Probability of picking an odd number from the set of naturals? I know there's no uniform distribution for a countably infinite set, but I'm wondering if there's still a way to determine the probability of picking from a subset of a countably infinite set.

For example, what's the probability that I pick an odd number from the set of naturals, assuming I'm picking randomly? Is this even a coherent question? If so, is there a textbook approach to this problem in measure theory/probability theory/probability measure?

Thanks!

I'd like to give a semi-philosophical, semi-mathematical defense of the answer 1/2.

The received view in contemporary mathematical probability theory, due to Kolmogorov, is that probability functions are normalized measures: nonnegative, countably additive set functions that are defined on $\sigma$-fields and that assign the sure event measure $1$. This hasn't always been the received view, however.

Bruno DeFinetti argued, famously, that the concept of probability does not mandate countable additivity. His argument is based on the fact, pointed out by you and commenters, that countable additivity rules out the possibility of a uniform distribution on the natural numbers. DeFinetti, a subjectivist, thought there was nothing rationally incoherent about distributing one's credences uniformly over $\mathbb{N}$, and, on these grounds, argued that the countable additivity axiom be weakened. For DeFinetti, probabilities need only be finitely additive.

Working in the tradition of finitely additive probability, it is indeed possible to define a uniform distribution on $\mathbb{N}$. The needed extension results can be found in Rao and Rao's Theory of Charges and Hrbacek and Jech's Introduction to Set Theory. I will sketch a construction here so that this answer has some mathematical content (if needed or requested, I will edit later to include more details).

First, we define the natural density, as in comments above, to be the limit (if it exists) $$ \lim_{n \to \infty} \frac{|A \cap \{1,...,n \}|}{n} = :d(A) $$ for $A \subseteq \mathbb{N}$.

It is not difficult to show

Proposition. (i) $d(\emptyset) = 0$ and $d(\mathbb{N})=1$. (ii) $d(\{n\})=0$ for all $n \in \mathbb{N}$. (iii) The set of numbers divisible by $m$ has natural density $1/m$. (NB: Some of this has been asserted in comments above, but I include it here for completeness.)

We then have the following theorem.

Theorem. There exists a finitely additive probability measure $P$ on $\mathscr{P}(\mathbb{N})$ that extends the natural density $d$.

Proof sketch. The proof relies on the existence of the Frechet ultrafilter $\mathcal{U}$ (the ultrafilter of cofinite sets) on $\mathbb{N}$, and is therefore non-constructive. We say a sequence $(x_{n})$ of real numbers is convergent in $\mathcal{U}$ with limit $x$ and write $x = \lim_{\mathcal{U}}x_{n}$ if for all $\epsilon > 0$
$$\{n: |x_{n} - x| < \epsilon  \} \in \mathcal{U}.$$
It can be shown that every real sequence has a unique $\mathcal{U}$-limit. It can also be shown that
$$P(A) = \lim_{\mathcal{U}}\frac{|A \cap \{1,...,n \}|}{n}$$
is a finitely additive probability on $\mathscr{P}(\mathbb{N})$ extending $d$. $\square$

By the Theorem and the Proposition, $P$ is a finitely additive probability measure that assigns measure $1/2$ to the set of odd numbers.

Now, you may object that I've changed the subject by invoking merely finitely additive "probabilities". In response, I would remind you that the concept of probability predates the Kolmogorovian theory, and that there are good arguments (due to DeFinetti and other subjectivists) in favor of relaxing countable additivity in some contexts.

This is a great question that I as well don't know the answer to.

Also the probability cannot be 1. If the probability to choosing an odd number is 1, then the complement (the probability of choosing an even number is 0). However there are the same number of odd numbers and even numbers, therefore we can use the same proof to show that the probability choosing an even number is 1 leading to a contradiction.

I'm guessing the other approach is first defining our measure. We can use the normalized counting measure (density) as our probability measure on $(\mathbb{N}, 2^\mathbb{N})$ we just need the set to be finite. This measure would imply that the probability is the same for each natural number.

$$\mu(A) = \begin{cases}
|A| & \text{if $A$ is finite} \\
\infty & \text{if $A$ is infinite}
\end{cases}$$

The only problem that arises is that our set is infinite which will prevent us from having a our counting measure be a probability measure (we cannot divide by infinity). Maybe I'm mistaken, but it seems in order to solve this problem we can then cut up the natural numbers into a set of intervals, and prove it true for each interval. We then can extend it to projection maps.

We write the projection map as: $$\pi_k(\mathbb{N}) = \{k, k+1\} \text{ for any $k\in \mathbb{N}$ }$$ and define the new measure $\mu_k$ on $(\{k, k+1 \}, 2^{\{k, k+1\}})$ as $$\mu_k(A) =  \mu(A \cap \pi_k(\mathbb{N})) = \mathbb{P}(A)$$ for any $A \in 2^\mathbb{N}$. Now we solve what happens when we add union two spaces together to our probability. It seems that we must prove the probability is then the average the measures together, i.e., if we union the interval $\{k, k+1 \} \cup \{k+2, k+3\} = B$, then our measure would be $\mu(B) = \frac{\mu_k(B) + \mu_{k+2}(B)}{2}$ assuming we normalize each density in order for it to be a probability measure.

Let $O = \{$ odd natural numbers$\}$. Then we can extend the measure to a limit after noticing $\mu_k(O) = \mu_{k+2}(O) = \dots = \mu_{k+2n}(O) = \frac12$ for any $n\in\mathbb{N}$. Our limit
$$\mu(O) = \lim_{n\rightarrow\infty} \frac{\mu_1(O) + \mu_3(O) + \cdots + \mu_{1 + 2n}(O)}{n} $$
$$= \lim_{n\rightarrow\infty}\frac{\frac12 + \frac12 + \dots + \frac12}{n} = \lim_{n\rightarrow\infty} \frac{n\frac12}{n} = 1/2$$

Please correct me if you find any major errors.


How to find the length of an unfolded piece of paper which has been folded by its diagonal? The problem is as follows:

The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\textrm{68 cm}\\
2.&\textrm{130 cm}\\
3.&\textrm{63 cm}\\
4.&\textrm{112 cm}\\
\end{array}$
For this particular problem. I'm really stuck. It seems that there isn't any way to go further because the only thing which I'm getting is that as a consequence of the lenght of the diagonal the sum of the other sides must add up to 90 hence to be 50 cm.
But I cannot say for sure which side is greater or what?. Am I missing something?. Can someone help me here?.
For reference this problem was obtained from my Reason and logic book from 2000's. And it seems to be a reprinted copy of a Martin Gardner's 70's book Puzzle Carnival with some modifications.
Since I'm not good at this it could help me a lot if it would include some drawing to see what's happening?.

Draw a vertical from the apex of the blue region to the diagonal fold. You should be able to see that the two right angled triangles you create are congruent and can be solved using what you have. Furthermore, those triangles are similar to the triangle which is half the page.
I get the fourth option as my answer, and Its at least possible ie it doesn't seem crazy when you look at all the other lengths you know.

We can see that a $3-4-5$ right triangle is created here.
Knowing the hypotenuse of $3-4-5$, the lengths of the other two sides are $\frac{3+4}5=1.4$ times the length of the hypotenuse.
Since the triangle above and the triangle equal to half the piece of paper are similar, we can conclude that the length and width of the paper are $1.4$ times the length of the diagonal.  Thus, we can calculate:
$$40\times1.4\times2=112.$$


Integration of $\int\frac{1}{\cos x(5-4\sin x)}\, dx$ $$\int\dfrac{1}{\cos x(5-4\sin x)}\, dx$$

I have tried multiple methods like converting this into partial fractions or substitution but no positive!

Hint: One way is $\sin x=u$ then
$$\int\dfrac{1}{\cos x(5-4\sin x)}=\int\dfrac{\cos x\, dx}{(1-\sin^2x)(5-4\sin x)}=\int\dfrac{du}{(1-u^2)(5-4u)}$$

Let,
$$Y=\int\dfrac{  1}{cos(x)(5-4sin(x)}
  =\int \dfrac{  1}{\dfrac{1-tan^2(x/2)}{1+tan^2(x/2)}  ×  (5-4\dfrac{2tan(x/2)}{1+tan^2(x/2))})}$$
Now,simplify it and then put  $tan(x/2)=t$

On differentiate bhoth side with d(t)
I.e.  d(tan(x/2))/d(t)=d(t)/d(t)
      sec^2(x/2)/2   ×  d(x)/d(t)=1
      sec^2(x/2)d(x)=2d(t)


Matlab code to compute the smallest nonzero singular value of the matrix without using SVD I want to compute the smallest nonzero singular value of the matrix A, which is defined as follows.

Let $B =  rand(500, 250)$, $A = B*B^t$, where $t$ denotes the transpose of the matrix.

I found the following matlab code to compute singular values of the matrix A which is based on the Singular value decomposition of the matrix.

svds = svd(A);
s = min(svds);  % smallest singular value

I want to know is there any other efficient way to smallest singular value?

Thank you in advance

Since $A=BB^T$, it is symmetric and positive semidefinite, and its singular values are identical to its eigenvalues. Furthermore, its nonzero eigenvalues are identical to the nonzero eigenvalues of $B^TB$. To get the smallest eigenvalue of $B^TB$, use eigs(B'*B,1,'sm').

Thanks to all answers/comments, I suggest a method based on @Rahul's method (which has converted the problem from SVD to EVD).

Summary

The solution in short: s = 1/eigs(inv(B'*B),1,'lm');

Description

There exist a Power Method for calculation of the eigenvector corresponding to the largest eigenvalue of matrix $M$. It is to compute a sequence of unit vectors:

$\frac{Mv}{||Mv||}$, $\frac{M^2v}{||M^2v||}$, ..., $\frac{M^kv}{||M^kv||}$

starting from an initial random vector $v$ which almost always converges to the eigenvector with largest eigenvalue of $M$.

If Power Method requires few iterations to converge and also calculating the inverse matrix $(B^tB)^{-1}$ requires relatively lower computations, then we can use it to compute eigenvector $v$ with largest eigenvalue of $(B^tB)^{-1}$ which is the eigenvector with smallest eigenvalue of $B^tB$. finally use the inverse of largest eignvalue of $(B^tB)^{-1}$.

Having the eigenvector $v$ its corresponding eigenvalue can be obtained using $Mv=\lambda v$ (i.e: with $M = B^tB$ eigenvalue of v is equal to all elements of vector $M*v ./ v$ hence with less calculations $M(1,:)*v / v(1)$).

I have conducted a comparison of results and running times of all above methods in MATLAB.

n = 100;
t1 = 0; t2 = 0; t3 = 0; t4 = 0; t5 = 0; % for measuring running times

B = rand(500,250);
A = B*B';
for i = 1:n,
 tic; svds = svd(A); s1 = min(svds(svds > 1e-6)); t1 = t1 + toc;   % @srijan
 tic; s3 = eigs(B'*B,1,'sm'); t2 = t2 + toc;                       % @Rahul
 tic; svds = svd(B); s2 = min(svds(svds > 1e-6))^2; t3 = t3 + toc; % @Hans Engler
 tic; s4 = 1/eigs(inv(B'*B),1,'lm'); t4 = t4 + toc;                % @Alireza Kazemi(*)
end;
% Computed smallest non-zero signular value of A
[s1,s2, s3, s4]
%-> RESULT: 3.403670094444867  3.403670094444888  3.403670094444812  3.403670094444802
% average times
[t1/n , t2/n, t3/n, t4/n]
%-> RESULT: 0.1193    0.0410    0.0317    0.0202 (seconds)
% ration of average times to the initial question of @srijan
[t1/t1, t2/t1, t3/t1, t4/t1]
%-> RESULT: 1.0000    0.3436    0.2655    0.1693

It is clear that the proposed method is fastest among compared methods, even though it needs a matrix inversion.

I must also mention that I did not implemented the Power Method here but just guessed that Matlab $eigs(...,'lm')$ must have implemented it.


De Rham cohomology groups of projective real space I would like to calculate the de De Rham cohomology groups of projective real space $\mathbb{RP}^{n}$. Well, i know all groups of De Rham cohomology os $n$-sphere $\mathbb{S}^{n}$ and that the map $\pi:\mathbb{S}^{n}\to \mathbb{RP}^{n}$ (the restriction to the projection map on $\mathbb{S}^{n}$) is a smooth submersion. With these tools, can I compute the cohomology groups of real projective space? How can I do this?

The following approach has the benefit that it is elementary in the sense that it only uses basic (homological and linear) algebra and the tools you allude to. More on this later*. I will just expand on the relevant parts of what is in this link.

You can use the antipode $A:S^n \to S^n$ to decompose $\Omega^k(S^n)$ in a direct sum of two subspaces $\Omega^k(S^n)_+ \oplus \Omega^k(S^n)_-$, since the induced map $A^*$ is such that $A^*{}^2=\mathrm{Id}$ (the subspaces are the eigenspaces of $1$ and $-1$, respectively). By naturality of $A^*$, $d$ respects that decomposition. It follows that $H^k(S^n) \simeq H^k(\Omega(S^n)_+)\oplus H^k(\Omega(S^n)_-)$ (the right side is "algebraic" cohomology originated from the complex $\Omega(S^n)_+$, resp. $\Omega(S^n)_-$), and it is clear that each one is the eigenspace of $A^*$ (on the cohomology level) of eigenvalue $+1$ and $-1$, respectively.

Since $A$ has degree $(-1)^{n+1}$ (it is a diffeomorphism that reverses/preserves orientation accordingly with the parity of $n$) and $H^n(S^n;\mathbb{R})=\mathbb{R}$, we have that $H^n(S^n)=H^n(\Omega(S^n))_+$ if $n$ is odd, and $H^n(S^n)=H^n(\Omega(S^n))_-$ if $n$ is even.

