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I have not written in a while, but this blog is still alive, so I will quickly prove two of my favorite (among many) results of topology. These stem straight from my preparation for the coming preliminary examinations.

School all of a sudden snatched most of my time.

Let $X$ be a topological space.

Claim: $X$ is Hausdorff if and only if $\Delta:=\left\{\left(x,x\right):x\in X\right\}$ is closed.

Proof: Suppose that $X$ is Hausdorff, and let $\left(x_1,x_2\right)\in X\times X\setminus\Delta$. Then $x_1\neq x_2$, which implies that there exist disjoint neighborhoods $U$ and $V$ of $x_1$ and $x_2$, respectively, which implies that $U\times V\subseteq X\times X\setminus\Delta$, which implies that $X\times X\setminus\Delta$ is open, which implies that $\Delta$ is closed.

Suppose that $\Delta$ is closed, and let $x_1,x_2\in X$ such that $x_1\neq x_2$. Then $\left(x_1,x_2\right)\in X\times X\setminus\Delta$, which is open, which implies that there exists a neighborhood $U\times V$ of $\left(x_1,x_2\right)$ contained in $X\times X\setminus\Delta$, which implies that there exist disjoint neighborhoods $U$ and $V$ of $x_1$ and $x_2$, respectively, which implies that $X$ is Hausdorff. $\blacksquare$

Let $X$ and $Y$ be topological spaces, let $Y$ be Hausdorff, let $A\subseteq X$, let $f,g:\overline A\to Y$ be continuous, and let $f\left(x\right)=g\left(x\right)$ for every $x\in A$.

Claim: $f\left(x\right)=g\left(x\right)$ for every $x\in\overline A$.

Proof: Define $h:\overline A\to Y\times Y$ by $x\mapsto\left(f\left(x\right),g\left(x\right)\right)$. Then $h$ is continuous and $h\left(A\right)\subseteq\Delta$, which implies that $h(\overline A)\subseteq\overline{h\left(A\right)}\subseteq\overline\Delta$. Since $Y$ is Hausdorff, $\Delta$ is closed, which implies that $h(\overline A)\subseteq\Delta$, which implies that $f\left(x\right)=g\left(x\right)$ for every $x\in\overline A$. $\blacksquare$

I like these results because the first characterizes Hausdorff spaces in simple terms, and the second shows that no shenanigans happen at the continuous images in Hausdorff spaces of limit points.

Remark: It is implicit in the first claim that the topology on $X\times X$ is the product one, which is generated by the set of products of open subsets of $X$. Let $\left(x_1,x_2\right)\in X\times X$, and let $N$ be a neighborhood of $\left(x_1,x_2\right)$. Then
$$N=\bigcup_{\alpha\in I}U_\alpha\times V_\alpha,$$
where $U_\alpha$ and $V_\alpha$ are open, and $I$ is some index set. Therefore, there exists a $\beta\in I$ such that $\left(x_1,x_2\right)\in U_\beta\times V_\beta$.


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3 of 7 Try retyping first $2$ equations and add explanation that it may not show properly.

I'm adding a new answer instead of updating or adding to my original one since this is quite different based on the new information provided. The answers by Ethan Bolker and Fabio Somenzi involve using (note: the following $$2$$ equations may not display properly, although they appear to be written correctly & show fine in Edit mode)

$$a + 2n(b - n) = k(b - 2n) \tag{1}\label{eq1}$$

to create a quadratic equation in $$n$$, i.e.,

$$n^2 - (k + b)n + \frac{bk - a}{2} = 0 \tag{2}\label{eq2}$$

which has a discriminant of

$$D = (bk - a)^2 - 2(bk - a) = k^2 + b^2 + 2a \tag{3}\label{eq3}$$

The value of $$n$$ is

$$n = \frac{k + b \pm \sqrt{D}}{2} \tag{4}\label{eq4}$$

The smallest positive integer $$n$$ comes from subtracting $$\sqrt{D}$$ where $$D$$ is a perfect square of the same parity as $$k + b$$, and less than but as close as possible to $$(k + b)^2$$. Thus, from \eqref{eq3}, you want to start with the smallest $$k$$ such that $$bk - a \gt 0$$, say $$k_0$$, so

$$b \ge bk_0 - a > 0 \; \Rightarrow \; -2b \le -2(bk_0 - a) \lt 0 \tag{5}\label{eq5}$$

