\infty$.
Therefore, for all $c > 0$, $F(x) = c$ has a unique solution. |
H: N-binacci numbers and ratios generated by them
I was bored at home today and playing with "n-bonacci" numbers, numbers generated by $$x_0=0,x_1=1,...,x_k=k; x_n=\sum_{i=0}^{n-1}x_i$$
I made an assumption based upon the quadratic responsible for the golden ratio, that the ratio of sequential n-bonacci numbers could be found by solving the $n$-th degree polynomial
$$x^n-\sum_{i=0}^{n-1}x^i=0$$
For example, if $n=2$, then we get as a real root $\frac{1+\sqrt{5}}{2}$, approximately $1.6180339...$. If $n=3$, I let Maple generate the real root of $\frac{1}{3}\sqrt[3]{19+3\sqrt{33}}+\frac{4}{3}\frac{1}{\sqrt[3]{19+3\sqrt{33}}}+\frac{1}{3}$ approximately $1.83929...$
By using Maple, I started solving higher and higher degree polynomials and it seems that this ratio tends to 2 as $n$ gets large. Should I use an induction proof? Or is there a simpler way that I'm missing?
AI: The limiting case is the following series: $x_i=0$ for $i\le 0$, $x_n=\sum_{i\le n} x_i$ for all $n> k$, with $x_1,\ldots, x_k$ arbitrary initial conditions. This series has closed form $$x_{n}=2^{n-k-1}M ~~(\text{for }n>k)$$ where $M=x_1+\cdots+x_k$.
This series is an upper bound for all of yours and has ratio $2$. I provide this as evidence that 2 is the limit; a rigorous proof would require more care than I'm ready to spend at the moment. |
H: Evaluate $\oint_C\frac{dz}{z-2}$ around the square with vertices $3 \pm 3i, -3 \pm 3i$.
I'm having a tough time figuring out $\gamma(t)$ in $\oint_C f(z) dz = \int_a^b f(\gamma(t)) \gamma '(t) dt$. I think I have to split this into four parts, the four separate lines, but from there I'm trying to get things to cancel out so that I could arrive at the answer $2\pi i$. Clearly, we have to consider $3+3i \to -3+3i \to -3 - 3i \to 3-3i$, but I can't seem to find a good starting point.
AI: Cauchy's integral formula (and the homotopy invariance of the contour integral) tells you that for holomorphic $f$, $$\oint_C \frac{f(z)}{z-2} = 2\pi i f(2).$$ Apply this to the constant function $f = 1$. |
H: Is the Cantor set a subset of rational numbers, and is it countable or uncountable?
In Chapter 2 of Rudin's Priniciples of Mathematical Analysis, Rudin takes the Cantor set as an example of a perfect set in $\mathbb{R}^1$ which contains no segment. Here's the construction of the Cantor set and the proof:
2.44 The Cantor set The set which we are now going to construct shows
that there exist perfect sets in $\mathbb{R}^1$ which contain no segment.
Let $E_0$ be the interval $[0, 1]$. Remove the segment $(\frac13,\frac23)$, and let $E_1$ be
the union of the intervals
$$[0,\frac13], [\frac23,1].$$
Remove the middle thirds of these intervals, and let $E_2$ be the union of the
intervals
$$[0,\frac19], [\frac29,\frac39], [\frac69,\frac79],[\frac89,1]$$
Continuing in this way, we obtain a sequence of compact sets $E_n$, such that
(a) $E_1\supset E_2 \supset E_3 \dots $;
(b) $E_n$ is the union of $2^n$ intervals, each of length $3^{-n}$.
The set
$$P=\bigcap_{n=1}^\infty E_n$$
is called the Cantor set. $P$ is clearly compact, and Theorem 2.36 shows that $P$
is not empty.
No segment of the form
$$\left(\frac{3k+1}{3^m},\frac{3k+2}{3^m}\right)\tag{24},$$
where $k$ and $m$ are positive integers, has a point in common with $P$.
Since every segment $(\alpha,\beta)$ contains a segment of the form (24), if
$$3^{-m}<\frac{\beta-\alpha}6,$$
$P$ contains no segment.
To show that $P$ is perfect, it is enough to show that $P$ contains no isolated
point. Let $x \in P$, and let $S$ be any segment containing $x$. Let $I_n$ be that interval
of $E_n$ which contains $x$. Choose $n$ large enough, so that $I_n\subset S$. Let $x_n$ be an
endpoint of $I_n$, such that $x_n\ne x$.
It follows from the construction of $P$ that $x_n\in P$. Hence $x$ is a limit point
of $P$, and $P$ is perfect.
One of the most interesting properties of the Cantor set is that it provides
us with an example of an uncountable set of measure zero (the concept of
measure will be discussed in Chap. 11).
I can follow the proof with some effort, but in the end of this section Rudin claims that the Cantor set is an example of an uncountable set of measure zero. How can the Cantor set be uncountable? Corollary of Theorem 2.13 shows the set of all rational numbers is countable. Theorem 2.8 shows that every infinite subset of a countable set is countable. The elements in the Cantor set are the end points of all the intervals in $E_n$, it follows from the construction of the Cantor set that these end points are all rational numbers. Hence $P$ is a subset of the rational numbers and countable. Is there anything wrong with my reasoning here?
AI: "The elements in the Cantor set are the end points of all the intervals in $E_n$..." This is your mistake. This isn't true. In fact, written in ternary expansion, the elements of the Cantor set are precisely those elements in $[0,1]$ with a ternary expansion consisting of $0$'s and $2$'s (where we note $0.01=0.00\bar{2}\in\mathcal{C}$, but $0.0101\notin\mathcal{C}$, for example). Using this fact, it isn't hard to show that $\frac{1}{4}\in\mathcal{C}$ but $1/4$ is not an endpoint of any interval. |
H: Show that $f:[0,1] \to [0,1]$ is continuous if $f(x) = x^{1/k}$ for any $k \in \mathbb N$
I'm very confused right now and I want to apply the theorem that says " A mapping f of a metric space $X$ into a metric space $Y$ is continuous on $X$ if and only if $f^{-1}(V)$ is open in $X$ for every open set $V$ in $Y$". However, I'm not sure how to implement this, I'm considering an open set $V=(a,b) \in [0,1] \subset Y$ and showing that $f^{-1}(V)$ is also open in $X$ hence $f$ is continuous. Any tips would be appreciated!
AI: Hint: Show $f$ is increasing. Then $f^{-1}((a, b)) = (a^k, b^k)$ which is open.
Added Later: It is enough to show that $f^{-1}((a, b))$ is open as $\mathcal{B} = \{(a, b) \mid 0 \leq a < b \leq 1\}\cup\{\emptyset\}$ is a base for the topology on $[0, 1]$; that is, every open set in $[0, 1]$ can be written as a countable union of sets in $\mathcal{B}$. Now, for any open $V \subseteq [0, 1]$, there are sets $I_n \in \mathcal{B}$ such that $V = \bigcup\limits_{n=1}^{\infty}I_n$, then $$f^{-1}(V) = f^{-1}\left(\bigcup\limits_{n=1}^{\infty}I_n\right) = \bigcup\limits_{n=1}^{\infty}f^{-1}(I_n).$$ If you've shown that the preimage of every open interval is open, then $f^{-1}(V)$ is the countable union of open sets, and is therefore open. More generally, we have the following:
Theorem: Let $X$ and $Y$ be topological spaces, $f : X \to Y$, and $\mathcal{B}$ be a base for the topology on $Y$. If $f^{-1}(V)$ is open in $X$ for every $V \in \mathcal{B}$, then $f$ is continuous. |
H: having trouble finding the inverse.
Let $f(x) = \frac{1}{4}x^3 + x -1$ What is the value of $f^{-1}(x)$ when $x=3$?
First, check to make sure the fuction is stricly monotonic.
$f^{\prime}(x) = \frac{3}{4}x^2 + 1$
$f^{\prime \prime}(x) = \frac{3}{2}x$ Which is a linear function thus $f(x)$ is strictly monotic
Now, this is where I am having troubles finding the inverse. I plugged the formula into wolfram alpha, but the result was not understood by me, thus, I don't understand how the result came about.
I have tried the switch $x$ and $y$, but that is not very helpful.
$y = (1/4)x^3 + x -1$
$y +1 = (1/4)x^3 +x$
$y + 1 -x = (1/4)x^3$
$\sqrt[3]{4(y + 1 - x)} = x$
switch $x$ and $y$, then solve for y
$\sqrt[3]{4(x+1 - y)} = y$
$4(x+1-y) = y^3$
$x + 1 = \frac{y^3}{4} +y$
$x = \frac{y^3}{4} + y - 1$ IS THIS CORRECT?
$3 =\frac{y^3}{4} + y -1$
$4 = (1/4)y^3 + y$
$0 = (1/4)y^3 + y -4$
I have found equation to solve cubic polynomials
I think that is a bit much.
There has to be a better way, is there?
AI: Let $f^{-1}(3) = a$. Then, $3 = f(a)$.
So, this problem simplifies to solving the equation:
$$\frac{1}{4}a^3 + a -1 = 3$$
Rearranging:
$$\frac{1}{4}a^3 + a - 4 = 0$$
$$a^3 + 4a - 16 = 0$$
This is a polynomial with rational coefficients. The only possible (rational) roots are $\pm1, \pm2, \pm4, \pm8, \pm16$. With this, it is probably easiest to plug each one in, and see which is a root.
We can easily see that $a=2$ is a root to that equation. As $a$ is the value we seek, we have our answer. |
H: $0=a \cos (\phi + b) +c \cos (\phi +d)$
I am trying to find the maximum for a problem in continuum mechanics involving a 2D stress tensor. Now I arrived at a point where:
$0 = a \cos \left( \phi + b \right) + c \cos \left( \phi + d \right)$
My constants are $a$, $b$, $c$ and $d$. For which $\phi$ does the right side of the equation become $0$?
As a note on the side: The equation that I really want to solve is:
$0= \left(\sigma_y-\sigma_x\right)\cos\left(2\phi-\arctan \frac{1}{\mu} \right)+2\tau_{xy}\cos\left(2\phi-\arctan\mu\right)$
I'd be happy if someone could point me into the right direction. I've not found a fitting identity equation.
AI: Use the addition formula $\cos(\phi+b)=\cos\phi\cos b-\sin\phi\sin b$ then rearrange to get $p\cos\phi=q\sin\phi$. Then your answer is an arctangent. |
H: Regarding the question of compactness
My question is regarding Compactness and sequential compactness
How to show that the only compact subsets of $X$ are finite sets?
In case of specific sets like $\mathbb{N}$, we can choose cover containing open sets of the form $\mathbb N\setminus \{np: p$ is prime$, n\in \mathbb{N}\}$. Then no finite subcollection of it will cover $\mathbb N$.
But how to start the proof in case of arbitrary sets like $X$?
AI: Suppose that $X$ has the co-countable topology, and let $A$ be an infinite subset of $X$. Since $A$ is infinite, it has a countably infinite subset $A_0=\{a_n:n\in\Bbb N\}$. Let $A_1=A\setminus A_0$, and for each $n\in\Bbb N$ let $U_n=\{a_n\}\cup(X\setminus A_0)$; $X\setminus U_n=A_0\setminus\{a_n\}$, which is countable, so $U_n$ is open. Let $\mathscr{U}=\{U_n:n\in\Bbb N\}$. If $x\in A_1$, then $x\in U_0$, and if $x\in A_0$, then $x=a_n$ for some $n\in\Bbb N$, so $x\in U_n$. Thus, $\mathscr{U}$ is an open cover of $A$. However, if $n\in\Bbb N$, the only member of $\mathscr{U}$ that contains $a_n$ is $U_n$, so no finite subset of $\mathscr{U}$ covers $A$. In fact, no proper subset of $\mathscr{U}$, finite or infinite, covers $A$: if you remove $U_n$ from $\mathscr{U}$, the remaining sets do not cover the point $a_n$. |
H: Monotone Sequence Limit Question
Let $\{a_n\}$ be a monotone increasing sequence that converges to a finite limit. If a monotone subsequence $\{a_{n_k}\}$ (with $n_{k+1}>n_{k}$, and $n_k\rightarrow\infty$) converges to a finite limit as well, are both limits necessarily equal?