If we show that $\pi^*: \Omega^k(\mathbb{R}P^n) \to \Omega^k(S^n)_+$ is a isomorphism (the fact that the image indeed lies on $\Omega^k(S^n)_+$ follows from the fact that $\pi A=\pi$, because then $A^*\pi^*=\pi^*$), since it is natural it will follow that $H^k(\mathbb{R}P^n) \simeq H^k(\Omega(S^n))_+$, and hence we will have calculated the de Rham cohomology of $\mathbb{R}P^n$, since we know who are the $H^k(\Omega(S^n))_+$ (this is a good moment to see if you are following up: what is $H^k(\Omega(S^n))_+$, in terms of $n,k$? The only one to think about really is when $n=k$, the rest must be trivial - except $H^0$, of course).

To show that it is an isomorphism, you can check in the pdf I linked or use the answers by Mariano and user432847 here, which deal with a more general statement. Maybe seeing both would be best, since the general outlook that Mariano gives makes the fact that $\pi^*$ is an isomorphism with $\Omega^k(S^n)_+$ more intuitive.

*There are other methods to compute the cohomology of the projective space. Cellular homology + universal coefficients + deRham theorem is one example which you may want to check out.

Use the long exact sequence of a fibration and Hurewicz's theorem.


Calculate expectation of a geometric random variable When you download a file from a website, the file gets corrupted with probability 0.8. What is the expected number of downloads to get an uncorrupted file?

I have no idea how to do this. I only know the probability that a file isn't corrupted should be 0.2, but how do I get the expectation? Can anyone help me?

Your question essentially boils down to finding the expected value of a geometric random variable. That is, if $X$ is the number of trials needed to download one non-corrupt file then
$$X\sim Geo(0.2)$$
In general, if $X\sim Geo(p)$ then $$E(X)=\frac1 p$$
So in your case the expected number of trials to download an uncorrupted file is
$$E(X)=\frac{1}{0.2}=5$$
Addendum: Here is a derivation of the above mentioned result.

First note that $P(X=k)=p(1-p)^{k-1}$. The expected value is thus
$$\begin{align*}E(X)&=\sum_{k=1}^{\infty}kp(1-p)^{k-1} \\
&=p\sum_{k=1}^{\infty}k(1-p)^{k-1} \\
&=p\left(-\frac{d}{dp}\sum_{k=1}^{\infty}(1-p)^k\right) \\
&=p\left(-\frac{d}{dp}\frac{1-p}{p}\right) \\
&=p\left(\frac{d}{dp}\left(1-\frac{1}{p}\right)\right)=p\left(\frac{1}{p^2}\right)=\frac1p\end{align*}$$

Derivative step: (answer to comment)

Simple use of chain rule gives:
$$ -\frac{d}{dp}\sum_{k=1}^{\infty}(1-p)^k = \sum_{k=1}^{\infty}k(1-p)^{k-1} $$
It is clear that $$ -\frac{d}{dp}\sum_{k=1}^{\infty}(1-p)^k = -\frac{d}{dp}\left( \sum_{k=1}^{\infty}(1-p)^k \right)$$
given that $ 0 <1 - p < 1$ we can use the geometric series formula to obtain:
$$ \left( \sum_{k=1}^{\infty}(1-p)^k \right) = \frac{1 - p}{p} $$
then proof follows accordingly.

Expanding/complementing on what aflous said: your random variable is "number of non-corrupt files downloaded" , which has $p=1-0.8$......and expected number of non-corrupt files after _ trials is $\geq 1$....

EDIT: While it is true that the original question asks for a geometric random variable, one can look at the same problem from a different perspective, and still answer the question correctly. One can focus instead on whether a file is corrupt or not, and then define a new Binomial random variable to be the expect number of non-corrupt files in $n$ trials. Then this new random variable has mean $np$ for $n$ trials. We then want to find $n$ so that $np \geq 1$ , and , since $p=0.2$, we have $np \geq 1$ when $n \geq 5$.


Evaluate $\lim_{x \to 0+}{ (e^{1/\sin x}-e^{(1/x)})}$ I'm a high school student. For the past few hours, I was practicing the calculation of limits to improve my skills and I got stuck in a simple but complicated (for me) limit. The limit in question is

$$\lim_{x \to 0^+} \left( e^\frac{1}{\sin(x)} - e^\frac{1}{x} \right)$$

Calculating this limit I get zero, but Wolfram says it is infinite.
Could someone help me?

$$\lim_{x \to 0^+} \left( e^\frac{1}{\sin(x)} - e^\frac{1}{x} \right)= \lim_{x \to 0^+} e^{\frac{1}{x}} \left( e^\frac{x}{\sin(x)} - 1\right)$$
We know that $$\lim_{x \to 0^+} \frac{x}{\sin(x)}= \lim_{x \to 0^+} \left( \frac{\sin(x)}{x}   \right)^{-1}=1$$ and $$\lim_{x \to 0^+}\frac{1}{x}=+\infty$$
Hence:
$$ \lim_{x \to 0^+} e^{\frac{1}{x}} \left( e^\frac{x}{\sin(x)} - 1\right)=e^{+\infty}(e-1)=+\infty $$
So the initial limit is equal to:
$$ \lim_{x \to 0^+} \left( e^\frac{1}{\sin(x)} - e^\frac{1}{x} \right)=+\infty    $$


how to show that $a_{n+1}=\sqrt{6+a_{n}}$ is bounded? I have to show that the sequrnce $\{a_n\}$ with $a_{n+1}=\sqrt{6+a_{n}}$ is bounded, i know : $a_{n-1}<a_{n}$ because i demonstrated this. I tied to use this relation: $ab\leq \frac{a^2}{2}+\frac{b^2}{2}$. I don't kow how to increase $\sqrt{6+a_{n}}$

Here's a non-formal trick which can help you on these recurrence problems. But one must be careful and prove that $a_n > 0$ and monotonic first so that $b \geq 0$. You can prove $a_n > 0$ easily with induction and since you've already proven it is increasing let's get straight to the point.
$$\lim a_{n+1} = \lim \sqrt{6+a_n} = b = \lim a_n$$
So $b = \sqrt{6+b} \Leftrightarrow b^2-b-6 = 0 \Leftrightarrow (x+2)(x-3) = 0 \Leftrightarrow b = 3 \lor b = -2$. Since $b \geq 0$ then $b = 3$.
We conclude that the limit of this monotonic and positive recurrence sequence is $3$, so it becomes closer and closer to 3 from below. Meaning $a_n \rightarrow 3^-$.
This should not be your proof; just a side calculation that helps you get the supremum for $\{a_n\}_{n=1}^{\infty}$. We will use it to show that $a_n$ is bounded above.
Now we must demonstrate it with a formal proof (by induction) that for all $n \in \mathbb{N}, a_n < 3$. Since we are saying "for all $n$" it must be also true for $n+1 = n'$.
Our base case ($n=1$) is trivial since $a_1 =
\sqrt{6} < 3$.
Now assuming $p(n):= a_n < 3$ is true, we must show that $p(n+1) = a_{n+1} < 3$ is also true.
In fact,
$$a_n < 3 \Leftrightarrow a_n + 6 < 3 + 6 \Leftrightarrow \sqrt{6 + a_n} < \sqrt{3 + 6} = 3$$
But $\sqrt{6+a_n} = a_{n+1}$ for which $a_{n+1} < 3$ and the condition holds true for all $n$.
So we proved by induction that $0 < a_n < 3$ for all $n \in \mathbb{N}$.

Hint: Let $f(x)=\sqrt{6+x}$ for $x \ge -6$. Prove that

$f(x) \ge x$ when $ -6 \le x \le 3$

$f(x) \le x$ when $ 3 \le x$

Apply this result to $a_{n+1}=f(a_n)$ and conclude that the sequence is monotone.


Prove $\int_0^{\pi/2}\frac x {\sin x} \, \mathrm d x = 2\sum_{n\mathop = 0}^{\infty} \frac {(-1)^n}{(2n+1)^2}$ I would like to prove that

$$\int_0^{\pi/2}\frac x {\sin x} \, \mathrm d x = 2\sum_{n\mathop = 0}^{\infty} \frac {(-1)^n}{(2n+1)^2}$$

Any hints?

Both expressions equal $2G$, where $G$ is the Catalan constant


Proof $x$, $1+nx≤ (1+x)^n$ Prof using the binomial theorem: for all integers $n ≥0$ and for all nonnegative real numbers $x$, $1+nx ≤(1+x)^n$.

Don't have a idea to start this one. I don't know how to use math induction yet, so I need a answer without that.

Using the binomial theorem:

$$(1+x)^n=1+nx+\dots$$

Where $\dots$ is non-negative, as it is a polynomial in $x$ with non-negative coefficients and $x\geq 0$.

Equality holds for $x=1$. Then, taking the derivative,

$$n\le n(1+x)^{n-1}.$$


Interesting functional equation: $f(x)=\frac{x}{x+f\left(\frac{x}{x+f(x)}\right)}$ Solve for the function f(x):

$$f(x)=\frac{x}{x+f\left(\frac{x}{x+f(x)}\right)}$$
I'm not able to solve this.

[For instance, I tried solving for $f(\frac{x}{x+f(x)})$, but this doesn't lead me anywhere as the value obtained, when substituted into the original equation, just yields $f(x)=f(x)$]

Question: What is f(x) = ?

I suppose I need to mention that f(x) should be continuous, and that it's  a subset of $R$.

(Related : Functional equation $f(x)=\frac{1}{1+f(\frac{1}{1+f(x)})}$ , whose confirmed solutions are $-\phi$ and $\frac{1}{\phi}$)

If such an $f$ exists, plugging in $x=0$ we find that
$$f(0)=\frac{0}{0+f\left(\tfrac{0}{0+f(0)}\right)}=0,$$
and to avoid having to divide by $0$ this requires that
$$0+f(0)\neq0\qquad\text{ and }\qquad0+f\left(\frac{0}{0+f(0)}\right)\neq0.$$
But then $f(0)=0$ so $0+f(0)=0$, a contradiction. Hence no such $f$ exists.

take $g=\frac { x }{ (x+f) } $ (#)
by differentiating from two sides of this: $1=g+g'x+f'g+g'f$  $(*)$
by substituting $g=\frac { x }{ (x+f) } $ in the equation: $$\frac { x }{ g } -x=\frac { x }{ \left( x+f(g) \right)  } $$
differentiating from two sides of this and using (*) : $2x-gx=g-g^2$
find g in this quadratic equation.Then,find f by (#)


Embeddability of the cone of Klein bottle to $\mathbb R^4$ By using Alexander Duality, we can show that $K\not\hookrightarrow \mathbb R^3$ and we can also give an explicit formula for $K\hookrightarrow\mathbb R^4$ (here $K$ is the Klein bottle). How about the cone $CK$? Can it be embedded in $\mathbb R^4$?

It cannot be embedded in $\mathbb R^4$ 'nicely', which means if one considers any $\epsilon$-ball of the coning point, $K$ cannot be embedded into the boundary of the ball. I have also tried homology and cohomology exact sequences, and they did not give me any result.

The following lemma is an analogue of Corollary 3.46 in Hatcher’s “Algebraic Topology” and is proven by exactly the same method, using Alexander Duality (using cohomology with compact support instead of the ordinary cohomology as Hatcher does):

Lemma. Let $X$ be a locally contractible closed subset of $R^n$. Then $H_c^{n-2}(X)$ is torsion free.

Corollary. If $X$ is a nonorientable connected pseudomanifold of dimension $n-1$, it cannot be properly embedded in $R^n$.

Proof. Such $X$ will have $2$-torsion in $H_c^{n-2}(X,{\mathbb Z})$. qed

Now, turning to the open cone $CK$ over the Klein bottle, i.e.
$(K\times [0,1))/(K\times 0)$. This space is nonorientable connected pseudomanifold. Each open subset of $CK$ containing the apex $a$ of this cone (the projection of $K\times 0$) is also a nonorientable pseudomanifold.

Suppose that $CK$ embeds in $R^4$. Take the intersection $Y$ of $CK$ with a sufficiently small open ball $B(a,r)$ in $R^4$. Use the homeomorphism $B(a,r)\to R^4$ to identify $Y$ with a closed subset $X\subset R^4$.  We obtain a contradiction with Corollary above. Hence $CK$ does not embed in $R^4$.

Note that this argument is much more general applies to cones over compact nonorientable manifolds of any dimension.

Edit. Here is a better argument replacing Lemma and Corollary. I will use functoriality of the Alexander duality isomorphism AD: Given any closed subset $X\subset R^n$, we have a commutative square
$$
\begin{array}{ccc}
\check{H}^{n-1}_c(X; {\mathbb Z}) &\stackrel{AD}{\to} & \tilde{H}_0(R^n - X; {\mathbb Z}) \\
\alpha\downarrow &  & \beta\downarrow \\
\check{H}^{n-1}_c(X; {\mathbb Z}/2)&\stackrel{AD}{\to} & \tilde{H}_0(R^n - X; {\mathbb Z}/2)
\end{array}
$$
Here cohomology is the compactly supported Chech cohomology. In the case when $X$ is locally contractible, it is naturally isomorphic to the compactly supported singular cohomology.

Now, if $X$ is homeomorphic to a nonorientable connected pseudomanifold of dimension $n-1$, then the homomorphism $\alpha$ is not surjective (since it is a homomorphism $0\to {\mathbb Z}/2$). On the other hand, the homomorphism $\beta$ is always surjective (as a homomorphism of $0$-degree reduced homology groups). Therefore, a nonorientable connected pseudomanifold of dimension $n-1$ cannot be properly topologically embedded in $R^n$. The rest of the argument is as above.

I have a suggestion for your problem.

