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H: Use the definition of derivative to find $f'(1)$ for $f(x) = \frac{x}{\sqrt{x^2+1}}$
This is analysis.. So I am using the definition that
$$
f'(x)=\lim_{x\to c} \frac{f(x)-f(c)}{x-c}
$$
So far I have,
$$
\lim_{x\to 1} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{1^2+1}}}{x-1}=\lim_{x\to 1} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{2}}}{x-1}
$$
Now how do I make the correlation from there to the real derivative of $\displaystyle\frac{1}{\sqrt{(1+x^2)^3}}$
AI: The next few steps involve getting a common denominator on top, reducing it to a single fraction, and then multiplying top and bottom by the conjugate of the numerator. |
H: Are homomorphisms a finite group?
I have a question from my homework that says:
Find all the homomorphisms φ:ℂ->ℂ such that for any $x$ in ℝ: φ($x$) = $x$
I don't even know how to begin. The field of complex numbers is infinite. The number of homomorphisms must be huge, if it is even finite. How am I supposed to find them all??
AI: You would think there are infinitely many such homomorphisms but you would be wrong. In fact there are only two! This is essentially the beginning of Galois theory in that what you are being asked to find is the Galois group Gal$(\mathbb{C}/\mathbb{R})$.
Ok so let $\tau$ be such a homomorphism.
Then it's action on $\mathbb{C}$ is defined as follows:
$\tau(a+ib) = \tau(a) + \tau(b)\tau(i) = a + \tau(i)b$
since $a,b\in\mathbb{R}$ and $\tau$ is known to fix those.
So now it remains to find $\tau(i)$. What do we know about $i$? We know that $i^2=-1$ of course.
Aha but $-1$ is real, and $\tau$ fixes that!
So lets consider $\tau(i)^2$.
$\tau(i)^2 = \tau(i^2)=\tau(-1) = -1$.
Thus $\tau(i)=\pm i$.
So we get two possible homomorphisms, one being the identity ($\tau(i)=i$) and one being complex conjugation ($\tau(i)=-i$).
It is easy to check that both of these work. |
H: Use product rule and mathematical induction to show that $f^n$ is differentiable on $I$
Suppose that $f$ is differentiable on $I$. Use the product rule and mathematical induction to show that $f^n$ (the function f is raised to the nth power) is differentiable on $I$ for every positive integer $n$ and find a formula for $(f^n)’$.
AI: I think other replies to this thread are confused about what the question is asking. The question is not asking to prove that $f^{(n)}$ is differentiable, it is asking to prove that $f^n$ (the $n^{th}$ power of $f$) is differentiable.
We'll use strong induction to show that $(f^n)' = nf^{n-1}f'$.
Consider the base case, $n=1$. $f^1 = f$ is differentiable by assumption of the problem, and the formula for its derivative is given by $f' = 1f^{1-1} f'$ because $f^{1-1} = 1$. So clearly our property holds.
Now we'll assume that $f^i$ is differentiable for any $i < n$ and its derivative moreover is given by $(f^i)' = if^{i-1}f'$ and we'll prove the same holds for $f^{n}$. We note that $f^{n} = f^{n-1} f$. From here, we can invoke the $i=2$ case (which is simply a statement of the product rule) to get that $f^n$ is differentiable. We can also use the $i=2, i=1$, and $i = n-1$ cases to get that $$(f^{n})' = [f^{n-1} f]' = (f^{n-1})'f + f' f^{n-1} = (n-1)f^{n-2}f'f + f'f^{n-1} = (n-1)f^{n-1}f' + f^{n-1}f' = nf^{n-1}f'$$ which is the formula we posited. So in particular, for every natural number $n$, we've proved by strong induction that $$(f^n)' = nf^{n-1}f'.$$ |
H: $\epsilon, \epsilon/2 $ Proof
How is it proved that $$ \left[ |X_n - X|+|Y_n-Y| \geq \epsilon \right] \subset \{ \left[ |X_n -X|\geq \epsilon/2 \right] + \left[ |Y_n-Y| \geq \epsilon/2 \right] \} $$
Thank you.
AI: Intuitively, if you add two positive numbers and get at least $1$, at least one of them must be greater than $\frac 12$. To be more formal, it is probably easier to prove the contrapositive: assume $|X_n-X| \lt \epsilon/2$ and $|Y_n-Y| \lt \epsilon/2$. Then you can bound the left side of the sum by $\epsilon$. |
H: Can you simplify a expression with an exponent that is divided by a number?
As the title suggests, I have
$\;a^{(b/c)}.$
Is there any way to simplify this so that there is no dividing in the exponent?
AI: You can write it as $\sqrt[c]{a^b}=\sqrt[c]{a}^b=(a^{\frac{1}{c}})^b=a^{b/c}$. There isn't really any other way to simplify it, unless there's some context to your question. |
H: Does $H_0^1(\Omega)$ embed into $H_0^1(R^d)$?
Given a domain $\Omega$ in $\mathbb{R}^d$ and a function $f\in H_0^1(\Omega)$, the closure of the test functions on $\Omega$, does the extension of f by 0 to all of $\mathbb{R}^d$ necessarily lie in $H_0^1(\mathbb{R}^d)$?
AI: Yes. The big picture is that for we can glue two $H^1$-functions together using continuity.
Say let the extension be:
$$
\widetilde{f}(x) = \begin{cases}
f(x) & \text{ for } x\in \Omega,
\\[2mm]
0 &\text{ for } x\in \mathbb{R}^d\backslash\Omega.
\end{cases}
$$
Apparently $\widetilde{f}\in L^2(\mathbb{R}^d)$. Now we want to find $\nabla \widetilde{f} = \big(\frac{\partial \widetilde{f}}{\partial x_i}\big)$, we don't know if this thing agrees with the weak derivative of $f$ in $\Omega$, and is 0 outside $\Omega$. We have to use the definition of the weak derivative:
$$
\int_{\mathbb{R}^d} \frac{\partial \widetilde{f}}{\partial x_i} \phi = -\int_{\mathbb{R}^d} \widetilde{f}\frac{\partial\phi }{\partial x_i} \quad \text{for any }\phi\in C^{\infty}_c(\mathbb{R}^d),
$$
now because of the integrability we can split the right hand side into two parts:
$$
-\int_{\mathbb{R}^d} \widetilde{f}\frac{\partial\phi }{\partial x_i} = -\int_{\Omega} \widetilde{f}\frac{\partial\phi }{\partial x_i} -\int_{\mathbb{R}^d\backslash\Omega} \widetilde{f}\frac{\partial\phi }{\partial x_i}
= -\int_{\Omega} f\frac{\partial\phi }{\partial x_i}.
$$
Then use boundary condition of $f$:
$$
-\int_{\Omega} f\frac{\partial\phi }{\partial x_i} = \int_{\Omega} \frac{\partial f}{\partial x_i}\phi - \int_{\partial\Omega} f\phi \nu_{x_i}\,dS = \int_{\Omega} \frac{\partial f}{\partial x_i}\phi.
$$
Thus we have:
$$
\int_{\mathbb{R}^d} \frac{\partial \widetilde{f}}{\partial x_i} \phi = \int_{\Omega} \frac{\partial f}{\partial x_i}\phi \quad \text{for any }\phi\in C^{\infty}_c(\mathbb{R}^d),
$$
and this tells us that
$$
\nabla \widetilde{f}(x) = \begin{cases}
\nabla f(x) & \text{ for } x\in \Omega,
\\[2mm]
0 &\text{ for } x\in \mathbb{R}^d\backslash\Omega.
\end{cases}
$$
Now since $\nabla f\in L^2(\Omega)$, $\nabla \widetilde{f}\in L^2(\mathbb{R}^d)$. The extension is valid. |
H: Use the definition of derivative to find $f'(x)$ for $f(x) = x^{1/2}$.
This is an analysis question so we use the definition that the derivative of a function is, $$\lim_{x\to c} \frac{f(x)-f(c)}{x-c}$$ But I'm not really sure if I should just continue to say that $x=c$ for now as the second $x$ perhaps as if $f(x) = \sqrt c$. Obviously if we leave $x$ as is, $(f(x)-f(x)) = 0$ and the lim is $0$, but this is obviously not the case since $f'(x) = \frac{1}{2}\sqrt x$
AI: Hint: $$\frac{ f(x)-f(c) }{ x-c }= \frac{\sqrt x - \sqrt c}{x-c}$$
and remark that: $$x-c=(\sqrt x - \sqrt c)(\sqrt x + \sqrt c)$$ |
H: Writing a relatively simple MATLAB function
I'm new to MATLAB and I have been asked to write a MATLAB function whose input arguments are two integers $a$ and $b$; the output is the remainder of the integer division $a/b$ if $a>=b$ or of the integer division $b/a$ if $b>a$
Can someone help me out? Thanks.
EDIT:
This is all I can string together at the moment but I know that it is incorrect. I'm just trying to put together what I know.
function r=remainder(n,m);
if n>=m;
r=rem(m,n);
elseif m>n;
r=rem(n,m);
end
AI: In MatLab, in general, functions can be write as the followings:
function [y1,...,yN] = myfun(x1,...,xM)
function [y1,...,yN] = myfun(x1,...,xM) declares a function named myfun that accepts inputs x1,...,xM and returns outputs y1,...,yN. This declaration statement must be the first executable line of the function.
in you case you have one output and two inputs;
Function:
function y = compare(a,b)
if a>=b
y = a/b;
else
y = b/a;
end
end
Main Program:
close all; clear all; clc;
a=2;
b=3;
y=compare(a,b)
y =
1.5000
Note: Remember to save both function and your MatLab code in the same folder (same location). and the name of the function is also important to be related to what you want to do. |
H: Uniform convergence of a family of functions on $(0,1)$
Let the family of functions be
$$f_n(x) = \dfrac{x}{1+nx}.$$
Is the sequence $f_n$ uniformly convergent in the interval $(0,1)$?
AI: $\frac{x}{1 + nx} = \frac{1}{\frac{1}{x} + n} \leq \frac{1}{n}$ which doesn't depend on $x$ hence your sequence converges uniformly to $0$ |
H: Determine whether $\int _{1}^{\infty}\frac{x\sin\left(x\right)}{\sqrt{1+x^5}}\,{\rm d}x$ is convergent or divergent
My task is to determine if the following function is convergent or divergent.
$$
\int_{1}^{\infty}{x\sin\left(x\right) \over\sqrt{1 + x^{5}\,}\,}\,{\rm d}x$$
I have been trying to find a more easily integrable function which could be determined to belong in any of the two categories, but I have yet to succeed. Any help is appreciated!
AI: Hint:$$ \left\vert\frac{x\sin{x}}{\sqrt{1+x^5}}\right\vert \leq\frac{x}{\sqrt{1+x^5}} $$
What does this bound tell us about the convergence? |
H: Question involving Sylow theorems and characteristic subgroups
Let $G$ be a group of order $2\cdot 17\cdot 23$. Let $P$ be a $17$-Sylow subgroup and $Q$ a $23$-Sylow subgroup. Show that $PQ$ is normal in $G$ and that $P$ is characteristic in $PQ$.
What I know: both $P$ and $Q$ are normal in $G$ by the Sylow theorems, $PQ$ is normal in $G$ because it has index 2.
So all that remains is to show that $P$ is characteristic in $PQ$. Now, certainly any inner automorphism of $PQ$ will fix $P$ since $17$-Sylow subgroups are all conjugate and $P$ is normal in $G$ (and hence unique). But how do I know there isn't some outer automorphism that pushes $P$ around?
More generally, it seems like this would be true for groups of order $2pq$ for primes $2<p<q$, since I doubt $17$ and $23$ are particularly significant. Thoughts?
AI: Hint. Use Sylow #3 to show that there's only one Sylow $17$-subgroup of $PQ$. Then show that a unique Sylow subgroup is characteristic. |
H: Integral with bounded function
If $m$ is a finite measure on $X$, for which bounded measurable functions $g$ do we have the following: if $f$ is integrable on $X$ and $\int_{X}fdm = 0$ then $\int_{X} fg dm = 0$?
Maybe I'm not understanding something here, but if $g$ is bounded, let $|g| \leq M$ don't we have $$0 =(-M)\int_X fdm = \int_X f (-M)dm \leq \int_X fg dm \leq \int_{X} f M dm = M\int_Xfdm= 0$$
and therefore we don't need anything else for $g$? And the finite measure hypothesis also doesn't seem to be important. If someone sees why this is wrong please say...
AI: Let $X=[0,2\pi]$ with the Lebesgue measure and let $f(x)=\sin(x)$. It is easy to see that $g=f$ does not satisfies your claim. |
H: Modern Definition of the Real Numbers
I have been told:
"The real numbers are defined to be the set of equivalence classes of pairs of rational sequences $(a_i,b_i)$, where (1) $\{a_i\}$ is increasing, (2) $\{b_i\}$ is decreasing, (3) for each $i=1,2,..., \hspace{2mm} b_i-a_i>0$, and (4) $\lim_{x \to \infty} (b_i - a_i)=0$."
I have then been asked to (1) prove the distributive law of multiplication for numbers, and (2) describe the representation of $\pi$ in this sense.
However, I am struggling with the definition as given to me, and thus I have no clue where to start for either of the other parts.
Any insight into any, or all, of these parts would be very much appreciated.
AI: Let $x\in\mathbb{R}$. $\exists (a_i,b_i)\in\mathbb{Q}^2$ that define $x$. Let $y\in\mathbb{R}$ be similarly defined by $(c_i,d_i)$, and $z\in\mathbb{R}$ by $(e_i,f_i)$. We can say $x=\lim_{i\to\infty} \frac{1}{2}(a_i+b_i)$. $$x(y+z)=\left(\lim_{i\to\infty}\frac{1}{2}(a_i+b_i)\right)\left(\lim_{i\to\infty}\frac{1}{2}(c_i+d_i)+\frac{1}{2}(e_i+f_i)\right)$$
Since all the numbers in the expression are rational, we can distribute:
$$ x(y+z)=\lim_{i\to\infty}\frac{1}{4}(a_i+b_i)(c_i+d_i)+\frac{1}{4}(a_i+b_i)(e_i+f_i)=x\cdot y+x\cdot z $$
As for $\pi$, since $\pi=4\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}$, you can use the odd terms as an increasing rational series and the even terms as a decreasing one that converges on $\pi$. |
H: On $C^0 [0, 1]$, define $f \cdot g = \int_0^1 f(x) g(x) dx$. For $f(x) = x$.
a. find $||f||$
b. find all linear polynomials that are orthogonal to $x$
Okay, so I know that
$||f|| = \sqrt(f_1^2 + f_2^2 +... + f_n^2)$
and that linear polynomials are of the form $ax + b$
I am not sure however, how to apply these to the actual question..
AI: If you have a scalar product $(f,g)$ then the norm is defined by $\|f\| = \sqrt{(f,f)}$. In your case
$$ \|f\|^2 = \int_0^1 f(x)^2 dx$$
To find $\|f\|$ you just need to integrate $x^2$ from $0$ to $1$.
If you have a linear polynomial $ax+b$ which is orthogonal to $x$ then their scalar product is zero, so
$$ \int_0^1 x (ax+b)dx =0$$
and computing the integral you arrive at a necessary and sufficient condition for $a,b$. |
H: A Question on Digit Occurences
Here's a question I was thinking about:
For all positive integers n, list the decimal representation of the numbers 1, 2, 3, ..., n without any leading zeroes. Does there exist an n such that this list contains an equal number of each of the digits 0, 1, 2, 3, ..., 9? (For example, if n=15, the list is 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, and it contains eight 1's, one 0, and so on.)