From \eqref{eq3},

$$(k_0 + b)^2 - 2(bk_0 - a) \ge k_0^2 + 2bk_0 + b^2 - 2b \tag{6}\label{eq6}$$

However, note that

$$(k_0 + (b - 1))^2 = k_0^2 + 2bk_0 + b^2 - 2b + (1 - 2k_0) \tag{7}\label{eq7}$$

Thus, the next smaller perfect square of $$D$$ is $$(k_0 + (b - 1))^2$$. Note going from $$m^2$$ to $$(m + 1)^2$$ involves adding $$2m + 1$$, then to $$(m + 2)^2$$ involves adding $$2m + 3$$, then add $$2m + 5$$ to get $$(m + 3)^2$$, etc. In other words, even for very large integers, you can quickly and easily go from a perfect square to the next perfect square by adding a value which you increment by $$2$$ each time. Thus, starting at $$k = k_0$$ in \eqref{eq3}, and using \eqref{eq7}, you can quickly determine each next value for $$k$$ and compare it to the next possible perfect square, incrementing each value(s) as appropriate, until the values match, and the result gives an integer in \eqref{eq4}. This is relatively efficient, and is generally considerably faster than doing integer divisions (e.g., for factoring or to check modulo values). For example, https://agner.org/optimize/optimizing_cpp.pdf says at section 14.5 Integer division:

Integer division takes much longer time than addition, subtraction and multiplication (27 - 80 clock cycles for 32-bit integers, depending on the processor).

This speed issue will vary with things like the size of the integers, the compiler being used, etc., but I believe in basically all cases that integer division will be quite slow.

 answered May 12 at 8:39 John Omielan 14.9k22 gold badges88 silver badges3131 bronze badges

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# CTK Insights

• ## Pages

27 Oct

### That Divergent Harmonic Series

The harmonic series, i.e., the series of the reciprocals of integers is a prototypical divergent series: the sums

$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$

grow, albeit very slowly, without bound. There is no finite value S that could be ascribed as its sum. Assuming that such a value exists leads to a contradiction. A contradiction could be achieved in a multitude of ways. R. P. Boas once listed five of them, but there are more. Several has been listed by R. Honsberger and by W. Dunham. Kifowit and Stamps gave 20 proofs. More recently, the harmonic series attracted bloggers, at Let's Play Math and Random Walks.

Below I'll give a few proofs.

So assume that the series is convergent, i.e., for a finite S,

$S = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n} + \cdots.$

Let's mention the obvious, viz., that all the terms in the harmonic series are positive. It is well known that, for such series, the terms in the sum could be shuffled in any order without affecting the sum.

### Proof 1 (Johann Bernoulli)

$S = \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{1} + \frac{1}{3} + \cdots.$
$S = \frac{1}{2}(\frac{1}{1} + \frac{1}{2} + \cdots) + (\frac{1}{1} + \frac{1}{3} + \cdots).$
$S = \frac{1}{2}S + (\frac{1}{1} + \frac{1}{3} + \cdots).$
$\frac{1}{2}S = \frac{1}{2} + \frac{1}{4} + \cdots = \frac{1}{1} + \frac{1}{3} + \cdots,$

which is absurd as the sum of reciprocals of the even and odd integers could not be equal. To see that compare then term by term.

### Proof 2 (Johann Bernoulli)

$S - 1 = \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} + \cdots$
$S - 1 = \frac{1\times 1}{1\times 2} + \frac{2\times 1}{2\times 3} + \cdots + \frac{({n}-{1})\times 1}{({n}-{1})\times n} + \cdots$
$S - 1 =$
$(\frac{1}{1\times 2} + \frac{1}{2\times 3} + \frac{1}{3\times 4} + \cdots) +$
$(\frac{1}{2\times 3} + \frac{1}{3\times 4} + \frac{1}{4\times 5} + \cdots) +$
$(\frac{1}{3\times 4} + \frac{1}{4\times 5} + \cdots) + \cdots$

Now observe that every row in that infinite sum of infinite sums is a telescoping series. The first row sums to 1, the second to $\frac{1}{2}$, the third to $\frac{1}{3}$, and so on. We get

$S - 1 = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots = S,$

making $2S = 1,$ i.e., $S = \frac{1}{2}$ whereas already the first term in the series is 1. A contradiction.

### Proof 3

Combine terms in pairs and, in every pair, replace the largest term with the smallest:

$S = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots > (\frac{1}{2}+\frac{1}{2})+(\frac{1}{4}+\frac{1}{4})+(\frac{1}{6} +\frac{1}{6})+\cdots.$

$S = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots > {2} \times (\frac{1}{2}+\frac{1}{4}+\frac{1}{6})+\cdots.$

$S = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots > \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots = S.$

That is $S > S.$ A contradiction.

### Proof 3'

The terms in the series can be combined $m, m > 1,$ at a time with exactly the same result.

