Weird stuff happening with complex numbers on a ti-84 So, I'm trying to do some calculations for my Electrical engineering homework. This requires a bit of algebra with complex numbers. I have been finding that some of the calculations that my calculator has been making don't really make sense. This is especially apparent if I try to enter complex numbers into my calculator in their polar form. Correct me if I'm wrong, but it my understanding of switching back and forth between rectangular and polar forms of complex numbers is the following : A*e^(i*x) => A cos(x) + i sin(x)

However, if I input (while the calculator is in degrees mode, and it is set up to output complex quantities in rectangular form) e^(i*45) and press enter the output is : .525+.851 * i. My understanding of complex numbers leads me to believe that the output should have been: cos(45)+sin(45)*i = .707 + .707*i.  Also, if I put the calculator in polar mode and input e^(i*45) the output is: 1e^(58.31). Why is it giving me a different output? What is going wrong here? Also, it is saying that e^(i*100)/e^(i*50)= e^(i*-15.211). Also an input of 1+i yields an output of 1.414*e^(i*45)...which is correct. Any idea what's going on here? Is the ti-84 just garbage when it comes to complex algebra in polar form?

Your calculator won't work pleasantly with degrees in this case. It likely defines $e^{ix}$ as the power series $$e^{z} := \sum_{n=0}^\infty \frac {z^n}{n!}$$ In the case of $z=ix$, this simplifies to $\cos(x) + i\sin(x)$ by looking at the respective power series. These latter terms are, of course, in terms of radians; but changing 'radians' to 'degrees' doesn't change the above power series (or else your calculator would screw up, say, $e^2$ and instead output $e^{\pi/90}$ - trying to convert the entire exponent from 'radians' to 'degrees', even though they don't apply here!) So if you're doing things in degrees, you'll either need to plug into $\cos(x) + i\sin(x)$ directly, or first convert into radians.

The gentleman that suggests going to radians always is wrong. EEs use degrees many places, not radians. For instance when dealing with phasors. TI-84 is not good at handling that.

What you have to do if you want to express $45$ degrees as $\large e^{(\text{degree_angle}\cdot i)}$ is by entering this at your calculator: $\large e^{(\text{degree_angle}\cdot2\pi/360)}$

To check that it is true, try this on your $Ti-84$:
$e^{120i} = 0,814+0,580i$
Then to ehcek you result, use the angle from the CPX menu:
$\text{angle}(0,814+0,580i) = 35,49 $---- well thats plain WRONG!

What calculator is better at handling this? I messed up on the exam due to the unreliable handling of $e^{xi} $with the $TI-84$


Reference for "If $(a_n)$ is a sequence of real numbers with $\lim a_n^2=0$, then also $\lim a_n=0$"? Could you please give me a good reference as to the below-quoted real analysis result?

  If $(a_n)$ is a sequence of real numbers with $\lim a_n^2=0$, then also $\lim a_n=0$

By definition of a limit, for all $\varepsilon > 0$ there exists $N \in \mathbb N$ such that $|a_n^2 - 0| < \varepsilon$ for all $n > N$.

Then we can see that the same holds for $a_n$: for any $\varepsilon > 0$, we know that for $\varepsilon_1 = \varepsilon^2$ there exists $N \in \mathbb N$ such that $a_n^2 < \varepsilon_1\hspace{0.2cm}\forall n>N$, and this gives us $|a_n| < \varepsilon\hspace{0.2cm}\forall n>N$.

I am not sure, but if $\lim a_n=a,$ then if you divide $\lim a_n^2$ by $a,$ you get$$\frac{\lim a_n^2}{\lim a_n}=\lim\frac{a_n^2}{a_n}=\lim a_n=\frac{0}{a}=a.$$If you multiply both sides by $a,$ you get $a=0,$ but if $a=0,$ you can't multiply both sides by $a$. Since$$\lim_{\varepsilon\to0}\frac{\varepsilon}{a}=a\implies a^2=\lim_{\varepsilon\to0}\varepsilon,$$$a$ must converge to $0$.


How can I calculate this limit? ( I tried l'Hopital and failed ) I have to calculate this : $$ \lim_{x\to 0}\frac{2-x}{x^3}e^{(x-1)/x^2} $$ Can somebody help me?

Hint: It may be fruitful to substitute $\alpha = 1/x$, in which case you obtain the limit

$$ \lim_{ \alpha \rightarrow \infty} \left(2 - \frac{1}{\alpha} \right) \alpha^3 e^{\alpha - \alpha^2} $$

I should note that, here, I'm taking your limit to in fact be the limit as $x$ approaches $0$ from the positive direction. If you're intending for your limit to be two-sided, then you should think about why that would cause problems.

$$\lim_{x\to0}(f(x)g(x)) = \lim_{x\to0}(f(x)) \cdot \lim_{x\to0}(g(x))  $$

With that being said you can let $f(x) =  (2-x)/x^3 $ and $ g(x) = e^{(x-1)/x^2}  $

I hope this helps.


Determing the order when moving to the other side of the equation. Sorry if the title is vague. But I don't really know the terms I'm looking for ( Nor do I know any tags that go with it). I usually end up being confused with the little( and perhaps trivial things ) in math. The following for example:

to

What are the rules that demand it's a.b / ||a||||b|| and not ||a||||b|| / a.b ?

Perhaps this is a silly question. But, What are the rules when moving from one side of the equation to the other?

The general rule is, you can do anything to one side of an equation, so long as you do the same thing to the other side. So, if you have the equation, $$r=st$$ you can multiply both sides by $1/s$ (assuming that $s\ne0$) to get $$(r)(1/s)=(1/s)st$$ which becomes $${r\over s}=t$$ There's no way you can go from $r=st$ to $s/r=t$ by doing the same thing to both sides of the equation.

$x = y \cos \theta$

$x/y = \cos \theta$

$\arccos (x/y) = \theta$

Edit:

$x = y \times z$

$x \div y = z$


Prob 33, Ch. 3 - Two proofreaders reading a book with n typos - Discrete probability distributions, Blitzstein and Hwang I would like someone to verify my result to parts (a) and (b) of this problem.

  A book has $n$ typos. Two proofreaders Prue and Frida independently read the book. Prue catches each typo with probability $p_{1}$ and misses it with probability $q_{1}=1-p_{1}$, and likewise for Frida, who has probabilities $p_{2}$ if catching and $q_{2}=1-p_{2}$ of missing each typo.

  Let $X_{1}$ be the number of typos caught by Prue, $X_{2}$ be the number of typos caught by Frida and $X$, be the number of typos caught by atleast one of the proofreaders.

  (a) Find the distribution of $X$.

Solution.

Define an r.v. $I_{p}$.

$$\begin{align}
I_{p}&=1 \text{, if Prue catches an error}\\
&=0 \text{, if Prue misses to catch an error}
\end{align}$$

We have,

$$\begin{align}
P(I_{p}=1)&=p_{1}\\
P(I_{p}=0)&=1-p_{1}
\end{align}$$

On similar lines,

$$\begin{align}
P(I_{f}=1)&=p_{2}\\
P(I_{f}=0)&=1-p_{2}
\end{align}$$

We have,

$\begin{align}
P(I_{p}=1\cup{I_{f}}=1)&=P(I_{p}=1)+P(I_{f}=1)-P(I_{p}=1,I_{f}=1)\\
&= p_{1}+p_{2}-p_{1}p_{2}
\end{align}$

Thus, define success as, a typo is caught by atleast one of the proof-readers. Each of the $n$ typos can be caught or missed by atleast one proofreader. These can be conceived as $n$ independent Bernoulli trials. Therefore, $X\sim{Binomial(n,p_{1}+p_{2}-p_{1}p_{2})}$.

$\displaystyle{P(X=x)={{n}\choose{x}}(p_{1}+p_{2}-p_{1}p_{2})^{x}(1-p_{1}-p_{2}+p_{1}p_{2})^{n-x}}$

  (b) For this part only, assume $p_{1}=p_{2}$. Find the conditional distribution of $X_{1}$, given that $X_{1}+X_{2}=t$.

Solution.

$\begin{align}
P(X_{1}=x|X_{1}+X_{2}=t)&=\frac{P(X_{1}=x,X_{1}+X_{2}=t)}{P(X_{1}+X_{2}=t)}\\
&=\frac{{{n}\choose{x}}p^{x}(1-p)^{n-x}{{n}\choose{t-x}}p^{t-x}(1-p)^{n-(t-x)}}{{{2n}\choose{t}}p^{t}(1-p)^{2n-t}}\\
&=\frac{{{n}\choose{x}}{{n}\choose{t-x}}}{{{2n}\choose{t}}}
\end{align}$

Looks good to me. I think you may want to add

$\begin{align}
P(X_{1}=x|X_{1}+X_{2}=t)&=\frac{P(X_{1}=x,X_{1}+X_{2}=t)}{P(X_{1}+X_{2}=t)}\\
&=\frac{P(X_{1}=x,x+X_{2}=t)}{P(X_{1}+X_{2}=t)}\\
&=\frac{P(X_{1}=x)P(x+X_{2}=t)}{P(X_{1}+X_{2}=t)}\tag{*}\\
&=\frac{P(X_{1}=x)P(X_{2}=t-x)}{P(X_{1}+X_{2}=t)}\\
&=\frac{{{n}\choose{x}}p^{x}(1-p)^{n-x}{{n}\choose{t-x}}p^{t-x}(1-p)^{n-(t-x)}}{{{2n}\choose{t}}p^{t}(1-p)^{2n-t}} \tag{**}\\
&=\frac{{{n}\choose{x}}{{n}\choose{t-x}}}{{{2n}\choose{t}}}
\end{align}$

$(*)$ $X_1$ and $X_2$ are independent

$(**)$ This assumes $X_1$ and $X_2$ are binomial. Now, a binomial random variable is the sum of independent Bernoulli random variables. Are the Bernoulli random variables that make up each independent as well i.e. are the typo catching's for each person independent?

I mean

$$X_1 = B_1 + ... B_n$$
$$X_2 = C_1 + ... C_n$$

I guess $B_i$ and $C_j$ are independent for any i,j = 1, 2, ..., n, not necessarily distinct from the fact that $X_1$ and $X_2$ are independent, but are $B_1, B_2, ..., B_n$ independent? What about $C_1, C_2, ... C_n$?


11 divisiblity problem Given some digits $d_1, \dots, d_k$ is there a calculation that I can perform to determine if at least 1 permutation of $d_1, \dots, d_k$ is divisible by 11?

I am interested in all bases (radices) but a solution for base 10 should be sufficient for me to figure out the generalisation.

Generalised question:
Given a string of $k$ digits in base $b$, how can I test if at least 1 permutation of $d_1, \dots, d_k$ is divisible by $b+1$ (without computing them all)

Thanks in advance for any help!

To clarify:

I want to categorise a string $S$ of k digits into 1 of 2 sets.

Set1: S has no permutation $S'$ such that $S'$ divides by 11

Set2: S has at least 1 permutation $S'$ s.t. $S'$ divides by 11

An acceptable solution would be a finite list of criteria that applies to any combination of base and string length (b,k) i.e.
If $S$ satisfies any of the following:

Criteria 1
Criteria 2
etc

I believe the only information I can extract is that if the base is odd, the digit sum is odd if there are no permutations that divide 11.

If the base is even, no information can be extracted.

The usual divisibility by 11 test is to calculate the sum of the digits in the even positions minus the sum of the digits in the odd positions.  If that is divisible by 11 then the original number is.  From now on, I will say "even digit" to mean one in an even position, not necessarily an even value.  Similarly "odd digit".  I am counting from the right but it should work either way.

So, it is obvious that any permutation of the even digits or any permutation of the odd digits will not affect the divisibility by 11.  You could define a canonical form in which the even digits are in ascending order and so are the odd ones.

Also, any permutation can be expressed as a product of two element swaps.  From the above, any swaps of two even or two odd digits can be ignored.  For another two element swap, the effect on the test sum should be easy to calculate.  A swap of an odd digit and an even digit might affect the divisibility by 11.

Edit, due to comments, I have reworded the above to clarify the meaning of "even digit" as one in an even position and not necessarily an even value.


Is $\log(1+x)\sim x$ correct? This is my problem if $\log(1+x)\sim x$ when we are given $x\to 0$?

The assignment doesn't mention if $\log$ is $\log_{10}$ or $\log_{e}$? I assumed the first version and saw the claim is wrong. Am I on a right track?

I would definitely go with the taylor expansion around the origin:

$$\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+\mathcal O(n^4)\simeq x \ \ for \ |x|\ll 1$$

$$\lim_{x \to 0} \log_b(1+x) = \lim_{x \to 0}\log_b(1) = 0$$

Because $b^0 = 1 \space \forall b$


How to calculate limit in a quotient function How to calculate the limit of the following function:

$$\lim_{x \rightarrow 1}\frac{\sqrt[3]{2x^2-1} -\sqrt[2]{x} }{x-1}$$

I've tried to use the following formula: $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ without any success. Is there anything that I'm missing?

I can't use L'Hôpital rule.


Opposite of a function being bijective? A function is bijective if it is both surjective and injective. Is there a term for when a function is both not surjective and not injective?

As far as I know, there isn't.  The concept of a "non-surjective and non-injective function" just doesn't generally arise often enough to need a special term.

Of course not. This would be a useless notion (so there is no need for such terminology). Most functions are not surjective or injective (in the sense that if you take a "random" one, then you cannot expect it to be surjective or injective), mind you. We are mostly interested in the special class of mappings called injections and surjections, and for this reason the terminology "injective" and "surjetive" is useful.


Solving a Rubik's cube in 20 moves or less Most "cubers" solve a Rubik's cube in about 60 moves, using special algorithms and quick fingers. But any Rubik's cube can be solved in 20 moves or less (cube20.org). Is it possible that there is a way (without computers) to look at a cube and solve it in 20 moves or less, without trying to memorize several sextillion different positions?