I thought about it for quite a while, and intuitively it seems very unlikely, but I couldn't formulate a rigorous proof. Could you guys help me in discovering one?
Thanks!
AI: no, the number of times zero appears is always less than the number of times one appears.
proof:
set $k_i(n)$ equal to the number of times $1$ has appeared in the $10^i$'s place minus the number of times $0$ has appeared in the $10^i$'s place.
each $k_i(n)$ is non-negative and if the place of the first digit of n is $10^m$, $k_m(n)$ is positive, so the sum of all the $k_i(n)$ is positive. |
H: What is the additive inverse of $\langle x-2\rangle+(x^2-2)$ in $\mathbb{Q}[x]/\langle x-2\rangle$?
I'm not sure if I'm thinking about this correctly, but this is what I did:
$\langle x-2 \rangle +(x^2-2) = x^2-2$ (since here, $\langle x-2 \rangle$ is equivalent to $0$) whose additive inverse is $2-x^2$
The book (A First Course in Abstract Algebra 2nd ed. by Anderson and Feil) says the answer should be $\langle x-2 \rangle + (2-x^2)$. Do I need the $\langle x-2 \rangle$ part?
AI: It depends on the notation you are using. $\langle x-2\rangle + (2-x^2)$ means the coset of $(2-x^2)$ in the quotient group $\mathbb{Q}[x]/\langle x-2\rangle$. Some authors write this in short hand as $(2-x^2)$, where it is understood that this is an element of $\mathbb{Q}[x]/\langle x-2\rangle$, meaning that $(2-x^2)$ is well defined up to adding elements of the ideal $\langle x-2\rangle$. |
H: Why is $-\ln(\cos(x))$ equal to $\ln(\sec(x))$?
Why does the value $-\ln(|\cos(x)|)$ become $\ln(|\sec(x)|)?$
I was doing an integral and I got my final answer as that, but I don't understand how you can just send the negative sign inside and make it $\sec(x).$
AI: You have $$a^b = c \iff b = \log_a(c). $$
Use this to show that for any power $r$, $\log_b(a^r) = r\log_b(a)$
for any base $b$. |
H: Can the exponentiation of an integer by a rational be a non-integer rational?
Consider a strictly positive integer $n\in\mathbb{N^*}$ and a rational $r=\frac{p}{q}\in\mathbb{Q}$.
My question is the following: what is the nature of $n^r$?
My first guess is that $n^r$ is an integer or an irrational but cannot be a non-integer rational. Is it correct? And if not, can you provide an example where $n^r$ is a non-integer rational?
AI: Your intuition is correct: $n^r$ is a root of $x^q-n^p=0$. By the rational root theorem, $n^r$ is either an integer or irrational. |
H: Combinatorics homework question
The answer is 54912.
This is what I've tried so far: So first you have to pick a rank to occur 3 times so thats 13, now you gotta pick a suit that that rank has, which is now 13 * 4. Now you need to pick that same rank 2 more times. The second time it will be 13 * 3, then the third time it will be 13 * 2. So the expression so far is (13^3)*(4*3*2) but that number is already too high when I try to add the other two cards. What am I doing wrong?
AI: The key to this question is the terms 'combination' and 'subset'. You want to model this problem with combinations for each subset/property and then multiply these together.
That is: rank $\times$ suit $\times$ other_rank $\times$ other_suits.
One way to think about this is build an example: (4,{diamond,heart,spade},(2,5),{diamond,heart}).
There are 13 ranks so we choose one with $\binom{13}{1}$, choose our suits with $\binom{4}{3}$, chose our other two ranks $\binom{12}{2}$ and then the suits for these two $\binom{4}{1}^2$.
So the answer is $\binom{13}{1}\binom{4}{3}\binom{12}{2}\binom{4}{1}^2=54912$. |
H: Finding roots of a function in an interval
Does the equation $x^3-12x+2=0$ have three solutions in the interval $[-4,4]$?
We know that this is a continuous function because it's a polynomial, and so we can use the Intermediate Value Theorem to do this problem:
If $f(x)$ is continuous on $[a,b]$, let $M$ be any number between $f(a)$ and $f(b)$. Then there exists a number $c$ such that:
1) $a<c<b$
2) $f(c)=M$
If we define $M=0$, $a=-4$ and $b=4$, then $-4<c<4$ and $f(-4)<0<f(4)$.
Compute $f(-4)$ and $f(4)$ to show that $f(-4)<0<f(4)$
$\implies -14=f(-4)<0<f(4)<18$.
So I've shown that $M$ is between $f(-4)$ and $f(4)$, and thus there must be some number $c$ such that $f(c)=M=0$.
Is all of this correct?
How do I show that there are three solutions (roots) in this interval?
Thank you.
AI: No quadratic equation can have three roots in any interval.
I expect that the correct equation is $x^3-12x+2=0$, which does have three roots in $[-4,4]$ because $f(-4)<0$, $f(0)>0$, $f(1)<0$, $f(4)>0$. By the Intermediate Value Theorem, each sign variation gives you at least one root in that interval. |
H: Is the fraction of the irrational exponentiations of two coprime integers by a rational an irrational?
Consider two strictly positive integer coprimes $n, m\in\mathbb{N^*}$ and a rational $r=\frac{p}{q}\in\mathbb{Q}$. Consider furthermore that the three number statifies the following condition:
$x=n^r$ is an irrational number
$y=m^r$ is an irrational number
Question: is $\frac{x}{y}$ an irrational (neither rational or integer)?
AI: Hint: Yet another use of the rational root theorem, $x/y = (n/m)^r$ solves $z^q - (n^p/m^p) =0$, now with leading coefficient $\neq 1$. |
H: Let G be cyclic and let H be a subgroup of G. Show that G/H is cyclic.
Not really sure how I go about doing this. Abstract Algebra. I'm fairly certain it has to do with normal subgroups though.
AI: Here is a quick answer, and if you fill out the details you'll learn a lot from it.
1) Prove that a group $G$ is cyclic if, and only if, there exists a surjective homomorphism $\mathbb Z \to G$.
2) Recall that the quotient group $G/H$ always comes equipped with the canonical surjection $\pi:G\to G/H$.
3) Combine 1 and 2. |
H: Difference between the real projective plane and the complex projective plane
Well the title says it all. If we consider the $P^2(\Bbb R)$ and the $P^2(\Bbb C)$, and we compare them, my guess is that it will be like a round $\Bbb R^2$ versus a sphere. I don't have very good geometric intuition, so I really can't picture this well, and I'm usually wrong about these things.
AI: Although their construction are similar, these two spaces are actually quite different from the topological point of view.
Recall that $P^2(K)$ is the set of all one dimensional subspace in $K^3$, where $K = \mathbb R$ or $\mathbb C$ in this case.
For $P^2(\mathbb R)$, as all lines passes through the unit sphere $S^2 \subset \mathbb R^3$ at exactly two points $x$ and $-x$, we can also think $P^2(\mathbb R)$ as
$$P^2(\mathbb R) = \mathbb S^2 /\{\pm 1\}\ .$$
As a result, we see that $\mathbb S^2 $ is a two to one cover of $P^2(\mathbb R)$, and thus $\pi_1(P^2(\mathbb R)) = \{\pm 1\}$.
For $P^2(\mathbb C)$, one can also consider the unit sphere $\mathbb S^5$ in $\mathbb C^3 \cong \mathbb R^6$. However, now every complex one dimensional subspaces intersects the sphere in a $\mathbb S^1$. In particular, we have the following fibration
$$\mathbb S^1 \to \mathbb S^5 \to P^2(\mathbb C),$$
this shows that $P^2(\mathbb C)$ is real four dimensional (complex two dimension). |
H: Surjectivity for permutation representation of a group action
I am having trouble proving that my function is surjective. Here is the problem statement:
Problem statement: Let T be the tetrahedral rotation group. Use a suitable action of T on some set, and the permutation representation of this action, to show that T is isomorphic to a subgroup of $S_4$.
Outline of my attempt at a proof:
Let $S$ be the set of faces of the tetrahedron. There exists a homomorphism $f:T$ $\rightarrow$ (permutations of the faces). There are $4$ faces, so $|Perm(|4|)| = 24$. We want to show that $T$ is a bijection, and show the faithfulness of $f$. Well, $|Ker(f)|=1$, because no rotation fixes all faces. So $f$ is injective. Also, the class equation for $T$ is $12 = 1+3+4+4$, which admits no "subsums" to $6$, implying that there are no odd permutations because $T$ has no subgroup of index $2$. So $T \approx A_4$, which is a subgroup of $S_4$.
The thing that evades me is how to prove that $f$ is surjective?
AI: The easiest way to do this would be to use the Orbit-Stabilizer Theorem.
$$
|G|=|\text{orbit}||\text{stabilizer}|
$$
Think about the action of a permutation on the tetrahedron. Draw a particular tetrahedron. How many rotations of the tetrahedron fix one of its faces?
There are 3
Then what is the orbit for that particular tetrahedron?
There are 4
So then the group has order?
The group has order 12.
To show that the group is $A_4$. Since we have already have an action of permutations, we need to show that we can only have even permutations of the "correct" order. I'll leave you to flesh out the details. But first ask why there can be no element of order $12$ in the group
There can be no element of order $12$ as then the group would be cyclic--Clearly not the case!
Finally, why are there no elements of order $6$?
It is a permutation of the FOUR faces of the tetrahedron.
Then finally, ask yourself, why does this only leave the subgroup $A_4$ of $S_4$? |
H: Show that $|1-\varphi_X (u)|\leq E\{ |uX| \}$
Show that $|1-\exp\{ix\}|^{2}=2(1-\cos x) \leq x^{2}$ for all $x \in \mathbb{R}$. Use this to show that $|1-\varphi_X(u)|\leq E\{|uX|\}$, where $\varphi_X(u) =E\{\exp(i\langle u,X\rangle)\}$ is the characteristic function.
We were given the hint that the power series for $\cos x$ is an alternating series. However, I'm not sure where that comes into play. For the first part, by using the definition of a square of a modulus, Euler's formula, and a trig identity, I got that $|1-\exp\{ix\}|^{2}=2-\sin^{2}x$. Now, $\sin^{2}x \leq x^{2}$, right? And so, $2-\sin^{2}x \leq 2-x^{2} \leq x^{2}$, right? So, I'm not sure hiw it's supposed to come into play in the first part, if at all.
The second part, showing that $|1-\varphi_X(u)|\leq E\{|uX|\}$ I am having the most trouble with. Any help you could give would be most appreciated!
AI: $$
1-\frac{x^2}{2} < \cos x < 1 - \frac{x^2}{2} + \frac{x^4}{24} \text{ if }x\text{ is near }0.
$$
You couldn't draw this conclusion if the series didn't have terms alternating in sign and (if $x$ is near $0$) getting smaller as the degree grows.
Hence
$$
x^2 - \frac{x^4}{12}\le2(1-\cos x) \le x^2.
$$
You should carefully distinguish between $x$ and $X$. You're not doing that.
Before I say more, could you clarify whether you intend $X\sim N(0,1)$, as I'm guessing? |
H: Modified dirichlet function
Can Dirichlet's function be modified in such a way that it is continuous at some real number? For instance, as $xD(x)$ is continuous at $x=0$, is it possible that $(x-1)D(x)$ is continuous at $x=1$? Here $D(x)$ is the Dirichlet function.
Can it be modified to be continuous at finitely many points in $\mathbb{Q}$?
I am aware that there can't be a function continuous only on rational numbers, but with the same approach as $2$, if possible, why isn't it possible to construct such a function?
Thanks in advance!
AI: Given numbers $x_1,\ldots,x_n$, you can define an "accordion function" $g(x)$ by piecing together absolute values so that $g(x_j)=0$ and $g((x_j+x_{j+1})/2)=1$. Then $g(x)D(x)$ is continuous at $x_1,\ldots,x_n$ and discontinuous everywhere else.
The same approach wouldn't work for all the rationals because they are dense, and so there is no room for the function to "raise up" and "come down". |
H: Orthogonal Eigen-Basis Problem
In any euclidean space, will the eigenvectors be always orthogonal ?
As a result of spectral theorem ?
AI: For an arbitrary operator, no. For instance, on finite dimensional spaces, an operator is unitarily diagonalizable (i.e. has orthonormal eigenvectors) if and only if it is normal, that is it commutes with its adjoint. Take a look here. |
H: Solving an inequality
Name the property that justifies each statement.
If $-3x < 24$, then $x > -8$.
AI: The property is the Additive Inverse
If $a < b$ then $-a > -b$
If $a > b$ then $-a < -b$
This is really the same as multiplying by $(-1)$, and that is why it changes direction.
http://www.mathsisfun.com/algebra/inequality-properties.html
so in your case, it'd be (after dividing both sides by 3)
$-3x < 24 \implies -x < 8$
multiplying both sides by $(-1)$ and according to the Additive Inverse you'd change the sign..
$(-1)(-x) < (-1)8 \implies x >-8$ |
H: If I have the equation, $\frac{p}{K-p}=Ce^{rt}$, how can I solve for p?
There is where I'm at:
$$
\frac{p}{(K-p)}=Ce^{rt}
$$
$$
p=(K-p)(Ce^{rt})
$$
$$
p=Ce^{rt}K-Ce^{rt}p
$$
$$
p+Ce^{rt}p=Ce^{rt}K
$$
$$
p(1+Ce^{rt})=Ce^{rt}K
$$
$$
p=\frac{Ce^{rt}K}{1+Ce^{rt}}
$$
Is it possible to simplify this further?
AI: I'll include my comment as an answer just so I can demonstrate what I was referring to.
Comment:
Your answer is just fine. There are lots of things you could do but none that would make it look any better. Though many would write $p=\frac{KC e^{rt}}{1+Ce^{rt}}$ just to keep all constants in front of exponentials, but this is not an error and is simply aesthetic.
Ways you could 'simplfy', take out the $C$ from the top and from the bottom
$$
p=\frac{CK e^{rt}}{C(\frac{1}{C}+e^{rt})}=\frac{Ke^{rt}}{\frac{1}{C}+e^{rt}}
$$
or you could put the $e^{rt}$ into the denominator so only one exponential appears.
$$
p=\frac{KCe^{rt}}{1+Ce^{rt}}=\frac{KC}{e^{-rt}(1+Ce^{rt})}=\frac{KC}{1+Ce^{-rt}}
$$
The last one may be a bit cleaner, but many would argue that the exponential with a negative power should not appear in the denominator. So you see how thought there are things you can do, they don't necessarily simplify much (if anything). Your original answer seems clean enough as written. |
H: $2^{2x}-3 \cdot 2^{x+1}=16 \implies 2^x=-2$ or $2^x=8$
How did they get from $$2^{2x}-3 \cdot 2^{x+1}=16$$ to $2^x=-2$ or $2^x=8$?
I know that you could also write it as $2^{2x}-3 \cdot 2^{x+1}$
and $2^{2x}-6\cdot 2^x=16$
But I got stuck there...
