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H: Question about integration of functions
LEt $f$ be a measurable function. Define $\int\limits_E f dm = \sup Y(E,f)$ where $Y(E,f) = \{ \int\limits_E \phi dm : 0 \leq \phi \leq f \}$ and $\phi$ is simple function.
and $E$ is a lebesgue measurable set. I want to show that if $B \subseteq A$, $A$ and $B$ both lebesgue measurable, and $f$ measurable function then
$$ \int\limits_B f dm \leq \int\limits_A f dm $$
My try: Pick $b \in Y(B, f) \implies b = \int\limits_B \phi $ for some simple $\phi \leq f$. Since $\phi$ is simple, can put $\phi = \sum c_i 1_{C_i} $.
Therefore, by definition of integral of simple functions, we have
$$ \int\limits_B \phi dm = \sum c_i m( C_i \cap B) \leq_{ monotonocity} \sum c_i m(C_i \cap A) = \int\limits_A \phi dm $$
Hence, $ \int\limits_A \phi dm $ is an upper bound for $Y(B, f)$ So we have
$$ \sup Y(B,f) = \int\limits_B f dm \leq \int\limits_A \phi dm $$
Im stuck in this part, can we conclude from here the conclusion of the problem? Can someone help me? thanks a lot for your time.
AI: Suppose $a \le b$, then $Y(A,a) \subset Y(A,b)$, from which it follows that $\int_A a \le \int_A b$.
Now suppose $B \subset A$. Then let $a=f\cdot 1_B, b=f$. Then $a \le b$, and so $\int_A f\cdot 1_B \le \int_A f$. Since $\int_A f\cdot 1_B = \int_B f$ we have the desired result. |
H: If $(x-8)\cdot (x-10) = 2^y$, where $x,y\in \mathbb{Z}$. then $(x,y)$ is
(1) If $(x-8)\cdot (x-10) = 2^y$, where $x,y\in \mathbb{Z}$. Then the no. of ordered pairs of $(x,y)$
(2) If $x^4-6x^2+1 = 7\cdot 2^y$,where $x,y\in \mathbb{Z}$. Then the no. of ordered pairs of $(x,y)$
$\underline{\bf{My\; Try}}::$ for (1) one , Given $(x-8)\cdot (x-10) = 2^y$. and $x,y\in\mathbb{Z}$
So $(x-8)=2^m\Rightarrow x= 8+2^m$ and $(x-10) = 2^n\Rightarrow x=10+2^n$and $m+n=y$
Now $10+2^n-8-2^m = x-x = 0\Rightarrow 2^m-2^n = 2$
Now I did not understand How can i solve after that
Help me
Thanks
AI: for $2$, you have $x^4-6x^2+(1-7.2^y)=0 \Rightarrow x^2= \frac{6\pm\sqrt{32-28.2^y}}{2}$
do you now see what should be $y$??? |
H: how to prove $E(X|\mathcal G)(\omega)=n\int^{\frac{j}{n}}_{\frac{j-1}{n}}X(s)ds, \omega\in(\frac{j-1}{n},\frac{j}{n}].$
Suppose $\Omega=[0,1]$, and $\mathcal P=$lebesgue Measure , and $\mathcal F=\mathcal B([0,1])$
and also Suppose X is random variable and $\mathcal G$ is $\sigma-$algebra Produced With intervals $ (\frac{j-1}{n},\frac{j}{n}]$ $\forall j=1,2,...,n$ .
show that $E(X|\mathcal G)(\omega)=n\int^{\frac{j}{n}}_{\frac{j-1}{n}}X(s)ds, \omega\in(\frac{j-1}{n},\frac{j}{n}].$
thanks in advance
$\mathcal B$ means Borel $\sigma-$algebra.
AI: Minor technicality: We need $\{0\} $ in the collection of sets from which ${\cal G}$ is created, otherwise ${\cal G}$ would not be a sub-$\sigma$-field of ${ \cal F}$.
Let $I_0 = \{0\}$, and let $I_j$ be the intervals above. Then it is straightforward to show that ${\cal G} = \sigma \{I_j\} = \{ \cup_{j \in J} I_j | J \text{ is finite }\}$ (the empty union being $\emptyset$).
Let $Y$ be the formula to the right of $E( X | {\cal G})$ above, and note that $Y$ is simple. It is straightforward to verify that $Y$ is ${\cal G}$ measurable and $\int_{I_j} Y = \int_{I_j} X$ for all $I_j$. (Since $P\{0\} = 0$, we have $\int_{I_0} Y = \int_{I_0} X$.)
Let ${\cal C} = \{ C \in {\cal G} | \int_{C} Y = \int_{C} X \}$. Then we have $I_j \in {\cal C}$ for all $I_j$, and since the $I_j$ are disjoint, it follows immediately that $\cup_{j \in J} I_j \in {\cal G}$ for any finite $J$. Hence ${\cal C} = {\cal G}$.
It follows that $Y$ is a version of $E( X | {\cal G})$.
(Since $\{0\}$ is the only null set other than $\emptyset$, we can see that all versions of $E( X | {\cal G})$ have the form $Y+c\cdot 1_{\{0\}}$, where $c$ is some constant.) |
H: prove that metric and series
Denote $E$ the set of all real sequences $\{a_n\}$ such that $|a_n| \leq 1$ for every positive integers $n$.Let $\{a_n\},\{b_n\} \in E$
Prove that $$d(\{a_n\},\{b_n\})=\sum_{n=1}^\infty\dfrac{|a_n-b_n|}{2^n}$$
defines a metric on E.
AI: $d$ will be a metric if it is a finite non-negative, non-degenerate symmetric function from $E \times E$ to $\mathbf{R}$ verifying the triangle inequality. It's clear from the definition that $d(\{a_n\}, \{b_n\}) = d(\{b_n\}, \{a_n\})$, since sum is commutative. Also, it's non-negative because it's the sum of non-negative factors (due to the absolute value). Let's check that it's finite: remember that the geometric $\sum q^{n}$ converges for $| q | < 1$. Now, if $|a_{n}| \leq 1 \forall n \in \mathbf{N}$, we have:
$$
|a_{n} - b_{n}| \leq |a_{n}| + |b_{n}| \leq 2
$$
where I used the triangle inequality. Diving each term like the above by $2^{n}$ and summing up we get:
$$
d(\{a_n\}, \{b_n\}) = \sum_{n=1}^{\infty} \frac{|a_{n} - b_{n}|}{2^{n}} \leq 1 + \sum_{n=1}^{\infty} \frac{1}{2^{n}}
$$
Because of the considerations about geometric series I made, the sum in the right hand side is convergent and dominates $d(\{a_n\}, \{b_n\})$, so it must converge as well. The only way such distance to be zero is if each summand is zero, since they're all non-negative, which in turn implies $a_{n} = b_{n}$ for every $n$. Finally, triangle inequality follows from $|a_n - b_n| \leq |a_n - c_n| + |c_n - b_n|$, dividing by $2^{n}$ and summing up as I did above. I suggest working through the details, because it's a routine verification. |
H: Complex conjugate to the power proof.
How can I proof that:
$$(z^n)^* = (z^*)^n$$
Where:
z is a complex number,
n is a positive whole number
* is the complex conjugate
AI: Write $z = re^{i\theta}$, then $\overline{z} = re^{-i\theta}$. So
$$
\overline{z}^n = r^n e^{-in\theta} = \overline{r^ne^{in\theta}} = \overline{z^n}
$$ |
H: Definition Weierstrass $\zeta$-function unclear
The Weierstrass $\zeta$-function is defined as follows for a lattice $\Lambda$, where a lattice is a discrete subgroup of $\mathbb{C}$ containing an $\mathbb{R}$-basis for $\mathbb{C}$.
$$\zeta(z) = \frac{1}{z} + \sum_{\omega\in\Lambda\setminus\{0\}}\left(\frac{1}{z-\omega}+\frac{1}{\omega}+\frac{z}{\omega^2}\right)$$
Doesn't $\sum_{\omega\in\Lambda\setminus\{0\}}\frac{1}{\omega}$ equal 0, because if $x \in \Lambda$ then $-x\in\Lambda$? I understand that the term appears when differentiating the logarithm of the Weierstrass $\sigma$-function, but why is it written anywhere if it could just as well be left out?
AI: You can'd do that rearrangement there because as written it is not absolutely convergent. On the other hand, with a small simplification,
$$\sum_{\omega\in\Lambda\setminus\{0\}}\left(\frac{z}{(z-\omega)\omega}+\frac{z}{\omega^2}\right)$$
is absolutely convergent, because $\omega$ is quadratic is both denominators.
On a related note, an addition of some constant is often required to construct holomorphic/meromorphic functions with prescribed zeroes and poles and residues. See Weierstrass theorem and Mittag-Leffler's theorem and their proofs on more along this line of thought. |
H: How is the Borel-Cantelli lemma used in this proof on $\psi$-approximable numbers?
I'm trying to understand a paper called "Almost no points on a Cantor set are very well approximable". In the proof the author uses the Borel-Cantelli Lemma (in the eighth line at the beginning of the proof on the second page). There are more than one version of it, but I think that the following is used:
It seems the lemma is used when we consider $(A_q:q\in\mathbb{Z}_+)$ given by $A_q=B(p/q,\psi(q)/q)$, however $p$ is not fixed. So could someone explain how is the Borel-Cantelli lemma used in this proof?
Thanks.
AI: Just reading (5) one sees that the authors consider the family $(B(p/q,\psi(q)/q))_{p,q}$ where $(p,q)$ runs over every pair such that $p\in\mathbb Z$, $q\in\mathbb Z_+$, $(p,q)=1$, $p/q\in I$. Since this collection of pairs is (at most) countable, one can order it as $\{B(p/q,\psi(q)/q)\mid p,q\}=\{A_n\mid n\geqslant1\}$ and apply simple Borel-Cantelli lemma to the sequence $(A_n)_{n\geqslant1}$. |
H: $g_n(z)=z^n$ uniformly on $D={z: |z|<1}$?
I am looking at this example:
$g_n(z)=z^n$ and domain
$D={z: |z|<1}$
I see that every $g_n$ will converge to 0 for $n \rightarrow \infty$. Thus it converges.
Now, how can I show that it is or is not uniformly?
AI: Try to compute the limit of $g_n\left(1-\frac1n\right)$ when $n\to\infty$. If the limit exists and is not zero, the convergence cannot be uniform on the disk $D$.
Once this is done, you might note that $1-\frac1n$ converges to $1$ which is not in the disk $D$ and try to imagine subsets of $D$ on which the convergence is uniform.
Edit: In view of the comments, let us recall the following facts:
For every $c$, $\left(1-\frac{c}n\right)^n\to\mathrm e^{-c}$ when $n\to\infty$.
For every $c$ and positive $a\gt1$, $\left(1-\frac{c}{n^a}\right)^n\to1$.
For every $c\ne0$ and positive $a\lt1$, $\left(1-\frac{c}{n^a}\right)^n\to0$.
Similar statements hold for the limit of $\left(1-x_n\right)^n$ for sequences $(x_n)$ such that $x_n\to0$, the limit depending on the asymptotics of $x_n$. |
H: What is the lower bound of the subset $2^n,\; n\in\mathbb{N}$
Let:
$$
A = \{2^n,\; n\in\mathbb{N}\},\quad A\subset \mathbb{R}
$$
Is the lower bound:
$(-\infty,0]$
$(-\infty,1]$
$(-\infty,1)$
?
I think it can be the first because $\min_{A}=1,\;\inf_{A}=1$, according to the definition of lower bounds.
AI: Let $a$ denote the sequence indexed by $\mathbb{N} = \{0,1,2,\cdots\}$, with defining property $$a_n = 2^n,$$
and define $A = \{a_n \mid n \in \mathbb{N}\}.$
If I'm not mistaken, the problem is to find the set of all lower bounds of $A$. Observe that the bigger $n$ gets, the bigger $a_n$ gets. Thus $a_0$ ($=1$) is the least element of $A$. This implies that $(-\infty,1]$ is the set of all lower bounds of $A$. |
H: how to compute this definite integral?
it is the second moment of logistic distribution, i.e.
$$
\int_{-\infty}^{\infty}\frac{x^2 e^x }{(1+e^x)^2}dx
$$
I've been struggling with it for hours. but i fail.
thanks first!
AI: $$\int_{-\infty}^{\infty} \frac{x^{2} e^{x}}{(1+e^{x})^{2}} \ dx = 2 \int_{0}^{\infty} \frac{x^{2} e^{x}}{(1+e^{x})^{2}} = 2 \int_{0}^{\infty} x^{2} \sum_{n=1}^{\infty} (-1)^{n+1} n e^{-nx} \ dx $$
$$ = 2 \sum_{n=1}^{\infty} (-1)^{n+1} n \int_{0}^{\infty} x^{2} e^{-nx} \ dx = 2 \sum_{n=1}^{\infty} (-1)^{n+1} n \Big(\frac{2}{n^{3}} \Big) $$
$$ = 4 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^{2}} =4 \Big(\frac{\pi^{2}}{12} \Big) = \frac{\pi^{2}}{3}$$ |
H: $ \{ x : f(x) > 0 \} = \bigcup_n f^{-1}([\frac{1}{n}, \infty )) $
$ \{ x : f(x) > 0 \} = \bigcup_n f^{-1}([\frac{1}{n}, \infty )) $
My try: Pick $$x \in \bigcup_n f^{-1}([\frac{1}{n}, \infty )) \implies x \in f^{-1}([\frac{1}{n}, \infty )) \text{ for some $n$ } $$
$$ \therefore f(x) \geq \frac{1}{n} \text{for some $n$ } \implies f > 0$$
Other direction, if $x \in \{ x : f(x) > 0 \} \implies f(x) > 0 $. We want to show $f \geq \frac{1}{n} $. If not, then $f < \frac{1}{n} \implies f = 0$ contradiction and hence $f \geq \frac{1}{n} \implies x \in f^{-1}([\frac{1}{n}, \infty )) \text{ for some $n$ } $ therefore, $$x \in \bigcup_n f^{-1}([\frac{1}{n}, \infty )) $$
Is this correct? Im a little unsure about last picture. I would greatly appreciate any feedback.
AI: Yes, its correct (as far as I can see). But personally - and this is just a stylistic thing - I would rather break it into smaller problems. Consider the following argument.
$$\bigcup_n f^{-1}([\frac{1}{n}, \infty )) = f^{-1}\left(\bigcup_n [\frac{1}{n}, \infty )\right)=f^{-1}(0,\infty) = \{ x : f(x) > 0 \}$$
All that remains is to verify the intermediate steps, by proving a few lemmas. Namely:
$\bigcup_n f^{-1}(A_n) = f^{-1}(\bigcup_n A_n)$ for an arbitrary function $f$
$\bigcup_n [\frac{1}{n}, \infty ) = (0,\infty)$
$f^{-1}(0,\infty) = \{ x : f(x) > 0 \}$
Since 1 is a standard result, you can probably omit the proof. Also, 3 is just definition chasing. So 2 is what you want to show. I suggest you break it up into the following results:
The sequence $a$ defined by $a_n = 1/n$ is strictly decreasing.
If $a$ is a strictly decreasing sequence, then $a$ has no least element.
If $a$ has no least element, then $\bigcup_n [a_n, \infty ) = (\mathrm{inf}(a),\infty).$
The sequence $a$ defined by $a_n = 1/n$ has the property that $\mathrm{inf}(a) = 0$
Hope that was helpful! |
H: Geometric intuition for partial derivatives with a single dependent variable
Having watched an integralCALC video lesson, given
$$w=xe^{y/z}, x = t^2, y = 1-t, z=1+2t$$
which could be rewritten as
$$w=t^2 e^\frac{1-t}{1+2t}$$
How does $dw/dt$ differ from $\partial w/\partial t$? Is there some intuition behind the partial derivative (such as a geometric interpretation)?
AI: There's no difference. It's just that we use straight $d$s to denote derivatives of functions with respect to one variable. It may seem like the function $w$ depends on more than one variable, but the other variables all depend on $t$, so at the end of the day all the values of the function $w$ are determined by the values of $t$, so we use the straight $d$s to denote its derivative. |
H: Maximmum and minimum values of function in interval
Hello I would like to learn the following from this site and members.
How to find the intervals ( Increasing and decreasing) of the function and its Maximum and minimum value, where the function: $f(x) = ax^2 + bx + c$.
Thanking you all.
AI: In quadratic functions,
$
\begin{array}{c|cc}
x & -\infty & & -\dfrac{b}{2a} & & +\infty \\ \hline
y \ (a>0) & +\infty & \searrow & \text{minimum} & \nearrow & +\infty \\ \hline
y \ (a<0) & -\infty & \nearrow & \text{maximum} & \searrow & -\infty
\end{array}
$ |
H: How to solve system of equations depending on parameter
\begin{cases} y=x^4 \\ y+8=a(x+5/4) \end{cases}
How many solutions does this sytem have depending on parameter a?
I need to solve it using derivatives somehow.
Thank you.
I solved using hint which bubba gave me.
Basically, we have this equation: $ x^4 + 8 = a(x + \frac54) $
Let's suppose that we have two functions: $ f(x) = x^4 + 8 $ and $g(x)=a(x+\frac54)$
$f(x)$ is a parabola looking thing while $g(x)$ is a line. So, we would have one solution when tangent line to $f(x)$ in some point $(x_0,f(x_0))$ derivative to $f(x)$ would be the same line as $g(x)$
So the derivative of $f(x)$ is $f'(x) = 4x^3$. So the tangent line would be $y = 4x_0^3(x-x_0) + f(x_0) = 4x_0^3x - 3x_0^4 + 8$. But we have that $y = g(x)$.
So \begin{cases} a=4x_0^3 \\ \frac54a=8-3x_0^4 \end{cases}.
Then we need to solve this equation:
$5x_0^3=8-3x_0^4$
The solutions are $x_0 = -2$ and $x_0=1$
So, we would have one solution when $a = 4$ and $a=-32$
And we would have no solutions if $0<=a<4$ or $0>=a>-32$.
And we would have two solutions if $a>4$ or $a<-32$
AI: Think about the graphs of the two curves. The curve $y=x^4$ is a parabola-like thing. It's not really a parabola, but it has a similar shape. The other curve is a straight line through the point $(-\tfrac54, -8)$, whose slope depends on the value of $a$. For some values of $a$, the line will intersect the first curve twice, and for other values, there will be no intersection. The crucial state is when the two intersections coincide, which means that the line is tangent to the first curve. So, the key is to find a line that's tangent to the curve $y=x^4$ and passes through the point $(-\tfrac54, -8)$. As you said, you can use derivatives to do this. |
H: Different implicit definitions
I'm Dutch, and my books are written in Dutch, so appologies if technical terms are incorrectly translated.