Simple answer is that there is no known algorithm that can solve the Rubiks Cube in 20 moves maximum for an arbitrary starting position. Nevertheless, it is known that all positions are solvable in 20 moves, thus some algorithm must exist. Whether or not this algorithm requires less work than memorizing all the possible moves is currently somewhat open. (not comoletely.... We have used patterns and some group theory to reduce the number of solutions we needed to check to prove the number 20.) If we ever do find a simpler algorithm the proof will be much easier!


Common factor $x^4+6x^2+25$ & $3x^4+4x^2+28x+5$ If $f(x)=x^2+bx+c$ where b,c are Real number and $f(x)$ is a factor of both $x^4+6x^2+25$ & $3x^4+4x^2+28x+5$ then find the value of $f(x)$.

I manage to get the answer by dividing the function by $f(x)$ and substituting the remainder as zero. I managed to get the following

$(x^2-2x+5)(x^2+2x+5)=x^4+6x^2+25$

$(x^2-2x+5)(3x^2+6x+1)=3x^4+4x^2+28x+5$

but i am not satisfied as it is a tedious process.

Is there any short-cut method

$$x^4+6x^2+25=x^4+10x^2+25-4x^2=(x^2+5)^2-(2x)^2=(x^2+2x+5)(x^2-2x+5).$$
Now, you can check that just $x^2-2x+5$ is a factor of the second polynomial.

If you want factoring of the second without this hint, I am ready to show, but it's not so easy.


Functional Equation (no. of solutions): $f(x+y) + f(x-y) = 2f(x) + 2f(y)$ Find all the functions $f\colon\mathbb{Q} \to \mathbb{Q}$ such that
$f(x+y) + f(x-y) = 2f(x) + 2f(y)$, for all rationals $x,y$.

$f(0)=0$ By choosing $x=y=0$. Also by choosing $x=y$ we can see $f(2x)=4f(x)$. Moreover choose $x=0$ and use $f(0)=0$, then you have $f(y)=f(-y)$.

Now by strong induction one can prove that $f(nx)=n^2f(x)$.

Then choose $x$ as $\frac{x}{n}$ in preceding equation, then you get:
$$
f(x)={n^2}f(\frac{x}{n})\rightarrow\frac{1}{n^2}f(x)=f(\frac{x}{n})
$$
And finally we have:
$$
f(\frac{mx}{n})=\frac{1}{n^2}f(mx)=\frac{m^2}{n^2}f(x).
$$
Now assume that $f(1)=a$ for some $a\in\mathbb{R}$, then you get
$$
f(x)=ax^2
$$
for all rational $x=\frac{m}{n}$.

Alternatively, one could, upon getting that $f(2x)=4f(x)$, keep differentiating both sides with respect to $x$;

$\quad 2f'(2x)=4f'(x), \\\quad 4f''(2x)=4f''(x), \\\quad 8f'''(2x)=4f'''(x),\\ \qquad \qquad \vdots \\ 2^nf^{(n)}(2x)=4f^{(n)}(x), \\ \qquad \qquad \vdots \\$

Then setting $x=0$, we have;

$\quad 2f'(0)=4f'(0), \\\quad 4f''(0)=4f''(0), \\\quad 8f'''(0)=4f'''(0),\\ \qquad \qquad \vdots \\ 2^nf^{(n)}(0)=4f^{(n)}(0),\qquad(1) \\ \qquad \qquad \vdots $

From this, it can be seen that $f''(0)$ can have an arbitrary value; with all the other derivatives being zero.

Using the Maclaurin's expansion for $f(x)$, we get the same answer as above;

$f(x)=ax^2\qquad$ (for some rational $a$)


Jordan canonical form of the transpose of a matrix Let $A \in M_{n \times n}(\mathbb C)$. Prove that $A$ and $A^\text{T}$ have the same Jordan canonical form, and conclude that $A$ and $A^\text{T}$ are similar.

Intuitively, I know that I need to flip each cycle of generalised eigenvectors of $A^\text{T}$, but how to prove it?

Moreover, I've tried Mathematical Induction but was stuck...

Transposed matrices have the same characteristic equation and minimal polynomial. Furthermore, they have the same generalised eigenspace. With all of this in mind, we can say that $A$ and $A ^ \text{T}$ have the same Jordan chains, and hence Jordan basis. So $A$ and $A ^ \text{T}$ have the same Jordan canonical form.

Finally, similar matrices have the same Jordan matrix because if $P, Q \in \text{GL}(n, \mathbb{C})$ and $J$ is the Jordan canonical form, then we can write
$$P^{-1}AP = J = Q^{-1}A^\text{T}Q$$
i.e.
$$QP^{-1}APQ^{-1} = A^\text{T}.$$
But $Q, P ^{-1}$ and $P, Q^{-1}$ are invertible, so $QP^{-1}$ and $PQ^{-1}$ are also invertible. So write $R = QP^{-1}$. Then we know that $R^{-1} = (QP^{-1})^{-1} = (P^{-1})^{-1}Q^{-1} = PQ^{-1}$. So

$$RAR^{-1} = A^\text{T}$$

and we can conclude that $A$ and $A^\text{T}$ are similar.

$A$ and $A^t$ have the same eigenvalues (why ?), hence the same Jordan canonical form.
It follows that they are similar.


Resources for learning mathematics for intelligent people? Could people recommend resources to help my wife learn more complicated mathematics? She had a really terrible maths education, and while she essentially OK with every day maths she keeps wanting to know more about topics that would be college and university level.

She often wants to dig much deeper into a subject than basic texts allow, but there are some fundamentals that she has never been taught, which means there is quite a lot of going back to basics needed.

I can find a lot of resources for remedial mathematics that are aimed at basic numeracy, but the questions she wants to ask are things like: What is set theory(which led us into questions on what numbers mean, and Peano axioms), How does cryptography work? What are imaginary numbers? Various stats problems.

I've got a maths and comp sci degree so I usually know what the answer is, but there is such a void of knowledge between us, we can spend hours getting deeper and deeper trying to resolve a side issue from the main question, and it can get frustrating for both of us. :)

General resources would be great, or advice on how best to approach it. Specific resources about Crypto, number/set theory, and Statistics would also be appreciated.

I would suggest going to Khan Academy.  They have many videos on math with a wide variety of skill levels.

Bourbaki's Algebra and Set Theory. The structure of the books makes knowledge outside the series irrelevant, and they clean past misconceptions and handwaving out of the mind. They are very dense and challenging, though. While the books are not considered good introductions by many, I personally used them to start learning mathematics, and while under the same teaching circumstances as the OP, having the student read Algebra was the only way I was able to teach any real facility, in spite of many exercizes. The first chapter of Algebra can take a while to sink in, but in spite of protests it does not require reading Set Theory first, only knowledge of basic set operations and notation and of the product of sets. However, after a definition has been read, illustrations will make sense of it, and the illustrations will themselves make sense, because the difficulty comes from having to realize that the definition of the thing is exactly what it takes for something to be that thing, and that the definition is so open that you can make things that satisfy it at will.

After getting used to Algebra, Rotman - "An introduction to the theory of groups" is number theory-minded, and its historical expositions and diagrams add depth and explain, to the same degree Bourbaki doesn't, what it is you're doing. Note however, that neither of these introduce Cayley graphs early, which if she's geometrically minded you'll want to show her how to make as soon as she has a handle on the ideas (you wouldn't want her to rely on them entirely, as they depend on the group axioms).

These lay considerable conceptual groundwork: Between Bourbaki's Algebra and Rotman, she would have enough familiarity with axiomatic constructions and the interest in solutions to polynomial equations that you could introduce the complex numbers axiomatically, by showing how the algebra of the reals adjoined with a token solution to $x^2+1$ is defined completely just by assuming certain axioms on multiplication and addition that apply to the real numbers also apply to the new algebra, the ring axioms. I think that does its job to such satisfaction that it justifies the whole language. It would also be a shortcut from the early chapters of Rotman to the idea of the Galois group, which comes up somewhat late in the book.


Expectation score of multiple choice test A multiple choice exam has 100 questions, each with 5 possible answers. One mark is awarded for a correct answer and 1/4 mark is deducted for an incorrect answer. A particular student has probability $p_i$ of knowing the correct answer to the $i$th question, independently of other questions.

a) Suppose that if a student does not know the correct answer, he or she guesses randomly. Show that his or her total mark has mean $\sum p_i$ and variance $\sum p_i (1-p_i)+\frac{(100-\sum p_i)}{4}$.

b) Show that the total mark for a student who refrains from guessing also has mean $\sum p_i$  but with variance $\sum p_i (1-p_i)$.

b) is pretty easy, since it's a simple application of the binomial distribution. I can't really get a) though, mainly since I can't come up with a good-looking expression for the expected score of question number $i$.


Applications of Operator Algebras to modern physics I think that recently I've started to lean in my interest more towards operator algebras and away from differential geometry, the latter having many applications to physics.  But while taking physics courses, it was also brought to my attention that operator theory is a very integral part of quantum mechanics.  Are there applications of operator algebras in particular to quantum or relativity, or other fields of modern physics?  What about applications of operator theory?

In a related vein, I've been trying to find some research problems at my level or only slightly higher so that I can get a flavor, either for these applications, or for the abstract subject itself, as it's studied today.  I was wondering if anybody knows either of any such problems, or a source where I could find such problems.  To be honest, being a first year graduate student going on second, I'm not even quite sure where I'd look to find research problems that I could definitely guarantee are open, let alone ones I could reach right now.  In fact, I'm told that sometimes professors even mistake solved questions as being open.  If you have good examples of solved problems that are recent and representative of what I might face in the future, those would be helpful too.  Thanks.

The only applications to general relativity that I know of (my field!) is via Connes' noncommutative geometry, which is...complicated. Connes' work is freely (and legally!) available online. See also the math overflow thread Applications of Noncommutative Geometry, which may be interesting.

Operator algebras pop-up in Algebraic Quantum Field Theory too, which may be fun to look at. Actually, trying to discuss quantum fields on curved spacetime requires operator algebras.

The Physics.SX thread "Why are von Neumann Algebras important in quantum physics?" is also relevant.

I would suggest the textbook

Theory of Operator Algebras I
from Takesaki
enter image description here
good explanations with many good examples

Thanks


Given two disjoint closed sets, prove that there exist disjoint open sets containing the closed sets Given two disjoint closed sets, $F_1$ and $F_2$ in $\mathbb{R}$, prove that there exist disjoint open sets $G_1$ and $G_2$ such that $F_1 \subset G_1$ and $F_2 \subset G_2$

The answer is not trivial in that the distance between $F_1$ and $F_2$ could be going towards zero.

I'm outlining a proof from Baby Rudin. Let $X$ be any metric space an $\delta_A:X\to \Bbb R$ be defined by
$$
\delta_A(x)=\inf\{d(x,a)\,|\,a\in A\}
$$
where $A$ is a subset of $X$. You can verifiy that

  $\delta_A$ is a continuous function on $X$ (actually, it's even uniformly continuous).
  If $A$ is a closed set, then $\delta_A(x)=0$ iff $\ x\in A$.

Now for any disjoint closed set $A,B \subset X$, we define $\mu:X\to[0,1]$ by
$$
\mu(x)=\frac{\delta_A(x)}{\delta_A(x)+\delta_B(x)}
$$
Verify that

  $\mu$ is well-defined on $X$, i.e. $\delta_A(x)+\delta_B(x)\ne 0$ all $x\in X$(Hint: Use that fact that $A$ and $B$ are closed and disjoint.)
  $0\le\mu(x)\le 1$ all $x\in X$.
  $\mu(a)=0$ for all $a\in A$ and that $\mu(b)=1$ for all $b\in B$.

Since $\delta_A,\delta_B$ are continuous, $\mu$ is also continuous so that $\mu^{-1}[0,\frac 12)$ and $\mu^{-1}(\frac 12,1]$ are both open sets in $X$ and that they are disjoint. It's not hard to see that $A\subset\mu^{-1}[0,\frac 12)$, $B\subset\mu^{-1}(\frac 12,1]$.

Now let $A=F_1, B=F_2, G_1=\mu^{-1}[0,\frac 12)$ and $G_2=\mu^{-1}(\frac 12,1]$ and we're done.

Such a space is called T4 space.
Not all space is T4 space, so there should be some assumtion about the space.


Two lifts of a local homeomorphism Just learning about sheaves.

Suppose I have a sheaf $\mathscr{F}$ on a topological space. (The sheaf can take values in sets, let's say.) Let $E \overset{\pi}{\to}X$ be the etale space of this sheaf. Then $\pi$ is a local homeomorphism, but usually not a covering map.

A covering map $E' \overset{\pi'}{\to}X$, as we know, has the unique lifting property (if we assume things are nice), and to prove it, we let there be a map $f: U \to X$, where $U$ is connected, and we show that the points where two lifts of $f$ agree are

Open, (for which we use that given $x \in X$, there is a
neighborhood $V$ of $x$ and a homeomorphism $V \to V' \subset E'$
which is a section of $\pi'$, and where $V'$ is open)
Closed, (for which we use that $E'$ is assumed Hausdorff, and for
this it suffices to assume $X$ is Hausdorff by the covering map
property)
Nonempty. (For which we assume that they agree at a single point.)

If we were to try this for $\pi$ instead of $\pi'$, definitely (2) would fail, because $E$ is usually not Hausdorff.

  My Question: Would (1) fail also?

Clearly we don't have the same type of section of $\pi$ as we do in (1) above for $\pi'$. At first I thought that the definition of a germ would mean that if two lifts of $\pi$ agreed at a point, then they'd have to agree in a neighborhood of that point. But now I'm not so sure.

It is true : let $f:U\rightarrow X$ be any map and $g_1,g_2:U\rightarrow E$ be two lifts. Assume that $g_1(x)=g_2(x)=y$ and let $V$ be an neighborhood of $y$ such that $\pi:V\rightarrow \pi(V)$ is an homeomorphism. Then on the subset $U'=g_1^{-1}(V)\cap g_2^{-1}(V)\subset U$ the maps $g_1,g_2$ are given by $U'\overset{f}\rightarrow\pi(V)\overset{\pi^{-1}}\rightarrow V$ and so $g_1$ and $g_2$ coincide on $U'$ which is a neighborhood of $x$.