AI: Hint : $$2^x=t\iff t^2-6t-16=0$$ |
H: Spaces sharing all higher homotopy groups
Is it possible that two topological spaces share all higher homotopy groups, but are not homeomorphic? I should note that I have not studied much in the way of the theory of higher homotopy groups; I merely know of the groups' existence, and was wondering what could be said, in broad terms, about such a case.
If such a case is possible, an example would be appreciated.
AI: Yes it can happen. A simple example is to take $X=S^1\vee S^2$ and $Y$ to be a circle with a two spheres attached each at a different point. Then $Y$ is a double cover of $X$, so it shares all higher homotopy groups $\pi_n$ $n\geq 2$. (Here we use the fact that a covering map induces an isomorphism on all higher homotopy groups.) Also $\pi_1(X)=\pi_1(Y)=\mathbb Z$, so $X$ and $Y$ have the same fundamental group as well. |
H: Variation of parameters for a linear second order nonhomogeneous equation
I'm using variation of parameters for this problem, and I'm not sure if I'm on the right track.
The question is
Find a function $v_1$ and $v_2$ such that $v_1(x)e^x+v_2(x)e^{2x}$ is a solution of $y''-3y'+2y=4x+4$ and
$v_1(x)y_1(x)=v_2(x)y_2(x)$
$y''-3y'+2y=4x+4$
First we need to find the roots.
$r^2-3r+2=0$
$(r-1)(r-2)=0$
$r=1$
$r=2$
So, the $Y_h$ = $c_1e^x+c_2e^{2x}$
Then we need to find the derivatives and the Wronskian. I am using Cramer's rule.
$y_1$ = $e^x$
$y_2$ = $e^{2x}$
$y'_1$ = $e^x$
$y'_2$ = $2e^{2x}$
W[$e^x$ $e^{2x}$]
W= $$ \left[
\begin{array}{ c c }
e^x & e^{2x} \\
e^x & 2e^{2x}
\end{array} \right]
$$
$2e^{2x} e^x -e^xe^{2x}$ = $e^{3x}$
$W_1$= $$ \left[
\begin{array}{ c c }
0 & e^{2x} \\
4x-4 & 2e^{2x}
\end{array} \right]
$$
$2e^{2x}(0)-(4x-4)e^{2x}$=$-(4x-4)e^{2x}$
$W_2$= $$ \left[
\begin{array}{ c c }
e^x & 0 \\
e^x & 4x-4
\end{array} \right]$$
$(4x-4)(e^x)-(0)(e^x)$=$(4x-4)e^{x}$
$\frac{W_1}{W}$ = $\frac{-(4x-4)e^{2x}}{e^{3x}}$ = $\frac{-(4x-4)}{e^x}$
$\frac{W_2}{W}$ = $\frac{(4x-4)e^{x}}{e^{3x}}$ = $\frac{(4x-4)}{e^{2x}}$
After I calculated the Wronskian through Cramer's Rule I have to find the anti-derivative
$\frac{W_1}{W}$ = $\frac{-(4x-4)}{e^x}$
$\int{-(4x-4)}{e^x}\,dx = 4xe^{-x} + C$
$\frac{W_2}{W}$ = $\frac{(4x-4)}{e^{2x}}$
$\int {(4x-4)}{e^{-2x}}\,dx = (1-2x)e^{-2x} + C$
The result from the anti-derivative along with
$y_1$ = $e^x$
$y_2$ = $e^{2x}$
should be in a formula
$u_1y_1+u_2y_2 = 0$
$u_1y_1+u_2y_2 = 4x-4$
but I think I'm getting the letters confused. So, how do I find the values that equal to each other and satisfy this condition?
Do I solve using the method of undetermined coefficients?
EDIT: I have one more piece to this problem.
The Theorem states that If $y_1...y_n$
are linearly independent solutions of the reduced form of
$y^{n}+P_{n+1}y^{n-1}+...+p_{1}y'+p_0y=q$
then there are functions
$v_1,...v_n$
which satisfy
$y_1v'_1+...+y_nv'_n=0$
I'm just a beginner at proofs. How do I tackle this problem?
I got it. It needed the derivatives of $v_1$ and $v_2$
Therefore,
$v_1(x)=4xe^{-x}$
$v'_1(x)=-4e^{-x}(x-1)$
$v_2(x)=4(-1/2x+1/4)e^{-2x}$
$v_2(x)=-(2x-1)e^{-2x}$
$v'_2(x)=4(x-1)e^{-2x}$
The problem for the theorem just needed the original $v_1(x)$ and $v_2(x)$
Something isn't right...
the answer is $v_1(x)$=$x$+$\frac{1}{2}e^{-x}$ because it needed to satisfy this situation $v_1(x)y_1(x)=v_2(x)y_2(x)$
since $v_1$ would normally represent a Wronskian in this problem, there must be something or some value...that would satisfy $v_1(x)y_1(x)=v_2(x)y_2(x)$
I've done this problem with a study buddy and we found the answer. We were supposed to take the derivatives of $Y_p$ twice and use the rules from linear equations.
My study buddy's method:
AI: We are asked to solve this using Variation of Parameters (VoP), given:
$$\tag 1 y''-3y'+2y=4x+4$$
Step 1
Find the homogenous solution to $(1)$, so we have:
$$\tag 2 y''-3 y'+ 2 y = 0$$
This yields:
$$y_h = c_1e^x + c_2 e^{2x}$$
Step 2
We are now going to make use of VoP, so we set: $y_1 = e^x$ and $y_2 = e^{2x}$ from $y_h$ and $f = 4(x+1)$ from $(1)$.
We calculate the Wronskian of $y_1$ and $y_2$, yielding $W(e^x, e^{2x}) = e^{3x}$.
Using VoP, we have:
$$u_1 = \int \dfrac{-y_2 f}{W(e^x,e^{2x})} dx = \int \dfrac{-e^{2x}~ 4(x+1)}{e^{3x}} dx = 4 e^{-x} (x+2)$$
$$u_2 = \int \dfrac{y_1 f}{W(e^x,e^{2x})} dx = \int \dfrac{e^x ~4(x+1)}{e^{3x}} dx = -e^{2x}(2x+3)$$
Now, $y_p$ is given by:
$$y_p = y_1 u_1 + y_2 u_2 = e^x(e^{-x} 4(x+2)) + e^{2x}(-e^{-2x}(2x+3)) = 4(x+2) -(2x+3) = 2x + 5$$
Step 3
Our final solution is given by:
$$y(x) = y_h(x) + y_p(x) = c_1e^x + c_2 e^{-2x} + 2x + 5$$ |
H: Fourier transform formula
How to prove that $f(x)=\sum_{n=-\infty}^\infty f(n)K(x-n)$ where $K(y)=\sin\pi y/\pi y$
Here, $f$ is moderate decrease and its fourier transform is supported in $[-1/2, 1/2]$.
I show that $\hat f(k)=\sum_{n=-\infty}^\infty f(n)e^{-2\pi ink}$ but no improvement...
Plz help
AI: This formula is called Whittaker–Shannon interpolation formula.
Take inverse Fourier transform both sides of the formula $$\hat{f}(\xi)=\sum f(n) e^{-2\pi in\xi}\operatorname{rect}(\xi)$$
where $\operatorname{rect}(\xi)$ is rectangular function. |
H: Simplifying logarithmic equation $0.1 = e ^{-(\ln 2)t/5700}$?
$$0.1 = e ^{-(\ln 2)t/5700}$$
How do I simplify this? I took the ln of both sides so does the $e^{\ln}$ cancel out?
$$\ln 0.1 = \ln e ^{-(\ln 2)t/5700}$$
$$-2.3 = \ln e ^{-(\ln 2)t/5700}$$
but the $\ln$ in the exponent can't just disappear?
AI: No, but using the fact that $$\ln{e^x} = x$$ for all $x$, we see that
$$\ln{0.1} = \ln e^{(-\ln{2}) t / 5700} = \frac{(-\ln{2}) t}{5700}$$
Hence
$$t = \frac{5700 \ln{0.1}}{-\ln{2}}$$ |
H: What is the solution to this problem
I'm sorry I can't be more specific in the title.
I have this exercise I can't solve by myself, I've tried many times and I never get the right answer:
$[(\frac{1}{a^2}-b^2):(\frac{1}{a}+b)]^{-1}$
This is the answer:
$\frac{a}{1-ab}$
Any help is appreciated.
AI: Hint: The difference of squares formula is: $x^2-y^2 =(x-y)(x+y)$. Don't forget that $\frac{1}{a^2} = \left( \frac{1}{a} \right)^2$. |
H: Simple way to write Cumulative distribution function of $\alpha X + \beta$?
Let $X$ be a discrete random variable with Cumulative Distribution Function $F$. What is the cumulative distribution function of the random variable $\alpha X + \beta$, where $\alpha, \beta$ are constants and $\alpha \ne 0$?
Let $Y = \alpha X + \beta$. We want to find the CDF for $Y$. Going completely by definition, I have
$$\mathrm{P}(Y \le a) = \sum_{\large y\ \in Y:\ y\ \le\ a\ \ \ } \sum_{\large x\ \in X:\ \ \alpha x + \beta = y} \mathrm{P}(x) $$
for some $a \in Y$. Now, because it can be proven that
$$\mathrm{E}[\alpha X + \beta] = \alpha\mathrm{E}[X] + \beta$$
Is there a way to apply this result, or one something similar, to simplify the expression I gave above for $\mathrm{P}(Y \le a)$?
AI: Write: $F_Y(y) = P(Y < y) = P(\alpha X + \beta < y)$ and turn this into an expression of the form $P(X < \;?)$. |
H: which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$
Which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$?
Squaring both sides will give me something but I could not go any further.
AI: As $(\sqrt7+\sqrt5)^2=12+2\sqrt{35}$ and $(\sqrt6+\sqrt6)^2=12+2\sqrt{36}$
$$(\sqrt7+\sqrt5)^2<(\sqrt6+\sqrt6)^2$$
$$\implies \sqrt7+\sqrt5<\sqrt6+\sqrt6\text{ as } a^2>b^2\iff a>b\text{ for }a,b>0$$
$$\implies \sqrt7-\sqrt6<\sqrt6-\sqrt5$$ |
H: Show $\exp(A)=\cos(\sqrt{\det(A)})I+\frac{\sin(\sqrt{\det(A)})}{\sqrt{\det(A)}}A,A\in M(2,\mathbb{C})$
Show $$\exp(A)=\cos(\sqrt{\det(A)})I+\frac{\sin(\sqrt{\det(A)})}{\sqrt{\det(A)}}A$$
for $A\in M(2,\mathbb{C})$. In addition, $\operatorname{trace}(A)=0$.
Can anyone give me a hint how this can connect with cosine and sine? Thanks!
AI: The caracteristic polynomial of $2\times 2$ matrix $A$ is
$$X^2-\mathrm{Tr}(A)X+\mathrm{det}(A),$$
so that a trace $0$ matrix satisfies the equation
$$A^2=-\mathrm{det}(A)I_2.$$
Let $\lambda\in\Bbb C$ be a square root of $\det(A)$. It follows from the equation above that for every integer $p$
$$A^{2p}=(-1)^p\lambda^{2p}I_2\qquad\text{and}\qquad A^{2p+1}=(-1)^p\lambda^{2p}A$$
From then definition of the exponential function,
$$\begin{align}\exp(A)&=\sum_{n\in\Bbb N}\frac{A^n}{n!}\\
&=\sum_{p=0}^{\infty}\frac{A^{2p}}{(2p)!}+\sum_{p=0}^{\infty}\frac{A^{2p+1}}{(2p+1)!}\\
&=\sum_{p=0}^{\infty}\frac{(-1)^p\lambda^{2p}}{(2p)!}I_2+\sum_{p=0}^{\infty}\frac{(-1)^p\lambda^{2p}}{(2p+1)!}A.
\end{align}$$
Thus, if $\lambda\neq 0$,
$$\exp(A)=\cos(\lambda)I_2+\frac{1}{\lambda}\sin(\lambda)A.$$ |
H: Find the limit of a fraction
So far I have,
$$
\lim_{x\to 1} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{1^2+1}}}{x-1}=\lim_{x\to 1} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{2}}}{x-1}
$$
I have no idea how to keep going with this, every way I try I get stuck and can't do anything with it.
AI: More generally,
consider
$$\lim_{x\to a} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{a^2+1}}}{x-a}.$$
I will use just algebra.
If $x \ne a$,
$\begin{align}
\frac{\frac{x}{\sqrt{x^2+1}} - \frac{a}{\sqrt{a^2+1}}}{x-a}
&=\frac{x\sqrt{a^2+1}-a\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{a^2+1}(x-a)}\\
&=\left(\frac{x\sqrt{a^2+1}-a\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{a^2+1}(x-a)}\right)
\left(\frac{x\sqrt{a^2+1}+a\sqrt{x^2+1}}{x\sqrt{a^2+1}+a\sqrt{x^2+1}}\right)\\
&=\frac{x^2(a^2+1)-a^2(x^2+1)}
{\sqrt{x^2+1}\sqrt{a^2+1}(x-a)(x\sqrt{a^2+1}+a\sqrt{x^2+1})}\\
&=\frac{x^2a^2+x^2-a^2x^2-a^2}
{\sqrt{x^2+1}\sqrt{a^2+1}(x-a)(x\sqrt{a^2+1}+a\sqrt{x^2+1})}\\
&=\frac{x^2-a^2}
{(x-a)\sqrt{x^2+1}\sqrt{a^2+1}(x\sqrt{a^2+1}+a\sqrt{x^2+1})}\\
&=\frac{x+a}
{\sqrt{x^2+1}\sqrt{a^2+1}(x\sqrt{a^2+1}+a\sqrt{x^2+1})}\\
\end{align}
$
Letting $x \to a$,
this becomes
$\frac{2a}
{(a^2+1)(2a\sqrt{a^2+1})}
=\frac{1}
{(a^2+1)^{3/2}}
$.
As a derivative,
$\begin{align}
\left(\frac{x}{\sqrt{x^2+1}}\right)'
&=\frac{\sqrt{x^2+1}-x(1/2)(2x)(x^2+1)^{-1/2}}{x^2+1}\\
&=\frac{\sqrt{x^2+1}-x^2(x^2+1)^{-1/2}}{x^2+1}\\
&=\frac{(x^2+1)-x^2}{(x^2+1)^{3/2}}\\
&=\frac{1}{(x^2+1)^{3/2}}\\
\end{align}
$
which is comforting (and much easier).
Note that the "$1$"
in $x^2+1$ and $a^2+1$
can be any value -
it is just carried along
and,
if the expression is
$\frac{x}{\sqrt{x^2+b}}$,
the result is
$\frac{b}{(x^2+b)^{3/2}}$. |
H: Why do we stop at double cosets and do not consider triple, ... n-ple cosets?
The obvious answer seems to be that a group has only two sides, and once you're done taking quotients on the left and on the right, you can't quotient by another subgroup from another side.
But if $G$ is a group with subgroups $H$ and $K$, why can we not consider a quotient like $G/H/\bar K$, where $\bar K$ is the image of $K$ in $G/H$ and the whole should be understood as the set of cosets $xHK$?
AI: That's because you cannot build an equivalence relation on a double quotient on one side. Note that you can't get transitivity:
$$\begin{aligned} &a = xhk \\ &b = xh'k' \\ &b = yh''k'' \\ &c = y'''k'''\end{aligned}$$ so that $a \sim b$ and $b \sim c$, but a bit of manipulation will show that you can get $a\sim c$ when you have at most one coset on each side. |
H: Probability problem with deck of cards
In a poker game each player is dealt five cards. What is probability that a player's hand has exactly four diamonds?