$\in$ means 'element of'
$\mathbb{Z}$ is the set of 'whole numbers'
$\leq$ is 'less than or equal to'
$V = \{x \in\mathbb{Z} ~~|~~x \text{ is even}\}$
$W = \{x \in\mathbb{Z} ~~|~~ x^3 < 0 \text{ and } -10 \leq -2x \leq 20\}$
$X = \{x ~~|~~ x = v + w \text{ for some } v \in V \text{ and } w \in W\}$
I am asked to give an explicit definition for $X$, or if that's impossible give a different implicit one. Since I'd have to go over an infinite amount of numbers in $V$ to give an explicit definition, I'm pretty sure that's impossible. But what would a different implicit definition look like?
On a side note (if that's ok) I find $W$ confusing, because it seems to me it can only contain $-10$ through $-1$ yet the implicit definition defines "$\leq 20$" - is that just to put me off or did I miss something?
This is my first question here so sorry if I broke some conventions.
AI: You are right that $W$ is defined in a silly way. But you figured it out correctly, so that is good :)
Suppose that $n$ is an even number. Then $n,~n+2,~n-2,~n+4,~n-4,\dots$ are all in $V$. Can you use this to show that $n$ is also in $X$?
Suppose that $m$ is an odd number. Then $m+1,~m-1,~m+3,~m-3, \dots$ are all in $V$. A similar trick will show that $m$ is in $X$. |
H: Solving equation with absolute value. Two solutions where only one is right.
I have solved this equation:
$$
\sqrt{3x}-\sqrt{4x-5}=0
$$
And I got that the solutions are:
$$
x=\frac{5}{7}, x=5
$$
My question is, did I do it right, because
$$
x=\frac{5}{7}
$$
dont fit in the equation. I mean, if I put 7/5 in equation, I will not get the correct result.
This is how I solved the equation:
$$
\sqrt{3x}-\sqrt{4x-5}=0\\
\sqrt{3x}=\sqrt{4x-5}\\
(\sqrt{3x})²=4x-5\\
|3x|-4x+5=0\\
for\ x<0\\
-3x-4x=-5\\
-7x=-5\\
7x=5\\
x=\frac{5}{7}\\
for\ x>0\\
3x-4x=-5\\
-x=-5\\
x=5\\
x=5\\
$$
Thank you very much!!!
AI: As long as we are working with real numbers, the expression $\sqrt a$ is by definition the unique nonnegative number $b$ with the property that $b^2=a$, provided it exists (that is, we need $a\ge 0$ in the first place).
Unfortunately you did not join your lines of derivation with either $\Rightarrow$ or $\Leftrightarrow$ to tell the reader if you think that what you did was an equivalence transformation or just a conclusion. However, at one step you went from
$$\tag1 \sqrt{3x} =\sqrt{4x-5}$$
to
$$\tag2 |3x| = 4x-5.$$
These lines, as they stand, can unfortunately only connected by "$\Rightarrow$", not by "$\Leftrightarrow$".
According to what I wrote in the first paragraph, you could have written simply
$$\tag{2a}3x=4x-5$$
instead of $(2)$ and it would still be a valid conclusion (but not an equivalence), saving you the trouble of working with absolute values and cases.
If the sqaure roots of two numbers are equal, then these numbres themselves are also equal!
In fact we could turn this step of the derivation into an equivalence by adding the conditions that the oriignal radicands are nonnegative:
$$\tag{2b}3x=4x-5\text{ and }3x\ge 0\text{ and }4x-5\ge0$$
or slicghtly simplified
$$\tag{2c}3x=4x-5\text{ and }\frac54\le x.$$
Then we would indeed have an equivalence $(1)\Leftrightarrow(2c)$.
The introduction of absolute value would be justified if root and squaring occured in different order:
$$ \sqrt{a^2}=|a|\qquad\text{for all }a\in\mathbb R$$
$$ (\sqrt a)^2=a\qquad\text{if }a\ge 0 \text{ (and the left side is undefined if }a<0\text{)}.$$
Of course, it is then still true that $(\sqrt a)^2=|a|$, but this makes things unnecessarily complicated if we have to consider only the case $a\ge0$ anyway.
This is what one could have done:
$$\begin{eqnarray}\sqrt{3x}-\sqrt{4x-5}&=&0\\
\iff\sqrt{3x}&=&\sqrt{4x-5}\\
\implies 3x&=&4x-5\\
\iff 0&=&x-5\\
\iff 5&=&x\end{eqnarray} $$
So we find the only possible solution is $x=5$. Since the derivation involved a mere implication in one step, we need to check: $\sqrt{3\cdot 5}-\sqrt{4\cdot 5-5}=\sqrt{15}-\sqrt{15}=0$, OK.
Alternatively, one could avoid the checking phase by enforcing equivalence in all steps:
$$\begin{eqnarray}\sqrt{3x}-\sqrt{4x-5}&=&0\\
\iff\sqrt{3x}&=&\sqrt{4x-5}\\
\iff 3x&=&4x-5\land 3x\ge 0\land 4x-5\ge0 \\
\iff 0&=&x-5\land \frac54\le x\\
\iff 5&=&x\land \frac54\le x\\
\iff 5&=&x.\end{eqnarray} $$
Then again, it is a good idea anyway to crosscheck all solutions one finds ;) |
H: Inequality $|z_1+z_2|^2 \le (1+|z_1|^2)(1+|z_2|^2)$
I have a problem to prove this inequality
$|z_1+z_2|^2 \le (1+|z_1|^2)(1+|z_2|^2)$ $\forall (z_1, z_2)\in \mathbb{C}$.
I tried to take the right hand set and subtract the lfs and after simplification I got this:
$1+(ax)^2+(by)^2 -2(ax+by)+(ay)^2+(bx)^2$ and I couldn't prove thqt this result is positive.
Any help please?
AI: $$
(1+|z_1|^2)(1+|z_2|^2)=|z_1|^2+|z_2|^2+1+|z_1|^2|z_2|^2
$$
Now using AM-GM inequality we have:
$$
1+|z_1|^2|z_2|^2\geq 2|z_1||z_2|
$$
Hence
$$
(1+|z_1|^2)(1+|z_2|^2)\geq |z_1|^2+|z_2|^2+ 2|z_1||z_2|= (|z_1|+|z_2|)^2 \ge |z_1+z_2|^2
$$ |
H: Can I approximate sine and cosine without derivatives?
Assuming I don't know derivatives (and Taylor series) can I manage to approximate sine and cosine of a generic given (rational) angle in radians?
AI: I will only use things that are known in basic trigonometry, i.e., the fundamental trigonometric identity ($\sin^2(x)+\cos^2(x)=1$), the sine/cosine of the sum or difference of two angles, the usual values at $0$, $\pi/4$ and $\pi/2$ and the formulas of the half angle that can be deduce for the previous formulas.
First of all, imagine that we want to calc the sine or cosine of the angle $\alpha$. After a translation by an entire multiple of $\pi$, something of the form $k\pi$ with $k\in\mathbb{Z}$, I can assume that $\alpha$ is in $[-\pi/2,\pi/2]$ due to the the "antiperiodicity" of sine and cosine of period $\pi$, due to
$$\cos(x+k\pi)=(-1)^k\cos(x)\text{ and }\sin(x+k\pi)=(-1)^k\sin(x)\text{.}$$
And here, after using the symmetry of the cosine or antisymmetry of the sine, i.e. $\cos(-x)=\cos(x)$ and $\sin(-x)=-\sin(x)$; we can, furthermore, suppose that $\alpha\in[0,\pi/2]$.
Secondly, we know that for $x,y\in[0,\pi/2]$, if $x\leq y$, then
$$\cos(x)\geq \cos(y)\text{ and }\sin(x)\leq \sin(y)\text{.}$$
For this, remember that
$$\cos(y)-\cos(x)=-2\sin\left(\frac{y-x}{2}\right)\sin\left(\frac{x+y}{2}\right)\leq 0$$
and
$$\sin(y)-\sin(x)=2\sin\left(\frac{y-x}{2}\right)\cos\left(\frac{x+y}{2}\right)\geq 0\text{.}$$
We have only use the formula for the sine/cosine of the sum/difference of two angles and the fact that sine and cosine for angles in $[0,\pi/2]$ are non-negative.
Third of all, remember that for $\alpha\in[0\pi/2]$,
$$\cos\left(\frac{\alpha}{2}\right)=\sqrt{\frac{1+cos(2\alpha)}{2}}\text{ and }\sin\left(\frac{\alpha}{2}\right)=\sqrt{\frac{1-cos(2\alpha)}{2}}\text{.}$$
Therefore, we can recursively calculate
$$c_n=\cos\left(\frac{\pi}{2^n}\right)\text{ and }s_n=\sin\left(\frac{\pi}{2^n}\right)\text{,}$$
using the formulas
$$c_{n+1}=\sqrt{\frac{1+c_n}{2}}\text{ and }s_{n+1}=\sqrt{\frac{1-c_n}{2}}\text{.}$$
Finally, given $\alpha\in [0,\pi/2]$, express it as
$$\alpha=\sum_{n=1}^\infty a_n\frac{\pi}{2^n}$$
with $a_n\in\{0,1\}$, by using the binary expansion of $\alpha/\pi$, and then you will have for all $m$ that
$$\cos\left(\sum_{n=1}^m a_n\frac{\pi}{2^n}+\frac{\pi}{2^{m}}\right) \leq\cos(\alpha)\leq \cos\left(\sum_{n=1}^m a_n\frac{\pi}{2^n}\right)$$
and
$$\sin\left(\sum_{n=1}^m a_n\frac{\pi}{2^n}\right)\leq \sin(\alpha)\leq \sin\left(\sum_{n=1}^m a_n\frac{\pi}{2^n}+\frac{\pi}{2^{m}}\right)\text{.}$$
You can easily check that the LHS and RHS in the both expression can be calculated as polynomials in the $c_n$ and $s_n$ by applying the formula for the sine/cosine of the sum of two angles. |
H: Non-recursive way to present $ p_{0}=0$, $p_{n+1}=(e+1)p_{n}+e$ for some $e>0 \in \mathbb{R}$.
Is there a non-recursive way to present this function:
$ p_{0}=0$
$p_{n+1}=(e+1)p_{n}+e$
for some $e>0 \in \mathbb{R}$.
Or at least some estimation from the top would satisfy me.
AI: Note that $p_{n+1}+1=(\mathrm e+1)(p_n+1)$ for every $n\geqslant0$ and that $p_0+1=1$ hence $p_n=(\mathrm e+1)^n-1$ for every $n\geqslant0$. Thus, for example, $p_n\lt(\mathrm e+1)^n$ for every $n\geqslant0$. |
H: How to prove $\gcd(a^m,b^m) = \gcd(a,b)^m$ using Bézout's Lemma
The problem is to prove the following
If $\gcd(a,b) = c$, then $\gcd(a^m, b^m) = c^m$
I know that this can be solved easily by proving that $c\mid a \implies c^m \mid a^m$ and $c\mid b \implies c^m \mid b^m$. So the greatest common divisor of $a^m$ and $b^m$ is $c^m$. But the problem is that I want to use Bézout's Lemma (Identity) to prove this.
I first introduce the following notation:
$$a = ca' \quad \quad b = cb', \quad \quad \text{where $(a',b') = 1$.}$$
Now from Bézout's Lemma, becaus $a'$ and $b'$ are coprime integers then we have integer solutions for the following equation:
$$a'x + b'y = 1$$
It's obvious that also the following equation has an integer solution:
$$a'^mx_1 + b'^my_1 = 1$$
Although I'm not able to prove this using Bézout's Lemma.
Now multiply both sides by $c^m$ we get:
$$(a'c)^mx + (b'c)^my = c^m$$
$$a^mx + b^my = c^m$$
But Bézout's Lemma states that for given $c^m$ is the greatest common multiplier of $a^m$ and $b^m$ or it is a multiplier of it.
But how to prove that it's not a multiplier of the greatest common multiper, and in fact is the greatest common multiplier. Is it possible to solve this problem with Bézout's Lemma?
AI: You need both parts of your work to get the proof.
Your argument "this can be solved easily" only amounts to a proof that that $c^m$ divides the GCD: it doesn't actually prove $c^m$ actually is the GCD.
The rest of your work provides the other half: that the GCD divides $c^m$.
The only way I see to do the first half of the proof with Bezout's lemma doesn't really amount to anything different: it observes that $c^m | a^m$ and $c^m | b^m$, and thus $c^m | a^m x + b^m y$ for all $x,y$.
Here's a useful trick to fill in your gap: if $c x + d y = 1$, then we also have $(c x + d y)^{2m-1} = 1$ (any power at least $2m-1$ would suffice for this trick). You can expand this and collect terms to find an identity $c^m u + d^m v = 1$. |
H: How to interpret involutory change of basis transformation?
Just working through an assignment and a change of basis matrix popped up which was involutory - its own inverse.
I am not quite sure how to think about this... Presumably it means that the transformation doesn't 'scale' the basis vectors from one basis to another - this makes sense in the finite dimensional case since the determinant of any matrix representing the transformation must be $\pm 1$.
It was a 3-dimensional case, so is there some geometric interpretation?
What about a more general case? Arbitrary finite or infinite-dimensional?
AI: Obviously, to be involutory (at list in the finite dimension case), it has to be diagonalizable, and -- as you have noticed -- its eigenvalues must be $\pm 1$. In other words, your transformation $T$ has to have a form
$$T = S \Lambda S^{-1}, \quad \Lambda = \operatorname{diag}(\lambda_1,\dots,\lambda_n), \quad \lambda_k \in \{-1, 1\}.$$
The interpretation would be that your operator is a reflector along some of the basis vectors in some (not necessarily orthogonal!) basis, defined by the columns of $S$. |
H: How to show that the limaçon has only two vertices.
Question: Show that the limaçon has only two vertices.
I researched what is limaçon. And I reached the following result;
Note that I only know that The limaçon is the parametrized curve
$$\gamma(t) = ((1 + 2\cos t)\cos t, (1 + 2\cos t)\sin t)$$ for $t \in \Bbb R$.
After this, I calculated $$\dot{\gamma(t)}= (-\sin t -4\cos t \sin t, \cos t +2\cos(2t)).$$
Now I need to find the points at which $\dot{\gamma (t)}=0$ .
How can I find these points? And after I find, what do I need to do in order to conclude that it has only two vertices.
I will be happy if someone show this.
AI: Given the definition Ted mentioned in the comments, the vertices of a parameterised curve are the points of a curve at which the derivative of the curvature is zero (ie a critical point of curvature), and so your first action should be to calculate the curvature of $\gamma(t)$. You can do this by one of several ways, either from the definition of curvature, by reparametising the curve in terms of arc length, or by using any of the various other formulae for curvature in terms of a parametisation. I would personally use $$\kappa(t)=\frac{\|\gamma'(t)\times\gamma''(t)\|}{\|\gamma'(t)\|^3}.$$
Now you need to show that $\kappa(t)$ has two distinct critical points $t_1$ and $t_2$, at which $\kappa'(t_i)=0$. |
H: increasing sequence of sets
suppose $A_n \subseteq \mathbb{R}$ for all $n$. Also if $A_{k} \subseteq A_{k+1} $, does it follow that $\bigcup_{k=1}^{\infty} A_k = \mathbb{R} $?. I know this is obvious but I dont know how to show the other inclusion: that is $ \mathbb{R}$ lives inside the bigcup
AI: Why is that obvious? It's not even true, which is why you're unable to show it.
Examples of countable unions of subsets of $\mathbb R$, which are not $\mathbb R$:
$A_k=\{0\}$ for all $k$. Then $A_k\subseteq A_{k+1}$ for all $k$, but $\bigcup_{k=1}^\infty A_k=\{0\}$.
Even if you require that the sequence be strictly increasing, consider $A_k=\{1,2,\dots,k\}$. Then $\bigcup_{k=1}^\infty A_k=\mathbb N$.
If we require that each of the $A_k$ be an interval, we can have $A_k=[0,1-1/k]$ (as in Prahlad's comment) - then $\bigcup_{k=1}^\infty A_k=[0,1)$.
Perhaps you could give some reasons why you think this is obvious. Then we can help you see where your mistake is. |
H: Convergence in product topology
Let $x_1,x_2, \ldots$ be a sequence of points of the product space $\prod X_\alpha$. Show that the sequence converges to the point $x$ if and only if the sequence $\pi_\alpha (x_1), \pi_\alpha (x_2)\ldots$ converges to $\pi_\alpha (x)$ for each $\alpha$. Is this fact true for box topology instead of product topology?
But what does it mean to converge in product topology? Is it that a common neighborhood exists for all points in the sequence?
AI: Let $U\subseteq X_\alpha$ be an open neighbourhood of $\pi_\alpha(x)$.
By definition of product topology, the set $V:=\prod U_i\subseteq \prod X_i$ with $U_\alpha=U$ and $U_i=X_i$ for $i\ne \alpha$ is an open neighbourhood of $x$. Hence almost all $x_n$ are in $V$, hence almost all $\pi_\alpha(x_n)$ are in $U$. This just says that $\pi_\alpha(x_n)\to \pi_\alpha(x)$.
For the other direction suppose that $\pi_\alpha(x_n)\to \pi_\alpha(x)$ for all $\alpha$.
Let $U$ be an open neighbourgood of $x$.
Wlog. we may assume that $U$ is the product of open sets $U_\alpha$ of $X_\alpha$ where $U_\alpha=X_\alpha$ for almost all $\alpha$ (as these special sets form a basis of the product topology).
Then for each $\alpha$, we have that almost all $\pi_\alpha(x_n)$ are in $U_\alpha$, that is there are $n_\alpha\in\mathbb N$ such that $\pi_\alpha(x_n)\in U_\alpha$ for all $n>n_\alpha$.
Since $U_\alpha=X_\alpha$ for almost all $\alpha$, we can choose $n_\alpha=0$ for almost all $\alpha$. Therefore we can consider $N=\max\{n_\alpha\mid \alpha\}=\max\{n_\alpha\mid U_\alpha=X_\alpha\}\in\mathbb N$ because we are taking the maximum only over finitely many natural numbers.
Then for $n>N$ we have that $\pi_\alpha(x_n)\in U_\alpha$ and hence $x_n\in U$.
This shows that $x_n\to x$.
With box topology, the above argument fails at the bolded step.