AI: Yes. You do not even need the sequence to be monotonic, and if $\{a_n\}$ converges then any subsequence will converge to the same limit. |
H: Big Oh Notation for a Recursive Algorithm
I have a question that I'm unsure of:
Express the complexity of the following method using big-O notation. You must explain how you arrived at your answer. What value is returned by the call fred(1,4)?
int fred(int x, int y)
{
if(y == 0)
return x;
else if((y % 2) == 0)
return fred(2 * x, y / 2);
else
return fred(x, y / 2);
}
It seems that this would never reach the base case and therefore gets stuck in an infinite loop. Am I correct here? Do functions like this have an answer for complexity like O(infinity)?
AI: An algorithm is called algorithm because it always converges. You will see in this simple function $y$ is always decreasing and eventually obtains zero and then the algorithm terminates. In this case,
fred(1,4)=fred(2,2)=fred(4,1)=fred(4,0)=4
To analyse the complexity of the algorithm, note that y reaches its half each iteration so it belongs to $O(\log y)$. |
H: Use the definition of a limit to prove..
Use the definition of a limit to prove the following:
$$\lim_{x\to -2}(x-3x^2)=-14.$$
Our definition: Let $L$ be a number and let $f(x)$ be a function which is defined on an open interval containing $c$, expect possibly not at $c$ itself. If for every $\varepsilon>0$ there exists a corresponding $\delta>0$ such that $o<|x-c|<\delta \Rightarrow |f(x)-L|<\varepsilon$ then we say $f(x)$ has a limit $L$ as $x$ approaches $c$.
Not really sure how I go about doing this.
AI: We want to show that for any $\epsilon\gt 0$, there is a $\delta\gt 0$ such that $|(x-3x^2)-(-14)|\lt \epsilon$ whenever $|x-(-2)|\lt \delta$. So suppose that we are given an $\epsilon$. We show how to produce a suitable $\delta$.
Note that
$$(x-3x^2)-(-14)=(x+2)-3(x^2-4)=(x+2)(7-3x).$$
Thus
$$|(x-3x^2)-(-14)|=|x+2||7-3x|.$$
First of all, we will make sure that $-3\lt x\lt -1$, by picking $\delta\lt 1$. Then $|7-3x|\lt 16$. Also make sure that $\delta\lt \frac{\epsilon}{16}$. So to sum up, we choose $\delta=\min(1,\frac{\epsilon}{16})$. If $|x+2|\lt \delta$, then $|(x-3x^2)-(-14)|\lt \epsilon$. |
H: Definition of $ 1 + \frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{\ddots}}}}$
Is there a definition of $ 1 + \frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{\ddots}}}}$? I am somewhat familiar with continued fractions; that is, I am aware that their convergence depends on whether our input is rational or not. This is not a homework problem, I am just curious about this topic which has gotten little attention in my current studies. If this expression is defined, what is it and does it converge?
AI: Let $A = [a_0; a_1, a_2, \ldots]$ be shorthand for the continued fraction $$a_0 + \cfrac1{a_1 + \cfrac1{a_2 + \cdots}}.$$ It is perhaps the most important theorem of continued fractions that $A$ is always well-defined when the $a_i$ are positive integers. This is an immediate consequence of a more general theorem which states that $A$ converges if and only if the series $$\sum_{i=0}^\infty a_i$$ diverges. (See Theorem 10 (p. 10) in A. Ya. Khinchin, Continued Fractions, University of Chicago Press 1964.)
Since the $a_i$ in this case increase fairly quickly, the partial quotients of the continued fraction itself converge fairly quickly, and a computer calculation of the truncated fraction $[1; 2,3,4,5,6,7,8,9,10,11]$ yields the approximation $$A\approx 1.43312742672231,$$ which is good to 14 decimal places. (I used the cf-evaluate program found here.) Plugging the sequence of digits into OEIS reveals that this is OEIS sequence A060997 and the claim that the value is exactly $$\frac{I_0(2)}{I_1(2)}$$ where $I_0$ and $I_1$ are modified Bessel functions of the first kind. (Unfortunately, OEIS gives neither a proof nor a reference.)
OEIS also claims (again unfortunately without proof; I suppose it follows from the Taylor series expansions for the Bessel functions) that $$A = \frac{\sum_{n=0}^\infty \frac{1}{n!^2}}{\sum_{n=0}^\infty \frac{n}{n!^2}}.$$
It does at least provide an email address for the author of the entry, so you could write and ask him for a reference.
[ Addendum 2016-08-20: Jack D'Aurizio explains the Bessel function thing in https://math.stackexchange.com/a/1871798 .] |
H: Calculating probability of difference of two distributions.
A has normal distribution of scores of students with $X \sim \mathcal N(625, 100)$
B has normal distribution of scores of students with $X \sim \mathcal N(600, 150)$
Now I have to calculate probablity of 2 A students's average marks to be greater than average of 3 B students,if 2 students from A and 3 from B appear in test.
Here what I supppose is i should calculate difference of averages from which i will get rrquired proabability.
$E(X-Y)=625-600=25$
$Var(X-Y)=Var(X)+Var(Y)-CoVar(X,Y)$
Now here i have $Var(x)$ and $Var(Y)$ but i don't have $CoVar(X,Y).
So now lets suppose somehow i calculated $Var(X-Y)
I will say $Z=(X-Y+25)/\sqrt{Var(X,Y)}$
I will say $Pr(X-Y>0)=Pr(Z>{some number i need to calculate after i have Var(X,Y)
}=some probability
So here i got simple probability of distribution of A minus B how will I calculatethe probability of 2 students of A average greater than 3 students of B average.
AI: Let the first group have mean $a$, variance $\sigma^2$, and let the second group have mean $b$, variance $\tau^2$.
We pick $m$ students from the first group, and $n$ from the second group. Let $\bar{X}$ be the average of the $m$ grades from the first group, and $\bar{Y}$ the average of the $n$ grades from the second group. Then $\bar{X}$ has mean $a$, variance $\frac{\sigma^2}{m}$. A similar result holds fpr $\bar{Y}$.
If $\bar{X}$ and $\bar{Y}$ are independent, then $W=\bar{X}-\bar{Y}$ has normal distribution, mean $a-b$, and variance $\frac{\sigma^2}{m}+\frac{\tau^2}{n}$. We want the probability that $W\gt 0$.
Put in the appropriate numbers, and calculate. |
H: RSA encryption without a calculator
I'm doing an RSA encryption and to get part of the solution I need to solve
$$C=18^{17} \pmod{55}$$
How would I solve this problem without a calculator
Thanks in advance
AI: Let's use successive squaring:
\begin{align*}
18^2 = 324 \equiv 49 &\equiv -6 \pmod{55} \\
18^4 \equiv (-6)^2 &\equiv 36 \pmod{55} \\
18^8 \equiv 36^2 \equiv 1296 &\equiv 31 \pmod{55} \\
18^{16} \equiv 31^2 \equiv 961 &\equiv 26 \pmod{55}
\end{align*}
Now we have
$$18^{17} = 18^{16 + 1} = 18^{16} \cdot 18 \equiv 26 \cdot 18 \equiv 28 \pmod{55}$$
This approach never necessitates using numbers larger than $55^2 = 3025$, and can be done by hand easily. In fact, by allowing negative results in our computations, we can only deal with numbers between about $-30$ and $30$, which is even easier. |
H: Gcd of every other Fibonacci number
Let $f_n$ be Fibonacci Sequence.
$$gcd(f_{n},f_{n+2})=1,\quad \forall\,n\in\mathbb{N}.$$
Prove
Could you help me with this one? I have done the base case, I just can't figure out the inductive step. Thanks!
AI: Suppose that for a particular $k$ we have $\gcd(f_k,f_{k+2})=1$. We will show that $\gcd(f_{k+1},f_{k+3})=1$.
Suppose that $d$ divides $f_{k+1}$ and $f_{k+3}$. Then since $f_{k+2}=f_{k+3}-f_{k+1}$, we have $d$ divides $f_{k+2}$. But then because $f_{k}=f_{k+2}-f_{k+1}$, it follows that $d$ divides $f_k$.
This forces $d=1$ by the induction assumption. |
H: Let $I$ be an ideal of a commutative ring $R$. Let $1\in I$. Prove $I=R$.
Let $I$ be an ideal of a commutative ring $R$. Let $1\in I$. Prove $I=R$.
I think if $1\in I$, then the ring $R$ has unity, but I'm not sure where to go from there.
Any help/hints would be extremely helpful. ^_^
AI: By the definition of an ideal,
$$RI \subseteq I$$
In particular,
$$I \subseteq R = R1 \subseteq RI \subseteq I$$ |
H: Fourier transform of a real function is real
I was trying to find the Fourier transform of the function
$$x \mapsto \frac1{x^2 - 2x +2}$$
and I keep getting something with non-zero imaginary part. But the Fourier transform of a real function should be real, right? So I must be making a mistake?
What is a proof that the Fourier transform of a real function is real?
AI: The following are equivalent:
$f(-x)=\overline{f(x)}$ for a.e. $x\in\mathbb R$
$\hat f(\xi)\in\mathbb R$ for a.e. $x\in\mathbb R$
The proof is immediate from
$$
\overline{\hat f(\xi)} = \int_{\mathbb R} \overline{f(x)}e^{2\pi i x \xi}\,dx
= \int_{\mathbb R} \overline{f(-x)}e^{-2\pi i x \xi}\,dx
$$ |
H: Let $f(x)$ be a 3rd degree polynomial such that $f(x^2)=0$ has exactly $4$ distinct real roots
Problem :
Let $f(x)$ be a 3rd degree polynomial such that $f(x^2)=0$ has exactly four distinct real roots, then which of the following options are correct :
(a) $f(x) =0$ has all three real roots
(b) $f(x) =0$ has exactly two real roots
(c) $f(x) =0$ has only one real root
(d) none of these
My approach :
Since $f(x^2)=0 $ has exactly four distinct real roots. Therefore the remaining two roots left [ as $ f(x^2)=0$ is a degree six polynomial].
How can we say that the remaining two roots will be real . This will not have one real root ( as non real roots comes in conjugate pairs).
So, option (c) is incorrect. I think the answer lies in option (a) or (b) . But I am not confirm which one is correct. Please suggest.. thanks..
AI: No requirement for $f$ to have real coefficients was stated. So you could have
e.g. $f(x) = (x-1)(x-4)(x+i)$ which has two real roots, and $f(x^2)$ has the four distinct real roots $\pm 1$ and $\pm 2$. |
H: Is the set (0, $\infty$) open?
A set is open if it doesn't contain any of its boundary points. I think 0 is a boundary point here and I think it's the only one. So is the set open?