The answer is .011, I just don't know how to get it.
AI: There are $\binom{52}{5}$ $5$-card hands. They are all equally likely.
Now we count the hands that have exactly $4$ diamonds. The diamonds can be chosen in $\binom{13}{4}$ ways. For each of these ways, the non-diamond can be chosen in $\binom{39}{1}$ ways, for a total of $\binom{13}{4}\binom{39}{1}$.
For the probability, divide the number $\binom{13}{4}\binom{39}{1}$of "good" hands by the total number $\binom{52}{5}$ of hands. |
H: Prove that $\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$ are linearly dependent.
Functions $f$ and $g$ are independent on an interval $D$ if $af(x) + bg(x) = 0$ implies that $a = 0$ and $b = 0$ $\forall x \in D$
let $\alpha$, $\beta$, $\gamma$ be real constants. Prove that
$\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$
are linearly dependent vectors in $C^0[0, 1]$. Be convincing in your reasoning (argument)
I was researching and found Wronskian. Using the Wronskian for three functions. The determinant of $f$, $g$ and $h$ is $W(f, g, h) = $
$$
\begin{vmatrix}
f & g & h \\
f' & g' & h' \\
f'' & g'' & h'' \\
\end{vmatrix}
$$
If $W(f, g, h) \neq 0$ then $f(x)$, $g(x)$ and $h(x)$ are linearly independent.
If $f(x)$, $g(x)$, and $h(x)$ are linearly dependent then $W(f, g, h) = 0$
My attempt
Let
$f(x) = \sin(x + \alpha)$, $g(x) = \sin(x + \beta)$ and $h(x) = \sin(x + \gamma)$
$W(f, g, h) =$
$$
\begin{vmatrix}
\sin(x + \alpha) & \sin(x + \beta) & \sin(x + \gamma) \\
\cos(x + \alpha) & \cos(x + \beta) & \cos(x + \gamma) \\
-\sin(x + \alpha) & -\sin(x + \beta) & -\sin(x + \gamma) \\
\end{vmatrix}
$$
$= \sin(x + \alpha)[-\sin(x + \gamma)\cos(x + \beta) + \cos(x + \gamma)\sin(x + \beta)] - sin(x + \beta)[-\sin(x + \gamma)\cos(x + \alpha) + \cos(x + \gamma)\sin(x + \alpha)] + \sin(x + \gamma)[-\sin(x+ \beta)\cos(x + \alpha) + \cos(x + \beta)\sin(x + \alpha)]$
$= -\sin(x + \alpha)[\sin((x + \gamma) +(x + \beta))] + \sin(x + \beta)[\sin((x + \gamma) + (x + \alpha))] - \sin(x + \gamma)[\sin((x + \beta) + (x + \alpha))] = 0$
By Wronskian, $f(x)$, $g(x)$ and $h(x)$ are linearly dependent since $W(f, g, h) = 0$
Not sure if this argument is sound?
AI: As Mohamed pointed out in his answer, we can use the angle addition formula for $\sin$, viz.:
$\sin (x + \alpha) = \cos \alpha \sin x + \sin \alpha \cos x, \tag{1}$
$\sin (x + \beta) = \cos \beta \sin x + \sin \beta \cos x, \tag{2}$
$\sin (x + \gamma) = \cos \gamma \sin x + \sin \gamma \cos x, \tag{3}$
to see that all three functions lie in the subspace generated by $\sin x$ and $\cos x$, which is of course $\text{Span} \{ \sin x, \cos x \}$, and since $\sin (x + \alpha)$, $\sin (x + \beta)$, $\sin (x + \gamma)$ are thus three vectors in a subspace of dimension two, a linear dependence between them must exist.
Going a step further, we can actually solve for $\sin x$, $\cos x$ in terms of $\sin (x + \alpha)$, $\sin(x + \beta)$ and thus find the linear dependence explicitly. We have, from
(1) and (2),
$\begin{bmatrix} \cos \alpha & \sin \alpha \\ \cos \beta & \sin \beta \end{bmatrix} \begin{pmatrix} \sin x \\ \cos x \end{pmatrix} = \begin{pmatrix} \sin (x + \alpha) \\ \sin(x + \beta) \end{pmatrix}; \tag{4}$
the inverse of
$\begin{bmatrix} \cos \alpha & \sin \alpha \\ \cos \beta & \sin \beta \end{bmatrix} \tag{5}$
is readily seen to be
$(\cos \alpha \sin \beta - \sin \alpha \cos \beta)^{-1}\begin{bmatrix} \sin \beta & -\sin \alpha \\ -\cos \beta & \cos \alpha \end{bmatrix} = (\sin(\beta - \alpha))^{-1}\begin{bmatrix} \sin \beta & -\sin \alpha \\ -\cos \beta & \cos \alpha \end{bmatrix}, \tag{6}$
from which it follows that
$\begin{pmatrix} \sin x \\ \cos x \end{pmatrix} = (\sin(\beta - \alpha))^{-1}\begin{bmatrix} \sin \beta & -\sin \alpha \\ -\cos \beta & \cos \alpha \end{bmatrix}\begin{pmatrix} \sin (x + \alpha) \\ \sin(x + \beta) \end{pmatrix}. \tag{7}$
(7) gives $\sin x$, $\cos x$ in terms of $\sin(x + \alpha)$, $\sin(x + \beta)$; the resulting formulas may then be inserted into (3) to express $\sin(x + \gamma)$ in terms of $\sin(x + \alpha)$, $\sin(x + \beta)$; I leave the necessary algebra to my readers. One must of course remain vigilant to the possibility that $\sin(\beta - \alpha) = 0$; but this
is a special case best handled on its own.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!! |
H: Find orthogonal matrices such that $P^TAP$ is diagonal (dim=2)
Let
$$A = \begin{bmatrix}
-5 & 12 \\ 12 & 5
\end{bmatrix}.$$
I found the eigenvalues $13$ and $-13$, and the eigenvectors $[2,3]^T$ and $[3,-2]^T$. However, the matrix with them as columns transposed, times $A$, times that matrix is not diagonal.
I'm not sure why I expected it to be (I am checking my answers) and I sense I may have a huge misconception about what I am doing. Please help me find where and why I've gone wrong. And also some resources (questions and notes) for practice, and that I may learn from.
Addendum
I may be less wrong that I thought, my matrix of eigenvectors can be symmetric if I place it in a certain way,if I place it the other way, I get negative my first answer.
Not sure what I got wrong, I'd still love an answer if one is going.
AI: Hints:
Find the eigenvalues and corresponding eigenvectors of the matrix.
Define the columns of $P$ as the eigenvectors from the previous step.
Since the matrix A is symmetric, you will find that $P^T = P^{-1}$.
Write $A = P \cdot J \cdot P^T$, or $J = P^T \cdot A \cdot P$
Spoiler
$$J = \begin{bmatrix}-13 & 0 \\ 0 & 13\end{bmatrix}; P = \begin{bmatrix}-3 & 2 \\ 2 & 3\end{bmatrix}$$ |
H: How to calculate the angle between two vectors, defined by 3 points on the earth?
I want to develop a formula to calculate the angle between two vectors. The vectors will be OX and OY (from point O to X , and Y), where the points are defined by their latitude and longitude values.
I know that there is a little problem when calculating the angle between 3D vectors, so it could be calculating the angles between OX' and OY' prime, where the points X' and Y' are the projections of X and Y, on the plane that "skews" earth on point O.
Any ideas how to formulize this ?
Thanks in advance :)
AI: If you project your vectors onto a plane, you can derive the angle from the dot product. $$a \cdot b = ||a||\,||b|| \cos \theta$$
Or you may want the
spherical angle, in which case you will also need the radius of your sphere. |
H: Taylor Series Approximation for degree k Taylor polynomial?
Let $T_k(x)$ be the degree $k$ Taylor polynomial of the function $f(x)=\sin(x)$ at $a=0$. Suppose you approximate $f(x)$ by $T_k(x)$. If $|x|\le 1$, how many terms are needed (that is, what is $k$) to obtain an error less than $\frac 1 {5040}$?
I don't really understand what they are asking or how to get the answer. I was thinking about using alternating series approximation. I am sure how to go about doing that, if that is true.
How do I do this?
Thanks!
AI: When $|x|\le 1$, the Maclaurin series for $\sin x$ is indeed an alternating series. The absolute value of the error when we truncate just after the term in $x^{2n+1}$ has absolute value $\le$ the absolute value of the first "neglected" term. And it is $\lt$ except in the case $x=0$.
Note that $5040=7!$. So the approximation obtained by stopping at the term $\frac{x^5}{5!}$ has error with absolute value $\lt \frac{1}{5040}$.
Remark: For $\sin x$, and $|x|\le 1$, you can get the same estimate of the error by using the Lagrange form of the remainder. Since it is likely that you will be asked similar questions where we do not have an alternating series, it is useful to know how to use the Lagrange form of the remainder, |
H: Is sum of $\sum\limits_{n = 1}^{p}{(n(n+1)}\mod{p})= \frac{p(p-1)}{2} $ where $p \equiv 7 \mod{8}$ correct?
$\sum\limits_{n = 1}^{p-1}{(n^{2}+n)}\mod{p})= \frac{p(p-1)}{2} $ where $p \equiv 7 \mod{8}$
I am not sure if this is correct, this is a part of my homework and is having trouble with it.
AI: Hint :
$\sum\limits_{n = 1}^{p-1}{(n^{2}+n)}\mod{p}= \sum\limits_{n = 1}^{p-1}{(n^{2}}\mod{p}) + \sum\limits_{n = 1}^{p-1}{(n}\mod{p})$
Sum the $ p-1$ squares followed by summing up $p-1$ terms in the next expression.
I do not think that the RHS is fully worked out. |
H: Converse of a well known result.
We all know the following result.
Let $\sum a_n$ is a series of alternative positive and negative terms. Consider $p_n = \frac{a_n + |a_n|}{2}$ and $q_n = \frac{a_n - |a_n|}{2}$. So $\sum p_n$ and $\sum q_n$ are series of positive and negative terms of $\sum a_n$. Now if $\sum a_n$ converges absolutely, then both the series of positive and negative terms $\sum p_n$ and $\sum q_n$ are convergent. Similarly if $\sum a_n$ converges conditionally, then both $\sum p_n$ and $\sum q_n$ diverges.
I want to know if its converse is true i.e. I am interested with the result.
If $\sum p_n$ and $\sum q_n$ converges then $\sum a_n$ is absolutely convergent.
If $\sum p_n$ and $\sum q_n$ diverges then $\sum a_n$ is conditionally convergent.
How to prove? I am not seeing its converse in my text book. If not please give some counterexamples.
Thank you for your help.
AI: (1) is true, because if two series converge absolutely, their termwise sum also converges absolutely.
(2) is false. Counterexample: $a_n = (-1)^n$. |
H: Prove that $N$ is normal
Let $H$ be a subgroup of $G$. Consider the set $N=\cap_{x\in G}xHx^{-1}$. Prove that $N$ is normal subgroup of $G$.
Using the fact that any (finite or infinite) intersection of subgroups is a subgroup I am able to prove $N$ is a subgroup of $G$ and even $H$.
AI: For each given $n\in N$, $\forall x\in G$ there exists an $h_x\in H$ s.t. $xh_xx^{-1}=n$. Now for each $g\in G$, we have $h_{g^{-1}x}\in H$ s.t. $n=g^-1xh_{g^{-1}x}x^{-1}g$, then$$gng^{-1}=g(g^-1xh_{g^{-1}x}x^{-1}g)g^{-1}=xh_{g^{-1}x}x^{-1}\in xHx^{-1},$$Thus $gng^{-1}\in N$.Hence, $N$ is a normal subgroup of $G$. |
H: Characterization of functional monoids
I define a functional monoid to be a monoid that is isomorphic to the set of all functions from a set $S$ to itself under the operation of composition. I want to know useful necessary and sufficient conditions for a monoid to be functional. Any help would be appreciated.
AI: The more common name for what you call a functional monoid is the full transformation monoid on the given set. A transformation monoid is then by definition a submonoid of the full transformation monoid.
Suppose now that $M$ is a transformation monoid on a set $X$. Clearly, $(m,x)\mapsto m(x)$ is a left action of $M$ on $X$. Moreover, this action is effective, in the sense that for all $m,m'\in M$ if $m(x)=m'(x)$ for all $x\in X$, then $m=m'$ (this property is just a monoid formulation of the definition of function). So, transformation monoids give rise to effective actions.
Conversely, an effective action of $M$ on some set $S$, allows one to associate with $m\in M$ the function $T_m:S\to S$ given by $T_m(s)=ms$. It is easy to see then that the function $m\mapsto T_m$ is a monoid homomorphism which moreover is injective precisely if $M$ is a transformation monoid, evidently isomorphic to the image of this homomorphism. So, transformation monoid are obtained from effective actions.
Now, you are interested in the full transformation monoid, which is severely restricted by size. In particular, there is, up to isomorphism, precisely one full transformation monoid of every cardinality of the form $k^k$.
I hope all of this information will help you answer what you actually are trying to solve. |
H: A set of point sets, of cardinality greater than that of the continuum.
What would be the examples of a Set of point sets in $\mathbb R^2$ having the cardinality greater than that of real numbers?
Any stipulations over these point sets themselves may please be specified.
EDIT: My mistake which i correct in bold face as an edit. Motivation here is purely for informative purposes.
This is also related to a previous question of mine asking for the cardinality of all planar curves.
AI: A natural example is the collection of Lebesgue measurable sets. The reason is that there are sets of measure $0$ and size continuum, and any subset of a measure $0$ set is also of measure $0$. This example has size $2^{|\mathbb R|}$. Note that, on the other hand, there are only $|\mathbb R|$ many Borel sets, we really needed that the measure space is complete to obtain a larger collection.