More concretely, note under the box topology the product of infinitely many discrete two-point spaces is discrete and find an explicite counterexample from that. |
H: Prove $A \cap B = A − (A − B)$
$A$ and $B$ are sets, how would I prove the following equality:
$A∩B = A-(A-B)$
Would I be correct in saying that if I take $x$ to be in $A-(A-B)$, it means that $x$ is in $A$, but not in $A-B$. Now, suppose temporarily that $x$ was not in $A∩B$, however, we already know that $x$ is in $A$, so that means $x$ cannot be in $B$. But then, since $x$ is in $A$, but not in $B$, $x$ must be in $A-B$, which contradicts what I just discovered. Therefore, $x$ must be in $A∩B$.
Suppose $x$ is in $A∩B$, then $x$ is in both $A$ and $B$. I know from this that $x$ cannot be a part of $A-B$ because that would imply that $x$ is part of $A$, but not part of $B$. Since $x$ is already part of $A$, but not part of $A-B$, it follows by definition that $x$ is in $A-(A-B)$.
Would I be correct in saying this? I can write these proofs out in words but my notation is poor, how would I write this in strict set notation?
AI: This can be done pretty mindlessly using De Morgan's laws:
$$A - (A - B) = A \cap (A - B)^c$$
$$= A \cap (A \cap B^c)^c$$
$$= A \cap (A^c \cup B)$$
$$= (A \cap A^c) \cup (A \cap B)$$
$$= A \cap B$$ |
H: Natural Deduction (FeedBack)
I am looking for feedback to three proofs (alternatively derivations) that I have constructed. The first is:
Theorem. Injectivity does not imply surjectivity.
Proof: Suppose $\{\phi\} \vdash \theta$. Then according to the soundness theorem we have $\{\phi\} \models \theta$, which is to say that for every valuation $v$, $[[\phi]]_{v}=1 \implies [[\theta]]_{v}=1$. However, consider the following structure $\langle \mathbb{N};; +, \cdot, 1 \rangle$ which gives rise to the following formula $\exists x_{1}(x_{0} \dot{=} f_{2}(f_{1}(\overline{c}_{0},\overline{c}_{0}),x_{1}))$.Now, from this formula we define a function as follows:
$$f:\mathbb{N} \to \mathbb{N}, $$
$$f(x)=2x$$
This function is clearly injective. However it is equally clear that it is not surjective, as there are elements in the range that the function does not map to, more specifically all odd numbers. Thus, we have a case where $[[\{\phi\}]]_{v}=1$ while $[[\{\theta\}]]_{v}=0$, which shows $\{\theta \}$ is not a logical consequence of $\{\phi\}$ ($\{\phi\} \not \models \{\theta\}$). Hence, by the completeness theorem we have $\{\phi\} \not \vdash \theta$, or in more informal terms, the injective property does not imply the surjective property. $\square$
The two other proofs are a bit more complicated to present. As far as I can tell (and believe me I have tried), Math Stack Exchange does not support bussproof, which is a LateX package that enables users to construct proof trees. Furthermore, I haven't been able to simply attach a PDF file of the derivations (again Math Stack Exchange does not support it). As a result, I have decided to post my codes, hoping you (the community) still are willing to provide some feedback.
AI: The counterexample to injectivity implying surjectivity is presented in a quite unnecessary laborious way. You will be marked down for obfuscation.
The universally quantified wffs at the top of both branches of the first displayed proof shouldn't be at the same level as the existential. Move the existentials down five lines, and then you need to repeat the conclusion of the sub proof. Thus ...
$\quad\quad\quad\quad\quad\quad\varphi\\
\quad\quad\quad\quad\quad\quad\ldots\\
\quad\quad\quad\quad\quad\quad\ldots\\
\quad\quad\exists x\varphi\quad\quad C\\
\quad\quad -----\\
\quad\quad\quad\quad C$
See van Dalen section 2.9, or other texts.
Here's the proof done Fitch-style: Natural Deduction and Identity .
[Yes, it would be good if we could use bussproofs here!] |
H: How to find a suitable orthogonal matrix
Assume $x,y \in \text{R}^n$ are two vectors of the same length, how to prove that there is an orthogonal matrix $A$ such that $Ax=y$?
Thanks for your help.
AI: Well, if they both have length zero, then take $A$ to be anything. If $\|x\| = \lambda = \|y\|$, then divide by $\lambda$ and assume that $\lambda = 1$.
Now note that the relation $x \sim y$ iff $\exists$ an orthogonal matrix $A$ such that $Ax = y$ is an equivalence relation. So it is enough to assume that $x = (1,0,0,\ldots, 0)$.
Now take $y$ and extend it to an orthonormal basis $\mathcal{B}$ of $\mathbb{R}^n$. Consider the change of basis matrix $A$ that sends the standard basis to $\mathcal{B}$. The columns of this matrix are precisely the vectors of $\mathcal{B}$ which are orthonormal, and hence $A$ is an orthogonal matrix. Furthermore,
$$
Ax = y
$$ |
H: Prove that this is an equivalence relation and give all the different equivalence classes
Let $R$ be a relation defined on real numbers by letting $a\mathrel R b$ iff $\cos (a) = \cos (-b)$. Prove that this is an equivalence relation and give all the different equivalence classes. Also show that this is all the classes and that they are different.
AI: $\cos(-b) = \cos b$ for all real $b$, hence, your equivalence relation is equivalent to $a\mathcal Rb \iff \cos a = \cos b$.
You need to show that $\mathcal R$ is reflexive. Since $\cos (a) = \cos(-a) $, then $a\mathcal{R} a$. Hence, $\mathcal R$ is reflexive.
You need to show that $\mathcal R$ is symmetric. Since $\cos(a) = \cos(-b) \iff \cos(a) = \cos (b) \iff \cos b = \cos a \iff \cos b = \cos(-a),$ we have that $a \mathcal{R} b \implies b \mathcal{R} a $, so $\mathcal R$ is symmetric.
You need to show that $\mathcal R$ is transitive. This means that if $a \mathcal{R} b$ and $b\mathcal{R} c$, then it must follow that $a \mathcal R c$. So supposing $\cos (a) = \cos (-b) = \cos(b) $ and also $\cos(b) = \cos( -c ) $, it certainly follows by transitivity of equality that $\cos (a) = \cos (-c).$ Hence $a \mathcal{R} c$.
Therefore, $\mathcal R$ is reflexive, transitive, and symmetric, and is therefore, an equivalence relation on the set of real numbers.
An equivalence class $[a]$ of $\mathcal R$ consists of all real numbers $x$ such that $x \mathcal R a$. Recall that an equivalence class partitions the set on which it is defined, meaning the every real number is in one and only one equivalence class, so that the union of the equivalence classes IS the set of reals, and so that the intersection of any two equivalence classes is empty.
Let's just explore $a \in \mathbb R$. $\cos a = \cos(-a) = \cos (2\pi + a) = \cos (2\pi -a) = \cos (2k \pi + a) = \cos (2k\pi - a)$, where $k$ is any integer. So for any given $a\in \mathbb R$, the equivalence class of $a$ is given by $$[a] = \{x\in \mathbb R\mid x=2k\pi \pm a,\;k \in \mathbb Z\}.$$ |
H: convergence of $\sum \frac {nx^{n-1}} {(1+x^n)^2}$
I want to examine the convergence of this function series $\sum \frac {nx^{n-1}} {(1+x^n)^2}$ for $x \in [2, +\infty)$. I showed pointwise convergence but I'm struggling with uniform convergence.
I tried to apply the Weierstrass M-test but it didn't work for me. I also tried to demonstrate that the series does not converge uniformly but that didn't work either. Does anybody have any hints?
AI: Hint: $$\frac {nx^{n-1}} {(1+x^n)^2} \le \frac {nx^{n-1}}{x^{2n}} = \frac {n} {x^{n+1}} \le \frac {n} {2^{n+1}}.$$ |
H: Knowing that $b\leq\frac{a}{1-a}$ and $a<0.01$ show that $b \leq 1.01a$
I've been solving a problem in numerical analysis and to finish one of the exercises I need the following result.
Knowing that $b\leq\frac{a}{1-a}$ and $a<0.01$ show that $b \leq 1.01a$.
Now I think that this might be BS, but everything that led up to this point seems proper to me, so maybe I'm just wrong.
I tried the following string of inequalities:
$b \leq \frac{a}{1-a} \leq \frac{a}{1-0.01}=1.(01)a $ but that leaves the possibility that $b=1.0101a$ for example and that goes against what I'm trying to prove...
AI: Hint aassuming $\;a>0\;$ :
$$\frac a{1-a}<\left(1+\frac1{100}\right)a\iff \frac1{1-a}<\frac{101}{100}\iff 1-a>\frac{100}{101}\iff a<\frac1{101}\left(<\frac1{100}\right)$$ |
H: Prove that there are only 2 solutions of $\frac{n+3}{n-1}=x$ for $n$ natural and $x$ natural.
I know that the solutions are $1) n=3$ and $2) n=5$ but I don't know how to prove that they are only the two.
AI: We have
$$\frac{n+3}{n-1} = 1 + \frac{4}{n-1}.$$
For that to be an integer, $n-1$ must be a divisor of $4$, so $n-1 \in \{1,2,4\}$, and $n \in \{2,3,5\}$. There are in fact three solutions. |
H: The series $\sum_{n=1}^{\infty}\frac{1}{n+1}$ diverge. What about $\sum_{n=1}^{\infty}\frac{1}{(n+1) \times (n+2)}$?
The series $\sum_{n=1}^{\infty}\frac{1}{n+1}$ diverge. What about $\sum_{n=1}^{\infty}\frac{1}{(n+1) \times (n+2)}$?
AI: Hint: $$\frac1{(n+1)(n+2)} = \frac1{n+1} - \frac1{n+2}.$$ |
H: Are these definitions of a prime ideal equivalent?
I just noticed I have three different definitions of a prime ideal in my notes. So are these definitions equivalent? Are they all correct...I have feeling I might have taken something down wrong.
Let $I$ be an ideal in a commutative ring $R$.
$I$ is prime if -
$a, b \in R$ with $a,b \notin I$ then $a \cdot b \notin I$
$a, b \in I$ then $a\cdot b \notin I$
$a, b \in I$ then either $a \in I$ or $b \in I$
AI: The usual definition for an ideal $I$ of the commutative ring $R$ to be prime is
for all $a,b\in R$, if $ab\in I$, then $a\in I$ or $b\in I$
The condition
for all $a,b\in R$, if $a\notin I$ and $b\notin I$, then $ab\notin I$
is just a reformulation of the above condition, using the fact that $P\implies Q$ is equivalent to $(\text{not }Q)\implies (\text{not }P)$.
Your second condition is false for any ideal: if $a,b\in I$, then clearly $ab\in I$: it suffices that one of $a$ and $b$ belongs to $I$ in order that $ab\in I$.
Your third condition is the first I gave provided we remove the comma; saying $a,b\in I$ is just an abbreviation for “$a\in I$ and $b\in I$”, so it wouldn't define anything special. |
H: How to prove these propositions?
This is an exercise I don't quite know how to write in a acceptable mathematical form. (No, I haven't found a solution.)
"Prove that if $3 \mid n$ then $3 \mid n^2$.
And that if $3 \nmid n$ then $3 \nmid n^2$. (For the second part consider $n=3a+1$ and $n=3a+2$ in turn.)"
Thank you in advance for your help and time.
J
AI: Assume that $3\mid n$. Then $n=3q$ for some integer $q$. Hence $n^2=(3q)^2=9q^2=3(3q^2)$. Since $3q^2$ is an integer, $3\mid n^2$. |
H: Prove that if $n \cdot 2^{-t} <0.01$ then $n \cdot 2^{-t} <\frac{1}{101}$
Is the following theorem true?
If $n \cdot 2^{-t} <0.01$ then $n \cdot 2^{-t} <\frac{1}{101}$ for $t,n \in \mathbb{N} $.
I've tried basic induction but that has led me nowhere, same with thinking of a counter-example.
AI: Let's find t such that if $2^{-t}$ is fixed and $n$ is increased one by one from $0$, $n2^{-t}$ will satisfy the inequality $1/101<n2^{-t}<1/100$. If such t and n exist, the theorem is clearly not true. It is obvious that if $n$ is increased one by one, $n2^{-t}$ satisfies the above inequality at some value of $n$ if $2^{-t}$ is smaller than
$1/100-1/101=1/10100$.
Since $ 2^{-13}>1/10100>2^{-14}$, if $t=14,$ there must be some $n$ such that $1/101<n2^{-t}<1/100$. This is enough as proof to disprove the theorem. |
H: Why are non-separated schemes schemes?
In "the old days", e.g. in the famous texts by Grothendieck and Mumford, a scheme was defined as what we now call a separated scheme. (i.e. a scheme where the image of the morphism $\Delta:X \to X \times X$ is closed)
Nowadays, schemes are usually allowed to be separated. The question is then: Do non-separated schemes naturally occur in nature?
AI: Moduli spaces and stacks tend to be non-separated. See for example this poster by David Rydh and Jack Hall. The example of the Picard scheme appears in FGA explained, Ex. 9.4.14. |
H: How would I evaluate this limit? $\lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}}$
How would I evaluate the following limit by hand?
$\lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}}$
Thanks in advance.
AI: $$\begin{align} \lim_{x\to 3+} \frac{x-3}{\sqrt{x^2-9}} & = \lim_{x \to 3+}\; \frac{\left(\sqrt{x - 3}\;\right)^2}{\sqrt{(x - 3)(x+3)}}\\ \\
& = \lim_{x \to 3+}\;\frac{\left(\sqrt{x-3}\;\right)^2}{(\sqrt{x-3}\;)(\sqrt{x+3}\;)}\\ \\
& = \lim_{x \to 3+}\;\frac{\sqrt{x - 3}}{\sqrt{x+3}}\\ \\
& = \frac 06\; = \;0\end{align}$$ |
H: Coupon selecting problem
Suppose that there are N distinct types of coupons and that each time one obtains a coupon, it is, independently of previous selections, equally likely to be any one of the N types. One random variable of interest is T3, the number of coupons that needs to be collected between the acquisitions of the third new type of coupon and the fourth new type of coupon. Find the pmf of T3. Also, find the average number of coupon you need to collect in this case.
I'm not sure how to solve this. If T3 is the number of coupons need to be selected between the third and fourth new coupon. Is it just N-3?
AI: You have just gotten your third different coupon. You draw coupons until you find a fourth one, different from the three you have. Note that these drawing form Bernoulli experiments.
What is the probability, in a single draw, to get a new coupon?
What random variable is used to model the time it takes to obtain a success in a series of Bernoulli trials?
Added
In the same sense,define $T0$ amd $T1$ to be the number of coupons needed to get respectively from $0$ to $1$ coupon and from $1$ to $2$. We start with the easy case $T0$.
Since you have no coupons at hand, you will always draw a new coupon on your next (first) draw. I will do something weird and say $T0$ is a geometric random variable with parameter $N/N=1$.
We go on with $T1$.
You have $1$ of the coupons, so there remains $N-1$ that you need for your collection to be complete. We will compute the probability that you get a new coupon after $k$ new draws, $k=1,2,3,\ldots$. To get a new coupon on the $k$-th draw, you need to first have had $k$ times your first coupon, and then draw a new one. Check that the probabilities this happen is
$$
\frac{1}{N^k}\frac{N-1}{N}.
$$
This is known has the geometric random variable (linked above) with parameter $p=(N-1)/N$.
You should now be able to derive $T2,T3$ and so on. |
H: Proving that a holder continuous function always has a smaller exponent.
According to wikipedia if we have $f:X \rightarrow Y$ which is $\alpha$-Holderian then for all $\beta < \alpha$ the function is also $\beta$-Holderian.
How do we prove this starting from the fact that we have $d_y(f(x),f(y)) \leq K d_x(x,y)^{\alpha}$?
It seems we would need to show that $K d_x(x,y)^{\alpha}\leq K d_x(x,y)^{\beta}$ but $d_x(x,y) \leq 1$ does not always hold.
AI: It depends on the definitions. If one wants a global Hölder constant, so that
$$d_Y(f(x),f(y)) \leqslant K\cdot d_X(x,y)^\beta$$
for all $x,y \in X$, then $\alpha$-Hölder continuity does in general not imply $\beta$-Hölder continuity for $\beta < \alpha$. The easy counterexample to that is $f(x) = \lvert x\rvert^\alpha$ on $\mathbb{R}$, which is $\alpha$-Hölder continuous, but doesn't admit a global Hölder constant for any exponent $\beta < \alpha$.
It is locally true, however, on every bounded subset $S$ of $X$ we have a $\beta$-Hölder constant when we have an $\alpha$-Hölder constant (the $\beta$-Hölder constant depends on the $\alpha$-Hölder constant and the diameter of $S$). Since Hölder continuity is mostly interesting for local behaviour, one can replace the metrics on $X$ and $Y$ by bounded uniformly equivalent metrics (for example $\bar{d}_X(x,y) = \min \{d_X(x,y),\, 1\}$) to obtain the implication. |
H: almost everywhere convergence vs uniform convergence
Let $(\mathbb{R},\mathcal{L},m)$
Can someone explain to me why $f_{n}(x)=\chi_{(0,\frac{1}{n}]}$ converges almost everywhere to $0$ but not uniformly...
also why does $f_{n}(x)=n^{-1}\chi_{(0,n)}$ converge uniformly on $\mathbb{R}$?
AI: (1) For any $x_0\in \mathbb{R}$, if $x_0\not\in (0,1]$ then $f_n(x_0)=0$ for all $n$; if $x_0\in (0,1]$ there exists $N$ such that when $n>N$ we have $x_0\not\in (0,\frac{1}{n}]$, and then $f_n(x_0)=0$. So $\lim_{n\to \infty}f_n(x)=0$ for all $x\in \mathbb{R}$.
(2) For $\epsilon_0=\frac{1}{2}$, for any $N\in \mathbb{N}$, we have that $\frac{1}{N+2}\in (0, \frac{1}{N+1}]$, hence $|\chi_{(0,\frac{1}{N+1}]}(\frac{1}{N+2})-0|=1\geq \epsilon_0$. This means that $\chi_{(0,\frac{1}{n}]}$ does not uniformly converge t0 $0$.
(3) Since $|n^{-1}\chi_{(0,n)}|\leq \frac{1}{n}$ for all $x\in \mathbb{R}$. Then $\frac{1}{n}\chi_{(0,n)}$ uniformly converges to $0$. |
H: Continuous Function
Let $ f: \Bbb R^2 \to \Bbb R $ such that :
$ \forall _{y_0 \in \Bbb R\ }: $ function $ x \to f(x,y_0) $ continuous function and increasing
$ \forall _{x_0 \in \Bbb R\ }: $ function $ y \to f(x_0,y) $ continuous function
I mean the continuity of one variable.