AI: You've hit the nail on the head. The only boundary point of $(0,\infty)$ is $0$, which is not in the set. Well done! |
H: The limit of $\sin(1/x)$ as $x\to 0$ does not exists
Prove that the following limit does not exist.
$$
\lim_{x\to 0} \sin\left(1 \over x\right)
$$
Our definition of a limit: Let $L$ be a number and let ${\rm f}\left(x\right)$ be a function which is defined on an open interval containing $c$, expect possibly not at $c$ itself. If for ever $\epsilon > 0$ there exists a corresponding $\delta > 0$ such that $0 < \left\vert\,x-c\,\right\vert \left\vert\,{\rm f}\left(x\right) - L\,\right\vert$
Not really sure how I go about doing this?
AI: Hint: Consider $x$ of the form
$$x_n = \frac{1}{2n\pi + \frac{\pi}{2}}$$
with $n$ a positive integer, and then consider
$$y_n = \frac{1}{2n\pi}$$
Then consider, say, $\epsilon = \frac{1}{2}$. Then for any $\delta$, there is an $n$ large enough that $x_n$ and $y_n$ are $\delta$-close to $0$. |
H: How to prove $\frac{1}{m+n+1}-\frac{1}{(m+1)(n+1)} ≤ \frac{4}{45}$?
If $m$ and $n$ are any two positive integers then how do we show that the inequality $$\frac{1}{m+n+1}-\frac{1}{(m+1)(n+1)} ≤ \frac{4}{45}$$ always holds ?
AI: let
$$f(m,n)=\dfrac{1}{m+n+1}-\dfrac{1}{(m+1)(n+1)}$$
then we easy to
$$f(1,1)=f(1,2)=f(2,1)=\dfrac{1}{12}\le\dfrac{4}{45}$$
so,when $m,n\ge 2,$,then we let $j=m+n+2\ge 6$
then
$$f(m,n)=\dfrac{1}{m+n+1}-\dfrac{1}{(m+1)(n+1)}\le\dfrac{1}{j-1}-\dfrac{4}{j^2}\le\dfrac{4}{45}$$
It is clearly,
because $$f(x)=\dfrac{1}{x-1}-\dfrac{4}{x^2}\Longrightarrow f'(x)=-\dfrac{1}{(x-1)^2}+\dfrac{8}{x^3}=\dfrac{8(x-1)^2-x^3}{x^3(x-1)^2}<0,x\ge 6$$let
where $j\ge 6$. |
H: Two soccer team players sit at a round table question.
11 players of Team A and 11 players of Team B sit at a round table, the players from the two teams alternate, the goalkeepers sit together, but the captains do not sit together. In how many different ways can they be seated?
Here is my attempt. assume the goalkeepers are sitting on the same chair (since they must be sitting together). There are 20 spots for them. If one of the captain is sitting next to the goalkeeper,(either left or right side, either from team A or team B) there are 2*2*9 ways for the other captain to sit. If captains are not sitting next to the goalkeepers, there are (8*8-(22-6-1)) ways. and the rest of the players have (8!*8!) ways. So totally 2*2*9*(8*8-(22-6-1))*(8!*8!) ways.
AI: The usual convention about round tables is that arrangements that differ only by a rotation are considered the same.
So we can assume that the goalkeeper from Team A sits in a particular chair. The other goalkeeper has $2$ choices.
Now count the arrangements, without worrying about the restriction on the captains. Fill the seats counterclockwise from the rightmost goalie. The first free chair can be filled in $10$ ways. For every such choice, the next free chair can be filled in $10$ ways. Then the next chair can be filled in $9$ ways, as can the next, and so on, giving a total of $(2)(10!)^2$.
Now we must count the forbidden configurations, in which the two captains (assumed to be non-goalies) sit next to each other.
Start again from the two ways the goalies can sit. The nearest captain (counterclockwise from the right-hand goalie) can be in any one of $19$ chairs. That determines the team membership of that captain, and the position of the other captain. Now count the number of ways to fill in the rest of the positions. This will be $(9!)^2$. So the number of "bad" seatings is $(2)(19)(9!)^2$. Subtract. |
H: For all $X \geq 1$,$\sum\limits_{1 \leq n \leq X} \mu(n) \left[\frac{X}{n}\right] = 1.$
I'm currently sitting with the following number theory problem:
Prove that for all $X \geq 1$, $$\sum_{1 \leq n \leq X} \mu(n) \left[\frac{X}{n}\right] = 1.$$
A few ideas I have tried: Using that $[x]=x - \{x\}$ which I couldn't make out to bring anything. Switching around on the summation variables, like $$\sum_{d \leq N} \mu(d) f\left( \frac{n}{d} \right) = \sum_{d \leq N} \mu \left( \frac{d}{n} \right) f(d)$$ but that didn't seem to bring anything.
I tried writing out the first few terms, which give $$\mu(1) [X] - \left[\frac{X}{2}\right] - \left[\frac{X}{3}\right] - \left[\frac{X}{5}\right] + \left[\frac{X}{6}\right] - \left[\frac{X}{7}\right] - \left[\frac{X}{8}\right]$$ but I can't seem to find a system where it simply reduces to one!
What do you think?
AI: Claim: if $f$ is any function, then
$$\sum_{n=1}^X f(n)\left\lfloor\frac{X}{n}\right\rfloor = \sum_{n=1}^X \sum_{d\mid n} f(d).$$
Proof: By rearrangement. How many times does $f(d)$ appear in the sum on the right-hand side?
Alternatively, proceed by induction on $X$ using the fact that $$\left\lfloor\frac{X+1}{n}\right\rfloor - \left\lfloor\frac{X}{n}\right\rfloor$$
equals $1$ if $n\mid (X+1)$ and $0$ otherwise.
Corollary: for all $X\geq 1$, $$\sum_{n=1}^X \mu(n)\left\lfloor\frac{X}{n}\right\rfloor = 1.$$
Indeed, $\sum_{d\mid n} \mu(d)$ is $1$ for $n=1$ and $0$ for $n>1$. |
H: Limits, find a limit that exist in absolute value but not outside the absolute value.
Let I be an open interval that contains the point c and let f be a function that is defined on I except possibly at the point c. Suppose that lim |f(x)| as x->c exists. Give an example to show that lim f(x) as x->c may not exist.
Not really sure of an example for this?
AI: $$f(x) = \left\{\begin{array}{lr} -1 & x < 0 \\ 1 & x > 0\end{array}\right.$$ |
H: Can objects repeat in commutative diagrams?
Are objects allowed to repeat in commutative diagrams? This seems to be necessary when representing endomorphisms such as the morphism $f : X \to X$ in the category $\mathbf{Set}$, such as when $f$ is a constant function? Or in the category of binary relations, the morphism $R = \{(x, x)\} \subseteq X \times X$?
That is, can commutative diagrams include parts that look like $X \xrightarrow{f} X$ or $X \xrightarrow{R} X$?
The formal definition of a diagram as a functor from an index category $J$ to a category $C$ seems to allow this if different objects in $J$ are mapped to $X$, but I'd like to confirm if my reasoning is correct.
AI: Yes, of course! A functor $J \to C$ need not be injective on objects.
Another example is when you want to take the product of an object with itself. |
H: Prove that $1<\frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{3n+1}$
Prove that $1<\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n+1}$.
By using the Mathematical induction. Suppose the statement holds for $n=k$.
Then for $n=k+1$. We have $\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}=(\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1})+(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$
we know $\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}>1$
What can we do for $(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$?
AI: You are almost done. Prove that
$$\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}-\frac{1}{k+1}$$ is positive.
To do this it is enough to show that $\frac{1}{3k+2}+\frac{1}{3k+4} \gt \frac{2}{3k+3}$. The left side can be written as $\frac{6k+6}{(3k+2)(3k+4}$. So we want to show that $(3k+3)^2\gt (3k+2)(3k+4)$. |
H: Prove this limit as $x\to\infty$
Let $f:(a,\infty)\to\mathbb{R}$ be a function. Suppose for each $b>a$, $f$ is bounded on $(a,b)$ and $\lim_{x\to\infty}f(x+1)-f(x)=A$. Prove
$$\lim_{x\to\infty}\frac{f(x)}{x}=A.$$
Here we assume $f$ to be arbitrary and no further conditions. $f$ could be continuous,discontinuous,as long as it fits all assumptions. I have some trouble with the intermediate steps.
AI: We have that $$\lim_{x\to\infty}f(x+1)-f(x)=A$$ which means that for any $\varepsilon\in\mathbb R^+$ there is an $n\in\mathbb N$ such that for each $x>n$ then $A-\varepsilon<f(x+1)-f(x)<A+\varepsilon$. (if we find some $n<a$ for this condition then we can chose $n=\lceil a\rceil$ for the remind of the proof.)
For simplicity I will define $g(x):=f(x)-Ax$, then we have that $g(x+1)-g(x)=f(x+1)-f(x)-A$ and for $x>n$: $|g(x+1)-g(x)|<\varepsilon$.
As $f$ is bounded in $(a,b)$ for any $b>a$, then $f$ is bounded in $(a,n+2)$ and as we have find $n\ge a$, then $f$ is bounded in $(n,n+1]$, let's call $\beta$ that boundary: $|f(x)|<\beta$ for any $x$ in $(n,n+1]$.
Also, let $\gamma=\beta+(n+1)|A|$, then $g$ is bounded and $|g(x)|<\gamma$ for $x\in(n,n+1]$.
Now, we can prove that
\begin{align}
|g(x)|&\le\gamma+(\lfloor x\rfloor-n)\varepsilon,&&\text{and}\\
\frac{|g(x)|}x &\le\frac{\gamma+(\lfloor x\rfloor-n)\varepsilon}x =\frac\gamma x+\frac{\lfloor x\rfloor-n}x\varepsilon
\end{align}
For any fixed positive $\varepsilon$, we have:
$$\lim_{x\to\infty}\frac\gamma x+\frac{\lfloor x\rfloor-n}x\varepsilon=\varepsilon$$
therefor
$$\lim_{x\to\infty}\left|\frac{g(x)}x\right|=\lim_{x\to\infty}\frac{|g(x)|}x\le\varepsilon$$
This should prove that $\lim_{x\to\infty}\frac{g(x)}x=0$, now the limit for $\frac{f(x)}x$ should follow by having $f(x)=Ax+g(x)$. |
H: Proof of a limit theorem.
Prove the following theorem. Let $I$ be an open interval that contains the point $c$ and suppose that $f$ is a function that is defined on $I$ except possibly at the point $c$. If $m \le f(x) \le M$ for all $x$ in $I \setminus \{c\}$ and $\lim_{x\to c} f(x) = L$, then $m \le L \le M$.
Really not sure here.
AI: Without loss of generality we will prove only the first inequality: $m \le L$. The other follows from a symmetric argument.
Suppose by way of contradiction that $L < m$. This means there is some $\epsilon > 0$ such that $L + \epsilon < m$. By the definition of a limit we know there is a $\delta > 0$ such that for all $x$ satisfying $|x-c| < \delta$ we have $|f(x) - L| < \epsilon$.
We can write the inequality as $$-\epsilon < f(x) - L < \epsilon.$$ Which gives $$f(x) < L + \epsilon < m.$$ However this is a violation of our hypothesis. Therefore $m \le L$ as desired. |
H: Isomorphism between $\mathbb{F}^{m \times n}$ and $\mathcal{L}(V, W)$
Let $\mathbb{F}^{m \times n}$ be the vector space of all $m \times n$ matrices and let $\mathcal{L}(V, W)$ be the vector space of all linear maps from a vector space $V$ to a vector space $W$ ($V$ and $W$ both over $\mathbb{F}$) with $\dim V = n$, $\dim W = m$. Is $M_{B_V, B_W}: \mathcal{L}(V, W) \rightarrow \mathbb{F}^{m \times n}$ an isomorphism for each choice of bases $B_V$ of $V$ and $B_W$ of $W$?