Using language from sheaf theory, assume that $U$ is an open subset of $X$, then $g_1,g_2$ are elements of $\mathcal{F}(U)$. Assume that their germs are equals at $x$, it exactly means that ${g_1}_{|U'}={g_2}_{|U'}$ for an open neighborhood $U'$ of $x$.

1) is also false. Consider the sheaf defined on $R$ which is the sheaf of sections of the trivial bundle $R\times R\rightarrow R$ a section is a continuous map defines by $f(x)=(x,h(x))$.

A section $f$ of the etale space here may be defined by a continuous map $g:U\rightarrow C(U,R)$ such that for $x\in U$, $f(x)=[g(x)]$ where $[g(x)]$ is the germ of $g$ at $x$.  you can find two sections $f,f'$ such that $f(x)=f'(x)$ on $U$ if and only if $x=0$.


The general nth derivative formula of $(1-x/4)^{-2}$ The question: Determine the nth general (derivative) formula for the function, $$f(x) = \frac{1}{(1-\frac{x}{4})^2}$$

My approach:

Determine $f(x), f'(x) ... f^4 $
Compare the patterns and determine a general formula.

$$f'(x) = \frac{-32}{(x-4)^3}$$
$$f''(x) = \frac{96}{(x-4)^4}$$
$$f'''(x) = \frac{-384}{(x-4)^5}$$

My thoughts: It seems as the the next numerator is a result of the previous numerator multiplied by $ (n+1)*(-1)^{(n+1)}, ex: 96 = -32 * 3, -384 = 96 * -4 $ and the denominator is simply $(x-4)^{n+1}$

I have tried a few different attempts as well but can't seem to complete my logic. This question is meant to help me solve a question regarding the Maclaurin series (where I am suppose to find the series using the definition of the Maclaurin series.

You are very close. $f(x)$ can be rewritten as:

$$f(x) = \frac{16}{(4-x)^2}$$

So, the factors of the derivative of order $n$ are:

$$16 \times (-1)^n \times (n+1)!$$

Edit: Hence the nth order derivative is:
$$f^{(n)} = 16 (-1)^n (n+1)! (x-4)^{-2-n}$$

Or, if we switch the order of the terms in the denominator, we can get rid of the $(-1)^n$ term:
$$f^{(n)} = 16 (n+1)! (4-x)^{-2-n}$$


Help with this Trigonometry integral I've to find this integral:

$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx$$

So far as I know to go:

$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(\frac{4^\pi}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx=$$

$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(4^\pi\cdot \frac{1}{\tan(x)}+\tanh^{-1}(x)-4\sec^2(x)\right)dx=$$

$$\int_{\frac{\pi}{365}}^{\frac{365} {\pi}}\left(4^\pi \cot(x)+\tanh^{-1}(x)-4\sec^2( x)\right)dx$$

Thanks a lot!

Hints:

$$\begin{align}&\int\frac{dx}{\tan x}=\int\frac{\cos x}{\sin x}dx=\log|\sin x|+C\\{}\\
&\int\arctan x\;dx\stackrel{\text{by parts}}=x\arctan x-\frac12\log\left(1+x^2\right)+C\\{}\\
&\int\sec^2x\;dx=\tan x+C\end{align}$$


Orthonormal Sets in Hilbert Spaces let $H$ be a Hilbert space and let M be a dense linear subspace of H.
Can we find a complete orthonormal set $\{u_{\alpha}: \alpha \in A\}$ for H
in M?

I think the answer is negative in general, but I cannot find a counterexample.
Thank you very much in advance for your help.

Your question is equivalent to the following: does every pre-Hilbert space (= inner product space) $M$ admit an orthonormal basis?

It turns out that the answer is "no". A counter-example can be found in N. Bourbaki's Topological Vector Spaces, Exercise V.2.2.

I tried to solve the exercise, but I'm not quite sure that I understood it.

Let $H_1=\ell_2(\mathbb N)$, and $H_2=\ell_2(I)$, where $I$ is a set with cardinality $\mathfrak c$ (the continuum). Choose a linearly independent family $(x_i)_{i\in I}\subset H_1$ and an orthonormal basis $(e_i)_{i\in I}$ of $H_2$. Now let $H:=H_1\oplus H_2$ and $M:={\rm span}\,\{ x_i\oplus e_i;\; i\in I\}\subset H$. Let us try to show that $M$ admits no orthonormal basis.

First, note that, identifying $H_2$ with $\{ 0\} \oplus H_2\subset H$, we have $M\cap H_2=\{ 0\}$. This is rather easy to check using that fact that the $x_i$ are linearly independent.

Now, let us show that any orthonormal family $\mathcal Z\subset M$ is countable. To see this, choose an orthonormal basis $(f_n)_{n\in\mathbb N}$ of $H_1=\ell_2(\mathbb N)\subset H$ and observe that for any $n\in\mathbb N$, we have $\langle f_n, z\rangle=0$ for all but countably many $z\in\mathcal Z$, because $\sum_{z\in\mathcal Z}\vert \langle f_n,z\rangle\vert^2<\infty$. Since by the previous remark, any $z\in\mathcal Z$ must satisfy $\langle f_n ,z\rangle\neq 0$ for at least one $n\in\mathbb N$ (otherwise we would have $z\in H_1^{\perp}=H_2$ and hence $z=0$), it follows that $\mathcal Z$ must be countable. Indeed, we have $\mathcal Z=\bigcup_{n\in\mathbb N} \mathcal Z_n$, where $\mathcal Z_n=\{ z;\; \langle f_n,z\rangle\neq 0\}$ is countable for each $n$.

It follows that if $M$ were to admit an orthonormal basis, then it would be separable. But this is not possible because
$\pi_2(M)$ is dense in $H_2$ (since it contains all the vectors $e_i$) and $H_2$ is not separable.

consider $X:=\left\{A\in\mathcal{P}\left(M\right): A \text{ is an ON-System }\right\}$ and $\subseteq$. Obviously each partially ordered chain $C$ has an upper bound in $X$: simply consider $\cup_{a\in C}a$.  By Zorn we get that there exists at least one maximal element of $X$. Take the maximal element, by definition it's an orthonormal Hilbert-basis.

  see comments. i will not delete this. (for documentation)


Number of ways of distributing $N$ balls into $M$ bins such that at least one bin has at least $n$ balls in it? What is the number of ways of distributing $N$ indistinguishable balls into $M$ bins such that at least one bin has at least $n$ balls in it?

My attempt:

The number of ways of of placing $N$ balls into $M$ bins is $\binom{N+M-1}{N}$.

I tried, by stars and bars, to calculate the number of ways of distributing $N$ indistinguishable balls into $M$ bins such that exactly one bin has exactly $n$ balls in it:

If the first or last bin contains the $n$ balls, we have used one partition, so there are $M-2$ left. This gives $2\binom{N-n+M-2}{N-n}$ ways.

If the second to $(M-1)$th bin contains the $n$ balls, we have used two partitions, which gives $(M-2)\binom{N-n+M-3}{N-n}$ ways.

So the total number of ways of distributing $N$ indistinguishable balls into $M$ bins such that exactly one bin has exactly $n$ balls in it is
$$2\binom{N-n+M-2}{N-n}+(M-2)\binom{N-n+M-3}{N-n}.$$

Is this correct? It seems like a strange result.

Then I thought about simply summing this expression for $n$ going from $n$ to $N$, i.e.,

$$\sum_{k=n}^N 2\binom{N-k+M-2}{N-k}+(M-2)\binom{N-k+M-3}{N-k},$$

in order to obtain the expression for "at least $n$ balls," but I feel like this would be over-counting somehow.

And then there is the issue of "at least one bin," which I am rather blunted by.

Any help is much appreciated!

Note that I'm looking for a closed form solution to the problem in the yellow box. Thanks.

This is a fairly ugly problem; probably the most straightforward approach is the calculate the number of distributions that have fewer than $n$ balls in each bin and subtract that number from $\binom{N+M-1}N$. Unfortunately, there does not seem to be a closed form for the subsidiary problem. By a combination of stars and bars and inclusion-exclusion one can show that there are

$$\sum_i(-1)^i\binom{M}i\binom{N+M-1-in}{M-1}$$

of these ‘bad’ distributions.

Note that the $i=0$ term is $\binom{N+M-1}{M-1}=\binom{N+M-1}N$; the other terms from the inclusion-exclusion calculation get rid of the distributions that have more than $n$ balls in some bin. Thus, the answer to your question is actually just the negative of the some of the terms with $i>0$, i.e.,

$$\sum_{i=1}^M(-1)^{i+1}\binom{M}i\binom{N+M-1-in}{M-1}\;.$$

The inclusion-exclusion argument itself is pretty straightforward; I can give it in detail if you wish, though I’m not sure how useful the result is.

In this answer Marc van Leeuwen uses generating functions to deal with the more general problem in which each bin has its own upper limit, but he starts with the simpler case in which all of the limits equal, as in your question.

Seems that it's not correct, since you going to have double counting for the case where cell 1 and cell 2 both have $n$ balls.

Try use inclusion-exclusion principle.

It's easy using the properties $p_i = \mathrm{The\space ith\space cell\space has\space at\space least\space n\space balls}$ to calculate $E(0)$ the case when no property holds.
For $r$ different properties to hold, we want that $r$ specific cells will have at least $n$ balls we first put $n$ balls in each of them and then the rest $N - rn$ we split however we want, so we get: $$ w(p_{i_1},...,p_{i_r}) = {{N + M - rn - 1}\choose N - rn}$$ and with the number of ways to pick $r$ different properties and the fact that the properties are symmetric we get $$ w(r) = {M \choose r} \cdot  {{N + M - rn - 1}\choose N - rn} $$

Now we have that $$E(0) = {{N+M-1}\choose{N}} - \sum_{r=1}^M (-1)^r\cdot w(r)$$

After calculating $E(0)$ use the fact that whats youre loking for is the complement of $E(0)$ which gives you that the desired number is: $${{N+M-1}\choose{N}} - E(0) = \sum_{r=1}^M (-1)^r\cdot w(r)$$


Intuition behind stationary points in a multivariable function Why should the local optima for a differentiable function $f(x,y)$ be at the stationary point(s)? What is the intuition behind setting the first partials equal to zero? Is it because if one of the partials is positive (negative), then we can move in (away from) that variable's direction and increase the function value?

From geometry, $f(x,y)$ is a 2D surface, $df$ is the vector field on this 2D surface, while its dual $\star df$ tangent to the contours on the surface.
From the aspect of vector field $df$. The stationary points of $f(x,y)$ are those points, the vectors at which are of zero length. In other words, all the vectors either converge to those point, or emit from them.
From the aspect of dual vector field $\star df$. The stationary points of $f(x,y)$ are those points, around which all the dual vectors are running.


Express Integer as Sum of Four Squares This is kind of a follow-up to the question I posted here about expressing integers as the sum of two squares. Is there a similar general method for expressing integers as the sum of four squares? I believe the Lagrange's Four-Square Theorem states that all positive integers are expressible as the sum of four squares of integers, but how do you find these numbers. As an example consider the value $1638$. How can we find the four squares?

Similar to the Brahmagupta-Fibonacci two-square identity. Euler has a four square identity which involves the sum of 4 squares:

$$(a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2) =\\
\quad(a_1b_1 - a_2b_2 - a_3b_3 - a_4b_4)^2 + (a_1b_2+a_2b_1+a_3b_4-a_4b_3)^2
+(a_1b_3 - a_2b_4 + a_3b_1 + a_4b_2)^2 + (a_1b_4 + a_2b_3 - a_3b_2 + a_4b_1)^2$$

Factor $1638$ as products of any small factors you know how to represent as sum of 4 squares. Repeat apply the formula will allow you to represent $1638$ itself as sum of 4 squares.

For example, let's say we have factored $1638$ as $2\cdot 3^2 \cdot 7 \cdot 13$, we have:

$$\begin{align}
  & 2\cdot 3^2 \cdot 7 \cdot 13\\
= & (1^2+1^2+0^2+0^2)(1^2+1^2+1^2+0^2)^2(2^2+1^2+1^2+1^2)(3^2+2^2+0^2+0^2)\\
= & (0^2 + 2^2 + 1^2 + 1^2)(1^2+1^2+1^2+0^2)(2^2+1^2+1^2+1^2)(3^2+2^2+0^2+0^2)\\
= & ((-3)^2 + 1^2 + 2^2 + 2^2)(2^2+1^2+1^2+1^2)(3^2+2^2+0^2+0^2)\\
= & ((-11)^2+(-1)^2+2^2 + 0^2)(3^2+2^2+0^2+0^2)\\
= & ((-31)^2 + (-25)^2 + 6^2 + 4^2)\\
\end{align}$$

This give you a non-trivial representation of $1638$ as $31^2 + 25^2 + 6^2 + 4^2$.

In general, there are many representations of a number as a sum of 4 squares.
There is a theorem:

  The total number of representations of a positive integer $n$ as the
  sum of four squares, representations that differ only in order and sign
  being counted as distinct, is eight times the sum of the divisors of
  $n$ that are not multiple of $4$.

The above representation is only $1$ out of $8 \sum_{d\mid 1638, 4 \nmid d} d = 34944$
ways of representing $1638$ as sum of 4 squares.

This formula seems wrong. Try letting $a_1=b_1$, $a_2=b_2$, $a_3=b_3$, $a_4=a_4$. The result should be $(a_1^2+a_2^2+a_3^2+a_4^2)$.