Prove the continuity of a function $f: \Bbb R^2 \to \Bbb R $
I know definition, but I can not do. I wanted to use the definitions of Cauchy.
AI: Let $\varepsilon > 0$. There is $\delta_1 > 0$ such that $$f(x_0, y_0) - \varepsilon/2 < f(x_0 - \delta_1, y_0) < f(x_0 + \delta_1, y_0) < f(x_0, y_0) + \varepsilon/2.$$
Then there is $\delta_2 > 0$ such that
$$f(x_0 - \delta_1, y_0) - \varepsilon/2 < f(x_0 - \delta_1, y) < f(x_0 - \delta_1, y_0) + \varepsilon/2$$
and
$$f(x_0 + \delta_1, y_0) - \varepsilon/2 < f(x_0 + \delta_1, y) < f(x_0 + \delta_1, y_0) + \varepsilon/2$$
for any $y \in (y_0 - \delta_2, y_0 + \delta_2)$.
Now for any $(x,y) \in (x_0 - \delta_1, x + \delta_1) \times (y_0 - \delta_2, y_0 + \delta_2)$ you have $|f(x,y) - f(x_0, y_0)| < \varepsilon$. |
H: Does there exist an infinite number string without any 'refrain'?
Let us consider an infinite or finite number string which consists of $0,1,2$. Then, let us call an adjacent pair of repeating number(s) 'a refrain'.
For example, we have three refrains in the following string :
$$01\overline{2}\ \overline{2}01202\overline{12}\ \overline{12}10\overline{201}\ \overline{201}02$$
Question : Does there exist an infinite number string without any refrain?
Motivation : I've known that there exists an infinite number string which consists of $0,1,2,3$ without any refrain. This got me interested in the above expectation, but I'm facing difficutly. Can anyone help?
AI: What you call "refrain" is called a square in the literature of combinatorics on words. There are many square-free words on an alphabet of three letters. An example is the sequence
$$1, 2, 3, 1, 3, 2, 1, 2, 3, 2, 1, 3, 1, 2, 3, 1, 3, 2, 1, 3, 1, 2, 3, 2, 1, 2, 3, 1, 3, \ldots$$
which can be obtained by starting with $1$ and then using the morphism $1\to 123$, $2\to 13$, $3\to 2$. (See http://oeis.org/A007413.) |
H: Help with an elementary set theory proof
"If $A$ and $B$ are two sets, then prove that $A$ is the union of a disjoint pair
of sets, one of which is contained in $B$ and one of which is disjoint from
$B$."
Can somebody help me understand what this question even wants me to prove?
As far as I can understand, it wants me to assume $A = C\cup D$, with $C \cap D$ being the empty set. How does $B$ even relate to this assumption, if I am not given any detail on what $B$ is? Am I supposed to prove that for ANY set $B$, $C$ would be a subset of $B$ and $D$ disjoint from $B$? I think that's a false statement, so I doubt I am reading the question correctly. Any help is greatly appreciated.
AI: Hint:
$$A=\left(A\cap B\right)\cup\left(A\setminus B\right)$$ |
H: Evaluate the limit $\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$
I need to evaluate the following limit:
$\lim_{x\to 2} \frac{x-2}{\sqrt{x^2+5}-3}$
I have multiplied both sides by the conjugate $\sqrt{x^2+5}+3$ but am getting $x^2-4$ as the denominator. Is this the correct way to go about it?
AI: Yes indeed, that's the way to go about it. Now, we have $$\begin{align} \lim_{x \to 2}\frac{x-2}{\sqrt{x^2+5}-3} & = \lim_{x \to 2}\dfrac{(x-2)(\sqrt{x^2 + 5} + 3)}{x^2 - 4} \\ \\ & = \lim_{x \to 2}\dfrac{(x-2)(\sqrt{x^2 + 5} + 3)}{(x - 2)(x+2)} \\ \\ & \overset{x\neq 2}{=} \lim_{x \to 2}\dfrac {\sqrt{x^2 + 5} + 3}{x+2} \\ \\
& = \dfrac 64 = \frac 32\end{align}$$ |
H: Clopen and open sets have the same measure
I was positive I had already asked this one but apparently not. It is simply this:
Show that the Lebesgue outer measure of $[a,b)$ is $b-a$.
Let $\mathscr{K}$ be the collection of open subsets of $\mathbb{R}$ and $\lambda$ a nonnegative setfunction defined on $\mathscr{K}$ by $\lambda(\emptyset)=0$ and $\lambda\{(a,b)\}=b-a$ for $(a,b) \in \mathscr{K}$. Then the Lebesgue outer measure is defined as $\mu^*(A)=\mathrm{inf} \{ \sum \lambda(E_n) \mid E_n \in \mathscr{K},\mathrm{ } \cup E_n \supset A \}$. So, how does one go about answering the question when the outer measure is defined on open subsets?
I figure one should work with limits in some way. Perhaps $[a,b)$ can be considered as the union of $(a,b)$ with $\cup (a-\varepsilon_n,a+\varepsilon_n)$ "where $\varepsilon_n$ goes to zero as $n$ goes to infinity", whatever that would mean in terms of the latter open sets. If we let the $E_n$ be $(a,b)$ and the $(a-\varepsilon_n,a+\varepsilon_n)$ then we have $\sum \lambda(E_n) = b-a+2 \sum \varepsilon_n$. For the moment ignoring that this is totally not rigorous and possibly nonsensical, the desired result follows if we can define $\varepsilon_n$ properly. Or maybe not...
Anyway those are my thoughts. Can you give me some hints on how to really do this? :P Thanks.
AI: Consider the sets $(a-\frac{1}{n},b)$.
We then have that $\mu^*([a,b)) \leq b - a +\frac{1}{n}$ for all $n$ since $[a,b) \subseteq (a- \frac{1}{n},b)$, for all $n$.
Now we also have that $(a,b) \subset [a,b)$ so it must be that $b-a = \mu^*((a,b)) \leq \mu^*([a,b))$.
So we have that $\mu^*([a,b)) = b-a$. |
H: $\forall$ ${i \in \{1,...n\} }$ $ a_{i}
Here's my exercise:
EDIT:
$v=nu$, not $\nu$ (same Latex code but one is without )
$\forall n \in \mathbb{N}$
$\forall$ ${i \in \{1,...n\} }$ $ |a_{i}|\leq u $ and $nu<0.01$ and $u=2^{-t-1}$ prove that there exists $\eta$ such that $\prod_{i=0}^{n}(1+a_{i})$=1+$\eta$ and $\eta \leq 1.01\nu$ .
In my classroom we have proven the following theorem:
$\forall n \in \mathbb{N}$ if
$\forall$ ${i \in \{1,...n\} } $ $ |a_{i}|\leq u, p_{i} \in \{1,-1\} $ and $\nu<1$ where $u=2^{-t-1}$ for some integer $t$
then
there exists $\theta$ such that $\prod_{i=0}^{n}(1+a_{i})^{p_{i}}$=1+$\theta$ and $|\theta| \leq \frac{\nu}{1-\nu}$
At first I thought that my exercise demanded nothing more that using the theorem we have proven, for the case where all the $p$'s were $1$'s. Using that I put $\eta=\theta$ , but then I got an estimation $|\eta|\leq \frac{\nu}{1-\nu} < 1.(01)\nu$, which is slightly more than I wanted, and I see no other way to approach this exercise... Can anyone help?
AI: While the "class theorem" can be proved simply by induction, this one can be a bit "tricky". Let $\prod_{i=1}^n (1+a_i)=1+\theta$, so that
$$
\theta=\prod_{i=1}^n(1+a_i)-1.
$$
Since $|a_i|\leq u$, we have
$$
|\theta|=\left|\prod_{i=1}^n(1+a_i)-1\right|\leq(1+u)^n-1.
$$
Use the binomial theorem to get
$$
\begin{split}
|\theta|&\leq\sum_{k=1}^n\begin{pmatrix}n \\k\end{pmatrix}u^k
\leq\sum_{k=1}^n\frac{(nu)^k}{k!}\leq nu\sum_{k=0}^n\left(\frac{nu}{2}\right)^k
\\&\leq nu\sum_{k=0}^{\infty}\left(\frac{nu}{2}\right)^k\leq\frac{nu}{1-nu/2}\leq\frac{nu}{0.995}<1.01nu.
\end{split}
$$ |
H: Showing Expected Value Does Not Exist
Let X be a uniform random variable on [0,1], and let $Y=\tan\left (\pi \left(x-\frac{1}{2}\right)\right)$. Calculate E(Y) if it exists.
After doing some research into this problem, I have discovered that Y has a Cauchy distribution (although I do not know how to prove this); therefore, E(Y) does not exist.
Also, I know that if I can show the improper integral does not absolutely converge - i.e., that $\int_{-\infty}^{\infty}|\tan\left(\pi\left(x-\frac{1}{2}\right)\right|dx$ diverges - I can show that E(Y) does not exist.
The problem is that I do not know how to evaluate this integral. Could someone please enlighten me on how to do so? Thanks in advance.
AI: $-\pi^{-1}\ln\cos\pi\left(x-\frac{1}{2}\right)$ serves as primitive
of $\tan\pi\left(x-\frac{1}{2}\right)$ on $\left(0,1\right)$ |
H: Initial value problem $x'=x^{2/3}$
I am just analyzing the IVP $x'=x^{\frac{2}{3}}$ with initial condition $x(0)=0$. It is obvious that there is a solution which is not unique, since it is not Lipschitz (not bounded near 0). My book says it has a solution that satifies $x(t)=0$ iff $t\in [t_1,t_2]$ for eached fixed $t_1<0<t_2$.
When I divide both sides by $x^{2/3}$ tae integrals etc. I get $x(t)=(\frac{t}{3})^3$. I do not see the the above equivalance.
AI: $$x(t) =
\begin{cases}
\left(\frac{t-t_1}{3}\right)^3 & \text{if } t<t_1 \\
0 & \text{if } t_1\leq t\leq t_2 \\
\left(\frac{t-t_2}{3}\right)^3 & \text{if } t>t_2
\end{cases}$$
works.
You divide by $x^{2/3}$, but you cannot divide by $0$, so the validity of that division only holds when you know $x\neq 0$. The above shows all possible solutions; $x$ can only be zero on an interval, because once positive it stays positive to the right, and once negative it stays negative to the left. The initial condition just tells you that $t_1\leq 0\leq t_2$. ($t_1$ could be $-\infty$ and $t_2$ could be $+\infty$.) |
H: Regarding countable sets
Please help to to prove that if a collection of sets $A$ is countable, then the set of all finite intersections of members of $A$ is also countable.
I couldn't find the bijection between them.
AI: Let $\mathcal{A} = \{A_1, A_2, \ldots\}$ be a countable collection. For each $k \in \mathbb{N}$, note that $\mathbb{N}^k$ is countable. For each $\alpha =(j_1, j_2, \ldots, j_k) \in \mathbb{N}^k$, define
$$
B_{k, \alpha} = A_{j_1}\cap A_{j_2}\cap \ldots\cap A_{j_k}
$$
and let
$$
\mathcal{B}_k := \{B_{k,\alpha} : \alpha \in \mathbb{N}^k\}
$$
Then $\mathcal{B}_k$ is countable, and the collection you are interested in is
$$
\bigcup_{k=1}^{\infty} \mathcal{B}_k
$$ |
H: Given $Re(f(z))$ and the fact that $f(z)$ is analytic, find $Im(f(z))$
The question I'm trying to answer:
Find an analytic function $f(z)$ whose real part $u(x,y)$ is:
$$\frac{y}{x^2+y^2}$$
An analytic function satisfies the Cauchy-Riemann relations. So I thought to differentiate the real function, and then integrate it, to find two expressions for $v(x,y) = Im(f(z))$. The partial integrals are:
$$ \frac{\partial u}{\partial x} = \frac{-2xy}{(x^2+y^2)^2} = \frac{\partial v}{\partial y} $$
$$ \frac{\partial u}{\partial y} = \frac{x^2-y^2}{(x^2+y^2)^2} = - \frac{\partial v}{\partial x} $$
The next step would be to integrate and find $v(x,y)$.
Is this the right track? Is it possible to integrate these expressions? If so, where would you start?
AI: You are on the right track. In order to integrate $$\frac{-2xy}{(x^2+y^2)^2}$$ for example you could substitute $p(y)=x^2+y^2$. |
H: Subtle question concerning intersection of convex sets
I am attempting to convince myself that if $$\{S_{\alpha}: \alpha \in \mathcal{A}\}$$ is any collection of convex sets, then $$\cap_{\alpha \in \mathcal{A}}S_{\alpha}$$ is convex. This is my proof so far:
Let $C = \cap_{\alpha \in \mathcal{A}}S_{\alpha}$ be an intersection of convex sets. If there is only one point $\textbf{v}$ in $C$ then we immediately know that $C$ is convex. Assume, then, that there are at least two points $\textbf{u}, \textbf{v} \in C$. Then $\textbf{u}, \textbf{v}$ are contained in every convex set in $\{S_{\alpha}: \alpha \in \mathcal{A}\}$. Because each of these sets are convex, it follows that the set $\{t\textbf{u} + (1-t)\textbf{v}, 0 \leq t \leq 1\}$ is contained in each of these sets as well and is therefore contained in $C$. Thus, $C$ is convex.
It is my belief, however, that the above proof is incomplete because we have not accounted for the possibility that $C = \emptyset$. Now, I perused Google and found several sources claiming that the null set is convex. I am having trouble believing them; considering that there are no points in the set with which to take linear combinations with nonnegative components summing to 1. Is it really just as simple as saying that because there are no points in the set that need to satisfy the above property, that is is vacuously true that the null set is convex?
AI: Yes, the traditional definition of convexity is that "for any two points", the convexity condition holds. When there are no points, then the condition is vacuously true "for any two points", because showing convexity fails would require showing the existence of two points for which the convexity condition fails, which cannot be done. |
H: Taylor evaluation in a product solving a limit
I have the following function, which I am supposed to evaluate:
$\lim_{x \to 0}{\frac{(e^{-x^2}-1)\sin x}{x \ln (1+x^2)}}$
My though is to replace sin x by its Maclaurin polynomial, as such:
$\lim_{x \to 0}{\frac{(e^{-x^2}-1)(x+ O(x^3))}{x \ln (1+x^2)}}$
From here I think I should be able to simplify the denominator, divide everything by x, derivate according to l'Hopital's rule, and get the final result of -1. I'm just not quite sure if this is allowed.
First of all, what happens when I multiply $e^{-x^2}$ by $O(x^3)$ in the denominator? And what happens when I derivate $O(x^3)$? Is there a better strategy?
AI: You can use the Taylor series for the other functions as
$$ {\frac{(e^{-x^2}-1)(x+ O(x^3))}{x \ln (1+x^2)}}= {\frac{({-x^2}+O(x^4))(x+ O(x^3))}{x (x^2+O(x^4))}}\sim \frac{(-x^2)(x)}{(x)(x^2)}=-1. $$ |
H: Consider the ring $R=ℂ[X,Y]$ and the ideal $I=(X^2-Y,X^2+Y)$. We find (??) that $R/I ≅ℂ[X]/(X^2)$.
I'm trying to understand a step in an example of my reader about rings.
Consider the ring $R=ℂ[X,Y]$ and the ideal $I=(X^2-Y,X^2+Y)$. We find
that $R/I ≅ℂ[X]/(X^2)$.
As the author doesn't give any more details of this step, I think the author thinks this step is obvious. However, it is not yet obvious to me.
I do know the following facts:
$$ℂ[X,Y] = (ℂ[X])[Y]$$ and $$R[X]/(X-a) ≅ R$$ where $a \in R$ and $R$ a commutative ring with idendity.
I've been trying for some time, but without succes. Any help would be appreciated.
AI: You can treat the polynomials in your ideal as equations identifying elements of $R$ with each other. Thus $X^2 = Y$ and $X^2 = -Y$. This means you can leave out $Y$ by substituting it with $X^2$ which in turn equals zero. |
H: Does order matter for the convergence of infinite products
Similar to infinite sums, does order matter in the convergence of infinite products? More specifically, I'm interested in the product of all rational numbers in the interval $(0,a]$.
For example, let $a=3$. I assert that the product converges to 0. Since all rationals in the interval $[\frac{1}{3},1)$ have a unique inverse in the interval $(1,3]$, we are left with the infinite product of all rationals in the interval $(0,\frac{1}{3})$, which converges to 0.
Now, consider the following product:
$$\prod _{i=0}^{n-1}{\frac{3n-3i}{n}}=3\cdot\frac{3n-3}{n}\cdot\frac{3n-6}{n}\cdot \ldots \cdot \frac{6}{n}\cdot\frac{3}{n}$$
As a quick example, when $n=18=3\cdot3!$, we get (in simplest form):
$$3\cdot\frac{17}{6}\cdot\frac{8}{3}\cdot\frac{5}{2}\cdot\frac{7}{3}\cdot\frac{13}{6}\cdot2\cdot\frac{11}{6}\cdot\frac{5}{3}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{7}{6}\cdot1\cdot\frac{5}{6}\cdot\frac{2}{3}\cdot\frac{1}{2}\cdot\frac{1}{3}\cdot\frac{1}{6}$$
The important thing to see here is that we have all rational numbers from 0 to 3 with denominators of 1,2, and 3 in our product (as well as some other rationals). In fact, if we let $n=3k!$ for some $k\in\mathbb{N}$, we will get all rational numbers from 0 to 3 with denominators of k. Therefore, if we let $n\to\infty$, then (unless I'm mistaken) we should get the product of all rationals from 0 to 3. This, however, diverges to infinity:
$$\lim_{n \to \infty}\prod _{i=0}^{n-1}{\frac{3n-3i}{n}}=\lim_{n \to \infty}\frac{3^n}{n^n} n!=\infty$$
With that being said, why does one of the products converge and the other diverge? My two guesses there's some condition on infinite products for convergence, or I am missing something in the second product.
AI: An infinte product $\prod_{n=1}^\infty a_n$ with $a_n>0$ is said to converge if and only if $\sum_{n=1}^\infty \ln a_n$ converges.
(And a product inovlving negative or zero factors is allowed only if there are at most finitely many of them)
Note that this is more restrictive than saying that $\lim_{m\to\infty}\prod_{n=1}^m a_n$ converges.