The idea is due, to my knowledge, to Hassler Whitney from the 1940's, and it was used in a paper of William Massey from the 1960's about an estimate on the signature of a closed four manifold  constructed from a surface in $S^4$.

Whitney observed an embedded closed surface in four space has a well-defined intersection number that does not require a global orientation for the surface.

Choose a local orientation near each point of transversal intersection of the surface with a pushed off copy of itself.

Both branches now have orientations which give a local orientation of four space which can be compared to a background orientation of four space. This gives a well-defined sign to the intersection point, note the order of branches does not affect the sign because they are even dimensional.

Changing the local orientation on the surface changes the orientation on each branch, so two minus signs.

It is a subtle and beautiful construction.

I am thinking if the algebraic sum of these is nonzero then it obstructs embedding a cone, because from a cone embedding one would get a embedding of the surface $\times$ $\epsilon$-interval, which would move the surface off of itself locally, forcing the Whitney sum to be zero.

  The following lemma is an analogue of Corollary 3.46 in Hatcher’s “Algebraic Topology” and is proven by exactly the same method, using Alexander Duality (using cohomology with compact support instead of the ordinary cohomology as Hatcher does):

  Lemma. Let $X$ be a locally contractible closed subset of $R^n$. Then $H_c^{n-2}(X)$ is torsion-free.

  Corollary. If $X$ is a nonorientable connected pseudomanifold of dimension $n-1$, it cannot be properly embedded in $R^n$.

  Proof. Such $X$ will have $2$-torsion in $H_c^{n-2}(X,{\mathbb Z})$. qed

  Now, turning to the open cone $CK$ over the Klein bottle, i.e.
  $(K\times [0,1))/(K\times 0)$. This space is nonorientable connected pseudomanifold. Each open subset of $CK$ containing the apex $a$ of this cone (the projection of $K\times 0$) is also a nonorientable pseudomanifold.

  Suppose that $CK$ embeds in $R^4$. Take the intersection $Y$ of $CK$ with a sufficiently small open ball $B(a,r)$ in $R^4$. Use the homeomorphism $B(a,r)\to R^4$ to identify $Y$ with a closed subset $X\subset R^4$.  We obtain a contradiction with Corollary above. Hence $CK$ does not embed in $R^4$.

  Note that this argument is much more general applies to cones over compact nonorientable manifolds of any dimension.

  Edit. Here is a better argument replacing Lemma and Corollary. I will use functoriality of the Alexander duality isomorphism AD: Given any closed subset $X\subset R^n$, we have a commutative square
  $$
\begin{array}{ccc}
\check{H}^{n-1}_c(X; {\mathbb Z}) &\stackrel{AD}{\to} & \tilde{H}_0(R^n - X; {\mathbb Z}) \\
\alpha\downarrow &  & \beta\downarrow \\
\check{H}^{n-1}_c(X; {\mathbb Z}/2)&\stackrel{AD}{\to} & \tilde{H}_0(R^n - X; {\mathbb Z}/2)
\end{array}
$$
  Here cohomology is the compactly supported Chech cohomology. In the case when $X$ is locally contractible, it is naturally isomorphic to the compactly supported singular cohomology.

  Now, if $X$ is homeomorphic to a nonorientable connected pseudomanifold of dimension $n-1$, then the homomorphism $\alpha$ is not surjective (since it is a homomorphism $0\to {\mathbb Z}/2$). On the other hand, the homomorphism $\beta$ is always surjective (as a homomorphism of $0$-degree reduced homology groups). Therefore, a nonorientable connected pseudomanifold of dimension $n-1$ cannot be properly topologically embedded in $R^n$. The rest of the argument is as above.

The cone on a manifold is not really a closed psuedomanifold.

Can you describe the embedding of the Klein bottle $\times$ interval?

With regards to the "better argument" that has been written, I like this proof. It shows the intersection with the link argument disproving nice embeddings, also works in spirit for all topological embeddings using cohomology etc.

It also shows $X$ locally separates $\mathbb{R}^n$, which is of course intuitive.


Roadway and book recommendations to math study. I had some calculus, linear algebra and complex analysis courses back in college. But it is not comprehensive. And I felt that my college math was not taught in a logical sequence (maybe because my major is engineering). And some of the confusing theories may  be better appreciated if taught in more detail and in a more logical sense.

Now I had some time and not pushed by so many exams. I'd like to re-study the mathematics by myself. This is all about interest and curiosity. And I believe I can figure out all my confusions so far. But I need some guidance.

So could you guys recommend some books and logical study sequence for me? My target is as below:

(1) To fully understand the integral transform, such as fourier, laplace
and wavelet transform.
(2) To fully understand some strange fuctions, such as Dirac delta function, etc.
(3) To fully understand the Euler formula: e^ix = cos(x) + i sin(x)
(4) To fully understand the complex number "i" and complex analysis.
(5) To fully understand the differential equation.
(6) To get a thorough understanding of modern algebra.
(7) To help my work in computer science.
(8) To get the general idea of modern mathematics.

Really appreciate your advice!

I should like to point out from the beginning that several parts in your question are unclear and in some sense unanswerable (such as 'the differential equation' because there is no such thing as 'the' differential equation). Also it is unclear to me how deep you want to go into these topics so that I restrict myself to what I consider to be to the minimum. I suggest:

(1) start with Stein's Complex analysis (Princeton lectures in analysis, vol. II), chapter 1 (this will answer part 2 and 3 of your question, apart from the complex analysis part, for which you also have to look at least at chapters 2-4).

(2) Part 1 of your question will be answered by Stein's Fourier analysis (same series, vol. I), chap. 5 and 6. The fourier transform comes up in increasing level of sophistication in Stein's series, such as in chap. 2 of vol. II and chap. 5. of vol. III. A good reference for integral transforms is Antimirov's applied integral transforms, chap. 2.

(3) Part 5: the basic material can be found in Arnold's ODE text chap. 1-4 (several sections could be omitted). But of course one could endlessly read about differential equations; differential equations come up also in several place of Stein's series (so I suggest you get these books, they are relatively expensive but they are excellent in any aspect, and also suitably elementary).

(4) Part 2: frankly speaking there is no such thing as the Dirac delta function, the road leads to distribution theory, which is dealt with in chapter 3 of Stein's functional analysis (vol. IV of the series). If you want to know something about special (gamma, zeta, elliptic and theta) functions, a good place to look is Stein's complex analysis, chap. 6-7, 9-10.

(5) Part 6: Lang's algebra book (GTM), as much as you can, at any rate chapters 1-6 (mainly group, ring, module and field/galois theory).

(6) Part 7-8: This is difficult to answer, but the more mathematics you understand the better your thinking. As far as 8 is concerned, Gower's Princeton companion to mathematics gives you a good overview of what mathematics is about. But again this is a very broad question; however if you get to know some category theory (Mac Lane's book), you will certainly understand many ideas underlying the algebra parts of mathematics.

By the way there is not much to be said about Euler's formula, some people find it special but it is in fact a triviality.

Functional Analysis - Kreyszig

Methods of Mathematical Physics- Courant & Hilbert or Arfken

Complex Analysis - Stein & Shakarchi or Rudin or Ahlfors

Differential Equations - G.F. Simmons

Algebra- Herstein and thereafter Lang/ Hungerford/ Jacobson

Concrete Mathematics - Ronald L. Graham, Knuth, Patashnik

Hope this is helpful.


Hello, can you help me with this inequality? for $x,y,u,v\in R$
$$|ux-yv|\leq\sqrt{(x^{2}+y^{2})\times(u^{2}+v^{2})}$$.
Thank you

If $\vec x=(y,x,0)$ and $\vec u=(u,v,0)$, and $\theta$ is the angle between them, then $|ux-yv|=|\vec x\times \vec u|=\|\vec x \|\|\vec u\||\sin \theta|\le \|\vec x \|\|\vec u\|=\sqrt{(x^{2}+y^{2})\cdot (u^{2}+v^{2})}$

$|ux - vy| = \sqrt {(ux - vy)^2}$ by definition.

$\sqrt {(ux - vy)^2} \le \sqrt{(x^2 + y^2)(u^2+v^2)}$

Squaring both sides

$(ux - vy)^2 \le (x^2 + y^2)(u^2+v^2)$

Expanding the binomials

$u^2x^2 - 2uvxy + v^2y^2 \le u^2x^2 + u^2y^2 + x^2v^2 + y^2v^2$

Subtract the common terms.

$- 2uvxy \le u^2y^2 + x^2v^2$

At which point we could invoke the AM-GM inequality.  Or we could say.

$0 \le u^2y^2 + 2uvxy + x^2v^2$

Which is a perfect square.

$0 \le (uy + xv)^2$

And square numbers are always positive.

Our proposition is true.


What is the difference between $\operatorname{Arcsin}$, $\operatorname{arcsin}$, $\operatorname{Sin}^{-1}$, and $\sin^{-1}$? This is a homework problem I have. I don't remember learning the difference, and searching hasn't helped explain the difference between the capitalization.

While I'm almost certain the other answers are correct about $\arcsin$ being equivalent to $\sin^{-1}$, the majority of sources I can find contradict the answers given by MarianD and Polygon:

The Math Forum's "Ask Dr. Math": Inverses of Trigonometric Functions
MathWords.com: Inverse Trigonometry
Dummies.com: How to distinguish between trigonometry functions and relations
MathWorld: Inverse Trigonometric Functions
SparkNotes: Inverse Trigonometric Functions

According to these sources
$$\arcsin x =\sin^{-1}(x) = \{ \theta \in\Bbb{R} : \sin \theta = x\} ,$$
whereas
$$\operatorname{Arcsin} x =\operatorname{Sin}^{-1}(x)= \theta,$$
where $\theta$ is the unique angle between $-\pi/2$ and $\pi/2$ (inclusive) such that $\sin\theta = x$, which is often called the principal value.

Side note

I'd like to comment that in practice, in most cases the lower case variants are used to denote the principal value, and the upper case variants are not used at all. However, when they are used, this does appear to be the correct answer regarding their meaning.

$\operatorname{Arcsin}$ is used when showing $\sin^{-1}x$ as a relation whereas $\arcsin$ is used when showing $\sin^{-1}x$ as a function. If you put the function sin^-1(x) into Desmos or a graphing calculator set to radians, you can see the graph of the arcsine function (inverse function of $\sin x$). You will also see that the line stops abruptly at $x=-1$ and $x=1$ and $y=-\pi/2$ and $y=\pi/2$. Since there cannot be two $y$ values for any one $x$ value, the graph stops. We denote this segment as the "arcsine". But if the graph continued as a relation, it would look like a sine function stretching from $y=-\infty$ to $y=\infty$. We denote this as "Arcsine".
I hope someone finds this useful.


Known proofs of Wirtinger's Inequality? I am looking for proofs of the (Poincare-) Wirtinger inequality which states that if $f:[0,\pi]\to \mathbb{C}$ is $C^1$ and $f(0)=f(\pi)=0$ then
\begin{equation}
\int_0^\pi |f(t)|^2 dt \leq \int_0^\pi |f'(t)|^2 dt.
\end{equation}
See link.

The proof that I know starts by proving that if
$$ \int_0^{2\pi} F(t) dt =0 $$
then
$$
\int_0^{2\pi} |F(t)|^2 dt \leq \int_0^{2\pi} |F'(t)|^2 dt.
$$
using Parseval's identity. From this, one proves the desired inequality for $f$ on $[0,\pi]$ by "extending" $f$ to an odd $C^1$ function on $[-\pi,\pi]$.

Are there other proofs? (Straightforward or otherwise...)

The following proof is found in section 7.7 of Hardy-Littlewood-Polya Inequalities, it is motivated by Hilbert's investigations into calculus of variations, especially Hilbert's method of invariant integrals.

Consider the expression
$$ (y'^2 - y^2) - (y' - y\cot x)^2 = -(1+ \cot^2 x) y^2 + 2y y' \cot x $$
So
$$ \left[(y'^2 - y^2) - (y' - y\cot x)^2 \right]\mathrm{d}x = -(\csc^2 x)y^2 \mathrm{d}x + 2y \mathrm{d}y \cot x = \mathrm{d} ( y^2 \cot x )$$

Now, since $y' \in L^2$, we have that
$$ y^2(x) = \left(\int_0^x y'(s) \mathrm{d}s\right)^2 \leq \int_0^x y'(s)^2 \mathrm{d}s \int_0^x 1\mathrm{d}s \leq x \int_0^x y'(s)^2 \mathrm{d}s $$
So we have that
$$ \frac{y^2(x)}{x} = o(1) $$
and hence $y = o(\sqrt{x})$. Similarly we have that $y^2$ approaches 0 superlinearly at $\pi$. This implies that $\lim_{x\to \{0,\pi\}} y^2 \cot x = 0$. Hence the exact integral

$$ \int_0^\pi (y'^2 - y^2) - (y' - y\cot x)^2 \mathrm{d}x = \int \mathrm{d}\left( y^2 \cot x\right) = 0 - 0 = 0 $$

Therefore we have

$$ \int_0^\pi (y'^2 - y^2) \mathrm{d}x = \int_0^\pi (y' - y\cot x)^2 \mathrm{d}x \geq 0 $$

with equality only if

$$ y' = y \cot x $$

which is when $y = k \sin x$.

You can find different proof in the book of B. Dacorogna 'Introduction to the calculus of variations'.


Is every simple ring a division ring? I know that every division ring is simple.

Is the converse true?

I think it isn't. But I can't find a counterexample.

Well, most of the time when two concepts are given names, as "division ring" and "simple ring" are, the ideas are different. If every simple ring were a division ring, then we would have created an extra unnecessary piece of terminology.

You could be forgiven for not finding a counterexample if you were only thinking of commutative rings though. It does turn out that a commutative simple ring is a field.

Anyhow, there are many good noncommutative rings which are simple. The most obvious one are, as has been mentioned, full $n\times n$ matrix rings over division rings (with $n>1$ .)