Euler's four-square identity says that the product of two numbers, each of which is a sum of four squares, is itself a sum of four squares. Specifically:

$$(a_1^2+a_2^2+a_3^2+a_4^2)(b_1^2+b_2^2+b_3^2+b_4^2)=
(a_1 b_1 + a_2 b_2 + a_3 b_3 + a_4 b_4)^2 +
(a_1 b_2 - a_2 b_1 + a_3 b_4 - a_4 b_3)^2 +
(a_1 b_3 - a_2 b_4 - a_3 b_1 + a_4 b_2)^2 +
(a_1 b_4 + a_2 b_3 - a_3 b_2 - a_4 b_1)^2$$


Determining if a function is strictly increasing for all natural numbers on a specific interval I am having some difficulties with tackling the following problem:
Show that $f(x)=x^n$ (where n ∈ $\mathbb{N}$) is strictly increasing on the interval $[0, ∞)$.
I am familiar with the following definition: Strictly increasing means if $x_1 < x_2$ then $f(x_1) < f(x_2)$. So, I started with this definition. However, I am not sure if this is supposed to be proved using the mathematical induction method. I tried to split it into two cases: an even and an odd exponent, but this didn't really lead me anywhere, unfortunately.

If you know how to use induction, then proceed like this: $f(x)=x^1=x$ is clearly strictly increasing, because if $0 \leq x<y$, then $f(x)=x<y=f(y)$. Now assume the result is true for $n=1,2,\ldots,k$. We need show this is true for $n=k+1$.

Suppose $0 \leq x<y$. Let $f(x)=x^{k+1}$ and $g(x)=x^k$. By assumption, we know that $g(x)$ is increasing so that $g(x)= x^k<y^k=g(y)$. But
$$
f(x)= x^{k+1}= x \cdot x^k= x \cdot g(x) \leq y \cdot g(x) < y \cdot g(y)= y \cdot y^k= y^{k+1}= f(y)
$$
But then $f(x)$ is increasing.

I just taught this today.

Let $0\leq x_1<x_2$.

Notice that

$x_2^n-x_1^n=(x_2-x_1)(\sum_{i=0}^{n-1}x_1^i x_2^{n-1-i})$ is positive since both factor are. Hence $x_2^n>x_1^n$ and the function is increasing.


An injective immersion that is not a topological embedding On page 86 of John Lee's Introduction to smooth manifolds there is an example of an injective immersion that is not a topological embedding:

$\beta : (-\pi, \pi) \to \mathbb{R}^2$, defined by $\beta(t) = (\sin{2t}, \sin{t})$, or pictorially:

It is explained that, although $\beta$ is an injective immersion, it is not a smooth embedding since the image is compact while the domain is not.
My understanding is that the image, while bounded in $\mathbb{R}^2$, is an open subset of the plane, whereas the statement claims that it is not.

Would anyone please explain why the image is compact? Thank you.

First proof: If $\beta(t_n)$ is a sequence of points in the image, the sequence $t_n$ is bounded in $\mathbb{R}$, hence there is a subsequence $t_{n_p}$ that converges to a $t \in [-\pi, +\pi]$. By continuity of the sine, $\beta(t_{n_p})$ converges to $(\sin(2t), \sin(t))$, which is equal to $\beta(t)$ if $t\in (-\pi, \pi)$ and to $\beta(0)$ otherwise. Thus, every sequence in the image has a subsequence that converges in the image, which is the definition of compactness.

Second proof: Let $\gamma$ be the map $t \mapsto (\sin(2t), \sin(t))$ from $[-\pi, \pi]$ into $\mathbb{R}^2$ The image of $\gamma$ is the same as $\beta$, hence it is the image of a compact set by a continuous map.

I would like to add a simple explanation. We can just use the fact that $\beta \left ( -\pi , \pi \right )$ is bounded (and closed because it has all its limit points) in $\mathbb{R}^{2}$ therefore it's compact in the topology of $\mathbb{R}^{2}$. It is a simple exercise to check that compactness in the bigger space ($\mathbb{R}^{2}$) implies compactness in the subspace topology of $\beta \left ( -\pi , \pi \right )$. Thus $\beta \left ( -\pi , \pi \right )$ is a compact space and also an image of a non-compact set. This shows that the map can not be a homeomorphism.


Determine type of set Given complement to set $M$ is recursively enumerable and recursive set $R$. What will be the type of the subset of M, elements of which are in R ?

I think they will be also recursively enumerable, but I'm not quite sure about it.

the complement of $M := \Bbb N$ is recursively enumerable; $R := \Bbb N$ is recursive; good luck classifying the subsets of $\Bbb N$.


Evaluate $(5-i)^4 (1+i)$ then prove that $ \pi/4=4\arctan(1/5)-\arctan(1/239)$ Evaluate $(5-i)^4 (1+i)$ then prove that $ \pi/4=4\arctan(1/5)-\arctan(1/239)$

I did the first part of evaluating it manually and I got 956-4i.  However I am having trouble seeing how I can use this result to calculate $\pi/4$ in that equation.  Can anyone help me here?

I think it is about the formula
$$e^{i\theta}=\cos\theta+i \sin\theta $$

$$(5-i)^4(1+i)=({e^{\arctan(-\frac{1}{5})i}})^4e^{\frac{\pi}{4}i}$$
$$956-4i=e^{\arctan(-\frac{4}{956}i)}=e^{-\frac{1}{239}i}$$
Since $(5-i)^4(1+i)=956-4i$,take log for both sides and you should get the answer.


Derivative of angle between two vectors singularity! I have been battling a problem of needing to know the derivative of the angle between two vectors, the vectors possibly being parallel at some points in time. I started off with:

$$\bf A \dot \bf B = \|A\|\|B\|\cos\theta \Rightarrow \theta=\arccos \left( \frac{\bf A \dot \bf B}{\|A\|\|B\|} \right)$$

Let's say that $\bf A$ and $\bf B$ are unit vectors for simplicity so:

$$
\theta=\arccos \left( \bf \hat A \dot \bf \hat B \right)
$$

To find the derivative of the angle $\theta$:

$$
\dot \theta = \frac{-1}{\sqrt{1-\left(\bf \hat A \dot \bf \hat B\right)^2}} \left[\bf \dot{\hat A}\dot \bf \hat B+\bf {\hat A}\dot \bf \dot{\hat B}\right]
$$

But, $\bf{\hat A} \dot \bf{\hat B}=$$1$ when $\bf \hat A \,\,\|\,\, \bf \hat B$ and the denominator of the expression for $\dot \theta$ becomes $0$. We hence get a division by $0$. Now I have tried other ways - by projecting $\bf{\hat B}$ onto $\bf{\hat A}$ and taking $\arctan$, by projecting $\bf{\hat B}$ onto a plane and forming a right-angle triangle and again taking $\arctan$. Every method I tried leads to the same singularity when $\bf \hat A \,\,\|\,\, \bf \hat B$.

Question: what method can I use such that I don't get a singularity for $\dot\theta$ when $\bf{\hat A}$ becomes parallel to $\bf{\hat B}$? Thank you!

You cannot get rid of the singularity. Note that in ${\mathbb R}^d$ with $d\geq3$ the angle between two vectors ${\bf a}$, ${\bf b}$ is nonoriented, implying that we always have $$\angle({\bf a},{\bf b})\in[0,\pi]\ .$$
Consider the following example in ${\mathbb R}^3$:
$${\bf a}(t):=(1,0,0),\quad{\bf b}(t):=(\cos t,\sin t,0)\qquad(-1<t<1)\ .$$
Then
$$\phi(t):=\angle\bigl({\bf a}(t),{\bf b}(t)\bigr)=|t|\ ,$$
and this is not differentiable at $t=0$.

Consider the function $F:\mathbb R^n \setminus\{0\} \to \mathbb S_{n-1}$, defined by $F(u) := u/\|u\|$. So you seek $\dot\theta(t)$, where $\theta(t):= \arccos(r(t))$, and $r(t): = a(t)\cdot b(t)$, where $a(t):= F(x(t))$ and $b(t) := F(y(t))$
Now, by chain rule, $\dot{r}(t) = a(t).\dot{b}(t) + \dot{a}(t)\cdot b(t)$. Using a algebraic fact (see lemma below) and the chain rule again, we get
$$
\begin{split}
\dot{a}(t) &= \nabla F(x(t))\cdot\dot{x}(t) = \frac{I_n-F(x(t))\otimes F(x(t))}{\|x(t)\|}F(y(t)),\\
\dot{b}(t) &= \nabla F(y(t))\cdot\dot{y}(t) = \frac{I_n-F(y(t))\otimes F(y(t))}{\|y(t)\|}F(x(t)),
\end{split}
$$
where $\otimes$ denotes the outer-product of vectors (produces a rank-1 matrix) and $I$ is the identity matrix of size $n$.
Putting everything together should give you the correct formula for $\dot{r}(t)$. Now by using the chain rule one last time to get $\dot{\theta} = -\dfrac{1}{\sqrt{1-r(t)^2}}\dot{r}(t)$.

Lemma. If $u,v \in \mathbb R^n$ with $u \ne 0$, then $\nabla_u (F(u)\cdot v) = \dfrac{I_n-F(u)\otimes F(u)}{\|u\|}v$.

Proof. Chain rule and basic matrix calculus.


Angle Between Two Faces of a Triangular Pyramid A regular triangular pyramid has all of edges of length 36. What is the angle between any two faces of this pyramid?

I have no idea what to do for this problem. Can someone explain to me what an angle between two faces is?

The lengths of the sides of the regular tetrahedron are irrelevant.

Hint:  Find the vectors from the tetrahedron's center to the center of two faces.  Then calculate the angle between them.

This is asking for the angle defect between the two faces. To find the angle defect or the answer to your question, do 360 degrees minus the sum of the interior angles meeting at the vertex of where the two faces meet. The answer of that will be the answer to your question.


Quadratic diophantine equations and fundamental solutions I posted a question about a class of ternary quadratic equations. However, after some searching, I found the problem is not so simple as I thought. Thus, I want to know just whether a property on the Pell equations can be generalized.

Let $a $ and $b $ be integers, $\ne 0, 1, -1$. Consider a quadranary diophantine equation $x^2-ay^2-bz^2+abw^2=1$. The "grand" problem is to classify "nontrivial" (that is, $x^2\ne 1$) solutions. The existence may follow from the classical theory of Pell's equations.

Then, I want to know whether there are always(or for almost all cases) a finite set of "fundamental" solutions, generating the other solutions by some method. In case of Pell's equations, there is a unique fundamental solution $x_1+\sqrt{n}y_1$ for $x^2-ny^2=1$ and the other solutions are given as $x_n+\sqrt{n}y_n = (x_1+\sqrt{n}y_1)^n$. To utilize a similar way, I chose the class of equations to lie in the unit group of a quaternion algebra $(a,b)_\mathbb {Q} $.

Suggest beginning with Carmichael, manuscript pages 30-35, pdf pages 42-47.

For any integer quadruple $(p,q,r,s)$ such that $p^2 - a q^2 - b r^2 + ab s^2 \neq 0,$ we get a rational solution
$$ (1 + tp, \; tq, \; tr, \; ts)  $$ where
$$ t = \frac{-2p}{p^2 - a q^2 - b r^2 + ab s^2}.  $$
This does include all integer solutions, but selecting those is likely to be a mess. The main problem is that you take a fixed right hand side, $1,$ rather than $w^2$ with another variable, so that the whole thing is the null set of an indefinite quadratic form. In this latter problem you could clear denominators by taking $w = p^2 - a q^2 - b r^2 + ab s^2,$ then divide out by the GCD of all five variables, giving primitive solutions. In some problems the set of possible GCD's to worry about is finite, once $\gcd(p,q,r,s) = 1.$ I have no idea whether such good fortune happens here.

From the language of integer (indefinite) lattices, we have an indefinite dot product, the bilinear form given by the evident 4 by 4 matrix. Over the rationals, a fixed vector $v$ with nonzero norm gives a reflection
$$  x \mapsto x - \frac{2 \, x \cdot v}{v \cdot v} v.  $$
This is what is called an "odd lattice," so this preserves all integer points only when $v \cdot v = \pm 1.$ i think it likely that the full automorphism group of your indefinite form is generated by reflections. This is somewhat circular, since you need to find vectors of norm $\pm 1$ to find reflections, and that is what you wanted in the first place. However, the thing sort of mushrooms; given a few solutions, you get a few reflections, these act on the solutions you have, and so on.

For quadratic forms, I recommend CASSELS  since it emphasizes $\mathbb Q$ and $\mathbb Z.$ Note that MAJID JAHANGIRI alludes to the book Fricke and Klein (1897) but that is not in the references. A fair amount of the relevant material is availbale, in English in MAGNUS.

This is nice, you can read several pages of CHALK

The easiest option is. Equation.

$$x^2-ay^2-bz^2+baw^2=f$$

To rewrite it.

$$x^2-ay^2=f+b(z^2-aw^2)$$

Find all possible solutions of the equation Pell.  They are made basic.

$$z^2-aw^2=t$$

It is necessary to consider all possible $t$.  And these decisions we find out whether the solution of this Pell equation.

$$x^2-ay^2=f+bt$$

And the problem is reduced to standard. Find out the solvability of the Pell equation and record his decision that much easier.

You can use either formula to find the solutions, or program to find the first solution. Knowing the first solution and the rest are easy.


Solving $8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$ I came up with this equation during my homework :
$8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$

My algebra is weak and I can't seem to find a way to solve for x nicely

Could someone please show me a decent way of doing this?