### Proof 4 (Nicole Oresme)

The terms of the series could be combined into growing groups of $2^{k}$ terms, $k=1,2,3,\cdots$:

$S = \frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots >$

$\frac{1}{1}+\frac{1}{2}+{2} \times \frac{1}{4}+{4} \times \frac{1}{8}+\cdots$

$\frac{1}{1}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots$

which clearly grows without bound.

### Proof 5

There are $9$ one-digit numbers; their reciprocals exceed $\frac{1}{10}$ each. The sum of the reciprocals exceeds $\frac{9}{10}$. There are 90 2-digit numbers; their reciprocals exceed $\frac{1}{100}$ each. The sum of the reciprocals of the 2-digit numbers exceeds $\frac{90}{100} = \frac{9}{10}.$ Similarly, the sum of the reciprocals of all 3-digit numbers exceeds $\frac{900}{1000} = \frac{9}{10},$ and so on. The sum of all the reciprocals of integers (i.e. the sum of the harmonic series) exceeds $k \times \frac{9}{10}$, for any positive integer $k$.

### Proof 6 (Pietro Mengolli)

The proof is based on the observation that, for $a > 1$,

$\frac{1}{a-1} + \frac{1}{a} + \frac{1}{a+1} > \frac{3}{a}.$

This suggests grouping the terms of the harmonic series by 3 starting with second one. The process leads to the inequality $S > 1 + 3S$, which appears to imply that $S < - \frac{1}{2}.$

The required inequality follows from $\frac{1}{n-1}+\frac{1}{n+1} = \frac{2n}{n^{2}-1} > \frac{2n}{n^2} = \frac{2}{n}.$

### Proof 7 (A. Cusumano)

$S = (\frac{1}{1} + \frac{1}{2}) + (\frac{1}{3} + (\frac{1}{4}) + (\frac{1}{5} + \frac{1}{6}) + \cdots =$

$(\frac{1}{1} + \frac{1}{2}) + (\frac{1}{2} + \frac{1}{12}) + (\frac{1}{3} + \frac{1}{30}) + \cdots =$
$S + (\frac{1}{2} + \frac{1}{12} + \frac{1}{30} + \cdots).$

### References

1. R. Honsberger, Mathematical Plums, MAA, 1979
2. R. Honsberger, Mathematical Gems II, MAA, 1976
3. W. Dunham, Journey through Genius, Penguin Books, 1991
4. A. Cusumano, The harmonic series diverges, American Mathematical Monthly 105(7), 608.
5. S. J. Kifowit, T. A. Stamps, The Harmonic Series Diverges Again and Again, The AMATYC Review, Vol. 27, No. 2, Spring 2006, pp. 31–43

1. 1
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# What are the main causes of Corrosion?

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## Corrosion is a phenomenon of oxidation of metal in the presence of oxygen when it is exposed to external environmental factors which include moisture, acid rain, etc., and chemicals like acids, bases, etc.For example - Rusting of Iron, tarnishing of Silver wares, green coating on Copper vessels.It can be prevented by a number of methods like paints and varnishes, alloy formation, galvanization, etc.The main causes of corrosion are given below.Weather conditions - Metals on exposure to environmental elements like water, wind, and moisture oxidize the metal surface and corrode it. For example formation of an Oxide layer on any metal surface like Fe, Al, etc forming Iron (III) oxide ( ${\mathrm{Fe}}_{2}{\mathrm{O}}_{3}$), Aluminium oxide ( ${\mathrm{Al}}_{2}{\mathrm{O}}_{3}$ ) etc.Harmful gases and chemicals - Corrosive substances like ${\mathrm{NH}}_{3},{\mathrm{Cl}}_{2},$Sulfur oxide, Hydrogen oxides, acids, bases, etc also promote corrosion of certain electrical equipment. For example - Oxidation of Copper wires.Biological Substances - Dirt and bacteria also cause corrosion in metals. For example - inland water pipes, waste pipes etc.

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How do you simplify 3(cos(pi/6)+isin(pi/6))div4(cos((2pi)/3)+isin((2pi)/3)) and express the result in rectangular form?