Another collection of examples is given by Weyl algebras, which in general aren't division rings but are nevertheless simple rings and additionally they don't have any zero divisiors other than $0$.

You can produce simple rings from noncommutative rings by forming their quotient by a maximal ideal of the ring. For example, if we take $R$ to be the ring of linear transformations of a vector space with countably infinite dimension, it is known that ring has exactly one proper ideal $J$ (which is obviously then maximal.) Then then ring $R/J$ is simple. It turns out that it has zero divisors but isn't Noetherian or Artinian, so this example is different from the first two I gave.

Note: I did not read the question carefully enough, as Prahlad Vaidyanathan points out in the comments.

Let $R$ be a division ring and let $L \subset R$ be a nonzero left ideal. Let $x \in L$ be nonzero. Since $R$ is a division ring, $x$ has a 2-sided inverse $x^{-1} \in R$. But then, $1 = x^{-1} \cdot x \in L$. Therefore, $y = 1 \cdot y \in L$ for every $y \in R$ and so $L = R$. So, $R$ does not even have a nontrivial (i.e. nonzero and proper) left ideal.

The whole point is that, as soon as an ideal contains a unit of a ring, it eats up the whole ring. Since division rings contains nothing but units (and zero), a nontrivial (nonzero and proper) ideal is impossible.


Logic Confusing Problem I Read one logic book, can my two conclusion are true?

1- Suppose

for each valuation v, we have such n that

    can we say we have such n that

2- Suppose

for each valuation v, we have such n that

can we say  set {φ1,φ1∨φ2,φ1∨ φ2∨φ3,...} is consistent?

i think the (2) is not always true?  for (1) and (2) any description?

This is a logic "confused" problem ...

We approach it "by steps".

Question 1

We assume that the meaning of the first part of the question is :

  for a valuation $v$ there is some $n$ for which : $v(\varphi_n)=1$.

Is so, due to the fact that : $v(\varphi_1 \lor \ldots \lor \varphi_n) = max_i \{ v(\varphi_i) \}$, we have that for the valuation $v$ :

  $v(\varphi_1 \lor \ldots \lor \varphi_n) = 1$.

Conclusion : if for a valuation $v$ there is a formula $\varphi_n$ such that $v(\varphi_n)=1$, then the conjunction $φ_1 \lor φ_2 \lor \ldots \lor φ_n$ is consistent (or satisfiable).

The symbol :

    $\vDash \varphi_1 \lor \ldots \lor \varphi_n$

means that the formula $\varphi_1 \lor \ldots \lor \varphi_n$ is valid (or a tautology).

The above formula is valid when we have $v(\varphi_1 \lor \ldots \lor \varphi_n) = 1$ for all valuation $v$.

Comclusion : the fact that for every valuation $v$ there is a $n$ such that ... does not implies that there is an $n$ such that for every valuation $v$ ...

In other terms, the condition :

  for a valuation $v$ there is some $n$ for which : $v(\varphi_n)=1$.

is not enough to conclude with $\vDash \varphi_1 \lor \ldots \lor \varphi_n$.

Question 2

Assuming a set $\Sigma = \{ φ_1, φ_2, \ldots \}$ of formulae, if $\Sigma$ is consistent, then :

  $\Gamma = \{ φ_1, φ_1∨φ_2, φ_1∨φ_2∨φ_3, \ldots \}$ is always consistent.

Being $\Sigma$ consistent, there is a valuation $v$ such that $v(\varphi_i)=1$, for all $i$.

Thus :

  $v(φ_1)=v(φ_1∨φ_2)=v(φ_1∨φ_2∨φ_3) = \ldots=1$

i.e. the valuation $v$ satisfy all the formulae in $\Gamma$ and thus $\Gamma$ is consistent.

If $\Sigma$ is not consistent, this means that for each valuation $v$ there is a formula $\varphi_i$ such that $v(\varphi_i)=0$; but it is enough that $i > 1$ and $\Gamma$ will be still consistent.

In fact, if for the valuation $v_0$, $v_0(\varphi_1)=1$, then again :

    $v_0(φ_1)=v_0(φ_1∨φ_2)=v_0(φ_1∨φ_2∨φ_3) = \ldots=1$.

If the formula $\varphi_1$ is unsatisfiable (e.g. a contradiction, like : $p \land \lnot p$) then for each valuation $v$, $v(\varphi_1)=0$; in this case, the set $\Gamma$ is not consistent.

By Compactness Theorem a set $\Gamma$ of formulae is consistent iff every finite subset of $\Gamma$ is consistent; thus, if $\{ \varphi_1 \} \subseteq \Gamma$ is not consistent, also $\Gamma$ is not.

i think 1 is true, by applying compactness theorem you can show that 1 is true


Reducing the Kronecker Delta I am not sure where my line of thinking for
$\delta_{ij} \delta_{ij} = \delta_{ii}$ is going wrong. Please help me find my error:

You cannot add the Kronecker indices before multiplying them fully out.


Large Deviations Counterexample in Durrett I am working on the following probability problem from Durrett:

  Suppose $(X_{i})_{i\geq 1}$ are i.i.d., $EX_{i}=0$, and $E\left(e^{\theta
X_{i}}\right)=\infty$ for all $\theta>0$. Show that $\frac{1}{n}\log
P\left(S_{n}\geq na\right)\to 0$ for all $a>0$.

I basically tried employing a similar strategy to one Durrett uses in one of his proofs by writing

$$
P\left(S_{n}\geq na\right)\geq P\left(S_{n-1}\geq -n\epsilon\right)\cdot P(X_{n}\geq (n+a)\epsilon).
$$
If I could conclude somehow that the right-hand side of the inequality above is
$$
\geq C P(X_{n}\geq (n+a)2\epsilon)\qquad(*)
$$
for some constant $C\leq 1$, or some similar $\epsilon$ modification, I'd be able to finish the proof since
$$
\limsup_{n\to\infty} P(X_{1}\geq (n+a)2\epsilon)^{n}=1
$$
by a Borel-Cantelli argument and the fact that $E(e^{\theta X_{i}})=\infty$ for all $\theta>0$. My problem is I don't know how to obtain a bound similar to $(*)$. In Durrett's proof, he states that it follows from the Weak Law of Large Numbers, but it is not clear to me how.

This indeed follows from the weak law of large numbers: note that $$\Pr\left\{S_{n-1}\geqslant -n\varepsilon\right\} \geqslant\Pr\left\{\left|S_{n-1}\right|/n\in \left[-\varepsilon,\varepsilon\right] \right\}=1-\Pr\left\{\left|S_{n-1}\right|/n \gt \varepsilon\right\}.$$
Since $\mathbb E\left[X_i\right]=0$ and $\left(X_i\right)_{i\geqslant 1}$ is an i.i.d. sequence, the sequence $\left(\left|S_{n}\right|/n\right)_{n\geqslant 1}$ goes to $0$ in probability and so does the sequence $\left(\left|S_{n-1}\right|/n\right)_{n\geqslant 1}$. Therefore, the quantity $\Pr\left\{S_{n-1}\geqslant -n\varepsilon\right\}$ is bigger than $1/2$ for $n$ large enough.


What does she mean by star? I was looking at this video http://www.youtube.com/watch?v=ygqIfLHGTu4&feature=g-all-f#t=06m33s33 and i wondered what she means by star. How is this number defined and where does it come up. Thanks in advance.

In the field of combinatorial game theory, certain two player games are assigned "values" that contain all of the information one might want to know about them. Some games' values are familiar numbers, which are similar to (but not quite the same as) points for one of the two players (traditionally named "Left" and "Right" but you can think of them as "white" and "black" in some games): 2 is similar to 2 points for Left, -3 is similar to 3 points for Right, etc. Technically, in the game represented by the value 2, Right can't move at all (so they would lose on their turn for having no legal moves), and Left has a move to the game represented by 1, where Left has a further move to "0", where neither player has a legal move.

"Star", often written "*", represents a particularly boring game where the player whose turn it is now (whether they're Left or Right) gets to move once, and then there are no legal moves anymore. Left isn't guaranteed a win, so it's not like a positive number, and Right isn't guaranteed a win, so it's not like a negative number (and more can be said along these lines), but unlike 0, the person to move now wins instead of loses, so it's not 0. It's a very weird value indeed.

She says that it is smaller than all positive numbers and larger than all negative numbers and not zero, so I would interpret it as an infinitesimal - something greater than zero that does not satisfy Archimede's axiom (for any two positive reals x and y, there is a positive integer n such that n times x is greater than y). In this case, for all positive integers n, n times "star" is less than one (or any other positive real).


Problem of Harmonic function. If H is a harmonic function on an unit disk; And $H=0$ on $R_1\cup R_2$, here $R_1, R_2$ are radius of $D(0,1)$. The angle between $R_1$ and $ R_2$ is $r\pi$; here $r\in (0,1]$. If $r$ is an irrational number then is $H$ identically zero on $D(0,1)$?

  I think if $r$ is irrational then $H\equiv0$ on unit disk $D$; and
  following lemma can help us to prove it.

Lemma:
If $a,b$ are two semilines starting at the same point $O$ there exists $u \not\equiv 0$ is a harmonic function vanishing on $a,b$ if and only if there exists $r \in \mathbb{Q}$ such that the angle between a and b is $r\pi.$

Proof of the Lemma: Using the following property of harmonic function we can prove this lemma.

A harmonic function vanishing on an open set is identically zero.
(Schwarz Reflection Principle) If a harmonic function is defined in a neighborhood of a line segment contained in one of the halfplanes determined by that segment and it is continued to zero on that segment, then it can be extended harmonically to the symmetric region by the given line segment.

Assuming a semiline need only exist within the domain of the function (I am unsure of the definition), we have the following proof:

Note that $R_1$ and $R_2$ are two semilines starting from the same point such that the angle between them is not a rational multiple of $\pi$, and that $H$ is harmonic on the unit disk, which is a neighborhood of these two segments. By our lemma, no non-zero harmonic function vanishes on $R_1 \cup R_2$.  We conclude that $H$ must be identically zero.

The Maximum and Minimum Principle of Harmonic functions imply that the answer of your question is positive.


How to define such a non-constant, continuous $f(x)$ on $[a,b] \subset \mathbb R$ .. such that for any $c$ on $f(x)$ 's image, there are infinite points $d_i \in [a, b]$ that $f(d_i) = c$?

It's a bit tricky as $[a, b]$ is closed, otherwise $\sin 1/x$ will help...

I will define such $f\colon [-2,2]\to [-2,2]$.
Let $$\tag1f_0(x)=\begin{cases}4x+6&-2\le x\le -1\\
-2x&-1\le x\le 1\\4x-6&1\le x\le 2\end{cases}$$
(which is continuous and attains each value $\in[-2,2]$ at least twice).
Now define recursively
$$\tag2f_{n+1}(x)=\begin{cases}4x+6&-2\le x\le -1\\
1-\frac12f_n(4x+2)&-1\le x\le 0\\
-1-\frac12f_n(4x-2)&0\le x\le 1\\
4x-6&1\le x\le 2\end{cases} $$
By induction on $n$ we see that

$f_n(\pm2)=\pm2$
$f_n$ is a continuous map $[-2,2]\to [-2,2]$
(Not needed, but hints that we are on the right track) For all $c\in [-2,2]$, the set $f_n^{-1}(c)$ has at least $n+2$ elements

Also, for the distance between two functions $f_n$,$f_m$, $n<m$, in this sequence, we find that $$\|f_{n+1}-f_{m+1}\|_\infty\le \frac12\|f_{n}-f_{m}\|_\infty$$ and consequently
$$\|f_n-f_m\|_\infty\le 2^{-n}\|f_0-f_{m-n}\|_\infty\le 4\cdot 2^{-n}. $$
Thus we have a uniformly converging sequence of continuous functions and let $$ f(x)=\lim_{n\to\infty} f_n(x).$$
Then $f$ is a continuous map $[-2,2]\to[-2,2]$ and it is a fixed point under the recursion $(2)$.

Assume that for some $c\in[-2,2]$ the set $f^{-1}(c)$ is finite. Pick $c$ for which this set is minimal. For $0\le c\le 2$ we have
$$f^{-1}(c)\supseteq\left\{\frac{c+6}4\right\}\cup \frac14\bigl(f^{-1}(2-2c)-2\bigr) $$
and the two sets on the right are disjoint because $\frac{c+6}4>0$ and $\frac14\bigl(f^{-1}(2-2c)-2\bigr)\subseteq [-1,0]$. We conclude that
$$|f^{-1}(2-2c)|\le |f^{-1}(c)|-1 $$
contradicting minimality. A similar argument works for $-2\le c\le 0$.
We conclude that there is no $c\in[-2,2]$ with finite pre-image.

$$
f(x)=
\begin{cases}
x\sin(1/x) & 0<x\leq 1\\
0 & x=0
\end{cases}
$$


For entire funtion $h$, $zh(z)$ cannot have a nonzero limit as $z\to \infty$ This is a question I encounter in learning complex variables.

Show that for any entire funtion $h$, $zh(z)$ cannot have a nonzero limit as $z\to \infty$.

Furthermore, use this statement to give an alternate proof of Liouville's theorem.

I am really having trouble proving the statement, and can't see any obvious connection between this and the Liouville's theorem. Any hints or suggestions are appreciated.