EDIT: Sorry if this question did not seem motivated - my main reason for asking is that I am confused by matrices in $\mathbb{F}^{m \times n}$ whose entries are all the same or in which there is a repetition of columns: in other words, how can such a map be surjective? How can an $a_{ij} \in \mathbb{F}^{m \times n}$ whose entries are all the same or in which there is a repetition of columns correspond to $B_V$ and $B_W$?
AI: $\mathcal{L}(V, W)$ is the set of all linear maps $V \to W$, they need not be injective or surjective. The map $M_{B_VB_W}$ takes a linear map $\varphi : V \to W$ and associates to it the unique matrix $A_{\varphi} \in \mathbb{F}^{m\times n}$ such that $[\varphi(v)]_{B_W} = A_{\varphi}[v]_{B_V}$; here $[\,\cdot\,]_{B_W}$ and $[\,\cdot\,]_{B_V}$ mean the components with respect to the bases $B_W$ and $B_V$ respectively. That is, $M_{B_VB_W}(\varphi) = A_{\varphi}$. If the matrix $A_{\varphi}$ has repeated columns, that just means the map $\varphi$ is not injective. |
H: Circles in Complex Planes
Points on the circle centre C and radius r are given by the equation $|Z-C|=r$ or $(Z-C)(\overline{Z}-\overline{C})=r^2$.
Where $Z = x + iy$.
When multiplied out, I understand that we have
$$Z\overline{Z}-C\overline{Z}-\overline{C}Z+C\overline{C}=r^2\;.$$
So the question I am stuck on states that
If $B$ lies on the circle centre $C$ and radius $r$, show that the equation of the tangent at $B$ is:
$$(\overline{B}-\overline{C})Z+(B-C)\overline{Z}={B}\overline{B}- {C}\overline{C}+r^2\;.$$
Can anyone give me pointers?
AI: First step
Show that $Z=B$ solves the equation, so $B$ lies on the assumed tangent line.
$$
\bigl(\bar B-\bar C\bigr)B + \bigl(B-C\bigr)\bar B = B\bar B - C\bar C+r^2 \\
\bar BB-\bar CB + B\bar B-C\bar B = B\bar B - C\bar C+r^2 \\
\bar CB + B\bar B-C\bar B + C\bar C = r^2
$$
So you arrive at the equation for the circle. If $B$ lies on that circle, then it will also lie on the line.
Second step
Show that the only $Z$ which satisfies both the line equation and the circle equation is $B$ itself. Assume $Z$ to lie on the circle, then plug it into the equation of the line.
\begin{gather*}
\bigl(\bar B-\bar C\bigr)Z + \bigl(B-C\bigr)\bar Z = B\bar B - C\bar C+r^2 \\
\bar BZ-\bar CZ + B\bar Z-C\bar Z = B\bar B - C\bar C+r^2 \\
\bigl(\bar BZ+B\bar Z-B\bar B-Z\bar Z\bigr)+\bigl(Z\bar Z-\bar CZ - C\bar Z+C\bar C\bigr) = r^2 \tag{$*$} \\
\bar BZ+B\bar Z-B\bar B-Z\bar Z = 0 \\
B = Z
\end{gather*}
Equation $(*)$ basically states that if $Z$ lies on the given circle around $C$, then $B$ lies on a circle of radius $0$ around $Z$, so the two must be identical. Therefore $Z=B$ is the only intersection of line and circle, and a line which intersects a circle in just one point is a tangent.
Alternative second step
You can even interpret $(*)$ as
$$\lvert Z-C\rvert^2 - \lvert Z-B\rvert^2 = r^2 = \lvert B-C\rvert^2$$
which you can see as the Pythagorean theorem, with $ZC$ the hypothenuse and $ZB$ and $BC$ the legs. So you can tell that $ZB$ is orthogonal to $BC$, which is another characterization of a tangent. |
H: How to draw a set $\{(x,y): y^2 \leq x^2\}$
I want to draw a set $\{(x,y): y^2 \leq x^2\}$. I know what the result should look like (the blue region):
But I don't really see why is that so. I have
$$ y^2 \leq x^2 $$
I am dividing the domain in two: for $x<0$ and $x \geq 0$ and after square root I get:
$$\pm y \leq \pm x$$
that is:
$$y \leq x$$
$$-y \leq x => y \geq -x$$
and that is fine, but I also get:
$$y \leq -x$$
which is below line for $x$ greater than zero. What am I doing wrong?
AI: Since $\sqrt{}$ is an increasing function, $y^2 < x^2$ is equivalent to $\sqrt{y^2} < \sqrt{x^2}$, i.e. $|y| < |x|$. Since $|y| = \max(y, -y)$, that
is equivalent to ($y < |x|$ and $-y < |x|$).
But $-y < |x|$ is equivalent to $y > -|x|$, so the condition is
$|x| > y > -|x|$. That is, for $x \ge 0$, $x > y > -x$, and for $x \le 0$,
$-x > y > x$. |
H: A $2 \times 2$ matrix $A$ such that $A^n$ is the identity matrix
So basically determine a $2 \times 2$ matrix $A$ such that $A^n$ is an identity matrix, but none of $A^1, A^2,..., A^{n-1}$ are the identity matrix. (Hint: Think geometric mappings)
I don't understand this question at all, can someone help please?
AI: Hint: Rotate through $\frac{2\pi}{n}$. |
H: Maximum cardinality
Let $X$ be some set and $P$ be some subset of ${\frak P}(X)$. We can define the smallest cardinal $\frak k$ such that any $A \in P$ has cardinal $\leq \frak k$. Indeed we can consider the set of all cardinals of elements of $P$ and pick the supremum. On the other hand, it is not clear to me that we can define the smallest cardinal $\frak k$ such that any $A \in P$ has cardinal $< \frak k$ (strictly less). Is it possible indeed?
AI: In the context of $\sf ZFC$ the cardinals are well-ordered. It suffices to show that there is a cardinal which is greater than any of the cardinals of the sets in $P$, in which case there is a least such cardinal.
Since $A\in P$ means $A\subseteq X$, this means that all those cardinals are strictly less than $|X|^+$. Therefore, there is a least cardinal $\kappa$ such that $|A|<\kappa$ for all $A\in P$. |
H: Prove $\binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a}$
I want to prove this equation,
$$
\binom{n}{a}\binom{n-a}{b-a} = \binom{n}{b}\binom{b}{a}
$$
I thought of proving this equation by prove that you are using different ways to count the same set of balls and get the same result. But I'm stuck. Help me please.
(Presumptive) Source: Theoretical Exercise 1.14(a), P19, A First Course in Pr, 8th Ed, by S Ross.
AI: First comes a bureaucratic answer.
We have a group of $n$ people. We want to select a committee of $b$ people, and choose $a$ of them to be a steering subcommittee.
The right-hand side counts the number of ways to pick the committee of $b$ people, and then choose the steering subcommittee from this committee.
The left-hand side picks the steering subcommittee first, then the rest of the committee.
Both sides count the same thing, so they are the same.
Or else we want to choose $b$ people from a class of $n$ to go on a trip. Of these $b$ people, $a$ will ride in the limousine, and the rest in an old bus. We can choose the $b$ people, and then choose the $a$ of them who will ride in the limousine.
Or else we can choose the $a$ people who will ride in the limousine , and then pick $b-a$ people from the remaining $n-a$ to ride in the bus. |
H: Mean value property implies harmonicity
It is fairly easy to show that harmonic functions satisfy the mean value property, but it seems harder to show the converse. I've seen the following theorem without proof:
If $u \in C(\Omega)$ satisfies $$u(z) = \frac{1}{|\partial
B_r(z)|}\int_{\partial B_r(z)} u\,dS$$ for all $z \in \Omega$ and
$B_r(z) \subset \Omega$, then $u \in C^\infty$ and $u$ is harmonic on
$\Omega$.
When I try to prove it myself, I got stuck. Could anyone kindly show me how to prove this? Thanks.
AI: Theorem: Let $\Omega \subset \mathbb{R}^N$, $u \in C(\Omega)$ be such that $$\frac{1}{|B(x_0,R)|}\int_{B(x_0,R)}u(y)\ dy = u(x_0) = \frac{1}{|\partial B(x_0,R)|}\int_{\partial B(x_0,R)}u\ dS$$ for every ball $\overline{B(x_0,R)} \subset \Omega$. Then $u \in C^{\infty}(\Omega)$ and it is harmonic
Proof: Consider the standard mollifier: $$\rho(x) := \begin{cases}Ce^{-\frac{1}{1 - \|x\|^2}} & \text{if $\|x\|$ < 1} \\0 & \text{otherwise.}
\end{cases}$$ Here $C$ is a constant such that $\|\rho\|_{L^1} = 1.$ Let $\epsilon > 0$ and consider $$\rho_{\epsilon}(x) = \epsilon^{-N}\rho(x\epsilon^{-N}).$$
Set $\Omega_{\epsilon} = \{x \in \Omega : \text{dist}(x,\partial \Omega) > \epsilon\}$ and define for $x \in \Omega_{\epsilon}$$$u_{\epsilon}(x) = \rho_{\epsilon} * u(x) = \int_{\Omega}\rho_{\epsilon}(x - y)u(y)\ dy.$$ The following is a well know theorem in analysis, if it is new to you you can look for a proof Analysis by Lieb and Loss or anywhere else.
**Theorem:***If $u \in C(\Omega)$, then $u_{\epsilon} \to u$ uniformly on compact subsets of $\Omega$, $u_{\epsilon} \in C^{\infty}(\Omega_{\epsilon})$ and for any multindex $\alpha$ we have $$\frac{\partial^{\alpha}u_{\epsilon}}{\partial x^{\alpha}}(x) = \int_{\Omega}\frac{\partial^{\alpha}\rho_{\epsilon}}{\partial x^{\alpha}}(x - y)u(y)\ dy.$$*
Finally we can proceed with the proof!
Fix $x_0 \in \Omega_{\epsilon}$.
$$u_{\epsilon}(x_0) = \int_{B(x_0,\epsilon)}\rho_{\epsilon}(x - y)u(y)\ dy = \int_{B(0,\epsilon)}\rho_{\epsilon}(z)u(x_0 - z)\ dz = $$ $$ = \int_0^{\epsilon}r^{N - 1}\int_{\partial B(0,1)}\rho_{\epsilon}(rw)u(x_0 - rw)\ dS(w)dr = $$ $$ \int_0^{\epsilon}r^{N - 1}\rho(r)\int_{\partial B(0,1)}u(x_0 - rw)\ dS(w)dr = \int_0^{\epsilon}r^{N-1}\rho_{\epsilon}(r)\frac{\alpha_N N}{|\partial B(x_0,r)|}\int_{\partial B(x_0,r)}u(y)\ dS(y)dr $$ $$ = u(x_0)\|\rho\|_{L^1} = u(x_0).$$
This proves that $u = u_{\epsilon}$ and hence $u \in C^{\infty}(\Omega_{\epsilon})$, for every $\epsilon$. We are left to prove that $u$ is harmonic. To this end consider $$f(r) = \frac{1}{|\partial B(x_0,r)|}\int_{\partial B(x_0,r)}u\ dS.$$
By assumption, $f$ is constant, hence $f' \equiv 0$. By the divergence theorem it is easy to show (and I am sure that you have already seen this result since you have proved that if $u$ is harmonic then it satisfies the mean value property) $$0 = f'(r) = \text{constant}\int_{B(x_0,r)}\Delta u(y)\ dy \to \Delta u(x_0),\ \text{if we let}\ r \to 0^+.$$ Thus $u$ is harmonic. QED |
H: tank and pipes problem
A tank of 425 liters capacity has been filled with water through two pipes, the first pipe having been opened 5 hours longer than the second. If the first pipe were open as long as the second pipe, the first pipe deliver half the amount of water delivered by second pipe; if the two pipes were open simultaneously, the tank would be filled up in 17 hours. How long was the second pipe open?
a. 10
b. 12
c. 15
d. 18
AI: x = first pipe rate
y = second pipe rate
a = hours of second pipe
$$y = 2x\\
17(x+y) = 425\\
x+y = 25\\
3x = 25\\
x = \frac{25}{3}, y = \frac{50}{3}\\
(a+5)\frac{25}{3} + a\frac{50}{3} = 425\\
(a+5) + 2a = 51\\
3a + 5= 51\\
3a = 46\\
a = 46/3$$
Are you allowed rounding? |
H: Order of $(\mathbb{Z}\oplus \mathbb{Z})/\langle (4,2)\rangle$?