Dec 14, 2016

The answer is $= - \frac{3}{4} i$

Explanation:

There are 2 ways for the simplification

$\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$

$\sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$

$\cos \left(2 \frac{\pi}{3}\right) = - \frac{1}{2}$

$\sin \left(2 \frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$

${i}^{2} = - 1$

$\frac{3 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right)}{4 \left(\cos \left(2 \frac{\pi}{3}\right) + i \sin \left(2 \frac{\pi}{3}\right)\right)}$

$= \frac{3}{4} \frac{\frac{\sqrt{3}}{2} + i \cdot \frac{1}{2}}{- \frac{1}{2} + i \frac{\sqrt{3}}{2}}$

$= \frac{3}{4} \frac{\left(\frac{\sqrt{3}}{2} + i \cdot \frac{1}{2}\right) \left(- \frac{1}{2} - i \frac{\sqrt{3}}{2}\right)}{\left(- \frac{1}{2} + i \frac{\sqrt{3}}{2}\right) \left(- \frac{1}{2} - i \frac{\sqrt{3}}{2}\right)}$

$= \frac{3}{4} \frac{- \frac{\sqrt{3}}{4} - i \frac{3}{4} - \frac{i}{4} + \frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{3}{4}}$

$= \frac{3}{4} \left(- i\right)$

We can also use $\cos \theta + i \sin \theta = {e}^{i \theta}$

$\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right) = {e}^{i \frac{\pi}{6}}$

$\cos \left(2 \frac{\pi}{3}\right) + i \sin \left(2 \frac{\pi}{3}\right) = {e}^{2 i \frac{\pi}{3}}$

$\therefore \frac{3 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right)}{4 \left(\cos \left(2 \frac{\pi}{3}\right) + i \sin \left(2 \frac{\pi}{3}\right)\right)} = \frac{3}{4} {e}^{i \frac{\pi}{6}} / {e}^{2 i \frac{\pi}{3}}$

$= \frac{3}{4} \left({e}^{i \pi \left(\frac{1}{6} - \frac{2}{3}\right)}\right)$

$= \frac{3}{4} {e}^{- i \frac{\pi}{2}}$

$= \frac{3}{4} \left(\cos \left(- \frac{\pi}{2}\right) + i \sin \left(- \frac{\pi}{2}\right)\right)$

$= \frac{3}{4} \cdot \left(0 - i\right)$

$= - \frac{3 i}{4}$

Dec 14, 2016

$3 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right) \div 4 \left(\cos \left(2 \frac{\pi}{3}\right) + i \sin \left(2 \frac{\pi}{3}\right)\right) = - \frac{3}{4} i$

Explanation:

Given two complex numbers ${z}_{1} = {r}_{1} \left(\cos \alpha + i \sin \alpha\right)$ and ${z}_{2} = {r}_{2} \left(\cos \beta + i \sin \beta\right)$

${z}_{1} \div {z}_{2} = {r}_{1} / {r}_{2} \left(\cos \left(\alpha - \beta\right) + i \in \left(\alpha - \beta\right)\right)$

$3 \left(\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right)\right) \div 4 \left(\cos \left(2 \frac{\pi}{3}\right) + i \sin \left(2 \frac{\pi}{3}\right)\right)$

= $\frac{3}{4} \left(\cos \left(\frac{\pi}{6} - \frac{2 \pi}{3}\right) + i \sin \left(\frac{\pi}{6} - \frac{2 \pi}{3}\right)\right)$

= $\frac{3}{4} \left(\cos \left(\frac{\pi}{6} - \frac{4 \pi}{6}\right) + i \sin \left(\frac{\pi}{6} - \frac{4 \pi}{6}\right)\right)$

= $\frac{3}{4} \left(\cos \left(- \frac{3 \pi}{6}\right) + i \sin \left(- \frac{3 \pi}{6}\right)\right)$

= $\frac{3}{4} \left(\cos \left(- \frac{\pi}{2}\right) + i \sin \left(- \frac{\pi}{2}\right)\right)$

= $\frac{3}{4} \left(\cos \left(\frac{\pi}{2}\right) - i \sin \left(\frac{\pi}{2}\right)\right)$

= $\frac{3}{4} \left(0 - i\right)$

= $- \frac{3}{4} i$


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Question

Write the following arithmetic series in summation notation.8+10+12+⋯+34

Series

Write the following arithmetic series in summation notation.
$$8+10+12+\cdots+34$$

2021-02-26

Well, let's see. This sum's terms increase by 2 each for each subsequent term, so our summand should be something like 2k where k is the summing index. If we just had 2k, with limits of say 11 to 1010, we'd get
10 $$\displaystyle∑{2}{k}={2}{\left({1}\right)}+{2}{\left({2}\right)}+{2}{\left({3}\right)}+⋯+{2}{\left({10}\right)}={2}+{4}+{6}+⋯+{20}$$ k=1
OK. That gets us the going up by two part, but not the fact that it should start at 8. It starts at 2 so what we really need is to add 6 to every term. That should shift up the starting point while keeping the spacing between numbers.
$$10\sum[2k+6]=[2(1)+6]+[2(2)+6]+[2(3)+6]+⋯+[2(10)+6] k=1 =8+10+12+⋯+26$$
Now the only thing left is make sure that the sum stops in the right place. It looks like we are 8 shy. So because we're skipping every other number, that should mean we need 4 more terms. So let's try stopping at the index$$k=14k=14$$.
$$14\sum[2k+6]=[2(1)+6]+[2(2)+6]+[2(3)+6]+⋯+[2(14)+6] k=1 =8+10+12+⋯+34$$
And there we go! We're all done.