Let us assume the opposite, i.e. that $\lim_{z\to \infty} zh(z)=k;k\neq 0$

Assuming Liouville theorem:
$$\lim_{z\to \infty}zh(z)=k\neq 0\\ \lim_{z\to \infty}h(z)=0\\
h(z)=a (\text{const})\\
\lim_{z\to \infty}zh(z)=\infty$$

This contradiction proves that the limit, if it exists, must be either $0$ or $\infty$

Without Liouville theorem:
By the Cauchy integral formula:

$$|h^{(n)}(0)|=\left|\frac{n!}{2\pi i}\oint_{c_r}\frac{h(z)}{z^{n+1}}dz\right|\le \frac{n!}{r^n}\underset{|z|=r}{\text{Max}}|h(z)|$$
Letting $r\to \infty$ we have, thanks to the assumption, that $\forall_{n>1}h^{(n)}(0)=0$. This implies that $h(z)=az+b$, which contradicts
$$\lim_{z\to \infty}zh(z)=k$$
Proving Liouville with this statement:
Given a bounded entire function $h(z)$, let us define $$f(z):=f\left(\frac{1}{z}\right)$$ This function is  holomorphic on $\mathbb{C}-\{0\}$, and is bounded in a neighbourhood of $0$; By  Riemann's theorem on removable singularities $\lim_{z\to 0}f(z)=l\in \mathbb{C}$ exists, and thus $\lim_{z\to \infty}h(z)=l\in \mathbb{C}$ too.
Defining $g(z)=\frac{h(z)-h(a)}{z-a}$, we obtain an entire function, for which $$\lim_{z\to \infty}g(z)z=l-h(a)$$
Thanks to our statement, $l-h(a)=0$. Since $a$ was generic, $h$ is constant

Your statement is wrong. What about $h(z)=1$ ? Then, for example, for $t>0$, $$\lim_{t\to \infty }t h(t)=+\infty.$$


Summation of nonnegative numbers over countably infinite set Suppose that $T_1,T_2\subset \mathbb{N}$ and $a_i\geq 0$. I need to prove that $$\sum \limits_{i\in T_1\cap T_2}a_i+\sum \limits_{i\in T_1\cup T_2}a_i=\sum \limits_{i\in T_1}a_i+\sum \limits_{I\in T_2}a_i.$$

I do not know how to prove this fact at all because I even don't know what is the definition of $\sum \limits_{i\in K}a_i$ for $K\subset \mathbb{N}$.

I will highly appreciate if someone will explain my questions, please!

EDIT: Also I assume that $\sum \limits_{I\in T_1}a_i<\infty,\sum \limits_{I\in T_2}a_i<\infty$ and as I said above $a_i\geq 0$.

One definition for summations of non negative numbers over an arbitrary index set is $\sum_{n \in I} a_n = \sup_{J \subset I, J \text{ finite}} \sum_{n \in J} a_n$.

Note that it follows immediately that if $I \subset I'$ then
$\sum_{n \in I} a_n  \le \sum_{n \in I'} a_n$.

Now suppose $A \subset I$, then I claim that
$\sum_{n \in I} a_n = \sum_{n \in A} a_n + \sum_{n \in A^c} a_n$.

Suppose $J_1 \subset A, J_2 \subset A^c$ are finite, then since $J_1 \cup J_2$ is finite we have
$\sum_{n \in I} a_n \ge \sum_{n \in J_1 \cup J_2} a_n = \sum_{n \in J_1} a_n + \sum_{n \in J_2} a_n$, taking the $\sup$s on the right hand side gives
$\sum_{n \in I} a_n \ge \sum_{n \in A} a_n + \sum_{n \in A^c} a_n$.

Similarly, if $J \subset I$ is finite, then
$\sum_{n \in J} a_n = \sum_{n \in J\cap A} a_n + \sum_{n \in J \cap A^c} a_n \le \sum_{n \in A} a_n + \sum_{n \in A^c} a_n$.

Now, note that $T_1 \cup T_2 = T_1 \setminus T_2 \cup (T_1 \cap T_2) \cup T_2 \setminus T_1$, a disjoint union, hence
$\sum_{n \in T_1 \cup T_2} a_n+ \sum_{n \in T_1 \cap T_2} a_n=  \sum_{n \in T_1 \setminus T_2} a_n + 2 \sum_{n \in T_1 \cap T_2} a_n  + \sum_{n \in T_2 \setminus T_1} a_n$
and
$\sum_{n \in T_1 } a_n + \sum_{n \in T_2}a_n =  \sum_{n \in T_1 \setminus T_2} a_n + 2 \sum_{n \in T_1 \cap T_2} a_n + \sum_{n \in T_2 \setminus T_1} a_n$.

Note that this result holds whether or not the sum is finite.

For the notation part:
For example, let $K$ be the even numbers. Then $\sum\limits_{i\in K}a_i$ is just $a_2+a_4+a_6+....$

For the proof part:

Lemma: If $A$ and $B$ are disjoint and the terms are always positive and the sums converge: $\sum\limits_{i\in A\cup B}(a_i) = \sum\limits_{i\in A}a_i + \sum\limits_{i\in B}a_i$

Hence, since $T_2= T_1 \cap (T2-T1)$ with this to sets being disjoint:
$\sum\limits_{i\in T_1}a_i + \sum\limits_{i\in T_2}a_i = \sum\limits_{i\in T_1}a_i + \sum\limits_{i\in T_2-T_1}a_i + \sum\limits_{i\in T_2\cap T_1}a_i$

And again by the same lemma. combining the first and the second terms from the right side of the equality (remember that $T_1 \cup T_2 = T_1 \cup (T_2 - T_1)$):

$\sum\limits_{i\in T_1}a_i + \sum\limits_{i\in T_2}a_i = \sum\limits_{i\in T1\cup T_2}a_i + \sum\limits_{i\in T_2\cap T_1}a_i$.


What are disasters with Axiom of Determinacy? It is well-known that Axiom of Choice has several consequences which might be viewed as counter-intuitive or undesirable. For example, existence of non-measurable sets or Banach-Tarski Paradox. H. Herrlich's book on AC has a chapter aptly named Disasters with Choice.

But omitting Axiom of Choice and working in ZF alone might lead to even more weird things. (Again, just have a glance at the table of contents of the same book and check various things mentioned in the chapter Disaster without Choice.)

Therefore there have been some suggestions on possible replacements of AC. For example, for some things working with ZF+countable choice is enough.

One of possible candidates is Axiom of Determinacy. For example, it is known that AD implies
countable choice for sets of real numbers. (See, for example Theorem 9.29 in L. Bukovský: The Structure of the Real Line.) Again, for some purposes this version of AC might be sufficient. AD also implies that every subset of $\mathbb R$ is Lebesgue measurable.

Let me quote from the book Pavol Zlatoš:
Ani matematika si nemôže byť istá sama sebou - Úvahy o množinách
nekonečne, paradoxoch a Gödelových vetách; (Even mathematics cannot be certain about itself, Essays on sets, infinity, paradoxes, and Gödel’s theorems, in Slovak), page 117.

The (not very good) translation is mine. This excerpt follows after mentioning French school represented by R. Baire, E. Borel and H. Lebesgue and Russian school lead by N. N. Luzin.

  Both these schools substantially used set-theoretical methods in analysis, topology, theory of functions, measure theory, etc.,
  but Zermelo's proof of well-orderability of continuum and some other "unpleasant" consequences of Axiom of Choice disconcerted them to the extent
  that they challenged this axiom and ineffective existential proofs in general.

  Let us mention that if Axiom of Determinacy had been known at the time, it is possible - or even probable - that these four great mathematicians would have given preference to this axiom over Axiom of Choice. Using Axiom of Determinacy it is possible to build nicer, more elegant and unified descriptive set theory, where various pathological (e.g., non-measurable) subsets of real numbers are impossible.

Or the quote at the beginning of the chapter on AD in H. Herrlich's book (which is attributed to U. Felgner and K. Schulz):

  Among all alternatives to the axiom of choice AC the axiom of determinateness AD undoubtedly is the most interesting.

I wonder whether working in ZF+AD would also lead to some consequences which some people might consider undesirable.

  What are consequences of Axiom of Determinacy which might seem antiintuitive/problematic/paradoxical?

It might not be a disaster, but I've always thought it counterintuitive that there is an equivalence relation on the real line with strictly more than continuum many equivalence classes.  The equivalence relation isn't even very complicated; it's just congruence modulo $\mathbb Q$ in the additive group of $\mathbb R$.  In other words: Vitali wants to give you his standard example of a non-measurable set, and, if you prevent him from doing so, he gets his revenge with this example of partitioning a set into strictly more pieces than it has elements.

You need AC to prove Gödel's completeness theorem, and that is a big deal.

See the wikipedia article on Completeness Theorem which I cannot make link.


If $\frac{r^2}{AP^2}$ can stated in the form $\frac{a}{b}$ where $\gcd(a, b) = 1$. Determine the value of $a + 10b.$ Given a circle $ω$ with center $O$ and length of radius $r$. There is
a point $A$, $B$ on the circle and point $C$ outside the circle so
that $BC$ tangent to the circle $ω$. Then $AC$ intersects the circle
$ω$ at $D$. Then $P$ is the midpoint of $AC$ and $AD = BP = BC$. If $\frac{r^2}{AP^2}$ can stated in the form $\frac{a}{b}$ where $gcd$ $(a, b) = 1$. Determine the value of $a + 10b.$

Suppose $AP = y$, $DP = x$
with the power of point theorem, i get :
$BC^2=CD.AC\Longrightarrow (x+y)^2=2y(y-x)\Longrightarrow x^2+4xy-y^2=0\Longrightarrow y=x(2+\sqrt{5})$
i need an equation containing $x$ and $r$,but I didn't find it, if someone can help, that's great!!!
There are the original question, Page 4, Number 5.

The ratio ${\large{\frac{r^2}{AP^2}}}$ can be computed by brute force as outlined below.

Assume $y=1$.

Since the ratio ${\large{\frac{r^2}{AP^2}}}$ is scale invariant, we can assume $y=1$.
Compute $x$.

By power of a point we get $(x+1)^2=(1-x)(2)$, so
$$x=\sqrt{5}-2$$
Compute $\cos\bigl(\angle{BCP}\bigr)$.

By the law of cosines in triangle $BCP$ we get
$$\cos\bigl(\angle{BCP}\bigr)=\frac{1}{2(x+1)}$$
Compute $BD$.

By the law of cosines in triangle $BCD$ we get
$$BD=\sqrt{3-7x}$$
Compute $\cos\bigl(\angle{CBD}\bigr)$.

By the law of cosines in triangle $CBD$ we get
$$\cos\bigl(\angle{CBD}\bigr)=\frac{3(1-x)}{2(1+x)\sqrt{3-7x}}$$
Compute $\cos\bigl(\angle{OBD}\bigr)$.

Since angles $OBD$ and $CBD$ are complementary we get
$$
\cos\bigl(\angle{OBD}\bigr)
=
\sin\bigl(\angle{CBD}\bigr)
=
\sqrt{1-\cos^2\bigl(\angle{CBD}\bigr)}
=
\frac{\sqrt{748-198x}}{44}
$$
Compute $r$.

By the law of cosines in triangle $OBD$ we get
$$r=\frac{22\sqrt{3-7x}}{\sqrt{748-198x}}$$
Compute ${\large{\frac{r^2}{AP^2}}}$.

Since $AP=y=1$ and $x=\sqrt{5}-2$ we get
$$
\frac{r^2}{AP^2}
=
r^2
=
\frac{22(3-7x)}{34-9x}
=
\frac{1138-422\sqrt{5}}{209}
$$

Thus we've computed ${\large{\frac{r^2}{AP^2}}}$, but contrary to the claim of the problem statement, it's not a rational number.

First we have to prove that points $AOB$ are collinear . For that you have to prove that $∠AOP+∠POD+∠DOC+∠BOC=180^o$ .
$\Rightarrow$ Now , as you can see that $ΔABC$ is a Right triangle . So apply pythagoras theorem . $$(x+y)^2+(2r)^2=(2y^2)$$ $$x^2-3y^2+2xy=-4r^2$$
$\rightarrow$ Now , subsitute $y$ with $x$ from $y=x(2+\sqrt5)$ $$4r^2=x^2(22+10\sqrt5)$$
$\rightarrow$ subsitute $x$ with $y$ . you'll get $$\frac{r^2}{AP^2}=\frac{11+5\sqrt5}{18+8\sqrt5}=\frac{a}{b}$$
$\Rightarrow$ so you'll get $hcf(a,b)=1$ . Now you can use the value of $a$ & $b$ . To get your answer .
$\Rightarrow$ I apologize for calculation mistakes if there is one , or any other mistake . Sorry for that !


Sum of $p\sum_{i = 1}^{\infty} i(1 - p)^{i - 1}$ Does anyone know how to find the exact sum of

$$
p\sum_{i = 1}^{\infty} i(1 - p)^{i - 1}
$$

Given that $|x|< 1$, $$\frac 1{(1-x)^2}=\sum_{k=0}^\infty (k+1)x^k$$

So $$p^{-1}=\frac{p}{(1-(1-p))^2}=p\sum_{k=0}^\infty (k+1)(1-p)^k$$

Reasoning: $(1-x)(1+x+x^2+\ldots)=1+x+x^2+\ldots-(x+x^2+\ldots)=\lim_{n\to\infty}1-x^n=1$ so $\frac{1}{1-x}=\sum_{k=0}^\infty x^k$.  Note that $$\frac{1}{(1-x)^2}=\sum_{k=0}^\infty \frac{x^k}{1-x}=\sum_{k=0}^\infty\left(\sum_{\ell=0}^\infty x^{k+\ell}\right)$$
For any $c$, there are $c+1$ pairs $(k,\ell)$ such that $k+\ell=c$, so the sum reduces to $\sum_{k=0}^\infty (k+1)x^k$


How many squares actually ARE in this picture? Is this a trick question with no right answer? This is one of those popular pictures on sites like Facebook. I always see a huge variation of answers such as $8, 9, 16, 17, 24, 28, 30, 40, 41, 52,$ etc., yet I've never seen a definitive answer on any website.