Thanks alot, Jason

Generally the best way is to just plough through the algebra (and algebra gets quite a bit more advanced than this!) :

$8=x(2(1-\sqrt{5}))+(1-x)(2(1+\sqrt{5}))$
$4=x(1-\sqrt{5})+(1-x)(1+\sqrt{5})$ (dividing through by $2$ simplifies a lot of the subsequent terms)
$4=x-x\sqrt{5}+1-x+\sqrt{5}-x\sqrt{5}$ (multiply out all the terms)
$4=x-x\sqrt{5}-x-x\sqrt{5}+1+\sqrt{5}$ (rearrange to get all the $x$ terms out front)
$4=x(1-\sqrt{5}-1-\sqrt{5})+1+\sqrt{5}$ (collect the x terms)
$4=x(-2\sqrt{5})+1+\sqrt{5}$ (simplify)
$(4-(1+\sqrt{5}))=x(-2\sqrt{5})$ (move the constant term to the left)
$3-\sqrt{5} = x(-2\sqrt{5})$ (simplify the left)
$x=(3-\sqrt{5})/(-2\sqrt{5})$ (divide both sides by $-2\sqrt{5}$)
$x=3/(-2\sqrt{5}) + {1\over2}$ (split out the terms)
$\displaystyle{x=-{3\sqrt{5}\over 10} + {1\over2}}$ (multiply the numerator and denominator of the first part through by $\sqrt{5}$)

Of course, I strongly recommend plugging this $x$ in to confirm that it satisfies your initial equation!

$$8 = x\left [ 2-2\sqrt{5} \right ] + \left ( 1 - x \right )\left [ 2 + 2\sqrt{5} \right ]$$

use foil

$$8 = 2x - 2x\sqrt{5} + 2 + 2\sqrt{5} - 2x - 2x\sqrt{5}$$

$$8 = -4x\sqrt{5} + 2\sqrt{5} + 2$$

subtract 2 and square both sides

$$36 = 16x^25 + 20$$

subtract 20 from both sides to obtain

$$80x^2 = 16$$

Now divide both sides by 80 and you get $$x^2 = 16/80$$

therefore $$x = \pm 1 / \sqrt{5}$$


What is theoretical convergence of formula for $\zeta(s)$? Assume the following definitions:
$\zeta(s)$ - Riemann zeta function
$\delta(x)$ - Dirac delta function

My question is:
What is the theoretical convergence of the following formula for $\zeta(s)$? Does this formula only converge for $Re(s)>1$?
$\zeta(s)=\int_{1-\epsilon}^{\infty}x^{-s}\sum_n \delta(x-n)\ dx$

This is nothing more than the definition of the Dirac delta distribution $\delta$ : $$\int_{-\infty}^\infty \delta(x-a)f(x)dx = \lim_{\epsilon \to 0} \int_{-\infty}^\infty \frac{1_{|x-a| < \epsilon}}{2\epsilon}f(x)dx = f(a)$$ whenever $f$ is continuous at $x=a$.

From this you can say that $$\int_{b}^c \delta(x-a)f(x)dx = \int_{-\infty}^\infty \delta(x-a)f(x)1_{|x-a| < \epsilon}dx=\begin{cases}f(a) \text{ if } a \in (b,c)\\ 0 \text{ if } a \not\in [b,c] \end{cases}$$
(since $f(x)1_{|x-a| < \epsilon}$ is continuous at $x=a$)

Finally by definition of the improper integral
$$\int_{1-\epsilon}^\infty \sum_{n=-\infty}^\infty \delta(x-n) x^{-s}dx = \lim_{N \to \infty}\int_{1-\epsilon}^{N+\epsilon} \sum_{n=-\infty}^\infty \delta(x-n) x^{-s}dx$$
$$ = \lim_{N \to \infty}\sum_{n=-\infty}^\infty\int_{1-\epsilon}^{N+\epsilon} \delta(x-n) x^{-s}dx=\lim_{N \to \infty} \sum_{n=1}^N n^{-s}$$
Converging of course to $\zeta(s)$ only when $Re(s) > 1$.

Now you are supposed to ask : so if this is "by definition" and "just a notation" then  what is it useful for ?

Well it is interesting because you can generalize this to many other Dirichlet series, you can integrate by parts and get the Abel summation formula, you can use the theory of Fourier series for writing that formally $$\sum_{n=-\infty}^\infty \delta(x-n)  = \sum_{n=-\infty}^\infty e^{2i \pi n x}$$
which leads to the functional equation $\zeta(s) = 2^s \pi^{s-1} \sin(\pi s/2) \Gamma(1-s) \zeta(1-s)$, and do many other things.

In other words : "$\zeta(s)$ is the Mellin transform of the Dirac comb $\sum_{n=-\infty}^\infty \delta(x-n)$" is important from the theoretical perspective.


Integral u-substitution of u=$\frac{2x}{1+x^2}$ I was curious about the validity of doing such a substitution in an integral:

(1)$$u=\frac{2x}{1+x^2}$$

Changing this around a bit

$$ux^2 - 2x + u = 0$$

Solving with the quadratic formula

with $a=u, b=-2, c=u$:

$$x=\frac{1 \pm \sqrt{1 - u^2}}{u}$$

I had discerned that both definitions for x (+ and -) are indeed true, plugging this into (1). However, I’m confused as to whether doing this substitution (with the intent to specifically replace parts in the integrand with x, rather than letting the differentials cancel things out) is even valid given there’s two possible definitions for x based on the original statement (1). Even if it is, I’m also not sure about how one would solve for the differentials.

For example, in this problem:

$$\int_0^1 \frac{1}{1+x^2}\sin^{-1}\left(\frac{2x}{1+x^2}\right)dx$$

The two substitutions are both valid. Note the reciprocal relationship between the two choices

$$\frac{1 + \sqrt{1 - u^2}}{u}
=\frac{u}{1- \sqrt{1 - u^2}}$$

to see that they are not equivalent, though, with one being the inverse of the other.


If $\lim\limits_{x \to \infty} f'(x) = L$ and $\lim\limits_{n \to \infty} f(n) = A$ exists, prove that $L = 0$. Here is the homework problem I am stuck on:

  Let $f$ be differentiable on $(0,\infty)$.  If $\lim\limits_{x \to \infty} f'(x) = L$ exists in $\mathbb{R}$ and $\lim\limits_{n \to \infty} f(n) = A$ exists in $\mathbb{R}$, prove that $L = 0$.

From the given information, I know that we get to assume:

$f$ is continuous at every point $s \in (0,\infty)$.
We can now apply the MVT.

So far this is what I'm thinking:
First, I think this describes a function which increases towards a horizontal asymptote.
I have created a strictly increasing sequence $\{x_n\} = \{x_1, x_2, \dots, x_n \}$ to serve as the $x$ values in $(0, \infty)$.  This gives me a sequence of intervals basically.

I can apply the MVT on each of these intervals, getting a sequence of $c_n \in (x_{n-1}, x_n)$.

But I don't see where to go from here.  I am guessing I will need the squeeze theorem later, but there's a gap in between.

Perhaps I'm doing something wrong?

Thanks for your help.

If we can show that $|L|\lt \epsilon$ for every $\epsilon \gt 0$ then we will have that $|L|\le 0\Rightarrow L=0$.

Let $\epsilon\gt 0$ be given. Since $$\lim_{x\to \infty} f^{\prime}(x)=L$$ we can find $M$ such that $x\ge M\Rightarrow |f^{\prime}(x) - L|\lt \frac{\epsilon}{2}$. Since $$\lim_{n\to\infty}f(n)=A$$ we can find $N$ such that $n\ge N\Rightarrow |f(n)-A|\lt \frac{\epsilon}{4}$.

Consider $m\gt max(M,N)$. The mean value theorem tells us we can find $c\in (m,m+2)$ such that $$f^{\prime}(c) = \frac{f(m+2) - f(m)}{2}$$ Then $|f^{\prime}(c)| = |\frac{f(m+2) - f(m)}{2}|\lt |f(m+2)-f(m)|\le |A+\frac{\epsilon}{4} - (A-\frac{\epsilon}{4})|=\frac{|\epsilon|}{2}=\frac{\epsilon}{2}$

Now, $|f^{\prime}(c)-L| = |L-f^{\prime}(c)|\lt\frac{\epsilon}{2}\Rightarrow -\frac{\epsilon}{2}\lt L-f^{\prime}(c)\lt \frac{\epsilon}{2}\Rightarrow f^{\prime}(c)-\frac{\epsilon}{2}\lt L\lt f^{\prime}(c)+\frac{\epsilon}{2}$.

But $-\frac{\epsilon}{2}\lt f^{\prime}(c)\lt \frac{\epsilon}{2}$ so $-\epsilon = \frac{-\epsilon}{2} + \frac{-\epsilon}{2}\lt f^{\prime}(c)+\frac{-\epsilon}{2}\lt L\lt f^{\prime}(c)+\frac{\epsilon}{2}\lt \frac{\epsilon}{2} + \frac{\epsilon}{2}=\epsilon$.

Therefore, $|L|\lt \epsilon$ and must be equal to $0.$

The proof can be done by contradiction.

Assume that $L\neq 0$, and without loss of generality you can assume that $L>0$ (otherwise use $g(x)=-f(x)$). Since $\lim_{x\to\infty}f(x)=L>0$, for small enough $\epsilon>0$ such that $L-\epsilon>0$, there exist $N>0$ such that for all $x>N$ $f'(x)>L-\epsilon>0$.

Now we have a differentiable function on an open interval $(N,\infty)$ such that $f'(x)>0$ we can conclude that $f$ is increasing, but this would imply that $\lim_{x\to\infty}f(x)=\infty$ obtaining a contradiction.


If $\operatorname{Im}(A)=\operatorname{Im}(A^2)$, what's the relation between $\operatorname{Im}(A^3)$ and $\operatorname{Im}(A^2)$? Let $A$ be a square matrix. If $\operatorname{Im}(A) = \operatorname{Im}(A^2)$, then what is the relation between $\operatorname{Im}(A^3)$ and $\operatorname{Im}(A^2)$? Here $im(A)$ denotes the columnspace of the matrix $A$.

The answer is $\operatorname{Im}(A^2) = \operatorname{Im}(A^3)$.

But I don't know how to obtain the answer. Please anyone help me solve it.  Thanks in advance.

If $v\in\operatorname{Im}(A^3)$, then $v=A^3(w)$ for some vector $w$. But then $v=A^2\bigl(A(w)\bigr)$, and therefore $v\in\operatorname{Im}(A^2)$.

And if $v\in\operatorname{Im}(A^2)$, then $v=A\bigl(A(w)\bigr)$ for some vector $w$. But $A(w)=A^2(u)$ for some vector $u$ (since $\operatorname{Im}(A^2)=\operatorname{Im}(A)$, and therefore $v=A^3(u)\in\operatorname{Im}(A^3)$.

We are given that  $\textrm{im}(A) = \textrm{im}(A^2)$. By definition, this means that $\forall$ $X$ $\in$ $\mathbb{R^n}$, we have $AX = A^2X$.
This equation implies that $A = A^2$ for any arbitrary $X$.
Hence, we have that $\forall$ $X$ $\in$ $\mathbb{R^n}$: $ A^3 X = (A A^2) (X) = (AA)(X) = A^2X$. Therefore, $\textrm{im}(A^2) = \textrm{im}(A^3)$, as desired.


Continuous one-to-one mapping of a compact space Prove that every continuous one-to-one mapping of a compact space is topological.

Does this problem statement refer to a mapping of a compact space to itself?

If so, suppose the mapping is $f:A\rightarrow A$, where $A$ is compact and $f$ is continuous and 1-1. We must show that $f$ is also onto (and hence bijective), and the inverse image $f^{-1}$ is continuous.

For any compact subset $B\subseteq A$, the set $f(B)$ is compact. Since closed subsets of a compact set are compact, for any closed subset $B\subseteq A$ , the set $f(B)$ is compact (and so, closed and bounded). How might that help?

This means that if $f:A\to B$ is continuous and one-to-one, and if $A$ is compact, then $f$ maps $A$ homeomorphically to $f[A]$. As written, though, it is incorrect. We need to know that $B$ has the property that compact subsets are closed. That's what will allow the proof to work.

It should be clear that such an $f$ maps $A$ bijectively and continuously to $f[A]$ (considered as a subspace of $B$). It remains to show, then, that $f$ maps closed subsets of $A$ to (relatively) closed subsets of $f[A],$ since this is equivalent (why?) to showing that $g:f[A]\to A$ given by $g(x)=f^{-1}(x)$ is a continuous function.

As a side note, "bounded" doesn't necessarily make sense for general spaces. It does suggest that you may be working with metric spaces, however, rather than arbitrary topological spaces. If that is the case, then there's no need to make the additional assumption that compact subsets of $B$ are closed, because this is always true in metric spaces.

This question comes from an exercise (Ex 3, pg 66) of Ahlfor "complex Analysis, 2nd edition".

In Rudin "Principle of Mathematical Analysis, 3rd ed" there is a theorem (Theorem 4.17, page 90) state that:

  Suppose $f$ is a continuous $1$-$1$ mapping of a compact metric space $X$ onto a metric space $Y$. Then the inverse mapping $f$ inverse is a continuous mapping of $Y$ onto $X$.

According to this theorem we need onto in the statement of the exercise or just show for the function $f\colon A \to f[A]$.


What's the expected number of coin tosses in order to get a sequence HHTTHH? Assume you have a fair coin. What's the expected number of coin tosses in order to get a sequence HHTTHH? (H=head,T=tail).

Assuming you start tossing and keep going until your last six tosses match the sequence.

I want to know if there is a general formula for this kinds of problems?

The answer is $70=2^{\color{red}{6}}+2^{\color{red}{2}}+2^{\color{red}{1}}$. The integers $6$, $2$ and $1$ are the lengthes of the prefixes of the word HHTTHH that are also its suffixes, here HHTTHH, HH and H.

For more details, see some previous posts on the site about this exact model, or the book DNA, Words and Models by Robin, Rodolphe, and Schbath (2005), or the survey Enumeration of strings (1985) by A. Odlyzko (see section 4, citing the paper A combinatorial identity and its application to the problem concerning the first occurrences of a rare event (1966) by A. D. Solov’ev).