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# American Institute of Mathematical Sciences

January  2019, 39(1): 585-606. doi: 10.3934/dcds.2019024

## Lower bound on the number of periodic solutions for asymptotically linear planar Hamiltonian systems

 1 Dipartimento di Matematica, Università degli Studi di Padova, Via Trieste 63, 35121 Padova, Italy 2 Centro de Matemática, Aplicações Fundamentais e Investigação Operacional (CMAF – CIO), Faculdade de Ciências da Universidade de Lisboa, Campo Grande, Edificio C6, 1749–016 Lisboa, Portugal 3 Departamento de Matemática and Centro de Matemática, Aplicações Fundamentais, e Investigação Operacional (CMAF – CIO), Faculdade de Ciências da Universidade de Lisboa, Campo Grande, Edificio C6, 1749–016 Lisboa, Portugal

* Corresponding author

Received  May 2018 Published  October 2018

In this work we prove the lower bound for the number of $T$-periodic solutions of an asymptotically linear planar Hamiltonian system. Precisely, we show that such a system, $T$-periodic in time, with $T$-Maslov indices $i_0, i_∞$ at the origin and at infinity, has at least $|i_∞-i_0|$ periodic solutions, and an additional one if $i_0$ is even. Our argument combines the Poincaré-Birkhoff Theorem with an application of topological degree. We illustrate the sharpness of our result, and extend it to the case of second orders ODEs with linear-like behaviour at zero and infinity.

Citation: Paolo Gidoni, Alessandro Margheri. Lower bound on the number of periodic solutions for asymptotically linear planar Hamiltonian systems. Discrete & Continuous Dynamical Systems - A, 2019, 39 (1) : 585-606. doi: 10.3934/dcds.2019024
##### References:

show all references

##### References:
An Hamiltonian flow corresponding to case ⅰ) of Remark 14. For small periods $T$, the system has $i_0=0$, $i_\infty=-1$, and the only non-zero $T$-periodic orbits are the fixed points $A$ and $B$
An Hamiltonian flow corresponding to case ⅱ) of Remark 14. For small periods $T$, the system has $i_0=1$, $i_\infty=-1$, and the only non-zero $T$-periodic orbits are the fixed points $M$ and $S$
An Hamiltonian flow corresponding to case ⅲ) of Remark 14. For small periods $T$, the system has $i_0=1$, $i_\infty=0$, and the only non-zero $T$-periodic orbit is the fixed point $S$
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2018 Impact Factor: 1.143


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# Quadrature of a circle solved

Discussion in 'Pseudoscience' started by Jason.Marshall, Jan 7, 2015.

1. ### rpennerFully WiredStaff Member

Messages:
4,833
You were wrong when you said "better". One is not better than the other. This was addressed above.

You were wrong when you said "calculate the square root of 2". You didn't obtain √2 by your manipulations, but a different number, 2.

The thought you expressed would be better written as: π/H² = 2 = 3.125/ה² which is a trivial rearrangement of the definitions of H and ה. Since the definitions use 2, this is not even calculating 2. This is recovering 2 from the definitions.

For x ≠ 0, y ≠ 0 the definition z = √(y/x) means y/z² = x and x z² = y, even if x and y are a complex numbers. It's not a mathematical mystery.

3. ### Jason.MarshallBannedBanned

Messages:
654
The thought you expressed would be better written as: π/H² = 2 = 3.125/ה² which is a trivial rearrangement of the definitions of H and ה. Since the definitions use 2, this is not even calculating 2. This is recovering 2 from the definitions.

You do realise that you have just confirmed my work as correct with that statement above?? and explain to me how 2 can have more than one quantity defining its position?? I was trying to express this above statement to you days ago hence the creation of my proof?? pi(1)^2= pi = 3.125(1)= 3.125.... π/H² = 2 = 3.125/ה².....π/H²*2 = 1 = 3.125/ה²*2...area = 2 for both equations Pi(1/H)^2=2=3.125(1/1.25)^2=π/H² = 2 = 3.125/ה²... so my question to you if they were not the exact same one defined rationally the other irrationally then how did the two different equations arrive at the area of a circle r=(1/H)...and r = (1/1.25)...the first area calculated by pi and the second area calculated by 3.125?? but the both areas = 2 ???