Just like the convergence and limit of $\sum_{n=1}^\infty \ln a_n$ may depend on the summation order, the same applies to the infinite product.
Only if the series converges absolutely, i.e. $\sum_{n=1}^\infty |\ln a_n|<\infty$, the order of summands (and hence also of factors) is irrelevant.
We can translate the condition for absolute convergence to products: If we replace all factors $<1$ by their reciprocals and then the sequence of partial products still converges, then the original product does not depend on the order of factors.
Since $|\ln x|$ becomes arbitrarily big when $x$ approaches $0$, your product of rationals in $(0,a]$ indeed depends on factor order. |
H: Finding Jordan Basis of a matrix
Having trouble finding the Jordan base (and hence $P$) for this matrix
$A = \begin{pmatrix}
15&-4\\ 49&-13
\end{pmatrix}$
I know that the eigenvalue is $1$, this gives an eigenvector $\begin{pmatrix}
2\\ 7
\end{pmatrix} $
Now to create the Jordan basis and find $P$ (of which its columns will consist of the two basis vectors) I'm aware that I need to find $v_1$ s.t $(A-I)v_1 = 0$, so $v_1 = \begin{pmatrix}
2\\ 7
\end{pmatrix}$. Now to find $v_2$ I need to do $ker(A-I)^2$ but $(A-I)^2 = 0$ so any non zero vector is in the kernel right$?$ So why doesn't it work if I choose $v_2 = \begin{pmatrix}
1\\ 0
\end{pmatrix}$ $?$
AI: Let
$$P=\left[ \begin{array}{cc}2 & a\\ 7 & b\end{array} \right]$$
You are meant to solve the equation $AP = PJ$, with
$$J=\left[ \begin{array}{cc}1 & 1\\ 0 & 1\end{array} \right].$$
$J$ has the previous form since its only eigenvalues $\lambda=1$ has an eigenspace of dimension $1$.
Expanding $AP = PJ$, you get the following non-trivial equations:
$$\left\{ \begin{array}{rcl}15a - 4b & = & a+2\\49a-13b & = & b+7\end{array} \right.$$
$$\left\{ \begin{array}{rcl}14a - 4b & = & 2\\49a-14b & = & 7\end{array} \right.$$
These equations are linearly dependent, so you can choose to solve the first one:
$$\left\{ \begin{array}{rcl}a & = & \frac{1+2k}{7}\\b & = & k\end{array} \right.$$
If you fix $k$, say $k=0$, then
$$P=\left[ \begin{array}{cc}2 & \frac{1}{7}\\ 7 & 0\end{array} \right]$$
Note that $$det(P(k))=\left[ \begin{array}{cc}2 & \frac{1+2k}{7}\\ 7 & k\end{array} \right] = -1 \neq 0 ~\forall k,$$ so the choice of $k$ is arbitrary. |
H: A subgroup of $S_n$ of odd order is contained in $A_n$
I saw this question. The original questioner asserted that the Lagrange's Theorem is sufficient to solve the problem, but I think that the theorem does just say that the order of $H$ divides the order of $A_n$, but not say that $H$ is a subgroup of $A_n$...
Is my opinion correct? If not, please point out my fault. Thank you in advance.
AI: Let $H$ be an odd subgroup of $S_n$.
If $x \in H$, then by Lagrange theorem $ord(x) | |H|$, thus $ord(x)$ is odd. Thus by definition $x \in A_n$. |
H: How find Number of integer solutions $x_{1}+x_{2}+\cdots+x_{n}=m$
Question:
let $m$ and $n$ be positive integers,The number of positive integer solutions to the equation
$$x_{1}+x_{2}+\cdots+x_{n}=m,m\ge n,x_{i}\ge 1,1\le x_{1}\le x_{2}\le\cdots\le x_{n},(i=1,2,\cdots,n)$$
is $f(m,n)$,How find this close form $f(m,n)?$
I know this therom:
Let $n$ and $m$ be positive integers,The number of positive integer solution to the equation
$$x_{1}+x_{2}+\cdots+x_{m}=n(n\ge m)$$ is
$$N=\binom{n-1}{m-1}$$
But for my problem,I can't prove it.Thank you
BY the way
In 2010 China math competition,there is this problem
find the total number of sets of positive integer $(x,y,z)$,where $x,y$ and $z$ are positive integers, with $x\le y\le z$
and such that
$$x+y+z=2010$$
this answer is $$336675$$
can see this PDF：(problem 8) http://wenku.baidu.com/view/9a59934ee518964bcf847cba.html
and I also Find this same problem is Sinapore Mathematical Olympiad (SMO)2012 Problem 4:
find the total number of sets of positive integer $(x,y,z)$,where $x,y$ and $z$ are positive integers, with $x\le y\le z$
and such that
$$x+y+z=203$$
This follow is office Solution(Maybe is wrong).
First note that there are $\binom{202}{2}=\dfrac{202(201)}{2}=20301$ positive integers sets$(x,y,z)$ which satisfy the given equation.These solution sets include those where two of the three values are equal.if $x=y$ then $2x+z=203$,By enumeerating,$z=1,3,5,\cdots,201$.There are thus $101$ solutions of the form $(x,x,z)$, similarly,there are $101$ solutions of the form $(x,y,x)$ and $(x,y,y)$,since $x<y<z$,the required answer is
$$\dfrac{1}{3!}\left(\binom{202}{2}-3(101)\right)=\dfrac{20301-303}{6}=3333$$
AI: Suppose we want to count the number of non-decreasing sequences of $n$ non-negative integers which sum to $m$. That is, $a_{k+1}\ge a_k$ and
$$
\sum_{k=1}^na_k=m\tag{1}
$$
Note that if we set $a_0=0$, then
$$
\begin{align}
m
&=\sum_{k=1}^na_k\\
&=\sum_{k=1}^n\sum_{j=1}^k(a_j-a_{j-1})\\
&=\sum_{j=1}^n\sum_{k=j}^n(a_j-a_{j-1})\\
&=\sum_{j=1}^n(n-j+1)(a_j-a_{j-1})\tag{2}
\end{align}
$$
The following diagram illustrates $(2)$ for $n=4$:
$\hspace{4.5cm}$
Consider the product
$$
\overbrace{(1+x^n+x^{2n}+\dots)}^{\large x^{(a_1-a_0)n}}
\overbrace{(1+x^{n-1}+x^{2(n-1)}+\dots)}^{\large x^{(a_2-a_1)(n-1)}}
\dots\overbrace{(1+x+x^2+\dots)}^{\large x^{a_n-a_{n-1}}}\tag{3}
$$
In the first factor, we choose $x^{(a_1-a_0)n}$. In the second factor, we choose $x^{(a_2-a_1)(n-1)}$. In the $k^\text{th}$ factor, we choose $x^{(a_k-a_{k-1})(n-k+1)}$. In the product, the coefficient of $x^m$ is the number of ways to make the sum in $(2)$.
The product in $(3)$ can be rewritten as
$$
\prod_{k=1}^n\frac1{1-x^k}\tag{4}
$$
The coefficient of $x^m$ in $(4)$ is the number of non-decreasing sequences of $n$ non-negative integers which sum to $m$.
The number of non-decreasing sequences of $n$ positive integers that sum to $m$ is the number of non-decreasing sequences of $n$ non-negative integers that sum to $m-n$. Just add $1$ to each element of the latter to get the former.
The number of increasing sequences of $n$ positive integers that sum to $m$ is the number of non-decreasing sequences of $n$ non-negative integers that sum to $m-n(n+1)/2$. Just add $k$ to the $k^\text{th}$ element of the latter to get the former.
The number of increasing sequences of $n$ non-negative integers that sum to $m$ is the number of non-decreasing sequences of $n$ non-negative integers that sum to $m-n(n-1)/2$. Just add $k-1$ to the $k^\text{th}$ element of the latter to get the former.
The number of non-decreasing sequences of $3$ positive integers that sum to $2010$ is the coefficient of $x^{2007}$ in
$$
\prod_{k=1}^3\frac1{1-x^k}\tag{5}
$$
which is $336675$. This matches the answer you give.
The number of increasing sequences of $3$ positive integers that sum to $203$ is the coefficient of $x^{197}$ in $(5)$, which is $3333$. This matches the official solution. However, the question asks for the number of non-decreasing sequences of $3$ positive integers that sum to $203$, which is the coefficient of $x^{200}$ in $(5)$, $3434$.
Closed Form for $(5)$
Since $(5)$ is the inverse of
$$
(1-x)(1-x^2)(1-x^3)=1-x-x^2+x^4+x^5-x^6\tag{6}
$$
the coefficients of $(5)$ are determined by
$$
a_n=a_{n-1}+a_{n-2}-a_{n-4}-a_{n-5}+a_{n-6}\\
(a_0,a_1,a_2,a_3,a_4,a_5,\dots)=(1,1,2,3,4,5,\dots)\tag{7}
$$
Since the roots of $(6)$ are $\left\{1,1,1,-1,e^{2\pi i/3},e^{-2\pi i/3}\right\}$, the solution of $(7)$ looks like
$$
a_n=c_0+c_1n+c_2n^2+c_3(-1)^n+c_4\cos(2\pi n/3)+c_5\sin(2\pi n/3)\tag{8}
$$
Using the initial values from $(7)$, we can compute the coefficients in $(8)$:
$$
a_n=\frac1{72}\left(47+36n+6n^2+9(-1)^n+16\cos(2\pi n/3)\right)\tag{9}
$$
Note that $9(-1)^n+16\cos(2\pi n/3)$ repeats mod $6$: $(25,-17,1,7,1,-17)$. Putting this together with $(9)$ yields
$$
a_n=\left\lfloor\frac{12+6n+n^2}{12}\right\rfloor\tag{10}
$$
Using $(10)$, we get $a_{2007}=336675$, $a_{197}=3333$, and $a_{200}=3434$, just as above. |
H: If $x=\prod^{27}_{n=1}(1+\frac{2}{n})$ then find $13x$ - Ramanujan Mathematics olmpiad 2013
I tried this: $$x=\prod^{27}_{n=1}(1+\frac{2}{n})=(1+\frac{2}{1})(1+\frac{2}{2})(1+\frac{2}{3})\ldots(1+\frac{2}{27})=\frac{3}{1}\cdot\frac{4}{2}\cdot\frac{5}{3}\cdots\frac{29}{27}$$ Then the terms cancel out. But I am not getting the correct answer.
AI: As you noted, $$x=\frac31\cdot\frac42\cdot\frac53\cdots\frac{29}{27}=\frac11\cdot\frac12\cdot\frac{28}1\cdot\frac{29}1=406$$hence $$13x=13\cdot406=5278$$ |
H: vector represented by standard basis
let us consider following problem
i want to clarify one thing,namely vector $v=(a,b,0)$ satisfy yes this condition?or is there any other thing asked? generally every $v$ in $R^3$ is represented by
$v=v_1*i+v_2*j+v_3*k$
i have tried figure out what combination of $m$ and $n$ cancel out $z$ component?only possible is this vector be parallel to $z$ axis right?so $m$ and $n$ must make vector parallel to $z$ axis right?
AI: The question is asking: Can every vector $(a,b,c)$ be written as combinations of just $i$ and $j$?
So what you have to ask yourself is "Can we get a $z$-component by just adding $i$ and $j$ vectors?". |
H: Is $f(0, y) = f(y) \implies f_{y} (0,y) = f'(y)$ true?
OK, I think I'm being a bit stupid here, but I need to check whether my reasoning here is correct, as I don't want to mess this up. So my question is:
Is $f(0, y) = f(y) \implies f_{y} (0,y) = f'(y)$ true?
It seems to me that it is, but, for some reason, I've gotten confused and am having some trouble getting out of this confusion, and figured I would ask here just to get this straight.
AI: This is my initial answer which is wrong
Nope you may have some $y$'s multiplying $x$'s consider $f(x,y) = xy$
$$f(y) = f(0,y)= 0 \Rightarrow f'(y)=0$$
$$ f_y(x,y)=y \Rightarrow f_y(0,y)=y $$
It is true which can be seen from the definition of the partial derivative.
$$ f_y(0,y)= \lim_{h\rightarrow 0} \frac{f(0,y+h)-f(0,y)}{h} = f'(y) $$ |
H: Induction on natural numbers
My textbook, Logic and Discrete Mathematics by Grassman and Tremblay, has an example which I can't wrap my head around (example 3.4; page 127).
It shows that for all $n$, $2(n+2)\le(n+2)^2$.
As the inductive base, we have $P(0): 2(0 + 2) \le (0 + 2)^2 = 4 \le 4$, which is true.
We assume $P(n): 2(n+2)\le(n+2)^2$ is true as the inductive hypothesis.
But for the inductive step, the author does:
$$\begin{align}2((n + 1) + 2) &= 2((n + 2) + 1) \\
&= 2(n + 2) + 2 \\
&\le (n + 2)^2 + 2 \\
& \le n^2 + 4n + 6 \\
& \le n^2 + 6n + 9 \tag 5\\
& = ((n+1)+2)^2\end{align}$$
The first four lines are straightforward enough. But on line 5, the $2n + 3$ is added and the author doesn't state where that came from. Can someone please explain what happened in that step?
AI: Because it's not specified that we're working with natural numbers, you should probably take induction on all integer. If it's so then you should take in account only the second part of the proof.
Note that the LHS is negative for $n < -2$ and the RHS is always positive as a square. So you the basis should start from $n = -2$ instead.
Note that for numbers $n \ge -\frac 32$ the following statment is always true:
$$2n + 3 \ge 0$$
Our inductive step starts from $n=-1$ so it this inequality holds for every number.
Now just add $n^2 + 4n + 6$ to both sides. It won't change nothing, beacuse we are adding it to both sides.
$$n^2 + 4n + 6 + 2n + 3 \ge n^2 + 4n + 6$$
$$n^2 + 6n + 9 \ge n^2 + 4n + 6$$ |
H: If n is such that every element $(\mathbb{Z}/n\mathbb{Z})^{\times}$ is a root of $x^2-1$. Prove that $n$ divides 24.
I have a hard time formulating proofs. For this problem, I can see that if $n$ is equal to $8,$ this statement is true. $(\mathbb{Z}/8\mathbb{Z})^{\times}$ includes elements: $1,3,5,7$, and all of these are roots of $1-x^2 \pmod 8.$ And obviously $8$ divides $24.$
But how do I prove this without depending on number calculations and only using theorems? Help Please? I need a step by step walk through of how to do this proof and what theorems would be appropriate to use.
AI: Suppose that $n=2^\ell 3^{m}p_1^{e_1}\cdots p_r^{e_r}$. We know that $$(\Bbb Z/n\Bbb Z)^\times\simeq (\Bbb Z/2^\ell\Bbb Z)^\times\times (\Bbb Z/3^{m}\Bbb Z)^\times\times (\Bbb Z/p_1^{e_1}\Bbb Z)^\times\times \cdots \times (\Bbb Z/p_r^{e_r}\Bbb Z)^\times$$
Suppose $p>3$. We know $(\Bbb Z/p_r^{e_r}\Bbb Z)^\times$ is cyclic of order $\geqslant 4$, so $x^2=1$ for each $x$ is impossible. Thus we necessarily need $n=2^\ell 3^m$, that is $$(\Bbb Z/n\Bbb Z)^\times\simeq (\Bbb Z/2^\ell\Bbb Z)^\times\times (\Bbb Z/3^{m}\Bbb Z)^\times$$
Suppose $m>1$. Since $(\Bbb Z/3^{m}\Bbb Z)^\times$ is cyclic of order $\geqslant 6$ we cannot have $m>1$. Thus we have
$$(\Bbb Z/n\Bbb Z)^\times\simeq (\Bbb Z/2^\ell\Bbb Z)^\times\times (\Bbb Z/3^{m}\Bbb Z)^\times$$
with $m=0,1$. It remains to show $\ell=0,1,2,3$. Finally, if $\ell \geqslant 3$, $$(\Bbb Z/2^\ell\Bbb Z)^\times\simeq C_2\times C_{2^{\ell-2}}$$
If $\ell >3$ we have $2^{\ell-2}\geqslant 4$, incompatible with $x^2=1$. Thus you know that $n$ must be of the form $n=2^\ell 3^m$ with $\ell=0,1,2,3$ and $m=0,1$. |
H: Show that there exists a subspace $W \subset \mathcal{P}(4)$ such that $\mathcal{P}(4) = \mathcal{U}(4) \oplus W.$
Let $\mathcal{U}(4)$ be a subspace of $\mathcal{P}(4)$ consisting of polynomials that are even functions. Show that there exists a subspace $W \subset \mathcal{P}(4)$ such that $$\mathcal{P}(4) = \mathcal{U}(4) \oplus W.$$
Additionally,
I know that a function $f:\mathbb{R} \mapsto \mathbb{R}$ is even if $f(x) = f(-x)$ for all $x$.
And that a function can also be expressed as the sum of an odd an even function $$f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}.$$
AI: $$\mathcal{U}(4)=Span(1,x^2, x^4), \mathcal{W}=Span(x,x^3)$$ |
H: How to prove that $S_n^2 − Var(S_n )$ is a martingale
I would be grateful for some help with the following exercise:
Let $(X_n ,n≥1)$ be a sequence of independent random variables with $E[X_i]=0$, and
$Var(X_i)=σ_i^2<\infty, ∀i ∈\mathbb{N}$. Prove that $S_n^2 − Var(S_n)$ is a martingale, where
$S_n:= \sum\limits_{i=1}^n X_i$
We have to show: $E[S_{n+1}^2-Var(S_{n+1})|\mathcal{F}_n]=S_n^2-Var(S_n)$.
I tried to do this by induction, but I'm already having trouble with the base case:
n=1: $E[S_2^2-Var(S_2)|\mathcal{F}_1]=E[X_1^2|\mathcal{F}_1]+2E[X_1|\mathcal{F}_1]E[X_2
\mathcal{F}_1]+E[X_2^2|\mathcal{F}_1]-E[E(X_1^2)+E(X_2^2)|\mathcal{F}_1]=...?$
I would be glad, if you could tell me, if I did something wrong so far, or, if not, how to continue.
Thanks for your help!
AI: Here I will use $E_n[\cdot]$ for $E[\cdot | \mathcal{F}_n]$.Note that since the $X_i$ are independent, you have $E_n(\sum_1^N X_i)^2 = \sum_1^N E_n X_i^2$, all of the cross terms vanish since for $i < j$ we have by iterated conditioning $$E_nX_iX_j = E_nE_i X_iX_j = E_n X_i E_iX_j = E_n X_i E X_j = E_n X_i 0 = 0.$$
Use this and the fact variances add for independent random variables as well. |
H: Does there exist a symmetric tridiagonal matrix with zero determinant?