The correct answer is 68. The expected value of any pattern W of length m, comprised of any symbols, W=w1w2..wm, is given by the following formula:

E[T]=[1/p(W)][1+Sum p(k) over all k such that k is a period of W],

where p(k)=p(w1)p(w2)...p(wk), and P(W)=p(m). In coin tossing experiments, only two symbols are involved, H and T. If W = HHTTHH, then W has only one period, k=4. Assuming a fair coin (which need not be assumed in using the formula),
p(k)= p(4)=1/(2^4) and 1/p(W) = 2^6. Substituting these values into the above formula gives

E[T] = 2^6[1 + 1/(2^4)] =  2^6[17/(2^4)] = (2^2)(17) = 68.


Is there a connection between uncountable sets and exponential growth? Let $S_n$ be the collection of all binary strings of length $n$.  It seems that as $n$ goes to infinity, $S_n$ becomes the set of all infinite binary strings.  Each time we increment $n$, the size of $S_n$ doubles, so $|S_n|$ is undergoing "exponential growth" with respect to $n$.

Similarly, let $C_0$ be the set $\{0, 1\}$, let $C_1$ be the set $\{0, \frac{1}{3}, \frac{2}{3}, 1\}$, let $C_3$ be the set $\{0, \frac{1}{9}, \frac{2}{9}, \frac{1}{3}, \frac{2}{3}, \frac{7}{9}, \frac{8}{9}, 1\}$, and so on.  It seems that as $n$ goes to infinity, $C_n$ becomes the Cantor set.  Again, each time we increment $n$, the size of $C_n$ doubles, so $|C_n|$ is undergoing "exponential growth" with respect to $n$.

In both of these cases, the "limiting result" of "exponential growth" is an uncountable set.

In contrast, let $K_n$ be the set of $k$-tuples of $\{1, 2, ..., n\}$.  The size of $K_n$ only undergoes "polynomial growth", and in this case, the "limiting result" (set of all $k$-tuples of positive integers) is a countable set.

  Is there a connection between uncountable sets and exponential growth?

In particular, I'm wondering if there any interesting results that generalize or extend the observations I made above.

You are mistaken in your first assessment. The limit case is countable.

The limit case of $S_n$ is not $2^\omega$, the set of infinite binary strings, but rather $2^{<\omega}$, the set of finite strings of arbitrary length. Similarly, $C_n$'s don't limit as the Cantor set, but rather as the set of endpoints of removed intervals, which is just a countable set.

The issue here is the same in all cases, $$\sup n^k\neq n^{\sup k}.$$

Here is an example of a connection between uncountable sets and exponential growth. Consider a binary tree of infinite depth.  Each path down the tree codes a binary string of infinite length, and all possible binary strings of infinite length are accounted for by the paths in the tree.  The Cantor Diagonal Argument shows that the set of all possible paths cannot be put into a 1-to-1 correspondence with the Natural Numbers, in other words not countable.  The branch/node count grows exponentially as one moves down the tree (it doubles at each subsequent level of the tree.)


Given the distribution of a random variable $R$, who do you get a uniform random variable $U$? Let us say you have a random variable $R$. How would one generate a uniform random variable $U$, with the maximum possible entropy (or infinite entropy, if $R$ has such)? (For simplicity, you may assume that $R$'s sample space is a subset of the real numbers.)

Note: $U$ is essentially gotten by applying a function to $R$. I believe what I am seeking is called a "randomness extractor", specifically one that preserves entropy, and works based a probability distribution.

Here's an idea, assuming that you don't know anything about the distribution of $R$.

Let $(A_i)_{i\in\mathbb{N}}$, $(B_i)_{i\in\mathbb{N}}$ be samples of $R$, i.e. i.i.d. random variables with the distribution of $R$, and set $$
  C_i = \begin{cases}
    -1 & \text{if $A_i < B_i$,} \\
    1 & \text{if $A_i > B_i$,} \\
    0 & \text{if $A_i = B_i$.} \
  \end{cases}
$$
Then let $D_i$ be the $i$-th value of $C_i$ which isn't zero.

By construction, the $C_i$ are mutually independent, and so are thus the $D_i$. By symmetry, $P(C_i = -1) = P(C_i = 1)$, so the $D_i$ are uniformy distributed on $\{-1,1\}$. You can then use the $D_i$ to produce uniformy distributed values on larger probability spaces.


Prove that the convex hull is the union of all the triangles determined by triples of points from X I'm trying to prove that the convex hull of a set X of three or more points in the plane is the union of all the triangles determined by triples of points from X, however I can't think of the meaningful approach to go with. And now I'm really interested what kind of theorems or rules would explain how is the convex hull a union of all the triangles determined by triples of points from X.

(1) Every triangle must be a subset of the convex hull.  (2) Suppose some edge of the convex hull doesn't belong to any triangle -> contradiction.


show that$ U=\{(x,y) \text{is an element of} \mathbb{R^2}:x^2+4y^2<4\}$ is open by finding a ball around each point which is contained in U I am trying to figure out how to do this one.I know that it is an open set but I don't know how to prove that it is open. I know how to show the ball around each point contained in U but not the overall graph of the ball and what I need to show that it is open.

For any point in U there is a minimum distance to the boundary of the ellipse which is > 0. Just make your neighborhood smaller than that value and it will be open.


If $f'$ is increasing and $f(0)=0$, then $f(x)/x$ is increasing Let $a>0$ and $f:[0,a] \to \mathbb{R}$ continuous function that is twice differentiable on $(0,a).$ Also $f(0)=0$ and $f'$ is strictly increasing function on $(0,a).$

I have to show that the function $g$ defined as $$g(x) = \frac{f(x)}{x}$$ is strictly increasing on $(0,a].$

Progress: I computed $g'$ and got $g'(x) = \frac{f'(x)x-f(x)}{x^2}.$ I have to show that $g'(x)>0 \ \ \forall x \in (0,a].$  Since $x^2$ is always positive for $x \in (0,a]$ I have to show that $f'(x)x-f(x) >0 \ \ \forall x \in (0,a].$ I know both terms are positive, but I don't know how to show that $f'(x)x>f(x) \ \ \forall x \in (0,a].$

By Mean Value Theorem we have for $0 < x < a$ $$\frac{f(x)}{x} = \frac{f(x) - f(0)}{x - 0} = f'(\xi) < f'(x)$$ because $f'$ is strictly increasing and $0 < \xi < x$. Thus $xf'(x) - f(x) > 0$ for all $x \in (0, a)$. And therefore $$g'(x) = \frac{xf'(x) - f(x)}{x^{2}}> 0$$ and hence $g(x) = f(x)/x$ is strictly increasing in $(0, a]$. There is no need to assume existence of $f''(x)$.

Hint: The function $g$ is strictly increasing because $f$ is strictly convex. Since $f'>0$, then...


Strange Dice probability I started a discussion with some friends trying to understand: if I roll 1 dice of 10(1d10) and 1 dice of 4(1d4) 20 times, and do it again 1000 times. Why the d4 have the probability to score the average(50) or more, more times than the d10???

I also wrote a java code to try to understand that take a look:

double totalDfour = 0;
        double plusDfour = 0;
        double avarageDfour = 50;
        int dfour = 0;

        double totalDten = 0;
        double plusDten = 0;
        double avarageDten = 110;
        int dten = 0;

        Random generator = new Random();

        for (int y = 1; y <= 1000; y++) {
            plusDten = 0;
            plusDfour = 0;
            for (int x = 1; x <= 20; x++) {

                dfour = generator.nextInt(4) + 1;
                Log.d("DADO", " d4: " + dfour);
                plusDfour = plusDfour + dfour;

                dten = generator.nextInt(10) + 1;
                Log.d("DADO", " d10: " + dten);
                plusDten = plusDten + dten;

            }

            if (plusDten >= avarageDten) {
                totalDten++;
            }

            if (plusDfour >= avarageDfour) {
                totalDfour++;
            }

        }
        Log.d(TAG, "Total d10= " + totalDten + " Total d4= " + totalDfour);

And always totalDfour is bigger than totalDten, and i would like to understand why if both have the same probability to score the average number or more.

The average of $1d10$ is $5.5$, so the average of $10d10$ is $55$. You have the average of $20d4$ correct.  So the chance $20d4 \gt 50$ is less than $\frac 12$ (it is $\frac 12$ less half the chance of exactly $20$ by symmetry), while the chance $10d10 \gt 50$ is greater than $\frac 12$

The probability that a number of dice rolls produces at least the average can be divided into two components:

The probability that the total is exactly the average. Call this probability $X_4$ for the D4 rolls and $X_{10}$ for the D10s.
The probability that the total is more than the average. Call this $Y_4$ for the D4 rolls and $Y_{10}$ for the D10s.

The probability that the total is more than the average is equal to the probability that the total is less than the average, by symmetry (each possibility that produces more than the average corresponds to a possibility that produces less, and vice versa). Thus, we have

$$X_4+2Y_4=X_{10}+2Y_{10}=1$$
$$X_4+Y_4=\frac{1+X_4}2$$
$$X_{10}+Y_{10}=\frac{1+X_{10}}2$$

$X_4$, the probability that the D4s roll exactly their average, is higher than $X_{10}$, the probability that the D10s roll exactly their average. Intuitively, you can think of the distribution being "chunkier"; there are fewer possibilities for the D4s, so the average happens more. (This isn't very precise, but that's intuition for you.) Thus, $X_4+Y_4$, the probability the D4s roll at least their average, is higher than $X_{10}+Y_{10}$, the corresponding probability for the D10s.

Note that all this depends on the expected value of the dice distribution actually being in the distribution. If you change the number of rolls to, say, 21, then it won't actually be possible to roll a result of 52.5 or 115.5, and the results will be more like you might have expected.


Solution to initial condition problem Let $y(t)=-\ln(1-e^{(t+c)})$.
  I'm trying to find the solution to the initial condition $y(0)=-\ln 2$.

Isolate $c$:
$$0=\ln(2)-\ln(1-e^c)=\ln\left({2\over1-e^c}\right)$$
therefore
$$-e^c=2-1 \leftrightarrow e^c=-1 \leftrightarrow c=0$$
I can't figure out where I'm going wrong

Hint: You should rewrite the equation by replacing $e^c=a$.

$$y=-\ln(1-ae^t)$$

$c=0$ is wrong. You cannot find a real solution for $c$ because $e^c\ge 0\,$ for real $c$. A complex solution is $c=\pi i,\,$ i.e. the principal value of $\ln(-1).$


Use L'Hopital's Rule to Prove Let $$f: \mathbb R\rightarrow \mathbb R$$ be differentiable, let a in $\mathbb R$. Suppose that $f''(a)$ exists. Prove that $$\lim_{h\rightarrow0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2}=f''(a) $$

Suppose further that $f''(x)$ exists for all $x$, and that $f'''(0)$ exists. Prove that $$\lim_{h\rightarrow0}\frac{4(f(h)-f(-h)-2(f(h/2)-f(-h/2)))}{h^3}=f'''(0)$$

Question 1
$$
\lim_{h\rightarrow0}\frac{f(a+h)-2f(a)+f(a-h)}{h^2} \\
$$
when $h = 0$ is substituted numerator and denominator reduce to $0$. So, applying L'Hopital's rule (differentiate wrt $h$)
$$
\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a-h)}{2h} \\
\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a)-f'(a-h)+f'(a)}{2h} \\
\lim_{h\rightarrow0}\frac{f'(a+h)-f'(a)}{2h}+\lim_{h\rightarrow0}\frac{f'(a)-f'(a-h)}{2h} \\
= \frac{2f''(a)}{2} = f''(a)\\
$$
Question 2
$$
\lim_{h\rightarrow0}\frac{4(f(h)-f(-h)-2(f(h/2)-f(-h/2)))}{h^3}
$$
when h=0 is substituted numerator and denominator reduce to 0. So, applying L'Hopital's rule (differentiate wrt h)
$$
\lim_{h\rightarrow0}\frac{4(f'(h)+f'(-h)-f'(h/2)-f'(-h/2))}{3h^2} \\
= \lim_{h\rightarrow0}\frac{4(f''(h)-f''(0)+f''(0)-f''(-h)+\frac{1}{2}f''(0)-\frac{1}{2}f''(h/2)-\frac{1}{2}f''(0)+\frac{1}{2}f''(-h/2))}{6h} \\
= \lim_{h\rightarrow0}\frac{4(f''(h)-f''(0)+f''(0)-f''(-h)+\frac{1}{2}f''(0)-\frac{1}{2}f''(h/2)-\frac{1}{2}f''(0)+\frac{1}{2}f''(-h/2))}{6h} \\
= \frac{2}{3}\left(\lim_{h\rightarrow0}\frac{f''(h)-f''(0)}{h}+\lim_{h\rightarrow0}\frac{f''(0)-f''(-h)}{h}\right)+\frac{1}{3}\left(\lim_{h\rightarrow0}\frac{f''(0)-f''(h/2)}{h}+\lim_{h\rightarrow0}\frac{f''(-h/2)-f''(0)}{h}\right) \\
= \frac{4f'''(0)}{3}-\frac{f'''(0)}{3} = f'''(0)
$$
PS: Thanks to @Paramanand Singh for pointing out that f was not twice and thrice differentiable in first and second questions respectively.


Finding the side length of an equilateral triangle having three inscribed $120^\circ$ sectors in a certain arrangement How do i even start this question? I thought of using length of tangent equal from exterior point but still of no use.

Let $G$ be the common centroid for the two triangles.

Pick an edge, say $AB$, let $H$ be the vertex on the small triangle which is the
apex for the circular sector touching edge $AB$. Let $G'$ and $H'$ be the foot of
$G$ and $H$ on edge $AB$.

It is clear the distance of $G$ to line $AB$ is

$$|GG'| = |HH'| - |GH|\cos(\theta + \frac{\pi}{6})$$

where $\theta$ is the angle illustrated in above diagram.