5. ### Jason.MarshallBannedBanned

Messages:
654
The thought you expressed would be better written as: π/H² = 2 = 3.125/ה² which is a trivial rearrangement of the definitions of H and ה. Since the definitions use 2, this is not even calculating 2. This is recovering 2 from the definitions.

You do realise that you have just confirmed my work as correct with that statement above?? and explain to me how 2 can have more than one quantity defining its position?? I was trying to express this above statement to you days ago hence the creation of my proof?? pi(1)^2= pi = 3.125(1)= 3.125.... π/H² = 2 = 3.125/ה².....π/H²*2 = 1 = 3.125/ה²*2...area = 2 for both equations Pi(1/H)^2=2=3.125(1/1.25)^2=π/H² = 2 = 3.125/ה²... so my question to you if they were not the exact same one defined rationally the other irrationally then how did the two different equations arrive at the exact same area of a circle r=(1/H)...and r = (1/1.25)...the first area calculated by pi and the second area calculated by 3.125?? but the both areas = 2 ??? and the sides are equal to "sqrt{2}" for 3.125 area was equal to exactly 2 without calculating to infinite irrational pi???
sqrt{pi}/sqrt{2}=H ....and sqrt{3.125}/sqrt{2}= 1.25...This is because 1.25 is the universal growth ratio identifiying the relationship between pi and 2

Last edited: Jan 11, 2015

7. ### Jason.MarshallBannedBanned

Messages:
654

[3.125(H/1.25)] (H/1.25)= pi

8. ### James RJust this guy, you know?Staff Member

Messages:
30,015
Jason.Marshall,

It's a pity that you ignored most of what rpenner told you. Probably you skimmed over most of it because you couldn't follow the mathematics.

Perhaps the most important point you need to take away about your algebra is that it is circular. You start by defining certain things, like H, in a particular way. Then, you manipulate expressions until to get a result that is, in the end, just a rearrangement of your initial definition. This kind of process doesn't prove anything. It's just shuffling around of terms in an equation.

Look. Suppose I define J = 2/17.5. Now watch some amazing arithmetic:

$(8 + 9.5)\times J = 2$, exactly!

Wow! Look! That J thing must be really important, like pi. Because I can make the number 2 using J.

Do you see what happened there?

Now look what you did. You say the result $\pi \times (1/H)^2 = 2$ is incredible. But look at your definition of H:

$H=\sqrt{\frac{\pi}{2}}$

Therefore
$\frac{1}{H} =\sqrt{\frac{2}{\pi}}$

Therefore
$\left(\frac{1}{H}\right)^2 =\frac{2}{\pi}$

Therefore
$\pi \times \left(\frac{1}{H}\right)^2 = \pi \times \frac{2}{\pi} = 2$

There's nothing amazing about that. It's just a rearrangement of your original definition of H.

Notice also that it doesn't matter whether, in that definition of H, you use $\pi$ or 3.125. Suppose you define
$H = \sqrt{\frac{3.125}{2}}$

Then it follows that
$3.125 \times (1/H)^2 = 2$ exactly - just as before.

Instead of 3.125 in the definition, you could use 163.7 or 14.356 or -83 and the result would still hold.

9. ### Jason.MarshallBannedBanned

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654
James H is not a variable and pi is not a variable but the incorrect quantities in between pi and 3.125 are variables only to be desribed by infinite sets because it is not a rational number it is undefined in a rational way that is the reason, I am not just using arbituary number I just slotted in. These are the solutions for pi*r^2 - a^2+b^2= 0
If a circle is defined as 360 degrees 360 is a constant and does not change so then this follows that the variable that must change is diameter which is 360/pi so I define this by doing this [(360/pi) (H/1.25)] (H/1.25) = 115.2 therefore this follows 3.125 * 115.2 = 360 exactly so tell me is there some random number you can substitute that will solve for this equation your thinking about it the wrong way you cannot focus on pi because that quatity is arbituary until its defined what you need to focus on is 360 divided by some number and that number must also be a factor of 180/pi for which is the variables number. Yo must think logically if 360 doesnot change then only the diameter can change that is why it's defined by 360/pi here in this definition 360 was used as a constant divided by and irrational number will give you another irrational number. So now this number is approximately 114...... So logic would dictate if you add an infinite sum of numbers to that quantity you will eventually arrive at some rational number. you can not just keep adding infinitely small sums to a finite number and not expect it to increase that is not logical. So this follows that the next important equivalent factor in line next starting from 180/pi...114....... Will be equal 115.2. Is that so hard to believe, thus it makes complete logical sense why do you think when you use 3.14 as a substitute for pi the accuracy will be very dim? That is because the more space is added to the diameter then pi must decrease to compensate, but doesn't mean it's a variable because the variable quantities are insignificant. so this is why the computer can never produce a rational pi because a computer operates in binary code similar to a one dimensional point constructing two dimensional constructs by traveling in with only one direction 1+1+1.... And so on an infinite set of natural numbers arithmetic it will need an eternity to solve that equation which is a contradiction because time is not infinite. So this is why humans exist to help the machine along and guide it because it is blind to our realm of existence the machine conscience can not comprehend our 3 dimensional reality because it was created to comprehend a maximum of 2,... +1, and -1 am speaking metaphorically here so you will have use your imagination until I complete my full mathematical theorems and desribe all of this mathematically. But if the machine was to define pi it would arrive at 3.125 there would be no infinite set of digits after .14.......... Likewise. The real problem here is a true understanding of a dimensions and how they correctly relate to each other because things that may appear as separate on a lower dimension is unified on a higher dimension explaining the paradox of quantum entanglement and probability.