This is what I did:
any element is of the form $(a+4\mathbb{Z},b+2\mathbb{Z})$, where $a=0,1,2,3$ and $b=0,1$. So it the order should be $8$. But the answer is given to be $\infty$.
What is the wrong in the way i did?
AI: It is possibly that you just misread the question, and thought it was about $\def\Z{\Bbb Z}(\Z\oplus\Z)/\langle(4,0),(0,2)\rangle$, where you are modding out be the lattice spanned by those two group elements; for that question the resulting group is indeed isomorphic to $(\Z/4\Z)\oplus(\Z/2\Z)$ and has order $8$. However in the question one is modding out only by the multiples of a single element $(4,2)$ which all lie on a single line. As a consequence, for two elements of $\Z\oplus\Z$ to become equivalent it is necessary that their difference $(a,b)$ satisfies $a-2b=0$ (and even this is not a sufficient condition). In particular any collection of points $(x,y)$ for which the expression $2x-y$ takes different values for all of them are mutually inequivalent. Then the quotient group is certainly infinite, as $2x-y$ can take infinitely many distinct values (even when taking only points with $x=0$). |
H: If $x^2+y^2=z^2$ has a solution then $5$ divides $xyz$
I tried to solve a question but I did not succeed yet...my question is about number theory. Here it is:
How can we show that if the equation $x^2+y^2=z^2$ has a solution then $5$ divides $xyz$?
Can we have a general method for solving questions like that?
I need the beginning of the solution.
I will be very glad if you help.
Thanks
AI: If $5$ divides $x$ and/or $y,$ we are done
Else
Observe that $$(\pm1)^2\equiv1,(\pm2)^2\equiv4\pmod 5$$
If both $x,y\equiv1\pmod 5, x^2+y^2\equiv2\pmod 5\not\equiv z^2$
If both $x,y\equiv2\pmod 5, x^2+y^2\equiv8\pmod 5\equiv3 \not\equiv z^2$
If $x\equiv\pm1,y\equiv\pm2,x^2+y^2\equiv0\pmod 5$
Now prove that if a prime $p$ divide $a^2,p$ must divide $a$
Here $p=5$ |
H: Undefined limit $(0/0)$ for a function of two variables
I am trying to calculate the following limit:
$$ \lim_{(x_1,x_2)\to(1,1)} \frac{c_1 \cdot (x_1^2-1)}{\frac{x_1^{10}}{x_2^2}-1}$$ where $c_1$is a constant.
This is giving $\frac{0}{0}$ and I can't seem a way to escape it.
Can anyone help, please?
AI: If you make a Taylor expansion around $x_1=1$ and $x_2=1$, you will see that you can set $x_2=1$ without any problem. What is left is $c_1(x_1^2-1)/(x_1^10-1)$. If $x_1$ goes to $1$, then the limit is $c_1/5$. Is this of any help?
As Michael said, there is no problem setting $x_2=1$ ane then the problem reduces to $c_1 (x_1^2-1) / (x_1^{10} - 1)$ which is $c_1 / (1+x_1^2+x_1^4+x_1^6+x_1^8)$ then $c_1/5$. |
H: What is the expected number of rounds that everyone get back their own ball?
Sppose there are $n$ people and each of them a unique ball. Suppose now they put the ball inside a box and the balls inside the box would then be mixed up. Then each of them will go out to draw a ball from the box.
Suppose that for each round , all people would draw a ball from the box. Those who draw their own balls can leave and the rest would go to the second rounds. What is the expected number of rounds that everyone get back their own ball?
AI: n.
For each person the expected value is 1/n. The expected value for all n people is therefore 1, by linearity of expectation. If the expected value of each round is 1, then the expected value of n rounds is n. |
H: Represent a Toeplitz matrix in an array
I need to represent a $n \times n$ Toeplitz matrix in a $2n - 1$ array. I need to create a function that takes a pair $(i,j)$ and returns the value in the $2n - 1$ array.
I am having a difficult time trying to calculate the index. Can you please help me?
AI: Store the top row first, then the left column without the first element. (e.g. {a,b,c,d,e,f,g,h,i) from the example in wiki)
Assuming (i, j) start at (0, 0) and end at (n-1, n-1),
if j >= i:
return array[j-i]
else:
return array[i-j+n] |
H: Do you know how can I see any image of graph?
If exist homepage or method, please tell me.
I want see some image of graph like $y=x^2\sin\left(1/x\right)$ or $y=\frac{\ln(x)}{x}$.
Of course, I know that they are not complex, but I want to see it sketched lovely.
I know mathematica, but I don't have it.
AI: There are plenty of options online if you google. One nice alternative is Wolfram Alpha. For one of the plots you want, see here. It should be straightforward to generalise. |
H: Field extension of quotient field
$1$. Let $ F $ be a field and $a,b$ be nonzero element in $L/F$.
How to show that a field extension $F(a,b)/F(a^{−1}b^{−1},a+b)$ is an algebraic extension?
I tried to show all element in $F(a,b)$ is an algeraic over $F(a^{−1}b^{−1},a+b)$ but
I failed....
$2$. Compute $[F(x):F(x^2/(x−1))]$ ($F$ is also a field)
$F(x)$ is a quotient field of $F[x]$, polynomial ring over $F$.
AI: Here are some hints.
Look at the roots of $T^2 - (a+b)T + ab$. Conclude that $a,b$ are algebraic over $F(ab,a+b) = F(a^{-1} b^{-1},a+b)$.
If $a=x^2/(x-1)$, then $x^2 - a (x-1) = 0$. Prove that $T^2-a(T-1)$ is the minimal polynomial of $x$ over $F(a)$. |
H: About proving an abelian group as a ring
This question is the following.
Consider the abelian group under addition of elements of the form $ar_0+br_1+cr_2$ where $a,b,c\in \mathbb Z$ and $r_0,r_1,r_2$ are variables.
1. Define a multiplication on this set that turns it into a ring with the property that $r_i*r_k=r_{i+k}$ where i,k works like elements of $\mathbb Z_3$
2. Show that this is a ring. 3. Does unity exist? If so, find it.
I defined the multiplication to be the following.
For $A,B \in \{ar_0+br_1+cr_2|a,b,c \in \mathbb Z\}$, $A=a_0r_0+a_1r_1+a_2r_2$ and $B=b_0r_0+b_1r_1+b_2r_2$.
$A*B=(a_0b_0+a_1b_2+a_2b_1)r_0+(a_0b_1+a_1b_0+a_2b_2)r_1+(a_0b_2+a_2b_0+a_1b_1)r_2$
Is this the right multiplication that the question asks to me?
And for question 2, Since addition is given as abelian group, I need to show that the multiplication I've defined is associative and the distribution law holds. Right?
Thank you.
AI: Yes, this is right multiplication. In fact, this ring is isomorphic to the integer ring of the group $\mathbb{Z}_3$. |
H: Is each paracompact space with countable cellularity always Lindelöf?
Is each paracompact space with countable cellularity always Lindelöf?
We recall that the cellularity of a space is the minimal infinite cardinal $\kappa$ such that every family of pairwise disjoint open sets has cardinality less than or equal to $\kappa$?
Thanks ahead:)
AI: Yes. This follows immediately from the following lemma.
Lemma: If $X$ is ccc, every locally finite open cover of $X$ is countable.
Proof: Let $\mathscr{U}$ be a locally finite open cover of $X$. For each $x\in X$ let $V_x$ be an open nbhd of $x$ that meets only finitely many members of $\mathscr{U}$. For each $U\in\mathscr{U}$ pick any $x\in U$ and let $V_U=V_x$. Let $\mathscr{V}=\{V_U:U\in\mathscr{U}\}$. Define an equivalence relation $\sim$ on $\mathscr{V}$ by $V\sim W$ iff there are $n\in\omega$ and $V_0=V,V_1,\ldots,V_n=W\in\mathscr{V}$ such that $V_k\cap V_{k+1}\ne\varnothing$ for $0\le k<n$. Each member of $\mathscr{V}$ meets only finitely many members of $\mathscr{U}$, so
each member of $\mathscr{V}$ meets only finitely many members of $\mathscr{V}$ as well, and each $\sim$-equivalence class is therefore countable. The union of each equivalence class is open, and these unions are pairwise disjoint, so there are only countably many equivalence classes. Thus, $\mathscr{V}$ is countable, and therefore $\mathscr{U}$ is countable. $\dashv$ |
H: Game Theory Problem with Dice
I need solution to this game theory problem. It seems impossible to me.
Two players (1 and 2) play the following game. Player 1 must write the numbers from 1 to 18 on the sides of 3 dice without repeating the numbers. Then player two chooses one die and throws it. After that player 1 chooses one of the remaining 2 dice and throws it. Is there a way for player 1 to write down the numbers in a way that he has more than 50% chance of winning.
AI: I believe so. What if the dice were as such?:
A{16, 13, 12, 9, 8, 5}
B{15, 14, 11, 10, 7, 6}
C{18, 17, 4, 3, 2, 1}
Dice A and B have equal chance against each other, but die C has 1/3 chance of winning.
If player 2 chooses C, then player 1 has 2/3 chances of winning, so that is a bad choice for player 2.
If player 2 chooses A and player 1 then chooses B no matter what, or the equivalent alternative B then A, then obviously the chance of winning is 50%. However, player 1 has the advantage of being able to choose die C in the case that player 2 rolls 16, 15, or 14. Therefore, he has a greater chance than 50%.
Note: Remember that 50% is the probability of winning before anyone rolls any die, not after player 2 rolls. There is no way to have a 50% chance of winning after player 2 rolls 18 for example. |
H: Proof that limits stay within the bounds
Prove the following theorem. Let $I$ be an open interval that contains the point $c$ and suppose that $f$ is a function that is defined on $I$ except possibly at the point $c$. If $m \le f(x) \le M$ for all $x \in I \setminus \{c\}$ and $\lim_{x \to c} f(x) = L$, then $m \le L \le M$.
I've got this far but now I am stuck.
Assume that $m \leq f(x) \leq M$ for all $x \in I \setminus \{c\}$ and $\lim_{x \to c} f(x) = L$.
We first prove $m \le L$. Assume, for the sake of contradiction, that $L < m$. This means there is some $\varepsilon > 0$ such that $L + \varepsilon < m$. By the definition of a limit we know there is a $\delta > 0$ such that for all $x$ satisfying $|x−c| < \delta$ we have $|f(x)−L| < \varepsilon$. We can write the inequality as $−\varepsilon < f(x) − L < \varepsilon$. Which gives $f(x) < L+\varepsilon < m$. However this is a false by our hypothesis, therefore $m \le L$, as desired.