For a different example: There are sets of reals of size $\aleph_1$, the first uncountable cardinal. For example, fix a ("natural") way of coding binary relations on $\mathbb N$, using reals (for example, a real $r$ can code a sequence $a_0,a_1,\dots$ of natural numbers, using the continued fraction of $r$, and we can identify each natural with a pair of naturals, so this gives us a binary relation; there are many other natural ways, of course). Now pick a representative from each equivalence class where two reals are equivalent iff either neither codes a well-ordering, or else both code a well-ordering with the same order type. The collection of all subsets of this set has size $2^{\aleph_1}$. Of course, if the continuum hypothesis holds, this example has the same size as the previous one, but it is consistent that the size is different: It is strictly larger than $|\mathbb R|$, and it may be strictly smaller than $2^{|\mathbb R|}$. |
H: How prove this isn't exist prime number $p>7$
show that: there isn't exsit prime number $p>7$,such
$p^{12}+5039\times 5041$ the factor number is less than $120$
I think maybe use Fermat theorem? How can solve it? Thank you
AI: Hint : $5040=7!$ , $120=5!$ , and $p=2k+1$. Our expression becomes: $$p^{12}+5039\cdot5041=p^{12}+(7!-1)(7!+1)=p^{12}+(7!^2-1)=(p^{12}-1)+7!^2=\\=(p^6-1)(p^6+1)+7!^2=(p^3-1)(p^3+1)\cdot(p^2+1)(p^4-p^2+1)+7!^2=\\=(p-1)(p^2+p+1)\cdot(p+1)(p^2-p+1)\cdot(p^2+1)(p^4-p^2+1)+5!^2\cdot42^2$$ $p>7>2\iff p=2k+1\iff(p-1),(p+1)$, and $p^2+1$ are all multiples of $2\iff$ the product is a multiple of $2^3=8$. Also, since $p>7>3\iff p=3k\pm1\iff p\mp1$ is a multiple of $3\iff$ the product is also a multiple of $3$. $p>7>5\iff p=5k\pm1$ or $p=5k\pm2$. In the former case, $p\mp1$ is a multiple of $5$, in the latter, $p^2+1$ is divisible through $5\iff$ the product is a multiple of $5$ as well, meaning that the entire sum divides through $5!=120$ , for all primes $p>7$
If however the total number of factors is what's actually meant, then let us first notice that the expression divides through $2$ once more, because $(p-1)(p+1)=2k(2k+2)=4k(k+1)$, and $k(k+1)$ is always even, generating an extra factor of $2$ ; through $3$ twice, because of $p^2\pm p+1$ ; as well as through $7$. $[\ p=7k\pm1\iff7|p\mp1$ ; $p=7k\pm2\iff7|p^2\pm p+1$ ; $p=7k\pm3$ $\iff7|p^2\mp p+1\ ]$. Thus, $N=2^4\cdot3^2\cdot5^1\cdot7^1\cdot n^1$ , implying a minimum of $(4+1)\cdot(2+1)\cdot$ $(1+1)\cdot(1+1)\cdot(1+1)=120$ factors. — Thanks to user Achille Hui for his invaluable help ! |
H: Complex numbers equation problem
I've been having some trouble with this complex question (not my best topic), and I was wondering if I could get any hints or explainations on how to do it.
Prove that all the roots of the equation $$z^n\cos(n\alpha)+z^{n-1}\cos((n-1)\alpha)+z^{n-2}\cos((n-2)\alpha)+\cdots+z\cos(\alpha)=1,$$ where $\alpha$ is real, lie outside the circle $|z|=\dfrac 12$.
Any help would be greatly appreciated
AI: Suppose that there exists a root $z_0$ of the equation such that $|z_0| \le \frac{1}{2}$. Now we have that $$1 = |z_0^n\cos(n\alpha) + z_0^{n-1}\cos((n-1)\alpha) + \dots + z_0\cos(\alpha)|\le$$
$$|z_0^n\cos(n\alpha)| + |z_0^{n-1}\cos((n-1)\alpha)| + \dots + |z_0\cos(\alpha)|\le|z_0^n| + |z_0^{n-1}| + \dots + |z_0|\le$$ $$\frac{1}{2^n} + \frac{1}{2^{n-1}} + \dots + \frac{1}{2} = 1 - \frac{1}{2^n}< 1 $$
Contradiction. |
H: Solve for a variable in the limit of integration
How do I solve for $x$ in an equation such as the following?
$$A = \int_0^x f(t) dt$$
I feel like I must of come across this at some point in my courses, but for the life of me I cannot remember. Is this something that is really easy, or just hard to do for a general function $f$?
Edit: A is a constant. I am looking to solve for $x$.
AI: There isn't necessarily a unique solution, however consider the following:
We know via basic calculus that if $F$ is a function such that $F' = f$, then $$\int_0^x f(t)dt = F(x) - F(0)$$ so we can use this to get that $$F(x) = A+F(0).$$ If $F$ is invertible, then we get that $$x = F^{-1}(A+F(0)).$$
To see that it is not necessarily a unique solution, consider solving $$0 = \int_0^x \cos(t)dt.$$ Well, we are essentially solving $$0 = \sin(x)$$ and we know from trigonometry that we can have $x = n\pi$ for any $n\in Z$ as solutions.
Does this answer your question? |
H: All pairs of (a,b) of positive integers satisfying a given condition.
I have to determine all pairs $(a,b)$ of positive integers satisfying the condition $a^{b^2}=b^a$. So far i have found only one, namely $(1,1)$. How can i do it? Thanks for any help.
AI: Hints:
Take logarithms
$\log_b(a)\in\mathbb Q\Leftrightarrow \exists n\in\mathbb N:a=b^n$
Juggle a little with the remaining expression. |
H: Normal subgroup where G has order prime
If $o(G) = p^n,$ p a prime number, and H is a subgroup of G, show that there exists an $x\in G$, but $x\notin H$ such that $x^{-1}Hx = H$
How can I prove this?
AI: You can induct on $n$ :
Now if $N_G(H) \neq H$, then any element of $N_G(H)$ will do, so assume $N_G(H) = H$. Since $Z(G) \subset N_G(H) = H$, you can now look at the subgroup $H/Z(G) < G/Z(G)$.
Since $|G| = p^n, Z(G) \neq \{e\}$, so you can apply induction to get $\overline{x} \notin H/Z(G)$ which normalizes $H/Z(G)$. Now check that $x$ normalizes $H$. |
H: Largest triangle to fit in a circle will be isosceles triangle?
Largest triangle to fit in a circle will be isosceles triangle?
Or some other type?
AI: Yes, what you say is true, but you can say more than that.
Given a particular chord of a circle, you can maximize the area of the triangle by having the third vertex as far away as possible (area is half base times perpendicular height), which means that it will be on the perpendicular bisector of the chord where it crosses the circle, the other side of the circle's centre. So with this chord as an edge, the area will be maximised if the other two edges are equal, i.e. if the triangle is isosceles.
But the same thing is true of the other two edges, and that implies ... (I will leave you to work this out) |
H: Help in a proof in basic Algebraic Geometry
I'm trying to understand this proof:
Theorem
Let $X$ and $Y$ be affine closed sets. If $f:X\to Y$ is a function such that $g\circ f\in k[X]$ for every $g\in K[Y]$, then $f$ is a morphism.
Proof
Suppose that $f:X\to Y$ is a function such that $g\circ f\in k[X]$, whenever $g\in k[Y]$. We write $f=(f_1,\ldots,f_m)$ and take $g=s_i\in k[Y]$ the regular function corresponding to the polynomial $S_i\in k[S_1,\ldots S_m]$, then $s_i\circ f=f_i\in k[X]$ for each $i=a,\ldots,m$, showing $f$ is a morphism.
Remark
$k[X]$ is the coordinate ring of $X$.
My doubts
Why can we write $f=(f_1,\ldots,f_m)$? Why $s_i\circ f=f_i\in k[X]$?
Thanks in advance.
AI: You can write $f$ in components because $Y$ is a subset of $\mathbb K^m$. The monomial $S_i$ is just the projection onto the $i$th coordinate. |
H: Wreath product of finitely generated groups is finitely generated
Let $G$ and $H$ be two groups generated by finite sets $\Sigma_G$ and $\Sigma_H$, and let $W=G \wr H$ be the wreath product of $G$ and $H$. Show that $W$ is finitely generated by $\Sigma_G \times \{1 \} \cup \{1 \} \times \Sigma_H$.
I cannot prove this. Please advice to me.
AI: By definition, $G \wr H= \left( \bigoplus\limits_{h \in H} G_h \right) \rtimes H$ where $G_h$ is a copy of $G$. Therefore, $W= G \wr H$ is generated by $\Sigma_H \cup \{ \Sigma_{G_h} \mid h\in H \}$, and so by $\Sigma_H \cup \bigcup\limits_{h \in H} h \Sigma_G$ or by $\Sigma_H \cup \bigcup\limits_{h \in \Sigma_H} h \Sigma_G$. Finally, you can deduce that $W$ is generated by $\Sigma_H \cup \Sigma_G$. |
H: Question straight from the SAT
If a coordinate system is devised so that the positive y-axis makes an angle of 60 degrees with the positive x-axis, what is the distance between the points with coordinates (4,-3) and (5,1)?
I'm sure you guys can get it without the multiple choice answers.
Keep in mind the college board expects 18 and 17 year old kids to be able to answer this in 62 seconds on average.
AI: One brute force way to solve this is to note that the change in $x$ values is $1$ and the change in $y$ values is $4$. By considering the triangle formed by the two given points and the point $(5,-3)$, we form a Side-Angle-Side triangle with side $1$, angle $120^\circ$, and side $4$. Applying Cosine Law, we obtain:
$$
c^2 = 1^2 + 4^2 - 2\cdot 1 \cdot 4 \cdot \cos(120^\circ) = 17 - 8(-1/2) = 21 \implies c = \sqrt{21}
$$ |
H: Criterion for proving flatness
I am trying to show that:
$M$ is a flat $A$ module iff for all ideals $I$ of $A$, which are finitely generated, the $A$-linear map
$I \otimes M\longrightarrow M$, taking $x \otimes y\longrightarrow xy$ is injective.
Please help.
AI: Here's a reference: Proposition 3.58 of Rotman's 'An Introduction to Homological Algebra', second edition. |
H: Rationalizing the denominator with 3 roots
Well, I can't find the example on how to solve this.
If I multiply
$$
\dfrac{2}{\sqrt[3]{9}+\sqrt[3]{15}+\sqrt[3]{25}}
$$
with
$$
\dfrac{\sqrt[3]{9}-\sqrt[3]{15}+\sqrt[3]{25}}{\sqrt[3]{9}-\sqrt[3]{15}+\sqrt[3]{25}}
$$
or similar, it just gets even more complicated, and I get 4 terms instead of 3 in the denominator, and more can't be better...
Can someone tell me some principle by which all kinds of expressions with 3 or more terms and with different roots could be rationalised? On a test, I haven't got much time to use appropriate formula, as I can for square difference and so on, I need a principle.
AI: The expression is $$ \frac{2}{\sqrt[3]{9}+\sqrt[3]{15}+\sqrt[3]{25}}$$
Note that $9=3^2,\ 15=3\cdot 5, \ 25=5^2$ so the denominator is of the form $a^2+ab+b^2$. Now you just use the formula $(a-b)(a^2+ab+b^2)=a^3-b^3$ for suitable $a,b$.
In general, when you have a $3$ term denominator, or more, if it does not have a special form (like above) it is unlikely you will be able to rationalize the fraction in one step. Note that when cubic or superior roots are involved, multiplying with an expression where you just change the sign (for example $(a+b+c)(a+b-c)$) of some term doesn't work, since squaring a cubic or superior root does not make the root dissapear.
For superior roots you may also use $a^n-b^n = (a-b)(a^{n-1}+a^{n-2}b+...+b^{n-1})$ and for $n$ odd $a^n+b^n = (a+b)(a^{n-1}-a^{n-2}b+..-ab^{n-2}+b^{n-1})$ (in the last one the signs are alternating). |
H: Finding vector length based on parallell and orthogonal vectors
Do anyone know a simple way of finding the length of vector a in my figure? The known values are $(x_0, y_0), (x_1, y_1), (x_2, y_2), (x_3, y_3)$. (If you look closely, you can see that $f$-vector is a force, and that I need to find the arm between force and point $3$, to calculate the momentum.)
AI: If you know $(x_0,y_0)$ and $(x_1,y_1)$ then you can find the equation of line passes through these points. Suppose you find its equation as $y=mx+n$ then you can also find equation of line passes through $(x_2,y_2)$ because you know its slope $\frac{-1}m$, suppose you found it as $y=\frac{-1}{m}x+k$. Now you know already the slope of line passes through $(x_3,y_3)$ so you can find the equation of it either, suppose it is $y=mx+l$. If you solve the equations $y=\frac{-1}{m}x+k$ and $y=mx+l$ you will find a point $(x_4,y_4)$. After that you can find the length of vector by the following formula $$\sqrt{(x_3-x_4)^2+(y_3-y_4)^2}$$ |
H: $e^z, \ \ z \in \mathbb{C}$ isn't invertible
Could you tell me why the function $\mathbb{R}^2 \ni z \rightarrow e^z \in \mathbb{R}^2$, complex exponential, is not invertible globally. On a horizontal strip $[\ i y, \ i(y+2 \pi))$ its inverse is $\ln z$.
I found this example on this forum, but there is no explanation if this fact.
Or at least it is not explanatory enough for me to understand.
I would really appreciate all your insight.
Thank you.
AI: I think you have to understand that a general function $f : A\to B$ to be invertible need first to be injective. Why that? Well, because recall for each $x\in A$ there must be unique $y\in B$ with $y=f(x)$. If $f$ is not injective, then there are $x_1,x_2\in A$, $x_1\neq x_2$ with $f(x_1)=f(x_2)$. Now, try defining the inverse, you can easily see that if $g : B \to A$ must be such that $g(f(x))=x$, then you have a problem, because you would have $g(f(x_1))=x_1$ and at the same time $g(f(x_2))=x_2$, but $f(x_1)=f(x_2)=y$, so for this $y$ there are two values in $A$, so $g$ isn't a function.
If a function $f: \Bbb C\to \Bbb C$ is periodic, that is there is $k\in \Bbb C$ with $f(z+k)=f(z)$, then it's clearly not injective, hence not invertible. As you can see
$$\exp(z+2\pi i)=\exp(x+iy+2\pi i)=e^x(\cos (y+2\pi) + i\sin(\theta+2\pi))=\exp(z),$$
so it can't be invertible. However all is not lost, when a function is not injective, you can restrict it's domain so that it becomes injective. You simply take away from it's domain the points on which it "duplicates". What this means in this case?
This means that since the function repeats when you add $2\pi i$ to the point, in every horizontal strip of vertical length less than $2\pi$ the function doesn't have duplicate values: it's injective, hence invertible. More precisely, let $\alpha \in \mathbb{R}$, then you let $S_{\alpha}=\{z \in \mathbb{C} : \alpha < \Im(z) < \alpha + 2\pi\}$, in this region $\exp$ is injective and invertible. |
H: Show that $(F^n)^A$ is cyclic as an $F[x]$-module
Problem statement:
Let $A$ be an $n \times n$ matrix over the field $F$ and let $A$ have $n$ distinct eigenvalues. Show that $(F^n)^A$ is cyclic as $F[x]$-module.
I'm not sure I understand the notation $(F^n)^A$. What does this refer to?
Thanks.
AI: I'm fairly sure that the notation means the following. We can turn the vector space $F^n$ into a module over the polynomial ring $F[x]$ by declaring that
$$
x\cdot v=Av
$$
for all vectors $v\in F^n$ (viewed as column vectors so that the matrix multiplication makes sense). Consequently we have to define $x^k\cdot v=A^kv$, and
if $p(x)=\sum_{i=0}^k p_ix^i,$ $p_i\in F$ for all $i=1,2,\ldots, k$, is an arbitrary polynomial, we are forced to define
$$p(x)\cdot v=\sum_{i=0}^k p_iA^iv$$
for all $v\in F^n$. If you have never done this, it is a useful exercise to convince yourself of the fact that this gives, indeed, the space $F^n$ a structure of an $F[x]$-module. This is sometimes stated by saying that $F[x]$ is
a free $F$-algebra. Basically meaning that we can let the monomial $x$ act any which way it wants, and simply extend the action in the obvious way to its powers and then $F$-linearly to all polynomials. The crux is that the powers of $x$ do not satisfy any hidden relations in the ring $F[x]$ (they are linearly independent).
Notice that above we were free to choose the matrix $A$ any which way we wanted.