I counted $16$ whole squares ($4$ going down on left side, $4$ going down on right side, $8$ small squares in the centre). With this method the answer could also be $17$ if you want to count the entire shape as one big square.

I can also see the answer being $24$ if you count the smaller squares separately, as if they were floating above the main grid instead of intersecting with it.

Is there a right answer, or is it possible that there is no 'right' answer and it just depends on your counting method?

Thanks to Ross Millikan for helping me to find the last squares. Here is an GIF animation (344 kb) showing 40 squares from biggest to smallest:

Taking the side of each small square as ' one ' ( excluding the two squares in the  middle ) ; the number of squares = 16 (1x1 squares)+9 (2x2 squares)+4(3x3 squares) +1(4x4 square) = 30 ................................The two square in the middle have small sqares , with each side having a value of 1/2 . So the total no. of squares of these two squares = 2x [4(1/2x1/2 squares)+1 (1x1 square)] =10 .......Thus the total number of squares , are 30+10 =40


Why do books say $\sqrt{xy}=\sqrt x\sqrt y$ My question is simple. I'm teaching in a college (pre-calculus course) and I'm asking myself why high school books say $\sqrt{xy}=\sqrt x\sqrt y$. This is false (counterexample is $x=y=-1$).

How can I state this rule in a more general and correct manner?

Thanks

$\sqrt{x}:\mathbb{R}_{\ge0}\to\mathbb{R}_{\ge0}$ is defined as the unique $y\in\mathbb{R}_{\ge0}$ so that $y^2=x$ where $\mathbb{R}_{\ge0}=\left\{x\in\mathbb{R}:x\ge0\right\}$.

For this function, and this should be the function in your book, $\sqrt{xy}=\sqrt{x}\sqrt{y}$.

Unfortunately, after complex numbers are introduced, sometimes writers define an inverse of $z\to z^2$ on $\mathbb{C}$ minus some branch cut and call it $\sqrt{z}$. However, for this function, it is not the case that $\sqrt{zw}=\sqrt{z}\sqrt{w}$.

Corrected:

If $x$ and $y$ are allowed to be complex, your proposed counterexample is no such thing:

When $x = -1$ and $y = -1$,

$$\sqrt{xy} = \sqrt 1 = \pm 1$$

and

$$\sqrt x\sqrt y = \sqrt {-1} \sqrt {-1} = (\pm i) (\pm i) = \pm (-1) = \pm 1 $$

So the high-school text books are more correct than they seem to be.

The above interprets $ \sqrt x $ as a multivalued function. If we take $\sqrt x$ as the principle value, the contention is, as you say, false.


Is there any branch of Mathematics which has no applications in any other field or in real world? Is there any branch of Mathematics which has no applications in any other field or in real world ?

for instance , maybe : number theory ? mathematical logic ?

is there something like this  ?

Lots of branches of mathematics currently have no application in any other field or the real world. As you get higher up the ivory tower, the object that you're studying becomes so esoteric, that there might not be relevance to other things.

However, that does not preclude the possibility that someone eventually finds a relevance for it. Before the 20th century, Number Theory was considered recreational, 'useless' math. It has since spawned a huge industry of security.

Of course, someone might come along and say "Hey, there's this connection between (this esoteric field) and (that esoteric field)", like what Andrew Wiles did (Andrew Wiles proved Fermat's last theorem using many techniques from algebraic geometry and number theory [Source:Wikipedia]).

Being a matter that deals more with quality and less with experiment than any other; maths is an avant-garde needing a delay for it to be applied. If i did not believe in the necessity of ethics i would say it´s one of the most aesthetical disciplines; but also when you prove a theorem you open a new road ar at least a path and make the task of discovering  new formal landscapes easier to successors; are truth and beauty and harmony and structures in the Universe immediately useful?


Finding the last two digits of $17^{198}$ Question 1: What are the last two digits of $17^{198}$?

My Attempt: The pdf hinted to reduce it via mod $100$. So my work is as follows:

Since $17^2\equiv289\equiv-11\mod100$, we have$$17^{198}\equiv (-11)^{99}\equiv-11^{99}\mod100$$
However, I'm not sure how to compute $-11^{99}$ without too much of a hassle.

  Questions:

How do you complete this problem?
Why use mod $100$? Why not mod $50$ or some other number?

An alternative approach, exploiting the Chinese remainder theorem and the fact that $2$ is a generator in $\mathbb{Z}/(25\mathbb{Z})^*$:

$$ 17^{198}\equiv (-8)^{198} \equiv 2^{3\cdot 198} \equiv 2^{14} \equiv 16\cdot 1024\equiv -16\equiv 9\pmod{25} $$
$$ 17^{198}\equiv 1\pmod{4} $$
together imply $17^{198}\equiv \color{red}{9}\pmod{100}$.

Let $G=(\mathbb{Z}/100\mathbb{Z})^{*}$. Since $|G|$=40 and $17\in G$, we have
$$ 17^{198} = 17^{40\cdot 5}\cdot 17^{-2}=1\cdot 17^{-2} \ mod\ 100$$
To find $17^{-2}\ mod\ 100$ we note first that $17\cdot 3 = 1 \ mod \ 10$. Then, we only have to solve
$$ 17\cdot (10x+3) = 1 \ mod\ 100$$
for $x\in[0,9]\cap\mathbb{Z}$, in order to find the inverse. Doing some computations, we find
$$ 70x=50 \ mod\ 100 $$
$x=5$ solves the equation and one can verify directly that $17\cdot 53=1 \ mod\ 100$. We deduce $17^{198}=53^2 = 9\ mod\ 100$


Colored card probabilities A box contains 3 cards. One card is red on both sides, one card is green on both sides, and one card is red on one side and green on the other. One card is selected from the box at random , and the color on one side is observed. If this side is green, what is the probability that the other side is also green?

My attempt:
3 cards
R R
G G
R G

3/6=1/2 chance of getting green

P(GG|G)=$\frac{1/2)(1/2)}{1/2}$
$=1/2$

Does this look ok?

Same question I jsut found: Conditional probability question with cards where the other side color is to be guessed

Now I am torn between 1/2 and 2/3

No, this scenario is counter intuitive.

The model is that a card is picked at random, and then a side is displayed at random.

There are six equally probable results of this -- one for each face.   Three of these result in displaying a green side and only one of those will have red on the other side.

The probability that the other side is also green given that a green side is displayed is: $2/3$.

Remark   The information that "a green side is displayed" is not the same as knowing "the card has a green side", or even that "the card isn't double red."

By listing all the outcomes of the card drawn and side displayed (first or second).

$$\{\overbrace{\underbrace{GG:1, GG:2, GR:1}_\text{you see a green side}, GR:2}^\text{the card isn't double red}, RR:1, RR:2\}$$

The likelihood of drawing each card is equal, but if a card with two green sides is drawn it is twice as likely to display a green side than if the card with one green side were drawn.   Therefore, if a green side is displayed the card is twice as likely to be a double green card than a single green card.

Let $C$ be the card drawn, taking values $\{0, 1, 2\}$ for the number of green sides. (So $C=0$ is the red-red card, et cetera).

Let $D$ be the colour displayed, taking values $\{0, 1\}$ for red or green respectively.

So we want to find: $\mathsf P(C=2 \mid D=1)$

$$\begin{align}
 & = \cfrac{\mathsf P(C=2)\cdot\mathsf P(D=1\mid C=2)}{\mathsf P(C=2)\cdot\mathsf P(D=1\mid C=2)+\mathsf P(C=1)\cdot\mathsf P(D=1\mid C=1)+\mathsf P(C=0)\cdot\mathsf P(D=1\mid C=0)}
\\[1ex] & = \cfrac{\cfrac 1 3 \cdot\cfrac{2}{2}}{\cfrac 1 3 \cdot\cfrac{2}{2}+\cfrac 1 3 \cdot\cfrac{1}{2}+\cfrac 1 3 \cdot\cfrac{0}{2}}
\\[1ex] & = \dfrac 2 3
\end{align}$$

Everything fine.

$$Pr (GG \mid G)=\frac{Pr GG}{Pr G}=Pr G=1/2.$$


Place a circle 'on top' of two other circles? I have two circles (their radii and position) given. I then have a third circle (only it's radius), and would like to calculate its position so it touches both other circles:

There are always two points where the third circle can be placed, but the third circle should be place so that when 'walking' from the first to the second circle, the third is on the left. So on my drawing, the first circle is the big one on the left, the second the smaller one on the right, and the third is the orange one. The arrow shows the 'walking' path.

How can I do this?

The first two circles have radius $R1$ and $R2$; your third circle will have radius $R3$.
The centre of the third circle is $R1+R3$ from one centre, and $R2+R3$ from the second.
If the centres are $C1,C2,C3$, then you want the cross-product $(C2-C1)\times(C3-C1)$ to be positive (I think).  Certainly, the sign of the cross-product is what you need.


Does $\Delta u = f(|x|)$ mean that $u=v(|x|)$ for some $v$? If $u$ solves $\Delta u = f(|x|)$ in $B_1 \subset \mathbb{R}^n$ for some continuous function $f: [0,\infty) \to \mathbb{R}$ does that mean that $u(x) = v(|x|)$ for some function $v$?

I think this is true result about radially symmetric solutions, but I may need need one more condition that $u = 0$ on $\partial B_1$. Anyway, I am having a difficult time showing this result.

Hint: This might not be true

Take $$u(x) = \sum_{i=1}^{n}x_i \implies \Delta u  = 0 $$

or
 $$u(x) =\sum_{i=1}^{n}x_i^4\implies \Delta u  = 12|x|^2$$
 $$u(x) =\sum_{i=1}^{n}(x_i^4 +x_i)\implies \Delta u  = 12|x|^2 $$

Hint What happens if you take $u : \Bbb R^n \to \Bbb R$ to be any nonradial harmonic function (e.g., the projection onto the first component)?


A riddle about a liar Say Alice says:'the probability that I'm lying is greater than p.'

What's the probability that Alice is lying?

It's interesting to try to make sense of this problem.  We'll suppose there's a "real" probability that Alice is lying (either in general, or with respect to the particular statement she is making); call that $q$.  Take a Bayesian approach, and assume a prior distribution on $q$.  (Maybe we'll be lucky, and the answer won't depend on that prior, but, no, we won't be that lucky.)

Let $X$ be the event that Alice claims $q>p$.  We are interested in the probability that Alice is lying given $X$, i.e. $P(q\le p \mid X)$.  Bayes' theorem gives:

$$\begin{align}
 &P(q\le p\mid X) =P(q\le p\,\, \&\,\,  X)/P(X)\\
 &= \frac{P(X\mid q\le p) P(q\le p)}{P(X\mid q\le p) P(q\le p)+P(X\mid q> p) P(q> p)}\\
\end{align}
$$
Now, $P(X\mid q\le p)=q$, since Alice would be lying there, and $P(X\mid q>p)=1-q$ since Alice would be telling the truth there.  Thus, given Alice's statement, the posterior probability that Alice is lying is
$$\frac{q\, P(q\le p)}{q\, P(q\le p)+(1-q)P(q>p)}$$
where the probabilities are our Bayesian priors.

In general this depends on the priors, but in the particular case $p=1$ the probability simplifies to $1$.  Even when $p=0$ this probability may be nonzero, depending on whether our prior has a point mass at $q=0$.

Let $P$(Alice lying$)=q$, then there are $(1-q)$ chance that $q>p$ and $q$ chance that $q\leq p$, so ${1-q\over q}={1-p\over p-0}$ which means $q=p$.

Some intuitive example: if $p=0$ then obiously alice is not lying. If $p=1$ then obviously alice is lying. If $p={1\over2}$ then there is  $1\over2$ chance she is lying.


Given $a_{n} = 3a_{n-1}-2a_{n-2}$, $a_0=0$, $a_1 = 2$. Show $a_n = 2(2^n-1)$ by induction This is my problem:

The recurrence relation is \begin{aligned} a_{n} = 3a_{n-1}-2a_{n-2} \end{aligned}
This is given that $a_{0}=0, a_{1}=2$.

From the above information I calculated
\begin{aligned} a_{2} &=3(a_{2 - 1}) - 2(a_{2 - 2}) = 3(a_{1}) - 2(a_{0}) = 3(2) - 2(0) = 6  \\
a_{3} &=3(a_{3 - 1}) - 2(a_{3 - 2}) = 3(a_{2}) - 2(a_{1}) = 3(6) - 2(2) = 14 \\
 a_{4} &=3(a_{4 - 1}) - 2(a_{4 - 2}) = 3(a_{3}) - 2(a_{2}) = 3(14) - 2(6) = 30 \end{aligned}
I have been trying to solve this recurrence relation for quite some time and have come up with the solution
\begin{aligned} a_{n} = 2 * (2^{n} - 1) \end{aligned}
I am having trouble proving this solution by induction.
My Attempt:
Base Cases:
$$a_0 = 0 = 2(2^0-1)\\a_1 = 2 = 2(2^1-1)$$
Inductive Hypothesis:
Assume that $a_k = 2(2^k-1)$ and $a_{k-1} = 2(2^{k-1}-1)$
Inductive Step:
$$a_{k+1} = 3a_{(k+1)-1}-2a_{(k+1)-2} = 3a_{k}-2a_{k-1}$$
By the inductive hypothesis  $a_k = 2(2^k-1)$ and $a_{k-1} = 2(2^{k-1}-1)$,
$$a_{k+1}= 3 ( 2 (2^k-1) - 2 (2(2^{k-1}-1)).$$
From here it is just simplifying but I cannot get it to simplify to the correct expression.
Any ideas?