Now we prove $L \le M$. Assume, for the sake of contradiction, that $M < L$. This means there is some $\varepsilon > 0$ such that $M + \varepsilon < L$.
AI: You're almost there. Now pick $x\in I$ so that $0\lt|x-c|\lt\delta$. Then you have $f(x)-L\le|f(x)-L|\lt\epsilon$. From $f(x)-L\lt\epsilon$ it follows that $f(x)\lt L+\epsilon\lt m$, contradicting the assumption that $m\le f(x)$ for all $x\in I\setminus\{c\}$. This proves that $m\le L$. Now prove $L\le M$ the same way.
We now prove $L\le M$. Suppse for the sake of contradiction that $L\gt M$. This means there is some $\epsilon\gt0$ such that $L-\epsilon\gt M$. By the definition of a limit we know there is a $\delta\gt0$ such that for all $x$ satisfying $0\lt|x-c|\lt\delta$ we have $|f(x)-L|\lt\epsilon$. Pick $x\in I$ so that $0\lt|x-c|\lt\delta$. Then $L-f(x)\lt|f(x)-L|\lt\epsilon$, so $f(x)\gt L-\epsilon\gt M$, contradicting the assumption that $f(x)\le M$ for all $x\in I\setminus\{c\}$.
Alternatively, it might be more efficient to prove the more general assertion: if $f(x)\le g(x)$ for all $x\in I$, and if $\lim_{x\to c}f(x)$ and $\lim_{x\to c}g(x)$ exist, then $\lim_{x\to c}f(x)\le\lim_{x\to c}g(x)$. |
H: When is a cyclotomic polynomial over a finite field a minimal polynomial?
When is the cyclotomic polynomial $f(x)$ over a finite field $\mathrm{F}_q$ also the minimal polynomial of some element $\alpha \in \mathrm{F}_q$?
AI: In general, a polynomial $f\in F[X]$ is the minimal polynomial of its roots over $F$ if and only if $f$ is irreducible over $F$. Cyclotomic polynomials provide no exception.
However, while cyclotomic polymomials are irreducible over $\mathbb Q$, not all of them are irreducible when considered as a polynomial over a finite field.
For example, over $\mathbb F_2$ the $7$th cyclotomic polynomial $\Phi_7 = X^6 + X^5 + X^4 + X^3 + X^2 + X + 1$ factors into two factors of degree $3$:
$$
\Phi_7 = (X^3 + X + 1)(X^3 + X^2 + 1)
$$
Hence $\Phi_7\in\mathbb F_2[X]$ is not irreducible and therefore not the minimal polynomial of its roots.
Note: This was written as a response to an earlier version of the question. |
H: Injective map from sigma algebras to partitions
I want to show that for every sigma-algebra $\mathfrak A$, you can define for all $x\in X$ the set of all $A_x:=\bigcap_{A\in \mathfrak A,x\in A}A$( a partition of the set $X$) and that this map is injective. Thus, for different sigma-algebras we get different partitions. Does anybody have a good hint how to start with this?
AI: This is not true. Look at the Borel sets on $[0,1]$ and the power set on $[0,1]$ At each point, this intersection is the point itself. They are clearly different $\sigma$-algebras. |
H: Related rates: Find dA/dt of triangle, given d(theta)/dt -- Can't come to textbook answer
The question:
ABC is a triangle in which the lines $\overline {AB} = 20cm$, $\overline {AC} = 32cm$ and $\angle BAC = \theta$. If $\theta$ is increasing at the rate of 2° per minute, determine the rate at which the triangle's area is changing when $\theta = 120°$.
Here's my attempt:
Let $t$ be time in minutes. We're given $\frac{d\theta}{dt} = 2°$, Area of triangle with two sides and included angle:
$A = \frac12\overline {AB}$ $\overline {AC}$ $sin\theta$, i.e.
$A = \frac12(20)(32)sin\theta$
$= 320sin\theta$
$\therefore \frac{dA}{dt} = \frac{dA}{d\theta}\frac{d\theta}{dt} = 320cos\theta\frac{d\theta}{dt}$
i.e. at $\theta = 120°$:
$\frac{dA}{dt} = 320cos(120) * (2)$
$= -320 cm^2/min$
The textbook answer:
Decreasing at $5.59cm^2/min$
The question is exactly as above, and I've double checked the units.
Can anyone please point out where I've gone wrong, or if I haven't? Thank you!
AI: You need to express angles in radians, i.e. $2^{\circ} =2 \pi/180$ radians. |
H: Prove that lim sin (1/x) as x-> 0 does not exist.
Prove that lim sin (1/x) as x-> 0 does not exist.
Not really sure where to go with this, do I approach from both the right and left?
AI: Hint: Try to find two sequences $x_n\to 0$ and $y_n\to 0$ such that, for instance, $\sin(1/x_n)=1$ and $\sin(1/y_n)=0$. |
H: Quantum Hermiticity Bra-Ket notation please
If $A$ and $B$ are Hermitian operators, show that $$C~:=~i[A,B]$$ is Hermitian too.
My work:
$$\begin{gather}
C=i(AB-BA) \\
\langle\psi\rvert C\lvert\phi\rangle = i\langle\psi\rvert AB\lvert\phi\rangle-i\langle\psi\rvert BA\lvert\phi\rangle
\end{gather}$$
$A$ and $B$ are Hermitian such that:
$$\begin{align}
\langle\psi\rvert A\lvert\phi\rangle &= \langle\phi\rvert A\lvert\psi\rangle^* \\
\langle\psi\rvert B\lvert\phi\rangle &= \langle\phi\rvert B\lvert\psi\rangle^*
\end{align}$$
I know a little about the identy operator, which I've seen used to do a similar trick, but I'm not that clear on its exact meaning hmm... $$1=\sum_n\lvert n\rangle\langle n\rvert$$
The definition of Hermiticity I learnt from lectures is the one I stated above for A, can you prove it in this way for C? ie can you use bra-ket notation.
AI: The bra-ket notation is very formal, but well, let's go through it.
We must prove that $\langle \psi | C | \phi \rangle = \langle \phi | C | \psi \rangle ^*$, where the $^*$ is complex conjugation.
First we prove that the adjoint of the composite operator $AB$ is $BA$, that is, that $\langle \psi | AB | \phi \rangle = \langle \phi | BA | \psi \rangle^*$.
For that, notice that
$\langle \psi | AB | \phi \rangle =
\langle \psi | A \, 1 \, B | \phi \rangle =
\sum_n \langle \psi | A | n \rangle\langle n | B | \phi \rangle =
\sum_n \langle n | A | \psi \rangle^* \langle \phi | B | n \rangle^* =
(\sum_n \langle \phi | B | n \rangle \langle n | A | \psi \rangle ) ^* =
(\langle \phi | B \, 1 \, A | \psi \rangle ) ^* =
\langle \phi | BA | \psi \rangle^*$.
By the linearity of the inner product, one has then
$\langle \psi | (iAB) | \phi \rangle =
i(\langle \psi | AB | \phi \rangle) =
(-i)^* (\langle \phi | BA | \psi \rangle)^* =
(\langle \phi | (-iBA) | \psi \rangle)^*$.
Switching $AB$ to $BA$ then gives you:
$\langle \psi | (iAB) | \phi \rangle = (\langle \phi | (-iBA) | \psi \rangle)^*$.
$\langle \psi | (iBA) | \phi \rangle = (\langle \phi | (-iAB) | \psi \rangle)^*$.
Now one has
$\langle \psi | C | \phi \rangle
= \langle \psi | (iAB) | \phi \rangle - \langle \psi | (iBA) | \phi \rangle
= (\langle \phi | (-iAB) | \psi \rangle)^* - (\langle \phi | (-iBA) | \psi \rangle)^*
= (\langle \phi | C | \psi \rangle)^*$.
That concludes the proof.
--
If you want to know what is going on, I recommend you find the definition of the adjoint of an operator. Mathematicians use the star $A^*$, and physicists use the dagger $A^\dagger$. If you know that $A\mapsto A^\dagger$ is antilinear and satisfies the rule $(AB)^\dagger = B^\dagger A^\dagger$, then your exercise can be solved by checking $C^\dagger = (iAB - iBA)^\dagger = -i BA + i AB = C$. |
H: Lebesgue integration on real line
we know that if $A$ is a Lebesgue measurable set, $f$ and $g$ nonnegative, then
$$ f \leq g \implies \int_A f dm \leq \int_A g dm $$
Does the result still follow if we change $\leq$ with $<$ ??
AI: If $A$ is a null set, then we have of course
$$\int_A f\,dm = \int_A g\,dm = 0$$
regardless of $f$ and $g$, so then the strict inequality does not follow. But if $m(A) > 0$, then we have
$$f < g \Rightarrow \int_A f\,dm < \int_A g\,dm.$$
Let $A_n = \{x \in A : g(x) > f(x) + 2^{-n}\}$. Then $A = \bigcup A_n$, the $A_n$ are increasing, so $m(A_n) > 0$ for all large enough $n$. Then
$$\begin{align}
\int_A g\,dm &= \int_{A_n} g\,dm + \int_{A\setminus A_n} g\,dm\\
&\geqslant \int_{A_n} f + 2^{-n}\,dm + \int_{A\setminus A_n} f\,dm\\
&= \int_{A_n} f\,dm + 2^{-n} m(A_n) + \int_{A\setminus A_n} f\,dm\\
&= \int_A f\,dm + 2^{-n}\cdot m(A_n)\\
&> \int_A f\,dm.
\end{align}$$ |
H: Given a vector $\vec x$, are all vectors perpendicular to it constructible from skew-symmetrix matrices multiplied by $\vec x$?
As noted elsewhere, given a skew-symmetric matrix $S$ the vector $\vec x^T S$ is orthogonal to $\vec x$ since
$$\vec x^T S\vec x = -\vec x^T S^T\vec x = -(\vec x^T S\vec x)^T = -\vec x^T S\vec x = 0.$$
But is the converse also true, i.e. $\forall\vec y\perp\vec x\exists S=-S^T: \vec y = S\vec x$?
AI: If the dimension of vector space is no less than 3, then for any $y\in\mathbb R^n (n\geq3)$ such that $\langle x,y\rangle=0$ there's always a skew-symmetric matrix $A$ such that $y=Ax$.
Consider linear equations $y=Ax$, since $A$ is skew-symmetric, it has $\frac{n(n-1)}{2}$ unknowns to determine. On the other hand, we have $n$ equations. So the solution exists if and only if
$$\frac{n(n-1)}{2}-n=\frac{n(n-3)}{2}\geq0$$
which implies $n\geq3$ |
H: Discrete structures exercise
I have this exercise in my worksheet I am a beginner.
Prove or disprove that if $A,B$ and $C$ are sets such that $A\times B = A \times C$ then $B = C$.
AI: The statement is true if $A\neq \emptyset$. If $x \in A$ and $y \in B$, then $(x,y)\in A\times B=A\times C \Rightarrow y\in C$, so $B\subset C$. By a similar way, we can show that $C\subset B$, so $B=C$. |
H: Given a vector $\vec x$, what is the maximum possible rank for a matrix $A$ such that $A\vec x=0$?
This is basically an inverse Eigenvector problem:
Given a vector $\vec x\in\mathbb R^n$ ($\mathbb C^n$), what is the highest rank $m$ a matrix $A\in\mathbb R^{m\times n}$ ($\mathbb C^{m\times n}$) can have such that $\vec x$ is a zero-Eigenvector (i.e. $A\vec x = \vec 0$)?