Last edited: Jan 12, 2015
10. ### Jason.MarshallBannedBanned

Messages:
654
and one more thing James I am here to formulate a correct proof so I was not intentionally expressing myself with circular logic since I am not a formal mathematician I need to discover a formal way to express and show my discovery is correct, I have my own methods of confirming my findings in reality and I already confirmed them correct its amazing what I discovered this appears to connect to our very tools we use to measure reality itself there seems to be a universal gap in all real quantities versus our measuring instruments this is inconclusive until further investigation so am not giving a conclusive statement at the moment. but these ratios of great interest... 1.38 ...1.038... and 1.25....1.02497. So this discussion is very important to me here I can determine how to actually prove I am correct beyond the shadow of a doubt illuminated in light. another thing I found interesting but its not really mathematical just an arbituary fact 322 the number of skull and bones correlates exactly to sqrt{3.125} which is 101.8 ...or nov 18 10,18 322 is the 322 day of the year, just something interesting I noticed not sure what to conclude about it though just saying.

11. ### Jason.MarshallBannedBanned

Messages:
654
So I guess this is what needs to be solved 360/360/pi = pi = 360/115.2... solve for 360/pi this here I guess you can say is circular logic but I guess my proof is not strong enough so I will have to keep thinking of the best way to present it because with my own methods I already confirmed this and it true correct and 100% accurate again circular logic of words but it gets trickier because it would seem that our measuring instruments may not represent reality in its exact nature as well so this calls for further investigation boy does it seem to have many bad alignments on this planet.

12. ### DywyddyrPenguinaciously duckalicious.Valued Senior Member

Messages:
18,474
Nonsense.

You don't have any "proof" and if "your methods have confirmed it" then your methods are deeply flawed.

Unmitigated drivel.

13. ### Jason.MarshallBannedBanned

Messages:
654
So Rpenner if the sqrt{2} = c and c^2 = 2 and r = 1/2 diameter of a circle = 1 radian then ..... 3.125 (1/1.25)^2 = 2 so then should this follows r^2+r^2= L^2...L= hypotenuse so then this follows (L*1.25)^2= 2... did I not show that c= L*1.25??? what is wrong with this proof if you can represent "c" as equal to "sqrt{2}"?? 3.125 (r/1.25)^2= Area so if I am able to derive the sqrt{2} from 3.125 (r/1.25)^2 when r = 1...what is wrong with this proof? I don't think this is any different from comparing g(x) = f(x) its the same concept am just comparing c^2= sqrt{2} these are two different formulas and I showed they are exactly equal and I never even used pi or H and reach an exact quantity? my first attempt I think I confused you when I said the sqrt{2} I meant 2. Area of a circle = 2 =3.125 (1/1.25)^2 ...
and area of square =2=(L*1.25)^2....and the can arrive at L by this r^2+r^2= L^2...sqrt {L^2}= L...(L*1.25)^2=c^2... a^2+b^2=c^2...
c= L*1.25 I know there is a function f(x) this is equal to I just cant remember which one??

So in conclusion what am saying is sqrt{2}=sqrt{pi}=sqrt{3.125} when r=1 is this not correct?
this proof may be weak it still may be circular logic I know there is a way this could be equal to a specific function of f(x) its not going to be straight forward because of sine(x)

Last edited: Jan 12, 2015
14. ### rpennerFully WiredStaff Member

Messages:
4,833
Mathematics on a Distant Planet”, American Math Monthly, vol 105 no 7
In standard real algebra, you are free to define whatever terms you like as long as you take responsibility to be consistent with the rules of algebra. More generally, you are free to define new types of math as long as you take responsibility to be logically consistent and honest enough not to pass them off as standard math. But you are not doing either, so you doomed your thread and maybe the remainder of your existence to reside in pseudo-mathematics.