We know $|HH'| = 20$, the radius of the circular sectors. Since the small triangle has side
$9$, we also know $|GH| = 3\sqrt{3}$. In the diagram above, we find

$$\cos\theta = \frac{|HH'|}{|DH|} = \frac{20}{29}\quad\implies\quad\sin\theta = \frac{21}{29}$$
Combine these, we get

$$\begin{align}|GG'|
&= 20 - 3\sqrt{3}\left(\cos\theta\cos\frac{\pi}{6} - \sin\theta\sin\frac{\pi}{6}\right)\\
&= 20 - 3\sqrt{3}\left(\frac{20}{29}\cdot\frac{\sqrt{3}}{2} - \frac{21}{29}\cdot\frac{1}{2}\right)\\
&= \frac{980+63\sqrt{3}}{58}
\approx 18.77791725649723
\end{align}
$$

As a corollary,

$$|AB| = |BC| = |CA| = 2\sqrt{3}|GG'| = \frac{189+980\sqrt{3}}{29} \approx 65.04861349715516$$

Slide the third circles along the sides they tangent to - you will see that they merge into the inscribed circle. So the radius of the inscribed circle is 20.

The side of the triangle is than - $AB=2\times \sqrt{3}\times 20 $


Find coordinates of a point on a square I have been given the following question.

I've got no problem calculating the area of the square. However, I don't understand how to find points $A$ or $D$ in order to identify the equation of the line $AD$, so that I can then get the coordinates of $S$ by plugging in the $y$ value.

Furthermore, is there an equation where you can input distance, gradient & the coordinates of one point to output the coordinates of another point?

Thanks.

it is easy to find the line AB since it is perpendicular to CB, you know the length of AB so you find A, than the Line parallel to CB cuts the x- Achse in S
next question: if you use vectors to describe lines  and the directionvektor a unit vector,
in your first question you also could find A bei adding the Vektor BA to A.

I think that the triangle SAB and EBC are the same where E is intersection of y-axes and AB. You can easily compute the length of [EC]. After you need only make a substraction.


Graph and derivative of y=x I'm having trouble understanding why the graph of $y=x$ is different from the graph of $y = \sqrt{ x^2 }$. Aren't both equations the same once you simplify the second one? And isn't the derivative of each the same?

It's different. $$y=\sqrt{x^2}=|x|$$ that is different from $y=x$
The derivate of $y=x$ is $y=1$ while the derivate of $y=|x|$ is $y=1 $ for $x>0$ and $y=-1$ for $x<0$

Consider $x=-1$:

$
\begin{align}
x&=-1 \\
\sqrt{x^2}&=\sqrt{(-1)^2} =\sqrt{1}=\pm1
\end{align}
$

Every root has two solutions: $\sqrt{a}=\pm a$, for example
$2 \cdot 2 = -2 \cdot -2 = 4$

Because a function can only have one $y$ for every $x$, one solution (the negative) has to be omitted. This causes the graph of $y=\sqrt{x^2}$ to always be positive.

If you would plot a graph with both the positive and negative solution of $y=\sqrt{x^2}$, it would look like $y=x$ and $y=-x$ united.


Prove that this determinant is $0$ $$
\begin{vmatrix}
\cos 2x & \cos 2y & \cos 2u \\
\sin^2 x & \sin^2 y & \sin^2 u \\
1 & 1 & 1
\end{vmatrix} = 0
$$

This is the conclusion where I got from another excercise and I want to prove me it is right. Please don't use the fact that $\cos(2x)=\cos^2x-\sin^2x$. We haven't learned it yet.

Here's how easy it is if you do use $\cos(2x) = 1-2\sin^2(x)$:

$$\begin{align}\left|\begin{matrix} \cos(2x) & \cos(2y) & \cos(2u) \\ \sin^2(x) & \sin^2(y) & \sin^2(u) \\ 1 & 1 & 1\end{matrix} \right| &= \left|\begin{matrix} 1-2\sin^2(x) & 1-2\sin^2(y) & 1-2\sin^2(u) \\ \sin^2(x) & \sin^2(y) & \sin^2(u) \\ 1 & 1 & 1\end{matrix} \right| \\ &= \left|\begin{matrix} 1 & 1 & 1 \\ \sin^2(x) & \sin^2(y) & \sin^2(u) \\ 1 & 1 & 1\end{matrix} \right| - 2\left|\begin{matrix} \sin^2(x) & \sin^2(y) & \sin^2(u) \\ \sin^2(x) & \sin^2(y) & \sin^2(u) \\ 1 & 1 & 1\end{matrix} \right|\end{align}$$

This equals zero because the determinant of any matrix with a repeated row (or column) is $0$.

So maybe it's time to learn some trig.

By using the equality $cos(2x) = 2cos^2(x)-1 =1-2sin^2(x)$,
-2 second line + third line = first line


Calculus proof for the area of a circle I was looking for proofs using Calculus for the area of a circle and come across this one
$$\int 2 \pi r \, dr = 2\pi \frac {r^2}{2} = \pi r^2$$
 and it struck me as being particularly easy. The only other proof I've seen was by a teacher and it involved integrating $x = \sqrt{r^2 - y^2}$ from $-1$ to $1$, using trig substitutions and then doubling the area to get $\pi r^2$ but the above proof seemed much more straight forward.

Is it a valid proof, or is it based on circular logic or some other kind of fallacy?

The above integral seems geometrically as below figure.

\begin{align}
\int_0^r 2\pi r\, dr& = 2\pi\int_0^r\ r\, dr \\
 & = 2\pi\bigg\lvert_0^r\ \frac{r^2}{2} \\
 & = \bigg(2\pi \frac{r^2}{2}\bigg)-\bigg(2\pi \frac{0^2}{2}\bigg) \\
 & = \frac{2\pi r^2}{2} \\
 & = \require{cancel} \frac{\cancel{2}\pi r^2}{\cancel{2}} \\
 & = \color{red}{\pi r^2}
\end{align}

Consider the standard equation of circle $$x^2 + y^2 = a^2$$ where $a$ is the radius of the circle.


Find the general solution of: $y^"+3y^{'}+2y=\sin(x)$ (I'm using the Annihilator Method) The characteristic eqn corresponding to LHS is $m^2+3m+2=0$. The solutions are: $m = -2$ and $m = -1$. So $y_c=Ae^{-2x}+Be^{-x}$.

The Annihilator of $sin x = D^2+1 $. Which makes $y_p=Ccosx +Dsinx$. Where $A, B, C, D$ are constants.

To find the constants $C$, and $D$. I found out $y_p^{'} =-Csinx+Dcosx$
    and $y_p^{''} =-Ccosx-Dsinx$. I put these values in the original eqn ($y^"+3y^{'}+2y=sinx$). I'm not getting the right values. C should be -3/10 and D should be 1/10.

Any clues as to where I'm going wrong?

Thanks.

sorry for bad quality but the answer is correct.


Find the least positive integer $n$ such $S(n^2)+7=S(n)$ This is contest math problem:

  For a positive integer $N=\overline{a_{n}a_{n-1}\cdots a_{0}}$ in decimal representation we denote by $S(N)$ the sum of its digits $a_{0}+a_{1}+\cdots+a_{n-1}+a_{n}$. then find the minimal $n$ such
  $$S(n^2)=S(n)-7$$

It is well known that
$$S(n)\equiv n\pmod 9$$

Then, the equation implies:
$$n^2\equiv n-7\pmod9$$
Note that this implication is not reversible, so we will need to check the solutions we can find.
The latter equation can be written so:
$$n^2-10n+25\equiv 0\pmod 9$$

So $(n-5)^2\equiv 0\pmod 9$. Therefore,
$$n-5\equiv 0\pmod 3$$
So $n\equiv 2\pmod 3$.

On the other hand, note that
$$S(n)=S(n^2)+7\ge 9$$
and since $n\equiv 2\pmod 3$, $S(n)\ge 11$.

Trying a bit yields the solution: $149$.

149
179
389
449
548
749
899

And many others

java code:

public class PrecalcMS {
public static void main(String[] args) {
    LongStream.range(0, 100000).parallel().filter(x->digitsum(x)-digitsum(x*x)==7).forEach(x->System.out.println(x));
}

private static int digitsum(long x) {
    final String pres = ""+x;
    return IntStream.range(0, pres.length()).map(i->pres.charAt(i)-'0').sum();
}

}


factorization of $x^{15}-1$ in $\mathbb{Z}/11\mathbb{Z}$ how can we find factorization of $x^{15}-1$ in $\mathbb{Z}/11\mathbb{Z}$?

I already factorize
$x^{15}-1=(x-1)(x^2+x+1)(x^4+x^3+x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)$ in  $\mathbb{Z}[x]$.

it was relatively easy factorizing in $\mathbb{Z}/2\mathbb{Z}$
$x^{15}-1=(x-1)(x^2+x+1)(x^4+x^3+x^2+x+1)(x^4+x^3+1)(x^4+x+1)$ in $(\mathbb{Z}/2\mathbb{Z})[x]$.

thank you very much in advance.

This can be done by hand with the aid of a bit of ad hoc -trickery.

As John Hughes and Gerry Myerson explained, the polynomial $x^5-1$ splits into linear factors over the field $F=\Bbb{Z}_{11}$. The multiplicative group $K^*$ is cyclic of order ten (we can easily verify that $2$ is a generator), so all the non-zero squares in $K$ are zeros of $x^5-1$. This gives us the linear factors
$$
x-1,x-4,x-9,x-16=x-5,\ \text{and}\ x-25= x-3
$$
of $x^{15}-1$.

As $3\mid 15$, the third cyclotomic polynomial $x^2+x+1=(x^3-1)/(x-1)$ is also a factor. The fun part is to factor the degree eight cyclotomic polynomial
$$
\Phi_{15}(x)=x^8-x^7+x^5-x^4+x^3-x+1.
$$
Its zeros (in some extension field of $K$) are roots of unity of order $15$, and the factors are the minimal polynomials of the said roots of unity. We recall that the multiplicative group of the quadratic extension field $K=\Bbb{F}_{11^2}$ is also cyclic. More precisely, $K^*\simeq C_{120}$. As $15\mid 120$, all the fifteenth roots of unity are in $K$. Because $[K:F]=2$, the minimal polynomials are all quadratic. So we know that $\Phi_{15}(x)$ factors into a product of four irreducible quadratics.

Let $\alpha$ be one of the zeros of $\Phi_{15}$ in $K$. By the Galois theory of extensions of finite fields (Frobenius automorphism) we know that the other zero of the minimal polynomial $m_\alpha(x)$ over the prime field is $\alpha^{11}$. So
$$
m_\alpha(x)=(x-\alpha)(x-\alpha^{11})=x^2-[\alpha+\alpha^{11}]x+\alpha^{12}\in F[x].
$$
Next we observe that the constant term $\beta=\alpha^{12}$ is a root of unity of order five. As $\alpha$ ranges over the set of roots of unity of order fifteen, $\beta$ will range over the roots of unity of order five (if $\alpha$ corresponds to $\beta=\alpha^{12}$ then $\alpha^k, k=2,4,8$, corresponds to $\beta^k$). So we have deduced that the constant terms of the four quadratic factors are $4,9,3$ and $5$ in some order.

The remaining task is to figure out the relation between the coefficient $\gamma=\alpha+\alpha^{11}$ of the linear term and that of the constant term $\beta=\alpha^{12}$. Up to this point we have been running on an autopilot, but now I needed trickery (but there may be a more general clever way that I have missed). Let us write
$\omega=\alpha^5$. It is a third root of unity in $K$. The other third root of unity is $\omega^2=\alpha^{10}$. They are roots of $\Phi_3(x)=1+x+x^2$, so we have the relation
$1+\omega+\omega^2=0$. This comes handy as it implies that
$$
\begin{aligned}
\gamma^2&=(\alpha+\alpha^{11})^2\\
&=\alpha^2+2\alpha^{12}+\alpha^{22}\\
&=\alpha^{12}+\alpha^2(1+\alpha^{10}+\alpha^{20})\\
&=\beta+\alpha^2(1+\omega^2+\omega)\\
&=\beta.
\end{aligned}
$$
So we know that $\gamma$ is one of the square roots of $\beta$. As $\beta^5=1$ we know that $(\beta^3)^2=\beta$, so the alternatives are $\gamma=\pm\beta^3$. Which one is it?
Of the candidates $\beta^3$ has order five but $-\beta^3$ is a tenth root of unity. We can settle this by calculating $\gamma^5$:
$$
\begin{aligned}
\gamma^5&=(\alpha+\alpha^{11})^5\\
&=\alpha^5+\binom 51\alpha^{15}+\binom 52\alpha^{25}+\binom53\alpha^{35}+\binom54\alpha^{45}+\alpha^{55}\\
&=\omega+5+10\omega^2+10\omega+5+\omega^2\\
&=10+11\omega+11\omega^2\\
&=-1.
\end{aligned}
$$
Therefore $\gamma$ always has order ten, and we can deduce that $\gamma=-\beta^3$. Thus
$$
m_\alpha(x)=x^2+\beta^3x+\beta.
$$
Plugging in $\beta=4,9,5,3$ gives the respective factors
$$
x^2+9x+4,x^2+3x+9,x^2+4x+5,\quad\text{and}\quad x^2+5x+3.
$$


Is $X$ reflexive if $X=\cup Y_i $ and $Y_i$ is reflexive for every $i$? Let $X$ be a Banach space. Assume that there are a family of closed (with respect to $\left\|\cdot\right\|_X$) subspaces $Y_i$ of $X$ such $X=\cup Y_i $ and $Y_i$ is reflexive for every $i$. Can we say that $X$ is reflexive? (otherwise, under what condition,  $X$ is reflexive).

The answer is (somewhat surprisingly) yes. The reason it's yes is because the assumption $X=\bigcup\limits_{i}Y_i$ is a very strong assumption which greatly reduces which sets $Y_i$ we can consider. Assume that $X = Y_1\cup Y_2$ as a simple example. Then, without loss of generality, $Y_2\subset Y_1$ since otherwise we can take $y_2\in Y_2\backslash Y_1$ and $y_1\in Y_1\backslash Y_2$ to have that $y_1+y_2\in X$ but $y_1+y_2\not\in Y_1\cup Y_2$.