It is clear that each of the $m$ rows of $A$ must be a vector $\vec y^T$ orthogonal to $\vec x$, and as noted elsewhere, given a skew-symmetric matrix $S$ the vector $\vec x^T S$ is orthogonal to $\vec x$ since
$$\vec x^T S\vec x = -\vec x^T S^T\vec x = -(\vec x^T S\vec x)^T = -\vec x^T S\vec x = 0.$$
Assuming the converse to be true, since there are $n(n-1)/2$ parameters for a skew-symmetric matrix, for $n\geqslant 4$ one would actually obtain a too large $n\times n(n-1)/2$ matrix $A$ to start with, the rank of which obviously cannot exceed $n$ (it is interesting to note that for $n=3$, $A$ itself can also be chosen skew-symmetric, $A=\begin{pmatrix}0&z&-y\\-z&0&x\\y&-x&0\end{pmatrix}$, meaning its rank must be 2 then (or zero, but that implied $\vec x=\vec 0$); whether this skew-symmetry of $A$ is also the case for $n>3$ I don't know). However I don't know how to determine the actual rank from this or whether there is a better way to determine it...
AI: By a change of basis, we can always assume that $x=(1,0,0,\ldots,0)^\top$. So, $Ax=0$ if and only if the first column of $A$ is zero. Hence the maximum possible rank of $A$ is $\min(m,n-1)$. |
H: Numbers of different ways to distribute $m$ balls into $n$ boxes?
So my question is this:
assuming I have $m$ balls how many ways there is to divide them into $n$ boxes (at least one ball for each box)?
For example if I have $7$ balls and I want to split them into $3$ boxes I can do:
$5, 1, 1 $
$4, 2, 1 $
$3, 3, 1 $
$3, 2, 2 $
meaning there is 4 ways (notice that $3,2,2$ is equal to $2,3,2$)
thanks in advanced !!
AI: You want the number of partitions of $m$ into $n$ positive parts. See OEIS A008284
You can use the generating function and find the co-efficient of $x^m$ in the expansion of $$\frac{x^n}{\displaystyle\prod_{j=1}^n (1-x^j)}$$ or my Java applet with "Partitions of $m$" with "Exact number of terms: $n$" |
H: sum of the series $\sum e^{-n}\sin nz$
I need to Find the sum of the series $\sum e^{-n}\sin nz$ and indicate where the series converges. Make an appropriate statement about its uniform convergence.
I was doing calculation like below, but did not get any right way about the series.
$f_n(z)=e^{-n}\sin nz$, Clearly $\lim_{n\to\infty} f_n(z)=0$, at $z=(2m+1)\pi/2n$ we see $f_n(z)$ has maxima or minima? and $f_n((2m+1)\pi/2n)=0$?
AI: Hint: Note that $\sin(z)=\frac{1}{2i}(e^{iz}-e^{-iz})$. Thus, you can write your sum $$\sum e^{-n}\sin(nz)=\frac{1}{2i}\sum e^{-n}(e^{inz}-e^{-inz})=\frac{1}{2i}\sum e^{(iz-1)n}-e^{-(iz+1)n}.$$Now note that the convergence of $\sum a_n+b_n$ is equivalent to $\sum a_n$ and $\sum b_n$ both converging or diverging; you can argue at least one of them always converges (although which one may depend on the region you are in), hence you can determine the constraints on the other one. |
H: Proof something with multivariate normal distributions
Suppose that $X \sim N(\theta, \Sigma), X\in \mathbb{R}^p $, I need to prove
$$p(x) \propto \exp \{-\frac1 2 x^T\Sigma^{-1}x + x^T\Sigma^{-1}\theta\} $$
I can I do this?
I know that the density should be
$$ p(x) = \left(\frac{1}{2\pi}\right)^{n/2} \frac{1}{\sqrt{\det \Sigma}} \exp \{(x-\theta)^T\Sigma^{-1}(x-\theta) \} $$
Please help me :)
AI: First we see that
$$
p(x)\propto \exp\left[(x-\theta)^\intercal\Sigma^{-1}(x-\theta)\right]
$$
and if we expand what's inside the brackets, we get
$$
\begin{align}
(x-\theta)^\intercal\Sigma^{-1}(x-\theta)&=x^\intercal\Sigma^{-1}x+\theta^\intercal\Sigma^{-1}\theta-x^\intercal\Sigma^{-1}\theta-\theta^\intercal\Sigma^{-1}x\\
&=x^\intercal \Sigma^{-1}x+\theta^\intercal\Sigma^{-1}\theta-2x^\intercal\Sigma^{-1}\theta
\end{align}
$$
since $x^\intercal\Sigma^{-1}\theta=(x^\intercal\Sigma^{-1}\theta)^\intercal=\theta^\intercal\Sigma^{-1}x$
(remember that $\Sigma$ is symmetric). Therefore, we conclude that
$$
p(x)\propto\exp\left[x^\intercal\Sigma^{-1}x-2x^\intercal\Sigma^{-1}\theta\right]\propto\exp\left[-\frac12 x^\intercal\Sigma^{-1}x+x^\intercal\Sigma^{-1}\theta\right].
$$ |
H: Adjoint of resolvent of self-adjoint, densely-defined operator on a Hilbert space
Let $H$ be a Hilbert space, $T=T^*$ a densely-defined linear operator on $H$. Denote the resolvent set of $T$ as $\rho(T)=\{\lambda\in\mathbb{C}~|~T-\lambda$ has bounded, everywhere-defined inverse}, and define the resolvent of $T$ at $\lambda$ to be the bounded, everywhere-defined operator $R_\lambda=(T-\lambda)^{-1}$ $(\lambda\in\rho(T))$.
How would I set about proving that $(R_\lambda)^* = R_{\smash{\overline{\lambda}}}$? I've been playing around with the inner product and the definition of $R_\lambda$ and I'm getting nowhere, so any hints of what sort of calculations to attempt would be greatly appreciated.
AI: Let $D_T$ be the domain of $T$. Then, since the resolvent $R_\lambda$ is invertible, we know that $R_\lambda^{-1}(D_T)$ is also dense. For $y \in R_\lambda^{-1}(D_T)$ and $x \in D_T$ we have
$$\begin{align}\newcommand{\inner}[2]{{\langle {#1}\mspace{-3mu}\mid\mspace{-3mu} {#2}\rangle}}
\inner{y}{x} &= \inner{(T-\lambda)R_\lambda y}{x}\\
&= \inner{R_\lambda y}{(T-\overline{\lambda})x}\\
&= \inner{y}{R_\lambda^\ast(T-\overline{\lambda})x}.
\end{align}$$
Since $R_\lambda^{-1}(D_T)$ is dense, we have $R_\lambda^\ast(T-\overline{\lambda})x = x$ for all $x\in D_T$.
Can you conclude $R_\lambda^\ast = R_{\overline{\lambda}}$ from that? |
H: $f, g$ are continuous mapping from a connected set $S$ onto $\mathbb{C}^*$
$f, g$ are continuous mapping from a connected set $S$ onto $\mathbb{C}^*$
If $f^n=g^n$ for some positive integer $n$,
1.what is the relation between $f,g$?
2.If $f(x)=g(x)$ for some $x\in S$, what can be said about $f,g$ Then?
3.Show that $\mathbb{C}^*$ can not be replaced by $\mathbb{C}$ in the hypothesis.
Well, my guess is $f=ge^{2k\pi i\over n},k=0,1,n-1,$ for $1$, but I have no idea for the other two.
AI: Yes, you are right: You can argue as follows:
(1) Let $h = f/g$ (note that $g(s) \ne 0$ for all $s \in S$, hence $h$ is well defined). Then $h$ is continuous and has $h^n = 1$. So $h[S] \subseteq \{\exp(2k\pi i/n)\mid 0 \le k < n\}$. As $h[S]$ is connected, we must have $h[S] =\{\exp(2k\pi i/n)\}$ for some $k$, so $f = g\cdot \exp(2\pi i k/n)$.
(2) Now, as there is $s \in S$ with $h(s) = 1$, arguing as in (1), we have $h[S] = \{1\}$, so $f= g$.
(3) Let $f,g \colon [-1,1] \to \mathbb C$ defined by $f(t) = t$, $g(t) = |t|$ for $t \in [-1,1]$. Then for each $t \in [-1,1]$
$$ f(t)^2 = t^2 = |t|^2 = g(t)^2 $$
so $f^2 = g^2$, but $f \ne \pm g$. |
H: Stuck on order of integration problem
I need to set up a double integral for both orders of integration and use the more convenient order to evaluate the integral $\int\int_{R}\frac{y}{1+x^2}dA$, where $R$ is the region bounded by $y=0, y=\sqrt{x}, x=4$.
I have tried setting it up, but am getting different answers so I must be wrong. I've been stuck on this for a while now!
My work: $\int_{0}^{2}\int_{0}^{y^2}\frac{y}{1+x^2}dxdy$, $\int_{0}^{4}\int_{0}^{\sqrt{x}}\frac{y}{1+x^2}dxdy$
They don't evaluate the same, but I can't tell where I'm going wrong!
AI: Draw a picture. To switch the order of integration, you need to integrate out from the curve to $x=4$:
$$\int_0^2 dy \, \int_{y^2}^4 dx \frac{y}{1+x^2}$$ |
H: Dimension reduction in the Lotka-Volterra model
I'm not sure if I can post this here, but check this out. This is some text from Boccara's Modeling Complex Systems. The thing which confuses me is that there is a dimension reduction from 4 parameters to just 1 (as stated in the text). This dazzles me, because I still see 4 parameters when looking at the resulting equations.
Could anyone shed some light?
AI: Original equations contain four parameters $b,d,s,e$ so that in order to discuss the behavior of dynamics, you have to fix three parameters and investigate the case varying the rest one. That's so complicated of course. After introducing $h,p,\tau,\rho$, equations contain only one parameter $\rho$ so that in $(h,p;\tau)$ space, no matter how $b,d,s,e$ change, which means you don't have to fix anything, if we know the behavior of dynamics varying $\rho$, everything seems clear.
Let me give a more intuitive example. I want to investigate the shape of the region $x^2+y^2\leq1$. Instead of fix each $x$ and find $y$, we set $r^2=x^2+y^2$ and the region becomes $r^2\leq1$, in which case we don't need to consider both $x$ and $y$ and instead focus on $r$. When solving equations, variables substitution usually gives simpler view. |
H: How come independence does not imply linear independence? (Statistics)
$$
\begin{array}{c|lcr|c}
\text{X/Y} & \text{0} & \text{100} & \text{200} &\text{Pr(X)}\\
\hline
100 & 0.2 & 0.1 & 0.2 & 0.5\\
200 & 0.05 & 0.15 & 0.3 & 0.5\\
\hline
\text {Pr(Y)} & 0.25 & 0.25 & 0.5 &1
\end{array}
$$
Question: Are X and Y independent random variables? Can you say that X and Y are correlated or not? (X is row, Y is column).
The answer given says that X and Y are not independent but they might or might not be linearly independent therefore the correlation might or might not be zero.
How come we cannot imply there is no linear independence if the joint probability $\neq$ the product of marginal probabilities?
AI: In general: if $X$ and $Y$ are independent, then they also are uncorrelated (zero correlation, or linear independence). The reverse is not true.
So, the title is wrong: independence does imply linear independence.
But the answer of the exercise does not say this, but the reverse, which is true. That is, $X$ and $Y$ could (in general) be uncorrelated (linear independent) even if they are not independent. |
H: Probability of two lamps failing within total 1200 hours when exponential density probability function is used for modeling failure
Question: Lamps are of a type with an average lifetime of 1200 hours. Assume that we can model the probability of failure of these bulbs by an exponential density function with mean μ = 1200.