The claim that you want to prove by induction is
$$a_0=0\land a_1=2\land\forall n>2:a_n=3a_{n-1}-2a_{n-2}\iff\forall n:a_n=2(2^n-1).$$
There is in fact no induction here, and you can use a direct proof,
$$2(2^n-1)=3\cdot2(2^{n-1}-1)-2\cdot2(2^{n-2}-1)=3\cdot2^n-6-2^n+4,$$ which is true.
You could force induction by stating that if the property holds for all $n\le m$, then it holds for all $n\le m+1$, but this is a little overkill.


Solving the functional equation $f(x)=f(1/x)$ for $x>0$ I am trying to solve the functional equation
$$f(x)=f\left(\frac{1}{x}\right),\quad x>0$$
for a real-valued function $f$, with $f(1)=2$. We may assume $f$ is continuous or infinitely differentiable. There is, obviously, the solution $f(x)=2$ for all $x>0$, but I have been unable to show that this is the only solution. By differentiating the original equation, one easily that $f^{(2n+1)}(1)=0$, where $f^{(k)}$ denotes the $k$th derivative of $f$, and $n$ is a non-negative integer. I cannot see this helping, though. By evaluating $f$ on $f(x)$ and applying the equation which we want it to satisfy, one gets that $f$ must satisfy
$$f\left(\frac{1}{f(x)}\right)=f\left(f\left(\frac{1}{x}\right)\right).$$
This doesn't seem to help either. Substituting values on the original equation yields nothing. Another observation is that
$$\lim_{x\to0+}f(x)=\lim_{x\to0+}f\left(\frac{1}{x}\right)=\lim_{x\to+\infty}f(x),$$
if the limit even exists. I know these are very superficial observations, but I am inexperienced with functional equations. Any hints?

Consider any function $g : [1,+\infty) \rightarrow \mathbb{R}$ such that $g(1)=2$.
Then the function $f : \mathbb{R}_+^* \rightarrow \mathbb{R}$ defined by
$$f(x)=g(1/x) \quad \text{if } 0<x<1 \quad \quad \quad \text{and} \quad \quad f(x)=g(x) \quad\text{if } x\geq 1  $$
is a solution to your problem. (So there are many solutions)
(If you want continuous solutions, just ask that $g$ is continuous, it will give a continuous $f$. If you want differentiable solutions, then try to see at which condition on $g$ the constructed function will be differentiable - hint : you will have a condition about the derivative of $g$ at $1$)

I think, there are infinitely many such functions.
For example, $$f(x)=\frac{(x^{4040}-1)\ln^{2021}x}{x^{2020}}.$$


Prove that if $A \sim I_n$ and $A \sim I_m$ then $n=m$ Some definitions:
Definition of equinumerous sets
Two sets are equinumerous if there exists a bijection $f: A \rightarrow B$. We write $A \sim B $ if $A$ is equinumerous to $B$
Definition of finite set and cardinality
A set $A$ is said finite if $A \sim I_n$  where $I_n=\{k \in \mathbb{N} | k \leq n\}$ and $\mathbb{N}=\{1,2...\}$
If $A$ is finite the unique number $n$ such that $A \sim I_n$ is called the cardinality of $A$
Now the question:
Prove the cardinality of A is well-defined (that is if $A \sim I_n$  and $A \sim I_m$  then $n=m$ )
Suggestion: prove first that if $n \neq m$ then there exist no bijection between $I_n$ and $I_m$.
So, if I assume by absurd that $m \neq n$, let's say $m < n$
There exists an injective but not surjective mapping  $i:I_m \rightarrow I_n$.
Now since this is so obvious I am having trouble going forward. Actually, I don't know if the previous statement, provided it is needed, is too far a step into the proof, given that what I want to prove is as much as obvious.
Can someone enlighten me?

Proving the statement in the suggestion is not trivial. What you’ve done isn’t nearly enough: the existence of an injection from one set to another that is not a surjection does not prove that there is no bijection between the two sets. For instance, the map $f:\Bbb N\to\Bbb N:n\mapsto n+1$ is a non-surjective injection from $\Bbb N$ to itself, but obviously there are bijections from $\Bbb N$ to $\Bbb N$! It’s only for finite sets that the existence of a non-surjective injection implies that there is no bijection, and that’s essentially what you’re supposed to be proving here.
Let $B$ be the set of $n\in\Bbb N$ such that there is a bijection from $I_n$ to some $I_m$ with $m<n$. Suppose that $B\ne\varnothing$; by the well-ordering principle we can let $n=\min B$. Let $f:I_n\to I_m$ be a bijection, where $m<n$, and let $k=f(n)$. Let $g$ be the restriction of $f$ to $I_{n-1}$; $g$ is a bijection from $I_{n-1}$ to $I_m\setminus\{k\}$.
Now define a function $h:I_m\setminus\{k\}\to I_{m-1}$ as follows:
$$h(i)=\begin{cases}
i,&\text{if }1\le i<k\\
i-1,&\text{if }k<i\le m\;.
\end{cases}$$
It’s easy to verify that $h$ is a bijection. But then $h\circ g$ is a bijection from $I_{n-1}$ to $I_{m-1}$, so $n-1\in B$, contradicting the choice of $n$ as the smallest member of $B$. This contradiction shows that $B$ must be empty and hence that no $I_n$ can be mapped bijectively to an $I_m$ with $m<n$.

Let $m<n$. Suppose that there is a bijection $f: I_m\to I_n$. Write
$$ I_n=f(I_m)\cup A, f(I_m)\cap A=\emptyset, A\neq\emptyset. $$
For $y\in A$, since $f: I_m\to I_n$ is bijection, there is $x\in I_m$ such that $f(x)=y\in f(I_m)$. From this one can see that $y\in f(I_m)\cap A$ which is absurd since $f(I_m)\cap A=\emptyset$.


{Proof Check} Let $x_n$ be a convergent sequence. if $x_n \geq 0$ $\forall n \in \mathbb{N}$ then $\lim x_n \geq 0$ Please check and point out if there are logical flaws in my argument. I'm learning analysis but I'm not that confident yet.
Proof: Suppose $x_n \geq 0 $ $\forall n \in \mathbb{N}$ and let $\lim x_n = x < 0$. Then $|x| > 0$. Take $\varepsilon = |x|$ then $\exists N \in \mathbb{N}$ s.t. $\forall n \geq N$,
$$|x_n - x| \leq |x_n| + |x| < \varepsilon = |x|$$
$$0 \leq |x_n| = x_n < |x| - |x| = 0$$
This is a contradiction, thus $x \geq 0$.

You have the right value of $\varepsilon$ (and any value smaller than that but still positive can also serve).
But how did you conclude that $|x_n| + |x|<\varepsilon \text{?}$
I'd write $|x_n-x| = x_n-x$ since $x_n\ge0$ and $-x\ge0,$ and then $x_n-x\ge -x = \varepsilon.$ Thus no matter how big $n$ gets, $|x_n-x|$ never gets less than $\varepsilon.$

The proof seems correct, you showed there is an $\varepsilon$ for which no $N$ exists. The only thing that seems a bit unorthodox is using $x$ as the variable for the limit instead of the more common $L$.


Convergence of the series $\sqrt[n]n-1$ Let $a_n=\sqrt[n]n-1$. Does $\sum_{n=1}^\infty a_n$ converge?

Hint: $$\sqrt[n]n  = e^{\frac{\log n}{n}} > 1+\frac{\log n}{n}$$

Note that for any $n\ge3$:
$$ n>e$$
and so,
$$ n^{1/n} > e^{1/n}$$

And since we know:
$$ e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$ we can say:
$$ e^{1/n} = \sum_{k=0}^\infty \frac{1}{k!\;n^k} = 1 + \frac{1}{n} + \frac{1}{2n^2}+\cdots$$
Which, for $n\in\Bbb N$, is surely larger than the truncated series $1+1/n$, and so follows the strict inequality:
$$n^{1/n} > e^{1/n} > 1 + \frac{1}{n}$$

Subtract one from both sides, yielding:
$$\sqrt[n]{n}-1 = n^{1/n} - 1>\frac1n$$

Which is true $\forall n\ge 3$. Evaluate the sum:
$$ \sum_{n\ge 3} (\sqrt[n]{n}-1) > \sum_{n\ge 3} \frac{1}{n}$$

Thus, $\sum a_n$ is larger than the divergent harmonic sum, and is divergent too.


A question about a surface This is a question in "baby do Carmo" 2.2.14. The question says

A half-line $[0, \infty)$ is perpendicular to a line $E$ and rotates about $E$ from a given initial position while its origin $0$ moves along $E$. The movement is such that when $[0, \infty)$ has rotated through an angle $\theta$, the origin is at a distance $\displaystyle d=\sin^2\bigg(\frac{\theta}{2}\bigg)$ from its initial position on $E$. Verify that by removing the line $E$ from the image of the rotating line we obtain a regular surface. If the movement were such that $\displaystyle d=\sin\bigg(\frac{\theta}{2}\bigg)$, what else would need to be excluded to have a regular surface?

My idea is the following, which could be wrong.
Let say the half line is the $x$ axis and the line $E$ is the $z$-axis. I think this surface is quite similar to helicoid, except it's finite because when rotating the $x$-axis about the $z$-axis once, the height varies from 0 to 1 and back to 0.

My doubt:

I think one parametrization $x:\mathbb{R}^+\times(0, 2\pi)\to\mathbb{R}^3$ can be given by
$$x(u, \theta)=\bigg(u\cos\theta, u\sin\theta, \sin^2\bigg(\frac{\theta}{2}
\bigg)\bigg).$$
The surface we want should be the image of this map. I am not sure whether this is correct or not.

My question:

The latter part of the question says when $\displaystyle d=\sin\bigg(\frac{\theta}{2}\bigg)$, what else has to be excluded. Besides the line $E$ is there any other thing should be excluded?

The hint is to think about the parametrized curve
$$\alpha(t)=(\sin t,\sin 2t)$$
in the plane. If you restrict the domain to $(-\pi+\epsilon,\pi-\epsilon)$, it is a fine curve. But what happens when we allow either of $\pm \pi$? We still have a regular parametrization (i.e., the derivative is everywhere non-zero), but the curve now intersects itself.

A similar phenomenon is occuring with doCarmo's surface. With the latter parametrization, when $\theta = \pm 2\pi$ the surface now intersects itself along the curve coming from $\theta = 0$. It is an example of an immersed surface, just as the curve $\alpha$ is an immersed curve.


Sum of digits of a square number. A positive integer n is said to be good if there exists a perfect square whose sum of digits in base 10 is equal to n. For instance, 13 is good because (7^2)=49 and 4+9=13. How many good numbers are among 1, 2, 3, …, 2007?

I have started to make a list of all square numbers and adding their digits in an excel file. ...   Until now, I have got good numbers as 1, 4, 7, 9, 10, 13, 16, 19.   But, this method takes very long. ...   Is there a shorter and smarter way to solve this problem.

To be formal about this note that for $r$ large enough and $s$ a single decimal digit
$$ (10^r-s)^2 = 10^{2r}-2s \cdot 10^r+r^2
= 9\cdot(10^{2r-1}+10^{2r-2}+\dots 10^{r+3})+(100-2s)\cdot10^r+s^2$$

So we begin with a string of $9$s which is $r-3$ long followed by two digits which are fixed (and may be zero or $9$), then a string of zeros followed by either one or two digits of $s^2$. Every time you increase $r$ by one, the digit sum increases by the addition of a single leading $9$.

Take $s=1$ to see what happens

$9^2=81$

$99^2=9801$

$999^2=998001$

This deals with $9,18, 27\dots$

With $s=2$

$8^2=64$

$98^2=9604$

$998^2=996004$

Which deals with $10,19,28\dots$

With $s=3$

$7^2=49$

$97^2=9409$

$997^2=994009$

Which deals with $13,22,31\dots$

$s=4$ deals with multiples of $9$ again.

$s=5$ gives

$5^2=25$

$95^2=9025$

$995^2=990025$

And this deals with $7,16,25\dots$

With the addition of a few early cases which these miss, you can show that you hit every positive element of every class of quadratic residue modulo $9$


How to prove $\mathbb{N}\subseteq\mathbb{R}$ given basic field axiomatisation. I understand that $\mathbb{R}$ can be defined axiomatically in a number of different ways, including as an extension from more basic number systems (e.g., $\mathbb{N}$, $\mathbb{Z}$, or $\mathbb{Q}$) using equivalence classes or Dedekind cuts. However, my question here assumes that $\mathbb{R}$ is defined as a complete ordered field directly from the following axioms:

Field Axioms (defining addition and multiplication with algebraic field properties -- in particular including a multiplicative constant $1\in\mathbb{R}$)
Order Axioms ($\mathbb{R}$ is totally ordered)
Completeness Axiom (the order is Dedekind-complete)

And now to the question: Given these axioms, how does one prove the existence of a Peano system $P\subseteq\mathbb{R}$ with $1_{\mathbb{R}}=1_P$ (i.e., I want to prove that $\mathbb{N}\subseteq\mathbb{R}$)?

With $S\colon \Bbb R\to\Bbb R$, $x\mapsto x+1_{\Bbb R}$ let $\Bbb N$ be the intersection of all subsets of $\Bbb R$ that contain $1_{\Bbb R}$ and are closed under $S$.