I will like to know whether there exists a symmetric tridiagonal matrix with zero determinant? I will refer the definition of a tridiagonal matrix to the one found in Wikipedia:
"A tridiagonal matrix is a matrix that has nonzero elements only on the main diagonal, the first diagonal below this, and the first diagonal above the main diagonal.".
If yes, kindly provide an example. If not, please illustrate the proof or idea.
Thank you very much.
AI: The trivial example is
$$
\pmatrix{0 & 0 \\ 0 & 0}.
$$
Or you could consider:
$$
\pmatrix{1 & 1 \\ 1 & 1} \quad\text{or} \quad\pmatrix{1 & 1 & 0 \\ 1 & 1 & 0 \\ 0 & 0 & 1}.
$$
Otherwise if you only want non-zero entries of the diagonals, then how about
$$
\pmatrix{1 & a & 0 \\ a & 1 & a \\ 0 & a & 1}
$$
where $a = \frac{1}{\sqrt{2}}$.
Since you ask, I don't know if this might work, but try to consider
$$
\pmatrix{1 & a & 0 & 0\\ a & 1 & a & 0\\ 0 & a & 1& a \\ 0 & 0 & a & 1}
$$
and find the determinant of this matrix. This will be an expression in $a$. Can this ever equal zero? If this doesn't work, maybe this will give you an idea to something that would. |
H: Eigenvectors of a linear operator
Let $A$ be $m \times m$ and $B$ be $n \times n$ complex matrices, and consider the linear operator $T$ on the space $\mathbb{C}^{m\times n}$ of all $m \times n$ complex matrices defined by $T(M)=AMB$.
(a) Show how to construct an eigenvalue for $T$ out of a pair of column vectors $X,Y$, where $X$ is an eigenvector for $A$ and $Y$ is an eigenvector for $B^t$.
What I've tried: I tried using the definitions: $X$ eigenvector of $A$: $X \neq 0$ and $( \exists \lambda \in \mathbb{C}, AX= \lambda X)$ and $Y$ eigenvector of $B^t$: $Y \neq 0$ and $( \exists \mu \in \mathbb{C}, B^tY= \mu Y)$
I tried combining the two in order to get an expression similar to $T(M)$: $AX(Y^tB)^t=\lambda\mu XY$ but it doesn't seem to lead be anywhere.
I haven't looked at the other questions, but I will still post them:
b) Determine the eigenvalues of $T$ in terms of those of $A$ and $B$.
c) Determine the trace of this operator.
AI: Suppose $Ax=\lambda x$, $B^* y = \mu y$. Then $T(x y^*) = Ax y^* B = (Ax) (B^*y)^* = (\lambda x) (\mu y)^* = \lambda \overline{\mu} x y^*$. Hence $x y^*$ is an eigenvector of $T$ corresponding to the eigenvalue $\lambda \overline{\mu}$.
(b) follows from this, but you will need to deal with the case when $A,B^*$ do not have a full set of eigenvectors. One way is rely on 'continuity' of eigenvalues and find 'nearby' $A_n,B_n$ that have a full set of eigenvectors. Some attention to detail is needed.
(c) follows from (b) using the fact that the trace is the sum of the eigenvalues.
Addendum: More detail for (b):
If $A,B^*$ both have distinct eigenvalues, then both have eigenvectors $u_1,...,u_m$ and $v_1,...,v_n$ that span $\mathbb{C}^m$ and $\mathbb{C}^n$ respectively. Then $u_i v_j^*$ form a basis for $\mathbb{C}^{m \times n}$. To see this, suppose $\sum \alpha_{ij} u_i v_j^* = 0$.
Let $U,V$ be the corresponding matrices formed by stacking the columns together. Then we can write this as $\sum \alpha_{ij} u_i v_j^* = \sum \alpha_{ij} U (e_i e_j^T) V^* = U(\sum \alpha_{ij} e_i e_j^T)V^* = 0$. Since both $U,V$ are non-singular, it follows that $\alpha_{ij} = 0$. In this basis, $T$ is diagonal, and the characteristic polynomial is given by $\chi_T(x) = \Pi_{\lambda \in \sigma(A), \mu \in \sigma(B^*)} (x-\lambda \overline{\mu}) =
\Pi_{\lambda \in \sigma(A), \mu \in \sigma(B)} (x-\lambda {\mu})$.
One way to deal with repeated eigenvalues is to approximate $A,B$ with matrices $A_i,B_i$ that have distinct eigenvalues.
Let $\chi_T$ be the characteristic polynomial of $T$. Suppose $A_i \to A$ and $B_i \to B$,and let $T_i(X) = A_i X B_i$. Let $\chi_{T_i}$ be the characteristic polynomial of $T_i$. The coefficients of the characteristic polynomial are continuous functions of the constituent matrix, hence we have $\chi_{T_i} \to \chi_T$ (the space $\mathbb{P}^{(nm+1)}$ of polynomials of degree $nm+1$ or less is finite dimensional, so we can use any suitable norm to characterize convergence).
To finish, we must find suitable $A_i,B_i$ and show that $\chi_{T_i}$ converges to a polynomial of the required form.
Let $\Delta_k=\operatorname{diag}(1,...,k)$ be a diagonal matrix. Let $U^{-1}AU =J_A$ be the Jordan form of $A$. Let $A_\epsilon = U(J_A+\epsilon \Delta_m)U^{-1}$.
We see that $\sigma(A_\epsilon) = \{ [J_A]_{ii}+i \epsilon\}_{i=1}^m$. Hence for some $\delta_A>0$, if $\epsilon \in (0,\delta_A)$, we have $|\sigma(A_\epsilon)| =m$.
A similar construction gives a $B_\epsilon$ such that for some
$\delta_B >0$, then $\sigma(B_\epsilon) = \{ [J_B]_{ii}+i \epsilon\}_{i=1}^n$ and $|\sigma(B_\epsilon)| =n$, for all $\epsilon \in (0,\delta_B)$.
Abusing notation slightly, we let $A_i = A_{\frac{1}{i}}, B_i = B_{\frac{1}{i}}$. Hence for sufficiently large $n$ we have
$\chi_{T_i}(x)= \Pi_{a=1}^m \Pi_{b=1}^n (x-([J_A]_{aa}+\frac{a}{i}) {([J_B]_{bb}+\frac{b}{i})})$, and we see $\chi_{T_i} \to \xi$, where $\xi(x) = \Pi_{a=1}^m \Pi_{b=1}^n (x-[J_A]_{aa} {[J_B]_{bb}}) = \Pi_{\lambda \in \sigma(A)} \Pi_{\mu \in \sigma(B)} (x-\lambda {\mu})^{\alpha_A(\lambda) \alpha_B(\mu)}$,
where $\alpha_A$ and $\alpha_B$ are the algebraic multiplicities of the respective eigenvalues.
Hence we have $X_T(x) = \Pi_{\lambda \in \sigma(A)} \Pi_{\mu \in \sigma(B)} (x-\lambda {\mu})^{\alpha_A(\lambda) \alpha_B(\mu)}$.
As an aside, note that the multiplicity of each eigenvalue $\nu = \lambda {\mu}$ is given by $\alpha_T(\nu) = \sum_{\substack{\lambda {\mu} = \nu \\ \lambda \in \sigma(A), \mu \in \sigma(B)}} \alpha_A(\lambda) \alpha_B(\mu)$.
(c) follows since the trace is the sum of the eigenvalues, hence
$\operatorname{tr}T = \sum_{\lambda \in \sigma(A)} \sum_{\mu \in \sigma(B)} \alpha_A(\lambda) \alpha_B(\mu) \lambda {\mu}$. |
H: How to prove this similarly but changing the hypothesis?
I'm trying to prove the followong statement, which includes already a proof, but changing the hypothesis $P^2 = P$ by dim Im $P^2= $ dim Im $P$ and assuming $V$ is finite dimensional.
I was trying to prove that dim Im $P^2 = $ dim Im $P$ implies $P^2=P$, but I don't know if that's true. Instaed, I have a proof that dim Im $P^2 = $ dim Im $P$ implies that Im $P ^2 =$ Im $P$ but I'm stuck there, because I cannot use the key step $P(v)=P(w)$ of the proof, instaed I get $P(P(v))=P(w)$, which implies that $P(v) \in$ Ker $P$, but not that $P(v)=0$.
Thanks for your help.
AI: The assumption $\dim \operatorname{im} P^2 = \dim \operatorname{im} P$ says that the restriction $P\lvert_{\operatorname{im} P}\colon \operatorname{im} P \to \operatorname{im} P$ is surjective, if $\dim\operatorname{im} P < \infty$. Since an endomorphism of a finite-dimensional vector space is surjective if and only if it is injective, we have
$$\{0\} = \ker P\lvert_{\operatorname{im} P} = \ker P \cap \operatorname{im} P$$
under the condition that $\dim\operatorname{im} P < \infty$, without needing the finite-dimensionality of $V$. |
H: no. of quadrilateral in 12 sided polygon
Find the number of quadrilaterals that can be made using the vertices of a polygon of 12 sides as their vertices and having
(1) exactly 1 sides common with the polygon.
(2) exactly 2 sides common with the polygon.
$\underline{\bf{My \; Try}}::$ Let $A_{1},A_{2},A_{3},................A_{12}$ points of a polygon of side $=12$.
(1) part:: Let We Select adjacents pairs $A_{1}A_{2}$, Then other two vertices are from $A_{4},A_{5},.........A_{11}$.
Here $A_{12}$ is not included because it is Left consecutive point corrosponding to $A_{1}$
So this can be done by $\displaystyle \binom{7}{2}$ similarly we can take another consecutive pairs $A_{2}A_{3}$.
So there are Total $12$ adjacents pairs in Anticlock-wise sence.
So Total no. of Quadrilateral in which one side common with $12$ sided polygon is
$\displaystyle = \binom{7}{2}\times 12 = 21\times 12 = 252$
(2) part :: If $2$ selected sides are consecutive:
Let we select $A_{1}A_{2}$ and $A_{2}A_{3}$. Then we select one points from the vertices $A_{4},A_{5},A_{6},.........A_{12}$
This can be done by $\displaystyle {9}{1}$ ways.
Now we select consecutive adjacents sides in Anti-clockwise sence by $(11)$ ways.
So Total ways in above case(for two adjacents sides) is $\displaystyle = \binom{9}{1}\times 11 = 99$
If $2$ selected sides are not consecutive:
Now I did not understand How can i calculate in that case
Help required
Thanks
AI: In the first case there are 12 different sides, so there are 12 different pairs of vertices to count. Your reasoning to leave vertices $A_3$ and $A_{12}$ is OK, but you need to make another restriction after the selection of the third vertex, because if we chose the vertices $A_1, A_2, A_5, A_6$, then we'll have two common sides, right?
Now we have to distinct cases. The first is when our third vertex is $A_{11}$ or $A_4$. Then we will have 6 other vertices to chose those are $A_4, A_5, A_6, A_7, A_8, A_9$ for $A_{11}$.
The number of combinations is $2 \times 6 = 12$.
In the second case the third vertex is one of the rest. So for example if we choose $A_5$ as our third vertex we'll have 5 other options for the fourth vertex. Those are: $A_7, A_8, A_9, A_{10}, A_{11}$. So there are $6 \times 5 = 30$ combinations.
Now add those together and divide by 2, because every polygon is counted twice. Then we'll have:
$$\frac{12 + 30}{2} = \frac{42}{2} = 21 = \binom {7}{2} \text{ quadrilaterial with one common side}$$
Multiply by 12, because we said there are 12 different ways to chose a pair of consecutive vertices and you'll get $252$ distinct quadrilaterials.
For the second problem you should look in two cases. The one is when the two common sides are consecutive (we use 3 consecutive vertices). Again there are 12 such triples.
Note that if we select 3 consecutive vertices then there are 7 available vertices to choose.
There are total of $12 \times 7 = 84$ quadrilaterials.
For the second cas you need again to chose pair of adjacent vertices and anothe pair of adjacent vertices. Because for every pair of vertices there are 8 available vertices, there are 7 other pairs so for selected pair $A_1, A_2$ we can chose the other pair as $(A_4,A_5)$, $(A_5,A_6)$, $(A_6,A_7)$, $(A_7,A_8)$, $(A_8,A_9)$, $(A_9,A_{10})$, $(A_{10},A_{11})$
So there are total of $\frac{12 \times 7}{2} = 42$ quadrilaterial.
We are dividing by two, because the same quadrilaterial can be reached starting from $A_1, A_2$ and $A_5, A_6$, so every quadrilaterial is calcualted twice.
Add those two numbers and you'll end up with: $84 + 42 = 126$ quadrilaterials with two common sides. |
H: Equation of a circle given two points and tangent line
I am given that $P(-3,-1)$ and $Q(5,3)$ are points of the circle. Also, the line $L:0=x+2y-13$ is tangent to said circle. The objective is to find the equation of the circle.
I thought of a way for solving this, but it doesn't seem to be the best one. The option was to find the points equidistant to $P$ and $Q$. Afterwards, using any point $(x,y)$ of the line of points equidistant to $P$ and $Q$, the next step would be to make the distance from $(x,y)$ to $L$ the same as to the distance to $P$ or $Q$.
Equations:
$$
2a + b = 3 \\
5a^2-10a+25 = r^2\\
\sqrt{5}r = a+2b-13
$$
AI: HINT:
If $(a,b)$ is the center of the required circle
we have $$(a+3)^2+(b+1)^2=(a-5)^2+(b-3)^2=r^2$$ where $r$ is the radius
Now, the perpendicular distance of the tangent form the center $(a,b)$ is again equal to the radius
Do you know how to calculate the perpendicular distance? |
H: Show $a^2=6k+3 \Rightarrow a = 6n + 3$
Show that if $a^2=6k+3$, for some integer $k$, then also $a = 6n + 3$ for an integer n.
Or in in other words: $a^2=6k+3 \Rightarrow a = 6n + 3$.
Taking the square root, $a=\sqrt{6k+3}$, does not help. I've also looked after factorizations, but I haven't find anything useful.
AI: A more pedestrian approach is to note that a must have a factor of 3 in it (since the square of a is divisible by 3) and a must also be odd. If you use these two observations, you can also deduce that a must have the form 6n+3 for some natural number n. |
H: Definition of Ring Homomorphism
I am using a text right now for abstract algebra ("A Concrete Introduction to Abstract Algebra" by Lindsay Childs) that seems to use a non-standard defn of ring homomorphism. I want to see if others agree and if the difference is a significant one or not.
Let (R,+,$*$) and (R',+',$*'$) be rings and f:R->R'.
The standard defn for f to be a ring homomorphism (given in Birkhoff and McLane, Dummit and Foote, and Herstein) is that f preserves addition and mutliplication in the two rings.
Childs, however also requires that f satisfy a third condition: if R and R' have multiplicative identity elts 1 and 1', resp, then f(1)=1'.
To me this seems like a non-trivial distinction b/c some mappings that would be considered as homomorphisms under the standard defn would not be a homomorphism under the definition given by Childs. Intuitively, to me this seems non-trivial b/c there are some slippery things that happens when f doesn't map 1 to 1' (i.e. f(1') is the unit in the subring f(R), but not necessarily a unit in R', not to mention the fact that the map of the multiplicative inverse of a unit is the multiplicative inverse of the map of that element, again in f(R), etc)
Not having much experience in abstract algebra, this naturally leads me to wonder if this is a difference which isn't really important and I should just ignore, so I was wondering what others (much more experienced and intelligent than myself) thought.
Thank-you,
Matt
AI: If you’re talking about his book A Concrete Introduction to Higher Algebra, the definition in the third edition, as displayed at Google Books, is the definition of a homomorphism of unital rings, i.e., of rings with identity. He doesn’t say that a homomorphism $h:R\to S$ must satisfy $h(1_R)=1_S$ if the rings have multiplicative identities: he defines the notion of homomorphism only for rings that do have multiplicative identities. I looked back to see how he defines ring, and it appears that he requires a ring to be unital. |
H: If the matrix of $f$ is diagonal, why is $f$ equal to scalar multiple of identity map?
Suppose that $V$ be a finite dimensional vector space, and $f:V \longrightarrow V$ be a non zero linear map. If the matrix of $f$ with respect to any basis of $V$ gives a diagonal matrix, why is $f= \lambda Id. $ where Id is an identity map ?
Attempts:
I am trying to show by contradiction. Let $f \neq \lambda Id.$ This implies that for any $v \in V$, $fv \neq \lambda Id v $ which is same as $fv \neq \lambda v. $ Thus, for all $v \in V$, $v$ and $fv$ are linearly independent. I don't know where this leads now.
AI: You are on right path :
Suppose $f\neq \lambda Id$ for any $\lambda \in F$.
Fix some basis $\{v_1,v_2,\dots v_n\}$ for $V$ over $F$.
As $f$ is non zero linear map, I have for some $v_i \in \{v_1,v_2,\dots v_n\}$ : $f(v_i)\neq 0$ and $f(v_i)\neq \lambda v_i$.
So, by writing corresponding matrix with $\{v_1,v_2,\dots v_n\}$ as basis, you get some non zero element in $i^{th}$ column different from $a_{ii}$..
Does this say anything about matrix being diagonal?
EDIT :how do you write matrix corresponding to linear map????
take a basis vector $v_i$, and write $f(v_i)$ as linear combination of $\{v_1,v_2,\dots v_n\}$
and write coefficients as corresponding column to $v_i$....
you have $f(v_i)\neq \lambda v_i$ and $f(v_i)\neq 0$.
so, there does exists some non zero coefficient to some basis vector $v_j$ for $i\neq j$ which gives some non zero element which is out of diagonal i mean which is not in the diagonal.
i.e., you have some non zero element in a non diagonal entry...
Thus matrix is not diagonal.... |
H: Prove formally that |N| = | N union a finite set |.
I'd like to show that the cardinality of $\mathbb{N}$ is the same as the cardinality of $\mathbb{N}$ union some other finite set (disjoint from $\mathbb{N}$). For example show that:
$|\mathbb{N}|= | \mathbb{N} \cup \lbrace \sqrt{2},\sqrt{3} \rbrace |$.
To prove these two terms have the same cardinality, we must find a bijection from one term to the other.
I think we can prove it using the Hilbert's Grand Hotel paradox, i.e. making two new "free places" for the two values $\sqrt{2},\sqrt{3}$ by adding $2$ to the other elements.