So if $X$ can be written in this way, there must be a "biggest" $Y_k$ which contains all the other $Y_i$. In this case, however, we have that $\bigcup\limits_{i} Y_i = Y_k$ which is reflexive by assumption and so $X=Y_k$ means that $X$ is also reflexive.

To make sense out of this question you have to replace the condition $X=\cup_i Y_i$ by the condition that $X$ is the closed subspace generated by reflexive subspaces $Y_i$. But then $X$ need not be reflexive as seen  by considering the collection of all one dimensional subspaces. Of course, there is the trivial case where there is only one $Y_i$ and $X$ is of course, reflexive in that case.


Why the extension of distribution T in $(D(\Omega))^{'}$ to dual of sobolev space $W^{k,p}(\Omega)$ not unique? I am trying to understand section about dual of sobolev spaces  from Adam's book "Sobolev Spaces". As I understand it he states that every element in the dual of $W^{k,p}(\Omega)$, $\Omega \subset \mathbb{R}^{n}$ open,  can be represented by a vector $v=(v_1,v_2,....,v_N)$ where each $v_i$ is in $L^{q}$ and q is the conjugate exponent of p.N is the number of multi derivate indices that depend on k and n. This also represents an element in the space of distributions $(D(\Omega))^{'}$. But however he says that this representation need not be unique. Can you give an example where there are more than one such representation for elements in the dual $(W^{k,p}(\Omega))^{'}$?

Edited:

Is it possible that there are two different elements in $(W^{k,p}(\Omega))^{'}$, $T_1$ and $T_2$ such that they agree on $C_0^{\infty}(\Omega)$. ie is it possible that $T_1{\phi}=T_2{\phi}$ for all $\phi$ in $C_0^{\infty}(\Omega)$ but $T_1{u} \neq T_2{u}$ for some $u$ in $W^{k,p}(\Omega)$.

The relevant section can be found in chapter 3 of the book Adams, R.A. and Fournier, J.J.F, Sobolev Spaces, 2nd ed., Academic Press, 2003.

The non-uniqueness refers to the vector $v$. Note that for $f\in W^{1,p}(\Omega)$ and $\phi \in C_0^1(\Omega)$ we have
$$\int_\Omega \nabla f \cdot \phi d x = - \int_\Omega f \mathop{div} \phi d x.$$
In other words, if you find a function $\phi \in C_0^1(\Omega)$ with $\mathop{div} \phi = 0$, (and there are many of those) you can add it to the respective components of $v$ without changing the operator. The same of course also can be done in the higher derivatives.


how to prove $G$ is an abelian group under $*$ (called the real numbers mod 1) Let $G = \{x \in \mathbb{R}~|~0\leq x < 1\}$ and for $x,y \in G$ let $x*y$ be the fractional part of $x+y$ i.e $x*y = x + y - [x + y]$ where $[a]$ is the greatest integer less than or equal to $a$. I need help proving $*$ is a well defined binary operation on $G$ and that $G$ is an abelian group under $*$.

To prove that $*$ is well defined, I rely on the assumption that +,- is well defined in the set of real numbers. evaluating the brackets [] is also well-defined in the set of real numbers. (is this an okay assumption?)

To prove $*$ is associative, (I imagine because +,- is associative, so will *?) I show that $(x*y)*z = x*(y*z)$ $$(x*y)*z =(x+y-[x+y])*z = x+y-[x+y]+z - [x+y-[x+y]+z]$$ $$x*(y*z) = x*(y+z - [y+z]) = x + y + z - [y+z] - [x+y+z - [y+z]]$$ rearranging $$(x*y)*z = x+y+z-[x+y]-[x+y+z-[x+y]]$$ $$x*(y*z) = x+y+z-[y+z]-[x+y+z-[y+z]]$$ I am not so sure that these two are equal. I don't see how distributing the $-$ to remove the terms $[x+y]$ and $[y+z]$ is fair (ex: $x = y = \frac{1}{2}$ so $x*y = .5+.5-[.5]-[.5]$ yields a different answer than $x*y = .5+.5-[.5+.5]$

The identity element would be $0$, the inverse would have to be $-x$ which isn't in $G$. What am I doing wrong?

edit: commutativity would be proven by showing $x*y = y*x$ which is easy to show $x+y - [x+y] = y+x - [y+x]$ right?

To prove $∗$ is associative

Let $a,b,c\in G$.
\begin{align*}
(a*b)*c &= (a+b-[a+b])*c \\
   &= a+b+c-[a+b]-\left[a+b+c-[a+b]\right]
\end{align*}
Similarly,
\begin{align*}
a*(b*c) &= a+b+c-[b+c]-\left[a+b+c-[b+c]\right]
\end{align*}
Given $0\le a,b,c < 1, [a+b]$ and $[b+c]$ must be either zero or one individually, for a total of four possible cases.  In the cases when both are zeros or ones, clearly $(a*b)*c = a*(b*c)$.
Consider the case when $[a+b]=1$ and $[b+c]=0$.  Necessarily $1 \le a+b < 2, b+c< 1$ and $1\le a+b+c< 2\implies [a+b+c] = 1$ and $[a+b+c-1] = 0$. Therefore, $(a*b)* c=a+b+c-1-[a+b+c-1]=a+b+c-1$, whereas $a*(b* c)=a+b+c-[a+b+c]=a+b+c-1$; i.e. $(a* b)* c = a*(b* c)$.
Similar argument applies when $[a+b]=0$ and $[b+c]=1$.  This covers all four cases.
Hence $G$ is associative over $*$.

You can think of your group $(G, *)$ as $(\mathbb{R}, +)/\mathbb{Z}$: That is, the reals under addition modded out by the integers.

This clearly works because we are essentially saying that is $x, y \in \mathbb{R}$, $x - y \in \mathbb{Z}$, then $x = \mathbb{Z} + y$ in $\mathbb{R} /\mathbb{Z}$.
This has the same effect as “truncation”, but is now a nice subgroup
Hence,

$$
(G, *) \simeq (\mathbb{R}/\mathbb{Z}, +)
$$

To imagine this, realize that $\mathbb{R}$ just contains shifted copies of $[0, 1)$ all laid down one next to the other. So, we simply “collapse” all these copies together for this construction to work out.

Now, for this to be a group, we need the kernel of the map $$\phi: (\mathbb{R}, +) \to (\mathbb{R}, +) /\mathbb{Z}$$

to be a normal subgroup. Since the kernel of a quotient map is the quotienting subgroup itself, we simply need to
show that $\mathbb{Z}$ is a normal subgroup of $\mathbb{R}$. This is obvious in abelian groups.


Find the remainder when $p(x)$ is divided by $x^2-a^2$ if $p(x)$ leaves remainders $a, -a$ when divided by $x+a, x-a$ Let $a\neq0$ and $p(x)$ be a polynomial of degree greater than $2$. If $p(x)$ leaves remainders $a$ and $-a$ when divided respectively by $x+a$ and $x-a$. Find the remainder when $p(x)$ is divided by $x^2-a^2$

$$
p(x)=q(x).(x+a)+r_1=q(x).(x+a)+a\quad\big[r_1=p(-a)=a\big]\\
p(x)=s(x).(x-a)+r_2=s(x).(x-a)-a\quad\big[r_2=p(a)=-a\big]\\
p(x)=t(x).(x^2-a^2)+r=t(x).(x^2-a^2)+Ax+B\\
p(a)=aA+B=-a\\
p(-a)=-aA+B=a\\
B=0,\;A=-1\implies r=Ax+B=-x
$$
I was wondering Is there another way to solve this problem ?

Hint $\ \overbrace{ x\!-\!a,x\!+\!a\,\mid\, p\!+\!x\!}^{\large p(a)+a\ =\ 0\ =\ p(-a)-a\!\!\!\!\!}\!\iff\! (x\!-\!a)(x\!+\!a)\mid p\!+\!x\,$ over a field where $\,\color{#c00}{2a\neq 0}$

since then we have the Bezout GCD equation $\ x+a - (x-a)\, =\, \color{#c00}{2a}$

which implies that $\,x\!+\!a,\, x\!-\!a\,$ are coprime, so their lcm = product.

Alternatively, $ $ apply: $\ fg\bmod fh = f(g\bmod h) = $ mod Distributive Law (a form of CRT)

$x\pm a\mid p\!+\!x\,\Rightarrow\, p\!+\!x\bmod x^2\!-\!a^2 = (x\!-\!a)\left[\dfrac{p\!+\!x}{x\!-\!a}\bmod x\!+\!a\right] = (x\!-\!a)[0] = 0$

There are two of many ways to solve CCRT = Constant-case of CRT [Chinese Remainder Theorem]

The other answers give good elementary methods. Here's an approach from abstract algebra.

Assume that the allowable coefficients of $p$ are a field $F$ (e.g., $\mathbb{R}$ or $\mathbb{C}$). Define $S = \{ p \in F[x]: p(\pm a) = \mp a\}$. This set includes every  polynomial of degree $3$ or higher that leaves the required remainders. Define $I = \{ p \in F[x]: p(-a) = p(a) = 0\}$. Note that $S = I - x$ (i.e., adding $x$ to any element of $S$ gives an element of $I$, and vice versa), and that $I$ is an ideal. $F[x]$ is a PID, so $I$ must contain all multiples of some primitive element. Linear primitive elements are easy to rule out, and clearly $x^2 - a^2 \in I$, so $I = \langle x^2 - a^2\rangle$. Thus, every element of $S$ is $-x$ plus a multiple of $x^2 - a^2$, and you're done.

This approach gives an obvious generalization:

  Every polynomial $p(x)$ that satisfies $p(x_1) = y_1, p(x_2) = y_2, \ldots, p(x_n) = y_n$ for some fixed quantities $x_i$ and $y_i$ has the form $p(x) = L(x) + \prod_{i=1}^n (x - x_i)$, where $L(x)$ is the Lagrange interpolation polynomial.


Is V a vector space over the field of real numbers with the following operations? $V=R^2$, with operations:

addition ($\oplus$): $(x_1, y_1) \oplus (x_2, y_2) = (x_1 + x_2, y_1 + y_2)$

multiplication($\circ$): $c \circ (x, y) = (cx, y)$

This is exercise II.1.4 from Hoffman-Kunze's Linear Algebra book.  I was able to show that all the axioms are satisfied. Can someone confirm if that is correct?

$\textbf{EDIT}$: I made a mistake in checking one of the distributive properties. It is not a vector space.

The operation $\oplus$ has $(0,0)$ as neutral element, because
$$
(x,y)\oplus(0,0)=(0,0)\oplus(x,y)=(x,y)
$$
If we had a vector space, it would be true that
$$
0\circ(x,y)=(0,0)
$$
for every $(x,y)$. What about $0\circ(1,1)$?

Distributivity of multiplication over addition requires that

$2c\circ(x,y) = (c+c)\circ(x,y) = c\circ(x,y) + c\circ(x,y)$

but

$c\circ(x,y) = (cx,y)$

and

$c\circ(x,y) + c\circ(x,y) = (cx,y) + (cx,y) = (2cx,2y) \ne 2c\circ(x,y)$


Simulation Dickey-Fuller distribution I tried to simulate the Dickey- Fuller statistics with a short Matlab routine.
I set Y0=0 and generated
$$
Y_{t}=Y_{t-1} + \varepsilon_t,
$$
where $\epsilon_t$ is  $\mathcal{N}(0,1)$-distributed. Then I estimate the model
$$
\Delta Y_t = \beta Y_{t-1} + \varepsilon_t
$$
and computed the Dickey-Fuller statistic in order to estimate the quantile.

 function [Y] = genY(n)
 %generate random walk.
 Y=zeros(1,n);
 Y(1)=0;
 for i=2:n
     Y(i)=Y(i-1)+randn;
 end
 end

 function [dist] = estBeta(n)
 %estimate  Dickey-Fuller statistics in one step
 Y=genY(n);
 DeltaY=Y(2:n)-Y(1:(n-1));
 est=(sum(DeltaY(2:end).*DeltaY(1:(end-1))))/(sum((DeltaY(1:(end-1))).^2));
 dist=est/(std(DeltaY)/sqrt(n-1));
 end

 function [quantil] = perc(m)
 %calculation of quantile
 z=zeros(1,m);
 for i=1:m
    z(i)=estBeta(1000);
 end
 quantil=quantile(z,0.01);
 end

If I execute the program I will always end up with -2.3 at p=0.01.
Has anybody an idea how to fix this problem?

I have a solution for the no contant no trend case (you can adapt the code for further model specifications).

Set up the number of replications, I have done this in a for loop so that you can change it to meet your requirements.
Deriving the DF tests state see W.

%% Replicating Dickey and Fuller critical values

for q = [25,50,100,250,500]
    T = q;
    reps = 5000000;
    DFstats = zeros(1,reps);

for i = 1:reps
    u = normrnd(0,1,[1,T]);
    W = 1/sqrt(T)*cumsum(u);
    DFstats(i)=W(T)^2-1/(2*sqrt(mean(W.^2)));
end

CV = sort(DFstats);
ActualValues = CV([(0.01*reps),(0.05*reps),(0.1*reps)]);
ActualTable(:,q)=ActualValues;

end


Determining the formula for a linear map Determine the formula for the following linear map:

$L : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ with $L(1,2) = (0,-1)$ and $L(-1,-1) = (2,1)$.

Attempt at solution: On the basis of these examples I was able to guess the second element: everytime we have that $L(x,y) = (... , -x)$. But this method seems a bit random to me, and I don't know how to compute the first element. Is there some strategy to deal with these kind of problems? Maybe setting up some linear system? Help would be appreciated.

$$L(x,y)=(ax+by,cx+dy)$$
$$L(1,2)=(a+2b,c+2d)=(0,-1)$$
$$L(-1,-1)=(-a-b,-c-d)=(2,1)$$
This becomes two linear systems with two equations, yielding the solution $(a,b,c,d)=(-4,2,-1,0)$. That is, $L(x,y)=(-4x+2y,-x)$.