That is: With $e=1_{\Bbb R}$ (or $e=0_{\Bbb R}$ depending on taste) the definition
$$\Phi(A)\stackrel {\text{def}}\iff  e\in A\land \forall x\in A\colon S(x)\in A$$
we let
$$\Bbb N\stackrel {\text{def}}=\bigcap\{\,A\in\mathcal P(\Bbb R)\mid \Phi(A)\,\}.$$
(This definition makes sense because such subsets $A$ do exist, for example we trivially have $\Phi(\Bbb R)$).
One verifies that $\Phi(\Bbb N)$, so that $(\Bbb N,e,S)$ is a viable candidate: $\Bbb N$ is a set, $e\in\Bbb N$ is an element, and $S\colon \Bbb N\to\Bbb N$ is a map. The axioms are readily verified:

Assume $e=S(a)$ for some $a\in\Bbb N$. From $\Phi(\{\,x\in\Bbb R\mid x\ge e\,\})$ (which follows from $1_{\Bbb R}\ge 0_{\Bbb R}$) we conclude that $a\ge e$, hence $e=a+1_{\Bbb R}\ge e+1_{\Bbb R}$, contradiction
Assume $S(a)=S(b)$. Then $a+1_{\Bbb R} = b+1_{\Bbb R}$, hence $a=b$.
Let $A\subset\Bbb N$ be a set with $e\in A$ and $\forall x\in A\colon S(x)\in A$. In other words, $\Phi(A)$. Then by definition $\Bbb N\subseteq A$, hence $A=\Bbb N$.

Note that we didn't use completeness (so we find $\Bbb N$ also inside $\Bbb Q$) or Archimedean-ness of the order. Ordered field is enough.

Hint:

Along the lines of Hagen's answer, construct function $S$ such that:

$$\forall a\in R: S(a)=a+1$$

Construct $N$ such that:

$$\forall a:[a\in N \iff a\in R \land \forall b\subset R:[0 \in b \land \forall c\in b: [S(c)\in b] \implies a\in b]]$$

Then prove:

$0\in N$
$\forall a\in N: S(a) \in N$
$\forall a,b \in N:[S(a)=S(b)\implies a=b]$
$\forall a\in N: S(a)\neq 0$
$\forall P\subset N: [0\in P \land \forall b\in P: [S(b) \in P] \implies P=N]$


Continuous function problem on $\mathbb{R}$ Can somebody please help me with this problem?

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f([a,b])$ is an interval, for all $a,b\in\mathbb{R}$ such that $a < b$. Also, $f^{-1}(r)$ is closed for all $r\in\mathbb{Q}$. Show that $f$ is contiuous on $\mathbb{R}$.

Thanks in advance.

Since you’re asking for help, here is an intentionally incomplete solution. It suffices to show that $f$ is continuous on the left and on the right at every point.
 Let us show for example that $f$ is right-continuous  at a point $x\in{\mathbb R}$.

For $\delta \gt 0$, there are two nonnegative numbers (which we will denote by $A(\delta)$ and
$B(\delta)$) such that $f([x,x+\delta])=[f(x)-A(\delta),f(x)+B(\delta)]$ (why ?)
  The functions $A$ and $B$ are nonincreasing in $\delta$ (why?)
  So the limits $a=\lim_{\delta \to 0, \delta \gt 0} A(\delta)$ and $b=\lim_{\delta \to 0, \delta \gt 0} B(\delta)$ exist ; our job is to show that those limits are in fact zero.

Suppose for example that $a\gt 0$. Then there is a rational $r$ such that $f(x)-a \lt r \lt f(x)$. Show that the closed set $C=f^{-1}(r)$ must contain $x$, and derive a contradiction from that.


Prove that the function can be continued into a larger domain Prove that the function  $f(z)=\displaystyle\sum_{n=1}^{\infty}(-1)^{n+1}\frac{z^n}{n}$ can be continued into a larger domain by means of the series $$\ln2-\frac{1-z}{2}-\frac{(1-z)^2}{2\cdot 2^2}-\frac{(1-z)^3}{3\cdot2^3}-\dots$$

Attempt

First notice that $f(z)$ is similar to $\ln z=\displaystyle\sum_{n=1}^{\infty}(-1)^{n}\frac{(z-1)^{n+1}}{n+1}$

Thus I think we can say that $f(z)=\ln(z-1)$

I do not know what can I do from here

I don't understand

Where does $\ln 2$ come from?

How is f(z) and the series $\ln2-\frac{1-z}{2}-\frac{(1-z)^2}{2\cdot 2^2}-\frac{(1-z)^3}{3\cdot2^3}-\dots$ related?

Any help would be greatly appreciated.

The right formula of $f(z)$ is $\ln(1+z)$, with the domain $(-1,1]$.

The series converge only in $(-1,1]$, but the function $\ln(1+z)$ exist in $(-1,+\infty)$.

The series in your question is just the Taylor series of $f(z)$ at the point $z=1$

$$\ln(1+z)=\ln(2+(z-1))=\cdots$$


The equation of the bisector of the angle between two lines $3x+4y+12=0$ and $12x-5y+7=0$ which contains the point $(-1,4)$ The equation of the bisector of the angle between two lines $3x+4y+12=0$ and $12x-5y+7=0$ which contains the point $(-1,4)$

I know how to take out the equation of the 2 bisectors, but I am not able to figure out which angle would contain the point $(-1,4)$. Can someone explain how I can figure this out so as to identify the equation of the right bisector the question demands.

Once you know how to find those bisectors I think the problem is how to locate the point $P=(-1,4)$ in one of those for regions.

Note that you can distinguish those regions looking to up or down of each line:

For example, regions $2$ is above of line $s$ and below of line $r$. Taking as reference the axis $OY$.

First, consider the equation $(s)$ $3x+4y+12=0$. If you take $x=-1$ you get $y=-9/4$ which is below of $y_p=4$. So $P$ is above of the line $3x+4y+12=0$.

Samething for $(r)$ $12x-5y+7=0$. If you take $x=-1$ you get $y=1$ which is below of $y_p=4$. So $P$ is above of the line $12x-5y+7=0$.

So the point $P$ would be in the region $1$.

You have found the equations of the bisectors $b_1:a_1x+b_1y+c_1=0$ and $b_2:a_2x+b_2y+c_2=0$.

To determine which one of the bisectors contain a point $P=(x_0,y_0)$ you simply insert the point and check that satisfies the equations.


How to solve the equation $\exp(iz)=-e$ How do I solve the equation

$$\exp(iz)=-e$$

Can anyone please explain the procedure to solve this kind of question to me please?

Much appreatiate

Solution with comments:  $e^{iz -1} = -e$:

1.)  $e^{iz} = -e \Rightarrow e^{iz -1} = -1= e^{2\pi ni}e^{\pi i} = e^{(2n + 1) \pi i}$, where $n \in \Bbb Z$;

2.)  $e^{iz - 1} =  e^{(2n + 1) \pi i} \Rightarrow e^{iz - 1 - (2n + 1)\pi i} = 1$;

3.)  $e^{iz - 1 - (2n + 1)\pi i} = 1 \Rightarrow iz - 1 - (2n + 1)\pi i = 2m \pi i$, where $m \in \Bbb Z$;

4.) $iz - 1 - (2n + 1)\pi i = 2m \pi i \Rightarrow iz - 1 - 2(m + n + 1) \pi i = 0$
$\Rightarrow iz - 1 - (2k + 1) \pi i = 0,$ where $k = m + n \in \Bbb Z$;

5.) $iz - 1 - (2\pi k + 1) i = 0 \Rightarrow iz = 1 + (2\pi k + 1)  i \Rightarrow z = -i + (2k + 1) \pi$.

We check the solutions $z = -i + (2k + 1) \pi$ obtained in (5):

A.)  $z = -i + (2k + 1) \pi \Rightarrow iz = 1 + (2k + 1) \pi i \Rightarrow iz - 1 = (2k + 1)\pi i$;

B.)  $iz -1 = (2k + 1) \pi i \Rightarrow e^{iz - 1} = e^{(2k + 1) \pi i} = e^{2k \pi i} e^{\pi i} = -1$.

The above shows that all solutions of

$e^{iz - 1} = -e \tag{6}$

are of the form

$z = -i + (2k + 1)\pi, \tag{7}$

with $k \in \Bbb Z$, and that every $z$ as in (7) solves (6).

Note that we have directly used the fact that $e^w = 1 \Leftrightarrow w = 2 k \pi i, \;\; k \in \Bbb Z$, instead of invoking the logarithm function(s), $\ln$ or $\text{Ln}$, in an appropriate open set of $1$ in the punctured complex plane $\Bbb C \setminus \{0\}$.  This fact is elementary and easy to prove by writing $z = x + iy$ and then $e^z = e^x(\cos y + i \sin y) = 1$, and avoids the subtleties of the complex logarithm.  For these reasons, I chose this line of attack for my solution.

Hope this helps.  Cheers,

and as always,

Fiat Lux!!!

$e^{iz} = -e$

$e^{iz} = e^{\pi i}e$

$e^{iz} = e^{\pi i + 1}$

$ln(e^{iz}) = ln(e^{\pi i + 1})$

$iz\ \ln(e) = (\pi i + 1)ln(e)$

Can you take it from here?


Finding the $\cot\left(\sin^{-1}\left(-\frac12\right)\right)$ How can I calculate this value?

$$\cot\left(\sin^{-1}\left(-\frac12\right)\right)$$

Draw a right triangle (in the x>0,y<0 quadrant) with opposite edge -1 and hypotenuse 2.  Then the adjacent side is $\sqrt{2^2-1^2}=\sqrt{3}$.  cotangent is the ratio of adjacent side over opposite side.

Let arcsin(-1/2)=  x
Implies sinx=-1/2
Implies x= 120 degree
Now u need to find value of cot x
So  cot x= - underroot 3


Length of hypotenuse v/s change in height of the opposite I have always struggled to understand mathematical concepts, and have a very different way of thinking about problems.  I suspect this is a very simple problem, but its confusing me a great deal.

I made a right triangle where the "opposite" side was $10$ and the "adjacent side" was $60$.  I then kept increasing the height of the opposite side by $10$ and measuring the length of the hypotenuse.  This rate of change in the hypotenuse length is not constant and reminds me of a sine wave because of its non-linear acceleration.

However, the rate of change appears to become linear as the graph progresses.

I just want to have an intuition for why the rate of change the length of the hypotenuse is not linear, and if there is one or several different ways I could visualize this to have a better intuition for whats going on.

Graph & Triangle:

Imagine a right-angle triangle with opposite side of $x$, that you'll keep changing. An adjacent side $a$ a constant that you'll leave alone and fix. Then the length of the hypotenuse is given by $$\sqrt{x^2 + a}$$

Now, as $x$ grows very large, the constant $a$ becomes kind of irrelevant. Think of adding $10$ to $20,000,000,000,000$ it'll hardly make a difference to the calculuations, so we can say that $$\sqrt{x^2 + a} \approx \sqrt{x^2} = x$$
for large $x$. For the smaller values of $x$, that extra $+a$ will have quite an impact on the size of the $x^2 + a$ term.

This is a very good question and your diagram is helping me to understand a surveying problem im trying to solve, but anyhow i think i might be able to explain this a little bit, i hope haha.

Without getting into the math it might help to look at the problem by understanding the absolute limits of the triangle you are describing ( a right triangle with a fixed horizantal length) by finding the absolute limits of a geometric shape you will see that all real points that are obtainable fit within the absolute limits.  I hope this makes sense.  Absolute limits are 0 and infinty  both are un-attainable limits and everything must fit inside(or around in the case of zero) these limits.  Make sense?

So because of these two conditions you have created limits for the triangle.  When we take the two variables ( vertical length and hypotenuse length) to their limits ( infinity) we see that they both become infinite or in other words when the vertical length goes to infinity the hypotenuse goes to infinity at the same time.  This means that the vertical length is catching up to the hypotenuse length as the vertical gets longer ( hence not a linear ratio) until the vertical length reaches infinity and then it will equal the hypotenuse length of infinity.  Otherwise the vertical length is always less than the hypotenuse because infinty is un attainable.  This is why your graph stables out  vertical length is catching up to hypotenuse length the longer it gets.

Now, look at it another way.
If you take the vertical length to 0  you will see that the hypotenuse reaches a new limit of?.....  the adjacent side length.  And now when the vertical length is 0 it is no longer a triangle it is just a line.   As soon as the vertical length is anything greater than zero you will see that the hypotenuse becomes just greater than the adjacent length but it will never be less than or equal to the adjacent length or else you wont have a right triangle.

Another limit of this triangle exist within the angle between the hypotenuse and the adjacent side.  This angle can go from anything greater than 0 to anything less than 90 degrees.  It will never be 0 or 90 degrees or else again the triangle breaks down and becomes a line. Even as that vertical line goes onto infinty the angle between the adjacent and hypotenuse will only approach 90 degrees but will never be 90 degrees.

I hope this sheds some practical light on your mathmatical question.

Cheers!


Price Calculation based on probabilities This is a strange question, it might be easy for all you math wizards or it even may be impossible. If you don't understand what I mean let me know so I can change the way I post the question.

Product A can be built using Product B.

To get Product B I have to transform product C or Product D.

I have a 50% chance to get product B by transforming product C and 25% to get product B by transforming product D.

Is there any way to predict the cost of production for Product A knowing the price for Product C and D

Edit 1
There are no transformation costs, only the costs for the materials

Suppose you go with the strategy "buy C and attempt a transformation to B until you get B".  Then with probability 1/2, this will cost C, with probability $1/2(1-1/2)$, cost $2C$, with probability $1/2^n$ cost $nC$.  Thus the expected price will be

$C\sum_{n=1}^\infty \frac{n}{2^n}=2C$

Similarly, the strategy "buy D and attempt a transformation to B until you get B" will have expected price

$B_D+D\sum_{n=1}^\infty \frac{n3^{n-1}}{4^n}=4D$

It's no coincidence that the expected price is just the cost divided by the probability of success; this follows from the properties of the binomial distribution.

You'll want to compare these two and see which is minimal and run with that strategy.

Product A can be built using Product B at a 50% chance (costs $5)

To get Product B I have to transform Product D (costs $1400)

I have a 25% to get product B by transforming product D.

I want to produce 5 x Product A.

Could anyone help me determine the probability and expected costs to produce 5 of Product A? and what at what price would be better statistically to purchase Product B than to produce it?

An explanation and an answer would be great and much appreciated.