The bijection would be $\theta(x)$ such that:
$\theta(x) = \cases{
2+f(x) & for $x\ge2 $\\
\sqrt{2} & for $x=0 $\\
\sqrt{3} & for $x=1 $\\
}$
with $f(x)= x$ for $x\in\mathbb{N} $.
Is $\theta(x)$ well defined? How to show formally that $\theta(x)$ is a bijection? And does it prove formally that $|\mathbb{N}|= | \mathbb{N} \cup \lbrace \sqrt{2},\sqrt{3} \rbrace |$?
Thanks
Edit:
As Asaf Karagila pointed out, the right definition for $\theta(x)$ is
$\theta(x) = \cases{
x-2 & for $x\ge2 $\\
\sqrt{2} & for $x=0 $\\
\sqrt{3} & for $x=1 $\\
}$
for $x\in\mathbb{N} $.
To show it is a bijection see the explanation of the accepted answer.
AI: Yes, you need to argue why this is a well-defined function, and that it is a bijection; or at least an injection (in which case you will have to use the Cantor-Bernstein theorem).
To claim that it is well-defined you have to show that given $x$ there is only one "output" that the definition you have given can end up with; that this output is in the wanted codomain; and that every natural number appears in the domain of the function. But indeed either $x=0$ or $x=1$ or $x\geq 2$. And clearly $\theta(x)$ is in $\Bbb N\cup\{\sqrt2,\sqrt3\}$.
To show that it is a bijection you need to verify that the function is both injective and surjective. So you need to show that if $x\neq y$ then $\theta(x)\neq\theta(y)$. You have some cases to check, if $x=0$ and $y=1$; if $x,y\geq2$ and so on.
To show that it is surjective you need to show that given $y\in\Bbb N\cup\{\sqrt2,\sqrt3\}$ you can find $x\in\Bbb N$ such that $\theta(x)=y$. Here you have a slight problem, that you need to take $x-2$ rather than $2+x$. But apart of this, it should be fine.
And finally, how does that show the wanted conclusion holds? Well, $|A|=|B|$ if there is a bijection $f\colon A\to B$. Since you have written down such a bijection, it finishes the proof. |
H: Can you give me a good alternative to Rotman's Group Theory book?
I've been trying to learn out of Rotman's book "An Introduction To The Theory of Groups" for the last few months, and it's rough going. I've been studying Chapters 7, 10, and 11 in particular, and he's too wordy in many places and makes massive leaps of logic in others. I can't seem to learn from him.
I understand that this is a seminal book on group theory, so I don't want to just toss it out the window, but I think I need to go somewhere else to actually learn advanced group theory and then just use Rotman as a reference. I'd appreciate any help you can give!
AI: Rotman's book is quite old fashioned. Why not try one of the following:
John S. Rose, A Course on Group TheoryDerek J.S. Robinson, A Course in The Theory of GroupsMarty Isaacs, Finite Group Theory.Each of these books has a lot of good exercises! |
H: Why does Wolfram Alpha state that $-\infty/0 = +\infty$?
I ran into a scenario when practicing L'Hôpital's rule which yielded -infinity/0. I broke this down into $-1 \cdot \infty \cdot \frac 1 0$, which I assumed equaled $-1\cdot\infty\cdot\infty$, which simplified to $-1\cdot\infty$ which equals negative infinity. So where did I go wrong with my logic as Wolfram Alpha claims the answer is positive infinity?
Here is the limit problem:
$$\lim_{x\to0}\frac{\ln(\sin x)}{\ln(\cos x)}$$
AI: Wolfram Alpha does not say that $-\infty/-0 = \infty$ exactly, it says that this is equal to complex infinity. What's happening is that Wolfram Alpha is coming up with an interpretation for your inputs that makes the input sensible. Specifically, you can't divide infinity by zero in the context of real or complex numbers, but you can do this in the context of the Riemann sphere, which is usually treated as the union of the complex numbers with a single point at $\infty$. |
H: Show that $\lim_{x\to \infty}\left( 1-\frac{\lambda}{x} \right)^x = e^{-\lambda}$
Based on the definition of $e: = \lim_{x\to\infty} \left(1+\frac1x \right)^x$, how can we show that
$$\lim_{x\to \infty}\left( 1-\frac{\lambda}{x} \right)^x = e^{-\lambda}?$$
So far I've tried changing variables, $\eta = \frac{-x}{\lambda}$, so $=\lim_{\eta \to -\infty}\left( \left( 1 + \frac1\eta \right)^\eta \right)^{-\lambda}$. But then we would need to show $\lim_{\eta \to -\infty}\left( 1 + \frac1\eta \right)^\eta =e$.
AI: Hint:
$$\left(1+\frac1{\eta}\right)^\eta = \left(\frac1{\left(1+\frac1{-(\eta+1)}\right)^{-(\eta+1)}}\right)^{-\eta/(\eta+1)}$$
and $-(\eta+1) \to +\infty$ as $\eta \to -\infty$. |
H: Need help with proof: If R ⊆ S then R^(-1) ⊆ S^(-1)
Here is what I have so far:
If R ⊆ S then R^(-1) ⊆ S^(-1).
Suppose R ⊆ S and that (a, b) ∈ R^(-1). This means that (b, a) ∈ R. Since S is a relation from A to B, that means that (b, a) ∈ S. That would mean that (a, b) ∈ S^(-1), thus proving this statement true.
AI: The argument is fine, but for one thing (which is perhaps what your tutor was trying to get you to see).
The reason why $(b, a) \in S$, given the assumption $(b, a) \in R$, is that $S$ extends $R$, i.e. $R$ is contained in $S$. That's crucial -- it is evidently not enough just that $S$ is some relation or other from $A$ to $B$! So your "Since" clause is not right as it stands. |
H: Proving a special case of the Squeeze Theorem
Prove that if $\{a_n\}\to0$, $$0\leqslant b_n\leqslant a_n \implies \{b_n\}\to0,$$where $a_n,b_n$ are infinite sequences and $n\in\mathbb{N}$. Then, using the result $$\{a_n\}\to0\iff\{\left|a_n\right|\}\to0,$$prove that if $\{a_n\}\to0$, $$0\leqslant \left|b_n\right|\leqslant a_n \implies \{b_n\}\to0$$
My Attempt
We begin by proving that if $\{a_n\}\to0$, $$0\leqslant b_n\leqslant a_n \implies \{b_n\}\to0.$$
So we have
$$\{a_n\}\to0\iff\forall \epsilon > 0 \exists N\in \mathbb{N}\text{ s.t. }\left|a_n\right|<\epsilon \forall n>N.$$Now $$0\leqslant b_n\leqslant a_n \implies \left|b_n\right|\leqslant\left|a_n\right|$$and so$$\forall \epsilon > 0 \exists N\in \mathbb{N}\text{ s.t. }\left|b_n\right|\leqslant\left|a_n\right|<\epsilon \forall n>N\\\implies\forall \epsilon > 0 \exists N\in \mathbb{N}\text{ s.t. }\left|b_n\right|<\epsilon \forall n>N\\\iff \{b_n\}\to 0.$$
Now using the given result, we obtain that $\{\left|b_n\right|\}\to0$. After this point I am stuck so any hints will be appreciated.
What really confuses me about this question is that you can prove that if $\{a_n\}\to0$, $$0\leqslant \left|b_n\right|\leqslant a_n \implies \{b_n\}\to0$$ in a different order of the steps than the question requires. If you just write down the definition for $\{a_n\}$ being a null sequence and then use the fact that $0\leqslant \left|b_n\right|\leqslant a_n \implies \left|b_n\right|\leqslant\left|a_n\right|$ (since $\left|a_n\right|=a_n\geqslant0$) you can prove that $\{\left|b_n\right|\}\to 0$ and then using the given result it follows that $\{b_n\}\to 0$.
Again, any help will be appreciated.
Thank you in advance.
AI: You have:
1) If $\{a_n\}\to0$, $0\leqslant b_n\leqslant a_n \implies \{b_n\}\to0$
2) $\{a_n\}\to0\iff\{\left|a_n\right|\}\to0$
so from (1), by replacing $b_n$ by $|b_n|$
3) If $\{a_n\}\to0$, $0\leqslant |b_n| \leqslant a_n \implies \{|b_n|\}\to0$
and from (2), by replacing $a_n$ by $b_n$
4) $\{b_n\}\to0\iff\{\left|b_n\right|\}\to0$
and combining (3) and (4)
5) If $\{a_n\}\to0$, $0\leqslant |b_n| \leqslant a_n \implies \{b_n\}\to0$
quod erat demonstrandum. |
H: Problem on inequality
Prove that, $E|X|^p < \infty $ iff $\sum_{k=1}^{\infty}k^{p-1}P\{|X| \geq k\} < \infty$.
Where E is the expectation and P is the usual probability measure. There was a mistake one it's correct.
AI: Let $A_k:=\{\omega,k\leqslant |X(\omega)|\lt k+1\}$. Then switching the order or summation (allowed, as the series converge), we have that
$$\sum_{k=1}^\infty k^{p-1}\mathbb P\{|X|\geqslant k\}<\infty\Leftrightarrow\sum_{k=1}^\infty k^{p-1}\sum_{j\geqslant k}\mu(A_j)=\sum_{j=1}^\infty\mu(A_j)\sum_{k=1}^jk^{p-1}<\infty.$$
Since $\sum_{k=1}^jk^{p-1}\sim j^p$, the problem reduces to show that
$$\mathbb E|X|^p\lt\infty\Leftrightarrow\sum_{j=1}^\infty\mu(A_j)j^p\lt\infty.$$
To see this, notice that
$$j^p\chi_{A_j}\leqslant |X|^p\chi_{A_j}\leqslant (j+1)^p\chi_{A_j},$$
and $\bigcup_jA_j=\Omega$. |
H: Set difference and probability
Trying to work out the following question. If 80% of people like walking, 60% like biking, and everyone likes at least one of these, then what is the fraction of people who like biking, but not walking?
If $P(w)=0.8$, $P(B)=0.6$, and $P(W \cup B)=1$, then what is $P(B \backslash W)$?
AI: Use a Venn diagram!
A is walking and B is biking. Since
$$Pr(w)+Pr(B)-Pr(w \cap B)=1,$$ then $$Pr(w\cap B)=Pr(w)+Pr(B)-1=1.4-1=0.4.$$
The probability you are looking for is, in the diagram, the light blue part of the right circle only (i.e. not the intersection). So then you need to calculate: $$Pr(B)-Pr(w\cap B),$$ which should be clear from the diagram. And this is equal to...? |
H: Multiplication of linear maps $S$ and $T$
If $T \in \mathcal{L}(U,V)$ and $S \in \mathcal{L}(V,W)$ them we define $ST \in \mathcal{L}(U,W)$ by $$(ST)(v)= S(Tv) \text{ for } v \in U.$$ Verify that the composition $S \circ T$ of functions as defined above is linear.
Is the following verification correct?
$\begin{align} (ST)(v+w) &= S(Tv+Tw) \\&=(ST)(v) + (ST)(w) \\ (ST)(av) &= aS(Tv) \\&= a(ST)(v)\end{align}$
AI: To be a little more complete, insert a couple of intermediate equalities.
$$\begin{align} (ST)(v+w)
&= S(T(v + w))\\
&= S(Tv+Tw) \\
&=(ST)(v) + (ST)(w)
\end{align}
$$
\begin{align}
(ST)(av) &= S(T(av))\\&=S(aT(v))\\
&= aS(Tv) \\&= a(ST)(v)
\end{align}
Your logic looks just fine. |
H: Complex double integral
I'm having trouble calculating following (complex) integral.
$$\int_D z^n dA$$
where $D=\{ z \in \mathbb{C} \mid \lvert z \rvert \leq 1 \}$.
I know how to calculate complex (line) integrals and real double integrals but I'm having difficulties combining those.
AI: Hint: Using polar coordinates the integrand is $$z^n=r^n e^{in \theta} $$ and the area element becomes $$dA=rdrd \theta. $$
The limits of integration should be obvious. |
H: Sum of factorials equation
Could you explain why constant $c \gt 0$ can't satisfy equation $c + \frac{c \cdot n!}{\varphi } \le 1$ , where $\varphi = \sum_{i=0}^{n-1}i!$, where $n \to \infty$
AI: Hint:
$$\frac{n!}{0!+\ldots+(n-2)!+(n-1)!} \ge \frac{n!}{(n-2)!+\ldots+(n-2)!+(n-1)!} = \frac{n!}{(n-1)!+(n-1)!} = \frac{n}{2}.$$ |
H: Can't solve following limit: $\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right)$
Need to solve following problem:
$$\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right) $$ I've tried to do something like this: $$\lim_{x \to \infty} x\left(\sqrt[3]{5+8x^3} - 2x\right) =\lim_{x \to \infty}x\left( \sqrt[3]{ \left( \frac {5}{x^3}+8 \right)x^3 } - 2x\right) = \lim_{x \to \infty} x\left( \left ( \sqrt[3] { \frac{5}{x^3}+8} \right)x - 2x\right) $$
It seems to be the right way, but I can't do my next step.
AI: Hint:
$$\sqrt[3]{5+8x^3} - 2x = \sqrt[3]{5+8x^3} - \sqrt[3]{8x^3} = \frac{\left(5+8x^3\right)-8x^3}{\left(\sqrt[3]{5+8x^3}\right)^2 + \sqrt[3]{5+8x^3} \cdot \sqrt[3]{8x^3} + \left(\sqrt[3]{8x^3}\right)^2}.$$ |
H: exercise on the closed subspaces of an Hilbert spaces
I have a question regarding exercise 3.1.13 of "Analysis Now" by Pedersen volume 118 of the Springer GTM.
The exercise aim to show that any closed subspace $X$ of
$L^2([0,1])\cap L^{\infty}([0,1])]$ is finite dimensional.
I'd like to get whatever proof of this fact, however in the book there is a hint which is the following:
There is a constant $\alpha$ such that
$|| f||_\infty\leq \alpha\cdot ||f||_2$ for all $f\in X$.
If $\{f_1,\dots,f_n\}\subset X$ is a family orthonormal vectors,
there is a null set $N$ such that
$
|f(t)|\leq\alpha\cdot ||f||_2
$
for all $t\in [0,1]\setminus N$ and $f$ in the linear span of
$\{f_1,\dots,f_n\}$.
For all $t\in [0,1]\setminus N$
$
\Sigma_{k=1\dots n} |f_k(t)|^2\leq\alpha^2.
$
Conclude that there cannot be a family of orthonormal vectors in $X$ of size larger than than $\alpha^2$.
All items except item 3 are clear to me,
however I cannot really understand what is the reason why from item 2 one can get item 3.
AI: Hint: fix $t_0\in [0,1]\setminus N$ and consider the function $f(t):=\sum_{i=1}^nf_i(t_0)f_i(t)$. |
H: How to compute infinite series $\sum_{n=0}^{\infty} ne^{-n}$
I'm trying to compute the infinite series $\sum_{n=0}^{\infty} ne^{-n}$.
I know the answer is $e/(e-1)^2$, but I don't understand how to find this result.
Thanks for the help!
AI: You can differentiate the series $\sum_{n=0}^\infty e^{-nx} = \frac{1}{1-e^{-x}}$ to obtain $\frac{d}{dx} \sum_{n=0}^\infty e^{-nx} = -\sum_{n=0}^\infty n e^{-nx}$. Then just put $x = 1$. |
H: Limit Comparison Test Problem confusion
I have a problem where I'm supposed to find if the series converges. I'm looking at the solutions manual and I'm really confused. I know it uses the limit comparison test, but one moment it goes from $n+n^2+n^3$ to $1/n^2+1/n+1$ and I feel like I don't understand how they did that arithmetic.
AI: $$\large\frac{n+n^2+n^3}{\sqrt{1+n^2+n^6}}\cdot\frac{1/n^3}{1/n^3}=\frac{\frac1{n^2}+\frac1n+1}{\sqrt{\frac{1+n^2+n^6}{n^6}}}=\frac{\frac1{n^2}+\frac1n+1}{\sqrt{\frac1{n^6}+\frac1{n^4}+1}}$$
Remember that $a^3\sqrt{b}=\sqrt{a^6b}$. The point is to reduce both numerator and denominator to expressions with finite limits: those fractions with powers of $n$ in the denominator all tend to $0$ as $n\to\infty$, so the limit can be evaluated by eye. |
H: Chance of a double in three dice
What is the chance of rolling doubles in three six sided dice?
The answer I have is:
$$
\frac{1}{6}•\frac{1}{6}•\frac{3}{2} = \frac{1}{24}
$$
AI: The number of elements in the sample space = 216 or (6*6*6)
Of the three numbers two of them to be same is (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) = 6 doublets.
Every one of this can happen in 3 different ways, that is (1,1,x) (1,x,1) (x,1,1) = 3 ways.
Now x can be any of the remaining 5 numbers.
So total number of ways two of them can be same = 6 x 3 x 5 = 90
So probability of the event = 90/216 = 5/12 |
H: What is the cardinality of an element of an free ultrafilter?
Let $U$ be a free ultrafilter on a set $X$.
I want to prove that the cardinality of every element $u\in U$ is equipotent to $X$. Is that true? Or does it lack some hypothesis?
AI: No, this is not provable because it can be false.
Consider an ultrafilter on $\Bbb N$, say $\cal F$. You can show that $\{A\subseteq\Bbb R\mid A\cap\Bbb N\in\cal F\}$ is a free ultrafilter on the real numbers.
The property that you are looking for is called uniform ultrafilter. |
H: Prove the column span of $A$, an $m\times n$ matrix, is $F^m$ iff $\exists$ an $n\times m$ matrix $B$ such that $AB = I_m$
This problem has a second part, about the columns of $A$ being linearly independent in $F^m$ iff $\exists n \times m$ matrix $B$ such that $BA = I_n$ but I think if I understand the first part, I'll get the second.
I've tried making the matrix into a linear transformation $T \in \mathcal{L}(F^n,F^m)$ such that $A = M_T(B_{F^n},B_{F^m})$, and then $B$ would give us the right inverse of $A$, i.e. $\exists S \in \mathcal{L}(F^m,F^n)$ such that $T\circ S = I_{F^m}$, the identity transformation in $F^m$, but I don't know how to use the column span property or how to finish the proof.
Any advice would be greatly appreciated! Thanks!
AI: If the column span of $A$ is $F^m$, then every vector $F^m$ can be expressed as a linear combination of the columns of $A$, which means that the linear system
$$
Ax=c
$$
is solvable for every $c\in F^m$. In particular, the systems $Ax=e_i$ $(i=1,2,\dots,m)$ are solvable, where $\{e_1,e_2,\dots,e_m\}$ is the standard basis of $F^m$. Then, let $Ab_i=e_i$ $(i=1,2,\dots,m)$ and consider the matrix
$$
B=\begin{bmatrix} b_1 & b_2 & \dots & b_m\end{bmatrix}
$$
having those solutions as columns. By the definition of matrix product, you have $AB=I_m$.