If one bulb burns out and is replaced by a bulb of the same type, find the probability that the two bulbs fail within a total of 1200 hours.
I tried solving it by
$$\int_0^{1200}\int_0^{1200-x}\frac{1}{1440000}e^{-x/1200}e^{-y/1200}dy\, \, dx = 1-e^{-1}$$
but this seems not to be the correct answer, as I know that the value must be lower than this. Can anyone show where I got wrong?
AI: Since you are modeling $T_1 + T_2 = 1200$ where T ~ exp($\frac{1}{1200}$), you need to perform the convolution of two iid exponential variables. You may want to read about the Erlang distribution, or Gamma distribution which is the analytical solution to this problem in the time domain.
Another way to look at it is to use the connection to the Poisson(1) and ask what is the probability that N $\geq$ 2. This is not as strict a formulation, as we are not forcing the time of the second failure to be at 1200 hours.
As for your approach (integrating over the joint distribution):
$\int\limits_{0}^{1200} \int\limits_0^{1200-x}(\frac{1}{1200})^2 e^{\frac{-x}{1200}} e^{\frac{-y}{1200}} dydx = \int\limits_{0}^{1200}(\frac{1}{1200})e^{\frac{-x}{1200}} \int\limits_0^{1200-x}(\frac{1}{1200}) e^{\frac{-y}{1200}} dydx = \int\limits_{0}^{1200}(\frac{1}{1200})e^{\frac{-x}{1200}} [-e^{\frac{-y}{1200}}]^{1200-x}_0 dydx = \int\limits_{0}^{1200}(\frac{1}{1200})e^{\frac{-x}{1200}}[1-e^{-(1-\frac{x}{1200})}]dx = \int\limits_{0}^{1200}(\frac{1}{1200})e^{\frac{-x}{1200}}-\frac{1}{1200}e^{-1}dx = [1-e^{-1}]-e^{-1} =1-2e^{-1} = Poisson(N\geq2, \lambda = 1)$ |
H: Polynomial whose n no. of integrals are zero
Is it true that, say $y(x)$ be a polynomial of degree $\le$ n, such that $\int^1_0 x^i y(x) dx = 0$ for $0 \le i \le n$ then y(x) is zero polynomial ? Prove or disprove.
AI: Yes, by linearity, you get that $\int_0^1 y^2(x)dx = 0$. Since $y^2$ is continuous and non-negative, you get that $y^2 = 0$ on $[0,1]$.
Since a non-zero polynomial can have only finitely many zeroes, you are done. |
H: Why is the boolean closure of $F_{\sigma}$-sets not in $F_{\sigma}\cap G_{\delta}$?
In the Borel hierarchy, why is the boolean closure of $F_{\sigma}$ or $G_{\delta}$ equal to $F_{\sigma \delta} \cap G_{\delta \sigma}$? If I take the complement of an element in $F_{\sigma}$ I got an element of $G_{\delta}$, and finite unions and intersection of elements in $F_{\sigma}$ are still in $F_{\sigma}$, so as I see it the boolean closure should be contained in $F_{\sigma} \cap G_{\delta}$?
AI: For example, the union of an $F_\sigma$ and a $G_\delta$ might not be either $F_\sigma$ or $G_\delta$. |
H: How is this result obtained?
I am reading a paper, and having a hard time determining how a result was obtained. The paper states that: Since the total number of linear-extensions is initially $n!$ and probing an edge reduces the number of linear-extensions by a $1-1/e \sqrt{n}$, then only $O(n^{3/2}log(n))$ probes are necessary. I was under the impression that this meant that I had to solve for $x$ in the following: $$n! = (1-1/e\sqrt{n})^x $$ right? If so, I tried using Stirling's approximation on $n!$ and took the logs of both sides, but only get a dominating term of $O(nlog(n))$.
This gives that: $x = \frac{nln(n) - n + O(ln(n))}{ln(1-1/e \sqrt{n})}$, which just seems to be $O(nlog(n))$. The context given isn't really important, as I am just stuck on the algebra.
AI: You have $$\ln \left(1 - \frac{1}{e\sqrt{n}}\right) = -\frac{1}{e\sqrt{n}} + O(1/n),$$
since the Taylor expansion of the logarithm yields
$$\ln (1-x) = -\left(x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \dotsb \right).$$
So you get
$$x = \frac{\ln n!}{\ln\left(1-\frac{1}{e\sqrt{n}}\right)} = -e\cdot n^{3/2}\ln n + O(n^{3/2}).$$ |
H: Solution properties of $au''(t) + bu'(t) + u(t) = 0 $
Let $au''(t) + bu'(t) + u(t) = 0 $.
Find the values of $a$ and $b$ so that the above ODE has a solution $u$ that:
$|u| \rightarrow \infty$ as $t \rightarrow \infty$
$u$ is perodic
$u \rightarrow 0$ as $t \rightarrow \infty$
Is there a way to determine this WITHOUT actually solving the ODE?
AI: You need to solve the ode by assuming the solution as $y=e^{mt}$ then plugging back in the ode and solving the resulting polynomial in $m$. Once you consruct the general solution, you can see what conditions you need to impose on $a$ and $b$. |
H: Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$
This is for homework, and I could use a little help. The question asks
Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$.
Here is what I have done so far. Since $f$ is entire, I wrote
$$ f(z) = \sum_{n=0}^{\infty} a_n z^n = a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4 + \dotsb $$
for some $z \in \mathbb{C}$. Then
$$ f(2z) = a_0 + 2a_1z + 4a_2z^2 + 8a_3z^3 + 16a_4z^4 + \dotsb $$
and
$$ (1-2z)f(z) = a_0 + (a_1-2a_0)z + (a_2-2a_1)z^2+(a_3-2a_2)z^3 + (a_4-2a_3)z^4 + \dotsb. $$
Comparing coefficients, I find that
\begin{align*}
a_0 &= a_0 \\
a_1 &= -2a_0 \\
a_2 &= \frac{2^2}{3}a_0 \\
a_3 &= -\frac{2^3}{7 \cdot 3}a_0 \\
a_4 &= \frac{2^4}{15 \cdot 7 \cdot 3}a_0 \\
a_5 &= \frac{2^5}{31 \cdot 15 \cdot 7 \cdot 3} a_0 \\
&\vdots
\end{align*}
Now $f$ looks like
$$ f(z) = a_0 \left( 1 - 2z + \frac{2^2}{3}z^2 - \frac{2^3}{7 \cdot 3}z^3 + \frac{2^4}{15 \cdot 7 \cdot 3}z^4 - \frac{2^5}{31 \cdot 15 \cdot 7 \cdot 3}z^5 + \dotsb \right). $$
Does the series in parenthesis represent any elementary function? Besides the denominators, it looks like the Taylor expansion of $e^{-2z}$.
AI: Your function can be expressed as a convergent infinite product
$$f(z) = (1-z)(1-z/2)(1-z/4)(1-z/8)\dots$$
It's easy to see directly from this expression that it satisfies the functional equation $f(2z) = (1-2z)f(z)$.
I don't think that it can be expressed in terms of elementary functions, but I may be wrong.
Its values at the points $z=2^{-n}$ are related to the values of the Dedekind eta function. |
H: Solve $x^2 = I_2$ where x is a 2 by 2 matrix
I tried a basic approach and wrote x as a matrix of four unknown elements $\begin{pmatrix} a && b \\ c && d \end{pmatrix}$ and squared it when I obtained $\begin{pmatrix} a^2 + bc && ab + bd \\ ca + dc && cd + d^2\end{pmatrix}$ and by making it equal with $I_2$ I got the following system
$a^2 + bc = 1$
$ab +bd = 0$
$ca + dc = 0$
$cd + d^2 = 1$
I don't know how to proceed. (Also, if anyone knows of a better or simpler way of solving this matrix equation I'd be more than happy to know).
AI: The second and third equalities are
$$b(a+d)=0$$
$$c(a+d)=0$$
Split the problem in two cases:
Case 1: $a+d \neq 0$. Then $b=0, c=0$, and from the first and last equation you get $a,d$.
Case 2: $a+d=0$. Then $d=-a$ and $bc=1-a^2$. Show that any matrix satisfying this works. |
H: Reversing order of integration
I have to reverse the order of integration for the following problem, but the trig functions are tripping me up because I have to take domain and range restrictions into consideration.
Calculate
$$\int_0^1\!\int_0^{\cos^{-1}y}\!\!\sqrt{1+\sin x}\,dx\,dy$$
Does the region of integration look like a quarter circle in the first quadrant?
AI: Here is the change of order
$$ \int_0^1\!\int_0^{\cos^{-1}y}\!\!\sqrt{1+\sin x}\,dx\,dy= \int_0^{\pi/2}\!\int_0^{\cos x}\!\!\sqrt{1+\sin x}\,dy\,dx. $$
Here is the graph |
H: General linear group of a vector space
I am reading a text on Lie groups.
There is a whole chapter devoted to the group of invertible real or complex matrices of degree $n$, which are called the general linear groups (complex and real).
Further in the text, there is a reference to a general linear group of a vector space, in general. What is meant exactly by that? And why do we use the same terminology?
AI: The set of bijective linear maps from a finite-dimensional vectorspace $V$ to itself is isomorphic (as a group) to the group of invertible $n\times n$ matrices, where $n$ is the dimension of the vectorspace (to get an isomorphism, just pick a basis).
The first one is what is meant by the general linear group of a vectorspace. |
H: Integral Of $\int \frac{y}{e^{3y^2}}dy$
I want to integrate the following :
$$\int \frac{y}{e^{3y^2}}dy$$
what I did so far is:
set $t=e^{3y^2}$ and $dt=6y\cdot e^{3y^2}dy$ from here I get get the expression:
$\frac{dt}{t}=6ydy$, there is another way to do it? any suggestions? thanks.
AI: $$ I = \int ye^{-3y^2} \ dy $$
put $ -3y^2 = u \Rightarrow du = -6y dy $
$$ \Rightarrow I = -\frac{1}{6}\int e^u \ du = \frac{-e^u}{6} + c $$ |
H: Convert a matrix so it can be multiplied on the right side, instead of the left
Sometimes I come across examples of geometric transformations that multiply the transformation matrix on the left side of the vector and sometimes on the right.
How do you modify a matrix so that when multiplied on the opposite side it will yield the same result?
AI: Perhaps you're thinking about $(Ax)^T=x^TA^T$, where $^T$ denotes transpose. |
H: Prove by mathematical induction: $n < 2^n$
Step 1: prove for
$n = 1$
1 < 2
Step 2:
$n+1 < 2 \cdot 2^n$
$n < 2 \cdot 2^n - 1$
$n < 2^n + 2^n - 1$
The function $2^n + 2^n - 1$ is surely higher than $2^n - 1$ so if
$n < 2^n$ is true (induction step), $n < 2^n + 2^n - 1$ has to be true as well.
Is this valid argumentation?
AI: Yes. Your method works. Alternate: $$ n + 1 < 2n < 2 \cdot 2^n = 2^{n+1}, $$ as desired. |
H: Series Reduction
I'm not too familiar with sequences and series and I would like to show
$ n = \dfrac{n}{2^{n-1}}(\sum\limits_{i=0}^{n-2} 2^i + 1) $
I've been playing around with it on paper and tried expanding and rearranging but I'm sure there's some property I'm not familiar with that I could exploit.
If someone could walk me through some intuitive steps, I would appreciate it.
Thanks!
AI: Hint: $1+1 = 2$, $\ 2+2=2^2$, $\ 2^2+2^2=2^3$, $\ \ldots$, $\ 2^{n-2}+2^{n-2}=2^{n-1}$. |
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