Using a different matrix in place of $A$ gives a different gives a different module structure. This makes it necessary to tell the reader, which matrix was used. Hence the superscript.
But most likely the notation was explained somewhere. It may have been at an unexpected place. |
H: Some inequality; $(\sum a_i^p)^{1/p} \leq C \sum a_i$
I want to know whether or not the following inequality $$
(\sum_{i=1}^n a_i^p)^{1/p} \leq C \sum_{i=1}^n a_i$$
is satisfied for some $C$, where $0 \leq a_i,\ 1/2 \leq 1/p$.
For some constant $C$, the above inequality hold ?
Thank you.
AI: For $x =(x_i) \in \mathbb{R}^n$, define
$$
\|x\|_p = (\sum |x_i|^p)^{1/p} \text{ and } \|x\|_1 = \sum |x_i|
$$
Consider the unit ball in $\mathbb{R}^n$ given by
$$
B := \{x \in \mathbb{R}^n : \|x\|_1 = 1\}
$$
And define $f:B \to \mathbb{R}$ by
$$
f(x) = \frac{\|x\|_p}{\|x\|_1}
$$
Check that this function is continuous. Since $B$ is compact, it is bounded by some constant $C>0$. This constant does what you want. |
H: The limit at the end-point of an open interval, for a uniformly continuous function
If $f: (a,b)\to\mathbb R$ is uniformly continuous and $\{x_n\}$ in the domain tends to $b$, then why does $\{f(x_n)\}$ have a limit?
AI: Because $x_n$ is Cauchy sequence then $f(x_n)$ is Cauchy sequence due to uniform continuouity. And hence $f(x_n)$ is convergant. |
H: Can somebody explain why the interval $\left ( 0,1 \right )$ is not countable?
I cannot seem to understand the proof of why the interval $\left ( 0,1 \right )$ is not countable.
The proof that is written in my book using the method of Reductio ad absurdum.
It starts with the following statement:
We know that every real number can be written as a decimal.
Let $x_{1} = 0,x_{11}x_{12}x_{13}...$
$x_{2} = 0,x_{21}x_{22}x_{23}...$
$x_{3} = 0,x_{31}x_{32}x_{33}...$
Then by using the diagonial argument it constructs a $y = 0,y_{i}...$ but cannot understand the process that follows.
AI: For each $x_{ii}$, you pick $y_i$ such that $x_{ii}\neq y_i$, and of course, $y_i$ can only take values from $0,1,2,3,4,5,6,7,8,9$.
Then you set $y=0,y_1y_2y_3\dots$, and you'll note that $x_i\neq y$ for all $x_i$. So there is no bijection between $(0,1)$ and $\mathbb N$.
But of course this argument doesn't quite work, because what happens if $x_1=0.500\dots$, and $y_1=4$, $y_i=9$ for all $i\geq 2$. Then I would get $y=0.4999\dots=0.5=x_1$. So to fix this, simply don't let $y_i$ take the values $0$ or $9$. That's because a number has two decimal representations if and only if its decimal representation has repeating $9$ or repeating $0$ after some point. |
H: Let $\, g \,$ be a function defined on $(a,b)$ such that $\, a
I am stuck on the following problem that says :
What I guess option (A) is not possible. But,
I am not sure about the other options. Can someone explain? Thanks in advance for your time.
AI: If $g$ is a constant, then $g(x) = c$ for all $x\in (a,b)$. By hypothesis,
$$
a < c < x \quad\forall x \in (a,b)
$$
This is impossible. Hence $g$ is non-constant; so (B) is true.
(A) Not possible because $g(x) < b$ for all $x\in (a,b)$
(C) Not necessarily, unless $a\geq 0$
(D) No. The graph of $g$ can bounce up and down between $y=a$ and $y=x$.
(E) No. Same reason as (D) |
H: What precisely is the difference between Euclidean Geometry, and non-Euclidean Geometry?
I was wondering, what it is precisely which defines the difference between Euclidean and non-Euclidean Geometry, in a few words/equations/diagrams?
Would I be correct in understanding that non-Euclidean Geometry is just a more relaxed version of Euclidean Geometry?
AI: Let me make an example of something usual in our world, maybe you can find out a sort of differences better. That is the curvature. We are accustomed to think about the flat things on Earth,i.e. the curvature is zero. Making circles, triangles and many many shapes are drew on a piece of a blank paper. Indeed, the curvature of a paper when you put it on a table is $0$. And that is why we learn Ed. Geometry for students on blackboard. This is absolutely OK. But what is the real story? if you want to explain this concept as it is in the real world, you can consider an orange. A surface of an orange (its cover) inside it and outside of it can give us what is the main point about the curvature. Geometrically, when we are working with any spaces with positive curvature,i.e. outside the cover of an orange, if you draw two greater circle, they intersect themselves so, there are no lines parallel to a given line through an outside point (Rejecting the 5-th postulate of Euclidean Geometry).
On the other hand, while in a space of negative curvature, like the surface of an hyperbolic paraboloid or inside the cover an orange , we can draw many lines parallel to a given line through an outside point (Again rejecting the 5-th postulate of Euclidean Geometry). However, when sketching shapes on a blank paper, you are experiencing the zero curvature and inn this case you can easily draw just a single line parallel to a given line through an outside point.
The following figs are from Google's images: |
H: Prove that $[a,b]$ is connected space.
Prove that $[a,b]$ is connected space.
I know that $\mathbb{R}$ with euclidean metric is connected space. I would like find surjective function $f: \mathbb{R} \rightarrow [a,b]$. Because $\mathbb{R}$ is connected and $f$ is surjective function then $[a,b]$ is also connected space. Is it possible to find these function? I can't imagine how can look this function and how can I make a draw?
I tried with function $f(x) = \frac{1}{1+x^2}$ which is bounded ($0<f(x) \le 1$) and scale them such that $a<f(x) \le b$.
AI: Simply define it piecewise:
$$f ( x ) = \begin{cases}
a, &\text{if }x \leq a \\
x, &\text{if }a \leq x \leq b \\
b, &\text{if }x \geq b.
\end{cases}$$ |
H: Solving a first order non-linear ODE
I encountered a problem solving the following equation: $$t^2dy/dt+2ty-y^3=0$$
I have already tried the following steps, but to no avail:
1)Separation of variables
2.1)Integration Factor of 1 variable.
2.2)Integration Factor of 2 variables.
3)Transposing the equation and find a solution for x.
Thank you in advance.
AI: Rewrite the equation as
$$\frac{d}{dt} (t^2 y) = y^3$$
Multiply both sides by $t^6$:
$$t^6 \frac{d}{dt} (t^2 y) = (t^2 y)^3$$
or
$$\frac{d(t^2 y)}{(t^2 y)^3} = \frac{dt}{t^6}$$
Integrating,
$$-\frac12 \frac{1}{(t^2 y)^2} = -\frac15 \frac{1}{t^5} - C$$
Rearrange to get $y(t)$:
$$y(t) = \sqrt{\frac{5 t}{2+5 C t^5}}$$
where $C$ is a constant of integration. |
H: which one is larger $\sqrt[n]{x+\delta}-\sqrt[n]{x}$ or $\sqrt[n]{x}-\sqrt[n]{x-\delta}$?
Which is larger? $\sqrt[n]{x+\delta}-\sqrt[n]{x}$ or $\sqrt[n]{x}-\sqrt[n]{x-\delta}$?
Algebraic justilation does not help.
AI: $\sqrt[n]{x+\delta}-\sqrt[n]{x}\ \boxed{\phantom{A} }\sqrt[n]{x}-\sqrt[n]{x-\delta}$
$\sqrt[n]{x+\delta}+\sqrt[n]{x-\delta}\ \boxed{\phantom{A} }2\sqrt[n]{x}$
$\frac{\sqrt[n]{x+\delta}+\sqrt[n]{x-\delta}}{2}\ \boxed{\phantom{A} }\sqrt[n]{x}$
$\frac{\sqrt[n]{x+\delta}+\sqrt[n]{x-\delta}}{2}\ \boxed{\phantom{A} }\sqrt[n]{\frac{\left(\sqrt[n]{x+\delta}\right)^n+\left(\sqrt[n]{x-\delta}\right)^n}{2}}$
But $\frac{a+b}{2}\leq\sqrt[n]{\frac{a^n+b^n}{2}}$.
$\tiny{\text{some conditions apply}}$ |
H: using $\sin x$ to get a function with range $[a,b]$
$\sin x$ is a nice function on $\mathbb{R}$ whose range is $[-1,1]$. can we 'adjust' it so that its range will be $[a,b]$. By 'adjusting' I mean changing the argument $x$ to some other argument which is a function of $x$ , or multiplying by a constant or adding a constant. For example $2x$, $2x^2$ or $\sin x$ in which case we will have $\sin(2x)$, $\sin(2x^2)$, $\sin(\sin x)$.
Thanks in advance.
AI: First of all, the range of $\sin$ is $[-1, 1]$, not $[0, 1]$. Second, by adjusting the argument as you propose, i.e. composing $\sin$ with another function $f$ on the right and getting $\sin(f(x))$, you can only make the range smaller. You cannot turn it into anything larger, like $[-10, 10]$, because no matter what you put in the argument of $\sin$, the result will always be inside $[-1, 1]$.
UPDATE: Now I've also noticed that you allow multiplying by and adding constants, i.e. composing $\sin$ with an affine function on the left. That can indeed give you the desired range. For instance, the range of $$\frac{b+a}{2} + \frac{b-a}{2}\sin x$$ is $[a, b]$. |
H: Properties of the co-countable topology on $[0,1]$
I am trying to learn topology by myself. I have the following questions and my attempts of their proofs. Since I have no one to consult with, I am posting here. If anyone check my proofs that would be great. I know it is a bit long, my apologies for that. Many thanks for your time in advance!
Let $\tau$ be the system of sets consisting of $\emptyset$ and every subset of the the closed unit interval $[0,1]$ obtained by deleting a finite or countable number of points from $X$. Prove that
$T=(X, \tau)$ is topological space
$T$ satisfies neither the second nor the first axiom of countability
Only convergent sequences in $T$ are stationary sequences i.e., the sequences all of whose terms are the same starting from some index $n$
$M=(0,1]$ has the contact point $0$ as a contact point, but contains no sequence of points converging to $0$
So, for (1):
$\emptyset\in\tau$ and if we delete $0$ points (finite) from $[0,1]$ then $X\in\tau$.
Let $\cup_\alpha G_\alpha$ be open sets, then $G_\alpha=X-F_\alpha$ where $F_\alpha$ is at most countable. Then $\cup_\alpha G_\alpha=\cup (X-F_\alpha)=X-\cap_\alpha F_\alpha$, but $\cap\alpha F_\alpha \subset F_\alpha$ is countable (subset of a countable set is countable). Thus arbitrary union of open sets is in topology. I can similarly show "finite intersection property".
For (2): My idea is if I can show this topology does not satisfy first axiom of countability, then it also fails to satisfy second one (this idea might be wrong, though). Then, if $x\in T$, there is no countable neighbourhood base at $x$ since we have deleted countable points. The problem is I am not sure whether this explanation is true or not, also it does not seem rigorous enough to me.
For (3): I don't understand this claim. For any stationary sequence, I can delete them from my system so there would be no stationary sequence in this topology, but then I can create uncountably many stationary sequence (choose any $x\in T$ as members of the sequence). So a bit confusing, but it is clear that a stationary sequence is convergent. That is, any neighbourhood $G$ of $x$ contains every point after certain index. For contradiction, take any non-stationary sequence $\left\{x_n\right\}\to x$. Take neighbourhood $G$ of $x$ such that after certain index, say $N$, every member of $\left\{x_n\right\}$ deleted, except for $x_n=x$. Than $G$ does not contain any member of $\left\{x_n\right\}$, except $x$.
For (4): Suppose $G$ is neighbourhood of $0$ such that $G\cap M=\emptyset$. Then $G=\left\{0\right\}$ which cannot be an open set in this topology $T$. By using ii), if $x_n\to 0$ then, it should be stationary i.e., $x_n=0$ for $n>N$, which is impossible.
AI: The outline for (1) appears fine.
For (2) you must show for any $x \in X$ and any given family $\{ U_n : n \in \mathbb{N} \}$ of open neighbourhoods of $x$ you can find an open neighbourhood $V$ of $x$ such that $U_n \not\subseteq V$ for all $n \in \mathbb{N}$. As a hint, note that if $U$ is nonempty and $x \in U$, then $U \not\subseteq X \setminus \{ x \}$. (You are correct that if $X$ is not first-countable, then it cannot be second-countable.)
The point of (3) is to show that if a sequence $\langle x_n \rangle_{n \in \mathbb{N}}$ is not eventually constant, then it cannot converge to any point. As a hint, suppose that $\langle x_n \rangle_{n \in \mathbb{N}}$ is not eventually constant. Then the set $A = \{ x_n : n \in \mathbb{N} \}$ is countably infinite. For each $x \notin A$ we can easily construct an open set $U$ which contains no elements of the sequence (so the sequence cannot converge to any point not belonging to $A$). For each $x \in A$ we can make a small adjustment to $U$ to get an open neighbourhood $U^\prime$ of $x$ such that the sequence is not eventually always inside of $U^\prime$.
For (4), note that every open neighbourhood of $0$ must meet the set $M$, and so $0$ is a limit point (contact point) of $M$. By the idea of proving (3) we can show that no sequence of points in $M$ converges to $0$. (For any sequence of points in $M$, construct an open neighbourhood of $0$ containing no points of that sequence.) |
H: Probability: the expected value in a dice game.
If a dice is thrown till the sum of the numbers appearing on the top face of dice exceeds or equal to 100, what is the most likely sum?
AI: This has been computed several times on the site already so let us present some of the underlying theory.
Renewal theory deals with sums $(S_n)$ of i.i.d. nonnegative increments $(X_n)$ with common integrable distribution $\mu$ and asks for the occurrences of these sums just before and just after some given time.
Thus, one sets $S_0=0$, $S_n=X_1+\cdots+X_n$ for every $n\geqslant1$, $\mathcal S=\{S_n\mid n\geqslant0\}$ and, for every nonnegative $t$, $L_t=\max\mathcal S\cap[0,t]$ and $U_t=\min\mathcal S\cap(t,\infty)$.
A standard result of the theory is that, in continuous time, that is, when the distribution $\mu$ is continuous, $U_t-L_t$, $U_t-t$ and $t-L_t$ all converge in distribution. More precisely, $U_t-L_t$ converge in distribution to a size-biased version $\hat X$ of $X_1$, whose distribution has density $x/E[X_1]$ with respect to $\mu$. Furthermore,
$$
(U_t-L_t,U_t-t,t-L_t)\to(\hat X,Z\hat X,(1-Z)\hat X)\quad\text{in distribution},
$$
where $Z$ is uniform on $(0,1)$ and independent on $\hat X$. In particular, the so-called residual waiting time $U_t-t$ has density $g$ with respect to the Lebesgue measure, with
$$
g(x)=\frac{\mu([x,\infty))}{E[X_1]}.
$$
In particular, $g$ is maximum at $x=0$ and $g(0)=1/E[X_1]$.