Conversely, if $AB=I_m$ and $c\in F^m$, we can write
$$
A(Bc)=I_mc=c
$$
so $Bc$ is a solution of $Ax=c$, which means that every vector in $F^m$ is a linear combination of the columns of $A$. |
H: Discrete Distribution Help
If p(n) = c(5/8)^n, 3 <= n <= infinity is a p.m.f. for a discrete random variable X, find (a) c, (b) the probability P(6 <= X <= 16), (c) the mean and (d) the variance
Here's my work. I think got c wrong and haven't gone past it.
(a) Since the sum of p(n) = 1, I got c(5/8)^1 + c(5/8)^2 + the sum from 3 to infinity of c(5/8)^n = 1. (I figured that because p(n) started at 3 I need to add the first 2 terms). I got (5/8)c + (25/64)c + (1000/1536)c = 1. Which reduces to 1560/2560 = 39/64 = c.
(b) Would I just added up c(5/8)^6 + c(5/8)^7 + ... c(5/8)^16?
(c) Would I just find the expected value E(X) = (x)p(n) from 3 to infinity (using the geometric series?)
(d) Would I just find E(x^2) + ((E(X))^2)?
Sorry for the formatting, I haven't learned LaTeX
AI: a. $$\underset{n=3}{\overset{\infty }{\mathop \sum }}\,c{{\left( \frac{5}{8} \right)}^{n}}=1\Rightarrow c\cdot {{\left( \frac{5}{8} \right)}^{3}}\left( 1+\frac{5}{8}+\ldots \right)=1\Rightarrow c\cdot {{\left( \frac{5}{8} \right)}^{3}}\frac{1}{1-\frac{5}{8}}=1\Rightarrow c=\frac{192}{125}$$
b. $$\underset{n=6}{\overset{16}{\mathop \sum }}\,\frac{192}{125}{{\left( \frac{5}{8} \right)}^{n}}$$
c. $$E\left( N \right)=\underset{n=3}{\overset{\infty }{\mathop \sum }}\,\frac{192}{125}\cdot n\cdot {{\left( \frac{5}{8} \right)}^{n}}$$
d. $$E\left( {{N}^{2}} \right)=\underset{n=3}{\overset{\infty }{\mathop \sum }}\,\frac{192}{125}\cdot {{n}^{2}}\cdot {{\left( \frac{5}{8} \right)}^{n}}$$ and $$Var\left( N \right)=E\left( {{N}^{2}} \right)-{{\mu }^{2}}$$
for sums you can use
$$1+x+{{x}^{2}}+\ldots =\frac{1}{1-x}$$
$$1+2x+3{{x}^{2}}+\ldots =\frac{1}{{{\left( 1-x \right)}^{2}}}$$
(derivatives). |
H: Matrix calculation from an equation
I am trying to find matrix y from an equation
For example, I know (python)
>>> x = matrix([[5,1],[2,4]])
>>> y = matrix([[1,5],[3,3]])
>>> print x
[[5 1]
[2 4]]
>>> print y
[[1 5]
[3 3]]
>>> print x*y % 7
[[1 0]
[0 1]]
Assuming, I don't know what y is. But I know xy = I mod7
How do I calculate y?
AI: Let
$$
y=\begin{pmatrix} a & b\\ c& d\end{pmatrix}.
$$
Then $XY$ sets up some equations:
\begin{align}
5a+c &\equiv 1\,\, (\operatorname{mod} 7)\\
5b+d &\equiv 0\,\, (\operatorname{mod} 7)\\
2a+4c &\equiv 0\,\, (\operatorname{mod} 7)\\
2b+4d &\equiv 1\,\, (\operatorname{mod} 7).
\end{align}
You can solve this like you would solve any system of equations, only you have to remember to take every thing $\operatorname{mod} 7$. For example, take the second equation and multiply it by $4$ to get:
$$
20b+4d \equiv 0\,\, (\operatorname{mod} 7).
$$
But, $20\equiv 6\,\,(\operatorname{mod} 7)$. So, this becomes
$$
6b+4d \equiv 0\,\, (\operatorname{mod} 7).
$$
Then subtract fourth equation from this to get:
$$
4b \equiv -1\,\, (\operatorname{mod} 7).
$$
But, $-1\equiv 6\,\,(\operatorname{mod} 7)$. From above $b\equiv 5\,\,(\operatorname{mod} 7)$. Now substitute $b$ into those equations and just keep going.
I'm not sure what you are comfortable with, but the integers mod $7$ are a field. So, all this works. |
H: Reflection across a line?
The linear transformation matrix for a reflection across the line $y = mx$ is:
$$\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix} $$
My professor gave us the formula above with no explanation why it works. I am completely new to linear algebra so I have absolutely no idea how to go about deriving the formula. Could someone explain to me how the formula is derived? Thanks
AI: You can have (far) more elegant derivations of the matrix when you have some theory available. The low-tech way using barely more than matrix multiplication would be:
The line $y = mx$ is parametrised by $t \cdot \begin{pmatrix}1\\m\end{pmatrix}$. The line orthogonal to it is parametrised by $r \cdot \begin{pmatrix}-m\\1\end{pmatrix}$. The line $y = mx$ shall be fixed, the line orthogonal to it shall be reflected, so you want a matrix $R$ with
$$R \begin{pmatrix}1 & -m\\ m & 1\end{pmatrix} = \begin{pmatrix}1 & m\\ m & -1\end{pmatrix},$$
and that means
$$\begin{align}
R &= \begin{pmatrix}1 & m\\m&-1\end{pmatrix} \begin{pmatrix}1&-m\\m&1\end{pmatrix}^{-1}\\
& = \begin{pmatrix}1&m\\m&-1\end{pmatrix}\cdot \frac{1}{1+m^2}\begin{pmatrix}1&m\\-m&1\end{pmatrix}\\
&= \frac{1}{1+m^2} \begin{pmatrix}1 - m^2 & 2m\\2m &m^2-1\end{pmatrix}.
\end{align}$$ |
H: Please help me interpreting Elementary Set Theory question
I have to prove something for homework, but I can not for the life of me figure out what the theorem I'm supposed to prove is. Again, please: I am not asking for you to prove it for me (that's my hw and I want to do it myself), I am just asking you to help me understand what it is that I am trying to prove here.
What is this "theorem like 1.1.5 for the intersection and union of A" which I am supposed to prove?
AI: The book is hinting for you to show that
$$
\left(\bigcup_{Y\in\mathcal{A}}Y\right)^{c}=\left(\bigcap_{Y\in\mathcal{A}}Y^{c}\right)
$$
and
$$
\left(\bigcap_{Y\in\mathcal{A}}Y\right)^{c}=\left(\bigcup_{Y\in\mathcal{A}}Y^{c}\right).
$$
Note that when $\mathcal{A}=\left\{ A,B\right\} $ you get back your
original problem.
Edit:
We define $\bigcup_{Y\in\mathcal{A}}Y$ as the set of all $x\in X$ s.t. $x\in Y$ for some $Y\in\mathcal{A}$.
We define $\bigcap_{Y\in\mathcal{A}}Y$ as the set of all $x\in X$ s.t. $x\in Y$ for all $Y\in\mathcal{A}$. |
H: Checking understanding of concept
I want to check if I have understood a concept correctly.
Problem: Describe geometrically the action of an orthogonal $3$ x $3$ matrix with determinant -1.
My solution: The orthogonal $3$ x $3$ matrix with determinant $-1$ is an improper rotation, meaning it is a reflection combined with a proper rotation. In another sense, an improper rotation is an indirect isometry, which is an affine transformation with an orthogonal matrix with a determinant $-1$.
How do I prove the statement for an arbitrary $3$ x $3$ orthogonal matrix with determinant $-1$?
Any input is appreciated. Thanks.
AI: If you have that any orthogonal matrix is similar to
$$A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & * & * \\ 0 & * & *\end{bmatrix}$$
where the stars represent $2\times 2$ orthogonal matrix, then notice that $~\begin{bmatrix} 1 \\ 0 \\ 0\end{bmatrix}~$ is an eigenvector. This means that there is some change of basis such that the span of the new first basis element is an invariant subspace.
In fact, it is more than invariant, it is fixed pointwise. Why?
What does this mean geometrically?
How does the transformation act on a vector in the orthogonal complement?
Finally, use this to show how the transformation acts on an arbitrary vector. (You can do this only using the subspaces, with no need of either basis.)
Sanity check: Where did the $-1$ come in to play? |
H: Homeomorphism preserving partitions
Let $X$ and $Y$ are homeomorphic topological spaces.
Consider a equivalent relation $R_X$ and $R_Y$ that partition $X$ into $X_1,\ldots,X_n$ of and $Y$ into $Y_1,\ldots,Y_n$ respectively.
$X_i$ is homeomorphic to $Y_i$ for all $1\leq i\leq n$.
The quotient topologies $X/R_X$ and $Y/R_Y$ are homeomorphic.
Does there exist a homeomorphism $H$ such that $H(X_i) = Y_i$ for all $1\leq i\leq n$?
If not, what it we have the information that the homeomorphism of $X/R_X$ and $Y/R_Y$ maps $X_i/R_X$ to $Y_i/R_Y$?
If not, is it true on some special topological spaces? (I'm interested when $X$ and $Y$ are spheres with boundaries, the partitions are simply connected regions.)
AI: It’s certainly not true in general.
Let $X=Y=[0,2)$, $X_1=Y_2=[0,1)$, and $X_2=Y_1=[1,2)$. There is no homeomorphism of $[0,2)$ onto itself that interchanges $[0,1)$ and $[1,2)$.
It will be true if the partitions are clopen partitions.
Added: A stronger example: $X=Y=[0,2)\cup[3,4]\cup[5,6]$, $X_1=Y_2=[0,1)\cup[3,4]$, and $X_2=Y_1=[1,2)\cup[5,6]$. Now the map $H:X\to Y$ that is the identity on $[0,2)$ and interchanges $[3,4]$ and $[5,6]$ in the obvious way is a homeomorphism of $X$ onto $Y$ such that there are $x_1\in X_1$ and $x_2\in X_2$ such that $H(x_i)\in Y_i$ for $i=1,2$, but there is no homeomorphism of $X$ onto $Y$ that interchanges $[0,1)\cup[3,4]$ and $[1,2)\cup[5.6]$. |
H: determining abelian groups of a certain size up to isomorphism
Say the size is 360. My book uses this as an example. It says there are 6 distinct groups up to isomorphism:
$\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_5$
$\mathbb{Z}_2 \times \mathbb{Z}_4 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_5$
$\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_9 \times \mathbb{Z}_5$
$\mathbb{Z}_2 \times \mathbb{Z}_4 \times \mathbb{Z}_9 \times \mathbb{Z}_5$
$\mathbb{Z}_8 \times \mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_5$
$\mathbb{Z}_8 \times \mathbb{Z}_9 \times \mathbb{Z}_5$
The only result I have to use right now is that "$\mathbb{Z}_m \times \mathbb{Z}_n$ is isomorphic to $\mathbb{Z}_{mn}$ iff gcd($m$, $n$) = $1$". From this result, I know (6) is isomorphic to $\mathbb{Z}_{360}$ and that (1-5) are not. But how can I show that each group in (1-5) is not isomorphic to a different group in (1-5)?
AI: Let me do an easier example. What if the size were 60? Then we might have $\mathbb{Z}_3\times\mathbb{Z}_4\times\mathbb{Z}_5$ or $\mathbb{Z}_2\times\mathbb{Z}_3\times\mathbb{Z}_{10}$.
These cannot be isomorphic because $\mathbb{Z}_2\times\mathbb{Z}_{10}$ and $\mathbb{Z}_4\times\mathbb{Z}_5$ are not isomorphic (your theorem has iff, not just if), and taking the product of two non-isomorphic groups by the same group will not make them isomorphic.
The same idea works for size 360, but with a bit more legwork. |
H: Does $1/n$ satisfy the Cauchy criterion?
I have been told that if a series satisfies the Cauchy criterion, it converges. I was not quite convinced on this, however, since some of the posts on this website seemed to imply otherwise. I came up with what I thought was a counterexample: I argued that $1/n$ satisfies the Cauchy criterion but diverges. However, I've been told afterwards that $1/n$ doesn't satisfy the Cauchy criterion. Now, I've tried to show that the $1/n$ does not satisfy the Cauchy criterion but couldn't come up with a proof. Can someone provide me a proof that shows that the series $1/n$ does not satisfy the Cauchy criterion?
AI: The sequence $\{1/n : n=1,2,3,\ldots\}$ satisfies the Cauchy criterion and converges.
The series $\displaystyle\sum_{n=1}^\infty \frac1n$ diverges. That is the same as saying that its sequence of partial sums
$$
\left\{ \sum_{n=1}^m \frac1n : m = 1,2,3,\ldots \right\}
$$
diverges. The latter sequence is not a Cauchy sequence. To see that it's not a Cauchy sequence without first considering whether it converges, consider the difference between the $m_1$th term and the $m_2$th term, whree $m_2>m_1$:
\begin{align}
& \phantom{={}} \left| \sum_{n=1}^{m_1} \frac1n - \sum_{n=1}^{m_2} \frac1n \right| \\[6pt]
& =\left| \sum_{n=m_1+1}^{m_2} \frac1n \right| \ge \left| \int_{m_1+1}^{m_2} \frac1x\,dx \right| \\[6pt]
& = \left|\log\frac{m_2}{m_1+1}\right|\to\infty\text{ as }m_2\to\infty.
\end{align}
Don't confuse sequences with series. |
H: Need your help with the integral $\int_0^\infty\frac{dx}{e^{\,e^{-x}} \cdot e^{\,e^{x}}}$.
Is it possible to evaluate this integral in a closed form?
$$\int_0^\infty\frac{dx}{e^{\,e^{-x}} \cdot e^{\,e^{x}}}$$
AI: This is clearly equivalent to
$$\int_0^{\infty}e^{-2\cosh x}dx=K_0(2).$$
Here $K_0(x)$ denotes Macdonald function, which has integral representation $$K_{\nu}(r)=\int_0^{\infty}e^{-r\cosh x}\cosh\nu x\,dx.$$ |
H: When is the Cayley transform of a matrix $J$-orthogonal?
The (real) general linear group is defined $GL(n)=\{A \in \mathbb{R}^{n\times n} \mid \operatorname{det}(A) \neq 0\}$. It is a matrix Lie group. Let $J$ be a constant $n$-by-$n$ real matrix. The so-called $J$-orthogonal group is defined $O_J(n)=\{A \in GL(n) \mid A^T J A = J\}$. I'm pretty sure $O_J(n)$ is a matrix Lie group regardless of what $J$ is. The corresponding Lie algebra is $o_J(n) = \{B \in \mathbb{R}^n \mid B^T J + JB = 0\}.$ For $B \in o_J(n)$, the book I am reading defines the Cayley transform of $B$ as
$$A = (I-\alpha B)^{-1}(I+\alpha B). $$
I'm assuming $\alpha$ is some real constant. The book claims that $A$ is $J$-orthogonal, that is, $A^T J A = J$. In the book, it is unclear whether they claim this for $n=4$ and
$J = \operatorname{diag}(1, -1, -1, -1)$, or for all real square matrices $J$. I think I have verified that $A$ is $J$-orthogonal if $J$ is symmetric and $J^2 = I$ (which includes $J = \operatorname{diag}(1, -1, -1, -1)$. I guess I have two questions: first, is this correct? And second, are weaker assumptions on $J$ possible?
EDIT: The two factors defining $A$ above commute. If you use this fact and mimic the proof in the answer below, you can show that no assumptions on $J$ are necessary: $J$ can be any $n$-by-$n$ real matrix.
AI: To get some idea how this works, let us first assume only that $J$ is involutory, i.e., $J^2 = I$. Then $A$ is $J$-orthogonal if and only if $AJA^T = J$. We check that condition:
$$(I - \alpha B)^{-1} (I + \alpha B) J (I + \alpha B)^T (I - \alpha B)^{-T} = J$$
and see that it is equivalent to
$$(I + \alpha B) J (I + \alpha B)^T = (I - \alpha B) J (I - \alpha B)^T.$$
This is equivalent to
$$J + \alpha (BJ + JB^T) + \alpha^2 B J B^T = J - \alpha (BJ + JB^T) + \alpha^2 B J B^T,$$
which is equivalent to
$$\alpha (BJ + JB^T) = 0.$$
Assuming that $\alpha \ne 0$, this is equivalent to
\begin{equation}
BJ + JB^T = 0.\tag{*}
\end{equation}
If we premultiply and postmultiply $(*)$ with $J$, and keeping in mind that $J^2 = I$, the above is equivalent to
$$0 = J(BJ + JB^T)J = JB + B^TJ,$$
which is equivalent to saying that $B \in o_J(n)$.
Of course, for $\alpha = 0$, any $B$ gives as $J$-orthogonal matrix $A = I$, with no conditions on $J$.
Now, for the general case, let us also lose the involutority of $J$. Then we have to observe the condition $A^T J A = J$. This is equivalent to
$$(I + \alpha B)^T (I - \alpha B)^{-T} J (I - \alpha B)^{-1} (I + \alpha B) = J,$$
which is equivalent to
$$(I - \alpha B)^{-T} J (I - \alpha B)^{-1} = (I + \alpha B)^{-T} J (I + \alpha B)^{-1}.$$
Since $J$ is nonsingular, we see that this is equivalent to
$$\left( (I - \alpha B) J^{-1} (I - \alpha B)^T \right)^{-1} = \left( (I + \alpha B) J^{-1} (I + \alpha B)^T \right)^{-1}.$$
In other words,
$$(I - \alpha B) J^{-1} (I - \alpha B)^T = (I + \alpha B) J^{-1} (I + \alpha B)^T.$$
We have this above, with the only difference that we now have $J^{-1}$, instead of $J$. So, using the same steps, we obtain the following equivalent of $(*)$:
$$BJ^{-1} + J^{-1}B^T = 0.$$
Again, premultiplying and postmultiplying with $J$, we see that this is equivalent to
$$JB + B^T J = 0,$$
which means that the above is equivalent to $B \in o_{J^{\color{red}{-1}}}(n)$.
In other words, the first part of this proof is just an obvious special case ($J = J^{-1}$). This makes sense, since we've used involutority twice, which cancels itself out. |
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