This suggests that, in the present (discrete) case, the most probable overshoot when $t\to\infty$ is $0$, which happens with probability $6/21=2/7$. True, there are some subtle differences with the continuous setting since one asks about $V_n-n$, where $V_n=\min\mathcal S\cap[n,\infty)$ ($n$ included). But the result above probably carries through and time $t=100$ is probably already large enough for this asymptotics to apply. Anyway, the asymptotic distribution of the overshoot on $\{0,1,2,3,4,5\}$ is
$$
\frac6{21},\ \frac5{21},\ \frac4{21},\ \frac3{21},\ \frac2{21},\ \frac1{21}.
$$ |
H: integrate the following expression
Question:
integrate $\sqrt{t}$
My answer:
$t^{\frac{1}{2}} = \dfrac{t^{\frac{3}{2}}}{\frac{3}{2}} +c$
correct answer:
$\frac{2}{3}.t^{\frac{3}{2}} +c$
what am I doing wrong? thank's
AI: $$\int \sqrt{t} = \int t^{\frac{1}{2}} = \frac{t^{\frac{3}{2}}}{\frac{3}{2}} + C = \frac{2t^{\frac{3}{2}}}{3} + C$$
where $C \in \mathbb{R}$ |
H: Is the empty function always a bijection?
Let $f_A:\emptyset\to A$ be the empty function with range $A$. The definition of a bijection as applied to this function is:
$$\forall x,y \in \emptyset (x=y \implies f_A(x)=f_A(y))$$
negating you get:
$$\exists x,y \in \emptyset (x = y\land f_A(x) \neq f_A(y))$$
Which is obviously a false statement since there are no elements in $\emptyset$ at all.
I got troubled by this question when considering the empty set as an inital object of the category Set and the following theorem:
"if I is an initial object then any object isomorphic to I is also an initial object."
but since every empty function is a bijection and thus an isomorphism it follows that all the objects in Set are initial which is obviously false.
What did i miss?
AI: Your symbolic definition of bijectivity is incorrect. The condition you wrote holds because $f_A$ is a function (vacuously in the case of the empty set as domain). A function $f:A\rightarrow B$ is injective (resp. surjective) if and only if $f(x)=f(y)$ implies $x=y$ (resp. for each $z\in B$ there exists $x\in A$ with $f(x)=a$). The unique function from the empty set to any other set is injective, but can be surjective if and only if the target is empty as well. |
H: Family of sets which intersect isn't connected
Find any family of sets $A_n$ such that $A_n$ are connected sets and $A_{n+1} \subset A_n$ and $$\bigcap_{n=1}^{\infty}A_n$$ is not connected.
I tried find family of sets such that $$\bigcap_{n=1}^{\infty}A_n = (-1,0) \cup (0,1)$$ Of course it isn't connected but I don't have any idea how find family of sets.
AI: You can’t do it with subsets of $\Bbb R$: connected subsets of $\Bbb R$ are intervals, and the intersection of nested intervals is an interval or empty. You can, however, do it in the plane.
For $n\in\Bbb Z^+$ let $$A_n=\left([0,1]\times\left(0\frac1n\right]\right)\cup\{\langle 0,0\rangle,\langle 1,0\rangle\}\;;$$ each $A_n$ is connected, but $$\bigcap_{n\in\Bbb Z^+}A_n=\{\langle 0,0\rangle,\langle 1,0\rangle\}\;.$$ |
H: integrate the following expression again
Question:
$x - x^2 + 1$
My answer:
$\frac{x^2}{2} - \frac{x^3}{3} + {x} + C$
Correct answer:
$\frac{x^2}{2} - \frac{x^2}{3} - \frac{x^3}{3} + x + C$
What am I doing wrong? thanks
AI: The general formula is
$$-1\neq n\in\Bbb Z \implies \int x^ndx=\frac{x^{n+1}}{n+1}+C$$
Instead of $\;\Bbb Z\;$ we can in fact take $\;-1\neq n\in\Bbb R\;$ .
and , of course, using linearity of integral:
$$\int (f(x)+g(x))dx=\int f(x)dx+\int g(x)dx$$
and the above can be extended to any (finite) number of summands. |
H: Approximating continuous functions with growth condition by lipschitz functions
I wonder about the following. Is it possible to approximate $x^p$ for $x\in\mathbb{R}_+$ with lipschitz continuous functions? If so, is it possible to approximate it in a dominating way, i.e. $f_n\to x^p, f_n$ lipschitz and $f_n\ge x^p$? The convergence does not have to be uniform, pointwise would be fine.
If we need additional assumption, please let me know. Also any reference would be appreciated.
AI: Approximation by below ($f_n(x)\leqslant x^p$) can be obtained, taking $f_n(x):=\min\{n,x^p\}$.
But it's not possible for approximations by above:
$$\frac{f_n(n)-f_n(0)}{n-0}\geqslant \frac{n^p-f_n(0)}{n}\to \infty$$
if $p\gt 1$. |
H: Solving a probability equation.
One group of 30 people won a contest and as a reward they got a free vacation to Hawaii. The hotel they will be in has 10 rooms with 3-bed (bed for 3 people).
The question: How many ways can we deploy the students?
I used combinations because the students can change each other places, so I got:
$$
10 \cdot C_{30}^3,
$$
where $C_{30}^3 = \binom{30}{3}$.
Is it correct?
AI: If you’re trying to count the different ways in which the $30$ students can be distributed amongst the $10$ rooms, it’s not correct. There are $\binom{30}3$ ways to choose the $3$ students who are to occupy the first room. Once they’ve been chosen, only $27$ students remain, so there are $\binom{27}3$ ways to choose $3$ of them for the second room. Continuing in this fashion, we see that there are
$$\begin{align*}
\binom{30}3\binom{27}3\dots\binom33&=\frac{30!}{3!\,27!}\cdot\frac{27!}{3!\,24!}\cdot\ldots\cdot\frac{3!}{3!\,0!}\\
&=\frac{30!}{\underbrace{3!\cdot3!\cdot\ldots\cdot3!}_{10}}\\
&=\frac{30!}{3!^{10}}\\\\
&=\binom{30}{3,3,3,3,3,3,3,3,3,3}\;,
\end{align*}$$
a multinomial coefficient. |
H: What is the difference between these two given sums?
What is the difference between this: $\sum_{i=1}^n(x^i+9x*i)$ and $\sum_{i=0}^n x[x^i+9i+9]$ ?
So far I know that the first terms are not different.
The first term of the $\sum_{i=1}^n(x^i+9x*i)$ is $\ x+9x$.
And the first term for $\sum_{i=0}^n x[x^i+9i+9]$ is $\ x*x^0+9x*0+9x$ which equals $\ x+9x$.
What I'm unable to figure out is the difference between the sums still. I think that the last term of the first sum is $\ x^n+9x(n)$ and the last term of the second term is $\ x^{n+1}+9n(x)+9x$. But I don't know if that is correct, or even if it is correct where to go from this point.
AI: To get you started: The second sum is $\sum_{i=0}^n (x^{i+1}+9x(i+1)).$ Notice that if $0\leq i<n$, then the $i$th term of that sum equals the $(i+1)$st term of the first sum in your question. So some cancellation occurs in the difference...
Added: You already observed what I said in the case $i=0$: The $0$th term in the second sum is $x+9x$, and the $1$st term in the first sum is $x+9x$. So in the difference, these terms cancel. For $i=1$, the $1$st term in the second sum is $x^2+9x(2)$, while the $2$nd term in the first sum is $x^2+9x(2)$. So these terms cancel in the difference. For $i=2$, the $2$nd term in the second sum is $x^3+9x(3)$ while the $3$rd term in the first sum is $x^3+9x(3)$. So these cancel, etc. Of course this doesn't go on indefinitely. There is one term in the second sum leftover. |
H: Which is larger? $20!$ or $2^{40}$?
Someone asked me this question, and it bothers the hell out of me that I can't prove either way.
I've sort of come to the conclusion that 20! must be larger, because it has 36 prime factors, some of which are significantly larger than 2, whereas $2^{40}$ has only factors of 2.
Is there a way for me to formulate a proper, definitive answer from this?
Thanks in advance for any tips. I'm not really a huge proof-monster.
AI: It is probably easier to note that $2^{40} = 4^{20}$. The only ones of the 20 factors in $20!$ that are smaller than $4$ are $1$, $2$ and $3$. But, on the other hand, $18$, $19$ and $20$ are all larger than $4^2$, so we can see
$$ 20! = 1\cdot 2\cdot3\cdots 18\cdot19\cdot 20 > \underbrace{1\cdot1\cdot1}_{3\text{ ones}}\cdot\underbrace{4\cdot4\cdots 4\cdot 4}_{14\text{ fours}}\cdot\underbrace{16\cdot16\cdot 16}_{3\text{ sixteens}} = 2^{40} $$
by comparing factor by factor. |
H: Help needed with complements, partition and power sets
I'm working on some tasks which is listed below, and I'm trying to figure out if I've understood partition, power set, and complements correctly.
Here are the tasks:
Assume that $\{$$1, 2, 3, 4, 5$$\}$
What is the complement of the amount $\{$$1, 2, 3$$\}$?
What is $P(${$1, 4$}$)$ ie power set of the amount {$1, 4$$\}$?
Give a partition of the amount {$a, b, c, d, e, f$} so that {$a, b, c, d$} is included
Give a partition of the amount {$1, 2, 3, 4$} which has two elements
Why isn't $\{$$\{$$1, 2$$\}$, $\{$$2, 3$$\}$$\}$ a partition of {$1, 2, 3$$\}$?
Is it always the case that $X \in P(X)$, no matter what the $X$ is?
Here is my answers:
$\{$4, 5$\}$
$\{$$\emptyset$, $\{$1$\}$, $\{$4$\}$, $\{$1, 4$\}$$\}$
Not sure
Not sure
Not sure
Not sure
I would appreciate If someone could go through the tasks which I've done and see if they are correct. Would also appreciate help with $(3)$, $(4)$, $(5)$ and $(6)$.
Thank you.
AI: The first two are correct. The wording in (3) is very strange — I suspect that you’re translating — but I think that it’s asking for a partition of the set $\{a,b,c,d,e,f\}$ that has $\{a,b,c,d\}$ as one of its parts. There are two such partitions: $$\big\{\{a,b,c,d\},\{e,f\}\big\}\;,$$ and $$\big\{\{a,b,c,d\},\{e\},\{f\}\big\}\;.$$ Either is a correct answer; the crucial thing is that each member of $\{a,b,c,d,e,f\}$ must appear in exactly one member of the partition.
There are lots of two-element partitions of $\{1,2,3,4\}$; here are just a few of them:
$$\begin{align*}
&\big\{\{1\},\{2,3,4\}\big\}\\
&\big\{\{1,2\},\{3,4\}\big\}\\
&\big\{\{1,2,3\},\{4\}\big\}\\
&\big\{\{1,3,4\},\{2\}\big\}
\end{align*}$$
A partition of $\{1,2,3\}$ is a collection of non-empty subsets of $\{1,2,3\}$ such that each member of $\{1,2,3\}$ is in exactly one of the subsets. $\big\{\{1,2\},\{2,3\}\big\}$ is not a partition of $\{1,2,3\}$ because the element $2$ belongs to two members of the collection: $2\in\{1,2\}$ and $2\in\{2,3\}$. Another way to say it is that the members of a partition must be disjoint from one another, and $\{1,2\}$ and $\{2,3\}$ are not disjoint: they have $2$ in common.
The answer to the last one is yes: if $X$ is any set, then $X\subseteq X$, so by definition $X\in\wp(X)$. |
H: Can you get any irrational number using square roots?
Given an irrational number, is it possible to represent it using only rational numbers and square roots(or any root, if that makes a difference)?
That is, can you define the irrational numbers in square roots, or is it something much deeper than that? Can pi be represented with square roots?
AI: The smallest set of numbers closed under the ordinary arithmetic operations and square roots is the set of constructible numbers. The number $\sqrt[3]{2}$ is not constructible, and this was one of the famous greek problems: the duplication of the cube.
If you allow roots of all orders, then you're talking about equations that can be solved by radicals. Galois theory explains which equations can be solved in this way. In particular, $x^5-x+1=0$ cannot. But it clearly has a real root. |
H: Partial derivatives of a multivariable function
$f(tx,ty)=t^5f(x,y)$ for all values of $x, y, t$ where both functions are differentiable.
Show that
$$a)\ xf_x+yf_y=5f$$
$$b)\ \ x^2f_{xx}+2xyf_{xy}+y^2f_{yy}=20f$$
Clearly, there is differentiating of the initial equation in order to get to a. And a double differentiation to get to b. But, I am confused at differentiating it. I think it needs the chain rule but the composition of the different variables is confusing me.
a) $$\frac{\partial f(tx, ty)}{\partial t}=f_x*\frac{d(tx)}{dt}+f_y*\frac{d(ty)}{dt}=xf_x+yf_y$$
AI: In order to get the first equality differentiate the equality $f(tx, ty)=t^5f(x, y)$ with respect to $t$
and then put $t=1$. For the second one differentiate w.r. t. $t$ twice and put $t=1$.
Differentiate w.r.t. Let $u=tx$ and $v=ty$. Then differentiating either side of the inequality w.r.t. $t$ along with the use of chain rule gives: $f_u u'(t)+f_v v'(t)=5t^4f(x, y)=xf_x+yf_y=5f$. Notice that we put $t=1$ in the last equality and $u'(t)=x, v'(t)=y$. Differentiating $f_u u'(t)+f_v v'(t)=5t^4f(x, y)=xf_x+yf_y=5f$ w.r.t. $t$ we have $\big(f_{uu}u'(t)+f_{uv}v'(t)\big)x+\big(f_{vu}u'(t)+f_{vv}v'(t)\big)y=20t^3f(x, y)$. For $t=1$ you get the second equality. Observe that when $t=1$ you have $f_{uu}=f_{xx}$, $f_{uv}=f_{xy}=f_{yx}$ and $f_{vv}=f_{yy}$ and remember $u'(t)=x$ and $v'(t)=y$. |
H: Convert this scenario into algebra equation
Sales for the month minus the VAT @ 20% = (x).
20% of (x) is profit margin (y).
5% of (y) is commission earned (c).
How can I write an equation that demonstrates the above please?
I.e x - 20% of y - 95% of y = c. (Sorry, not explained well). Thanks.
AI: Why not : y = 0.2 x and c = 0.05 y. Then c = 0.05 (0.2 x) = 0.01 x |
H: For which t is the matrix invertible?
$$\begin{matrix} t&a_2&0&0&\cdots&0\\
0&t&a_3&0&\cdots&0&\\
\vdots&\vdots&\ddots&&\cdots&\vdots\\
0&0&\cdots&&t&a_n\\
a_1&0&\cdots&0&\cdots&t \end{matrix}$$
For what values of t is this matrix invertible? Thanks in advance for any help!
AI: Develop the determinant along the first column: you get
\begin{align}
&
t\det\,\begin{bmatrix}
t&a_3&\cdots&0&\\
\vdots&\ddots&\ddots&\vdots\\
0&\cdots&t&a_n\\
0&\cdots&\cdots&t \end{bmatrix}
+(-1)^{n+1}a_1\det\,
\begin{bmatrix}
a_2&0&\cdots&0\\
t&a_3&\cdots&0&\\
\vdots&\ddots&\ddots&\vdots\\
0&\cdots&t&a_n\\
\end{bmatrix}
\end{align} |
H: Combinatorics question about downsets
Prove that if
$\mathcal{A}$
is a downset then the average size
of sets in
$\mathcal{A}$
is at most
$\frac{n}{2}$
($\mathcal{A} ⊂ \mathcal{P}(n)$ is a
downset
if, for every
$A∈\mathcal{A}$
, every subset
of
$A$
belongs to
$\mathcal{A}$)
Have tried using induction but got stuck at the inductive step. Any help greatly appreciated!
AI: This might be using a cannon to kill a fly, but...
It can be proved that it is possible to partition $\mathcal{P}(n)$ into symmetric chains. Here, and chain $A_1,A_2,\dots, A_k$ is a sequence of sets such that $A_1 \subsetneq A_2 \subsetneq \dots$, and it is symmetric if $\# A_i + \# A_{k+1-i} = n $ for any $i$. The claim is not extremely difficult, and you can do it by induction on $n$. [For a proof of this, see e.g. these notes, p.78, Thm 8.3]
Knowing this, the solution is easy. Consider any chain $A_1,A_2,\dots, A_k$. Because $\mathcal{A}$ is assumed to be a downset, whenever $A_l \in \mathcal{A}$ then also $A_i \in \mathcal{A}$ for $i < l$. Hence, there is some $l$ such that the sets from the chain which belong to $\mathcal{A}$ are precisely $A_1,A_2,\dots, A_l$. Pairing $A_i$ with $A_{l+1-i}$ we see that on average, $A_i \in \mathcal{A} $ have at most $n/2$ elements. Averaging this over all chains in the partition (and using the fact that the chains are disjoint), we conclude that the average size of $A \in \mathcal{A} $ is at most $n/2$. |
H: Being Symmetrical in Limit
The question is given $f_n : \mathbb{R} \to \mathbb{R}$ a sequence of symmetrical functions around a constant point $c$. Is it true that the sequence converges to a symmetrical function around $c$?
I think the question can be modified in the following way. define $g_n(x)=f_n(x+c)$. Then $g_n$ will be the sequence of even functions. But still I don't know if this holds that for a sequence of even functions that have a limit, the limit is also even. Is this true?
AI: Some sequences of functions do not converge at all. But supposing $g(x)=\lim_{n\to\infty}g_n(x)$ exists for every value of $x$, and each $g_n$ is an even function, then the limit $g$ is also an even function.
$$
g(-x) = \lim_{n\to\infty} g_n(-x) = \lim_{n\to\infty} g_n(x) = g(x).
$$
The two sequences, $\lim_{n\to\infty} g_n(-x)$, and $\lim_{n\to\infty} g_n(x)$, are actually both the same sequence of numbers, so the limit has to be the same in both cases. |
H: What does $|A|$ denote in set notation?
What does $|A|$ of a set $A$ denote?
Also, what does $A\leftrightarrow B$ of sets $A, B$ mean?
I encountered this in one of my textbooks which said:
Of two sets $A, B$ we know $|B|$ but $|A|$ is unknown. If we succeed
in constructing a bijection $A\leftrightarrow B$, then$|A|=|B|$. A proof which shows
$|A|=|B|$ by such an explicit construction is called a bijective proof
or combinatorial proof.
I'm afraid I'm new to set notation. I would be grateful if someone could point me to a good resource on it.
AI: The notation $|A|$ often means the number of members of the set $A$, the "cardinality".
$A\leftrightarrow B$ might mean that there is a bijection --- a one-to-one correspondence --- between the two sets $A$ and $B$. |
H: The number of cyclic subgroups of order 15 in $\mathbb{Z}_{30} \oplus \mathbb{Z}_{20}$
What is the number of cyclic subgroup of order 15 in $\mathbb{Z}_{30} \oplus \mathbb{Z}_{20}$?
I have counted the number of elements with order 15 in $\mathbb{Z}_{30} \oplus \mathbb{Z}_{20}$ that is $48$, by counting all possibilities of $(a,b)$ where $a\in\mathbb{Z}_{30}$ , $b \in \mathbb{Z}_{20}$ and $\def\lcm{\operatorname{lcm}}\lcm(|a|,|b|)=15$.
My argument is that for an element that has order $15$, say $(x,y)$, then
$\langle(x,y)\rangle = \langle(x,y)^2\rangle = \langle (x,y)^4\rangle = \langle (x,y)^7\rangle = ...=\langle (x,y)^{14}\rangle$
So there are $8$ different elements with order $15$ such that the generate the same subrgroup. So my answer is that there are $\frac{48}{8}=6$ different cyclic subgroups of order 15,
am i correct? or maybe there are more simpler ways to solve this problem?
AI: Provided you correctly counted the elements of order$~15$, your answer is correct. You can indeed count cyclic subgroups by counting their generators (elements or order$~n$) and dividing by the number $\phi(n)$ of generators per cyclic subgroup, since every element of order$~n$ lies in exactly one cyclic subgroup of order$~n$ (the one that it generates).
Here is how I would count the elements of order$~15$. By the Chinese remainder theorem one has $\newcommand\Z[1]{\Bbb Z_{#1}}\Z{30}\cong\Z2\oplus\Z3\oplus\Z5$ and $\Z{20}\cong\Z4\oplus\Z5$, so all in all we are dealing with the group $\Z2\oplus\Z4\oplus\Z3\oplus\Z5^2$. To have order $15$, an element must have a trivial (zero) component in $\Z2$ and $\Z4$, in $\Z3$ it must have as component one of the $2$ generators, and it's component in $\Z5^2$ must be any one of the $24$ nonzero elements. Indeed you get $2\times24=48$ elements of order$~15$. |
H: Meaning of the set $\mathbb N^\mathbb N$
I came across a question which requires one to check if there's a bijection from the set $ \mathbb N^\mathbb N$ to another set. I've never seen a set defined this way and was wondering if this was just a typo. Could anyone clarify this for me? If this is correct, then what does the set actually mean? I mean, I can think of elements in $\mathbb R$ or in $\mathbb N$, but how do I think of elements in $\mathbb N^\mathbb N$?
AI: The elements of $A^B$ are function of the form $f\colon B\to A$. That is function whose domain is exactly $B$ and their values are elements from $A$.
If so, $\Bbb{N^N}$ is the set of all functions from $\Bbb N$ to itself. Or the set of infinite sequences of natural numbers (as a sequence is just a function).
As a footnote remark, I should add that this set is sometimes denoted by ${}^BA$ instead of $A^B$, especially in contexts were exponentiation is used in set theory. |
H: last three digits of $7^{100}-3^{100}$
How can I find the las three digits of $7^{100}-3^{100}$ ? I know one way is to use $7^{100}=(10-3)^{100}=\sum_{k=0}^n{100 \choose k}10^{100-k}(-3)^k$ but I'm totally stuck...
Thanks
AI: Well, note that for $0\le k\le 97,$ we have that $1000$ is readily a factor of $\binom{100}{k}10^{100-k}(-3)^k,$ so the last three digits of all of those numbers will be $0$s. Hence, the last three digits of $7^{100}$ will be the last three digits of $$\binom{100}{98}10^{2}(-3)^{98}+\binom{100}{99}10^1(-3)^{99}+\binom{100}{100}10^{0}(-3)^{100},$$ that is, of $$\binom{100}{98}10^{2}(-3)^{98}+\binom{100}{99}(-3)^{99}+3^{100}.$$ Hence, the last three digits of $7^{100}-3^{100}$ will be the last three digits of $$\binom{100}{98}10^{2}(-3)^{98}+\binom{100}{99}10^1(-3)^{99}.$$
Since $\binom{100}{98}=50\cdot99$ and $\binom{100}{99}=100$, the last three digits of this number are easily found. |
H: Finding a moment generating function given E(X) and E(x^2)
I am trying to find the moment generating function.
It takes values in the set {0,1,2} with moments
E(X) = 1 and E($X^{2}$) = $ \frac 3 2 $
I know then that M'(0) = 1 and M"(0) = $\frac 3 2 $
I have read through my course notes/ textbook and have found nothing.
If anyone has an idea where I can go with what I have, I would appreciate it.
AI: Hint:
Remember that the MGF of a random variable $X$ is defined by
$$
M_X(t):=\mathbb{E}[e^{itX}].
$$
In this particular case, if $X$ only takes values on $\{0,1,2\}$, then this can be rewritten as
$$
M_X(t)=1\cdot P(X=0)+e^{it}\cdot P(X=1)+e^{i2t}P(X=2).
$$
So, what you need to find are these three probabilities.
Now, you know that
$$
1=\mathbb{E}[X]=0\cdot P(X=0)+1\cdot P(X=1)+2\cdot P(X=2),
$$
and
$$
\frac{3}{2}=\mathbb{E}[X^2]=0^2\cdot P(X=0)+1^2\cdot P(X=1)+2^2\cdot P(X=2).
$$
Since we also know that $P(X=0)+P(X=1)+P(X=2)=1$, you have a system of three equations in three unknowns (namely, the probabilities you want to find). If you can find those, then you can substitute them in to your expression for the MGF to finish the problem. |
H: Computation of integral $\int_{0}^{1}\ln(p)\ln(1-p)p^{2}\,dp$
I want to compute this integral:
\begin{equation*}
J=\int_{0}^{1}\ln(p)\ln(1-p)p^{2}dp
\end{equation*}
It will be great if you can detail the proof.
I tried to do change of variable it does not work, and also integration by part.
Thanks.
AI: Change variables: $p=\mathrm{e}^{-t}$
$$
J = \int_0^1 \ln(p) \ln(1-p) p^2 \mathrm{d}p = \int_0^\infty t \mathrm{e}^{-3t} \ln\left(1-\mathrm{e}^{-t}\right) \mathrm{d}t = \int_0^\infty t \mathrm{e}^{-3t} \sum_{k=1}^\infty \left(- \frac{1}{k} \mathrm{e}^{-k t}\right) \mathrm{d}t
$$
Using
$$
\int_0^\infty t \mathrm{e}^{-(k+3) t} \mathrm{d}t = \left. - \frac{1+(k+3)t}{k+3}\mathrm{e}^{-(k+3)t} \right|_0^{\infty} = \frac{1}{(k+3)^2}
$$
Hence:
$$\begin{eqnarray}
J &=& - \sum_{k=1}^\infty \frac{1}{k} \frac{1}{(k+3)^2} = - \sum_{k=1}^\infty \left( \frac{1}{3} \frac{1}{(k+3)^2} - \frac{1}{9k} + \frac{1}{9(k+3)}\right) \\ &=& - \frac{1}{3} \sum_{k=1}^\infty \frac{1}{(k+3)^2} +\frac{1}{9} \sum_{k=1}^\infty \left( \frac{1}{k} - \frac{1}{k+3}\right) \\
&=& - \frac{1}{3} \left( \underbrace{\sum_{k=1}^\infty \frac{1}{k^2}}_{\zeta(2) = \pi^2/6} - \frac{1}{1^2} - \frac{1}{2^2} - \frac{1}{3^2} \right) +\frac{1}{9} \underbrace{\sum_{k=1}^\infty \left( \frac{1}{k} - \frac{1}{k+3}\right)}_{\frac{1}{1} + \frac{1}{2} + \frac{1}{3}} = \frac{71}{108} - \frac{\pi^2}{18}
\end{eqnarray}
$$ |
H: Why is $(2+\sqrt{3})^{50}$ so close to an integer?
I just worked out $(2+\sqrt{3})^{50}$ on my computer and got the answer
$39571031999226139563162735373.999999999999999999999999999974728\cdots$
Why is this so close to an integer?
AI: Let $x=(2+\sqrt{3})^{50} + (2-\sqrt{3})^{50}$
$x$ is clearly an integer, since all terms with $\sqrt{3}$ cancel.
Notice that $0<2-\sqrt{3}<1$, so $0<(2-\sqrt{3})^{50}\ll1$
So $(2+\sqrt{3})^{50} =x-(2-\sqrt{3})^{50}\approx x$. |
H: Can a vector of a-a, still be used as a point of direction even if it does equate to 0?
In my maths class we had a similar problem where we had to find the solution to get to Y from X. The answer is c however I argued that it could also be a-a, because technically you can go up to B from X by 1a and then down to Y by -1a, but apparently I was wrong, although I don't see why. Is X to Y simply c, or is my point (a-a) still technically correct, even if it does equate to 0.
AI: Bear in mind that $-1a$ will take you from $B$ back to $X,$ so $a-a$ takes from from $X$ to $X$.
We can get from $X$ to $Y$ directly by $c,$ or indirectly by $a+b$ (or many other, more complicated combinations, all of which simply amount to $c$).
Bear in mind that a vector is not just a magnitude (distance), but a direction. So, while it appears that $a$ and $b$ have the same magnitude by the drawing, the vector $a$ will not take us from $B$ to $Y$. The vector $b$ will do that, which is why $a+b$ does the trick (we could think of it as taking us from $X$ to $B$ to $Y$, or from $X$ to $Z$ to $Y$), but note that $a+b=c,$ so the simplest answer is $c$. |
H: Maximal elements in a set
Prove that if there are two maximal elements in a partially ordered set, then these maximal elements are not comparable.
I understand that I should show that if there are two maximal elements in a set, then they must be equal or not compareable. How should I write it down exactly?
Thanks
AI: If $m$ and $n$ are two maximal elements that are comparable then $m\leq n\wedge n\leq m$ leading to $m=n$. So if they are comparable then they cannot be distinct. |
H: Proving $\mathcal P(A-B) = \mathcal P (A) -\mathcal P (B)$
I started with trying to prove $\mathcal P(A-B) \subseteq \mathcal P (A) - \mathcal P (B)$
$x \in \mathcal P (A) - \mathcal P (B)$ that mean
$x \in \mathcal P (A) \wedge x \notin \mathcal P (B)$ means
$x \in A \wedge x \notin B$ meaning
$x \in A-B$ therefore $x \in \mathcal P (A) - \mathcal P (B)$
And using the same arguments for $\mathcal P (A) - \mathcal P (B) \subseteq \mathcal P(A-B) $
Am I on the right way?
Thanks
AI: I'm afraid your reasoning has gone awry. $x\in\mathcal P(A)\wedge x\notin\mathcal P(B)$ means that $x\subseteq A\wedge x\nsubseteq B$. From this, though, we cannot conclude that $x\subseteq A\setminus B$--to do that, we would need to know that $x$ and $B$ were disjoint--so we can't conclude that $x\in\mathcal P(A\setminus B)$.
Moreover, you're working in the wrong direction to show that $\mathcal P(A\setminus B)\subseteq\mathcal P(A)\setminus\mathcal P(B).$ You'd need to start with an element of $\mathcal P(A\setminus B).$
However, regardless of $A$ and $B$, we will always have $\mathcal P(A\setminus B)\nsubseteq\mathcal P(A)\setminus\mathcal P(B),$ since $\emptyset\in\mathcal P(A\setminus B),$ but $\emptyset\notin\mathcal P(A)\setminus\mathcal P(B).$
Moreover, it is rarely the case that $\mathcal P(A\setminus B)\supseteq\mathcal P(A)\setminus\mathcal P(B).$ Consider $A=\{1,2,3\}$ and $B=\{1,2\},$ for example. In general, if $A$ has more than two elements, then for any $B\not\supseteq A,$ we have that $\mathcal P(A)\setminus\mathcal P(B)$ has strictly more elements than $\mathcal P(A\setminus B),$ and so the latter cannot be a superset of the former. |
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