All the real numbers already are contained in standard algebra, so the most you can do is introduce new names for real numbers. An equals sign just says the name for the number on the left side refers to the same number as the name for the number on the right side, no matter how complicated the (legal) expression is.

You have defined $H \equiv \sqrt{\frac{\pi}{2}}, \; ה \equiv \frac{5}{4}$ which means both of these are just new names for real numbers that follow all the laws of algebra. So it follows that:
$1 = 1^2 = \frac{25}{16 ה^2} = \frac{16 ה^2}{25} = \frac{\pi}{2 H^2} = \frac{2 H^2}{\pi} = \frac{25 \pi}{32 H^2 ה^2} = \frac{25 H^2}{8 \pi ה^2} = \frac{32 H^2 ה^2}{25 \pi} = \frac{8 \pi ה^2}{25 H^2} \\ \frac{H^2}{ ה^2} = \frac{8 \pi}{25} \\ ה^2 = \frac{25}{16} \\ H^2 = \frac{\pi}{2} \\ 2 = 1 + 1 = \frac{\pi}{H^2} = \frac{\; \frac{25}{8} \; }{ה^2} = \frac{25}{8 ה^2} \\ H^2 ה^2 = \frac{25 H^2}{16} = \frac{\pi ה^2}{2} = \frac{25 \pi}{32} \\ 2 ה^2 = \frac{25 \pi}{16 H^2} = \frac{\pi ה^2}{H^2} = \frac{25}{8} \\ \pi = 2 H^2 = \frac{25 H^2}{8 ה^2} = \frac{25 \pi}{16 ה^2} \\ 4 = \frac{25 \pi}{8 H^2 ה^2} = \frac{25 \pi}{8 H^2 ה^2} = \frac{25}{4 ה^2} = \frac{2 \pi}{H^2} \\ 2 H^2 ה^2 = \frac{25 H^2}{8} = \pi ה^2 = \frac{25 \pi}{16} \\ 2 \pi = 4 H^2 = \frac{25 \pi}{8 ה^2} = \frac{25 H^2}{4 ה^2} \\ 4 H^2 ה^2 = \frac{25 \pi}{ 8}$
So $1 < \frac{8 \pi}{25} < \frac{25}{16} < \frac{\pi}{2} < 2 < \frac{25 \pi}{32} < \frac{25}{8} < \pi < 4 < \frac{25 \pi}{16} < 2 \pi < \frac{25 \pi}{ 8}$ and other facts about algebra and the real numbers remain true regardless of your new names.

You can't equate 1/2 diameter of a circle with 1 radian because they don't have the same units and don't measure the same thing.
That is true always, not just because of your assumptions.
You have that backwards. Here you introduce L for the first time, so therefore L is defined in terms of r. $L^2 = 2 r^2$.
That does not follow.
$\left( \frac{5}{4} L \right)^2= 2 \quad \Rightarrow \quad \frac{25}{16} L^2 = 2 \quad \Rightarrow \quad r = \frac{4}{5}$ follows. This constrains us to speak of a circle of a certain size.
No, you demonstrated the ratio r:L is the same as the ratio 1:c and since $r = \frac{4}{5} = 1/1.25$, you have left it for the reader to infer that L = c/1.25 or c = 1.25 L.
It lacks a statement of the givens. It lacks a statement of what is to be proven. It lacks a rigorous chain of explicitly justified statements that explain every step of the reasoning and connect the givens with the statement of what is to be proven. It has questions. It has false statements. There may be more problems.

Because r cannot be 1 ever since you asserted (L*1.25)^2 = 1.25 (2 r²)= 2. "[Y]ou are responsible for what you believe" and you are likewise responsible for what you communicate and assert.

You asserted that. You did not show that, because "c" didn't enter the conversation before.
The exact quantity you reached was $r = \frac{4}{5}$ which was neither your stated goal nor a conclusion you reached on your own.

You didn't confuse (just) me. You got it wrong for EVERYONE.

Only if the radius of the circle is $r = \sqrt{ \frac{2}{\pi} } \approx \frac{4}{5} - \frac{7}{3309}$
As above, this statement requires $L = \sqrt{\frac{32}{25}} = \frac{4}{5} \sqrt{2}$ (assuming L>0 as is typical in geometry).
As above, this statement requires $r = \frac{4}{5}$ which contradicts the implication of asserting the area of the circle is 2. Since you contradict yourself, you have not meaningfully related circles and squares.

As above, this relationship is incompatible with asserting a relationship between L and the two definitions of r. You have not related circles and squares.

r=1 contradicts both $r = \sqrt{ \frac{2}{\pi} }$ and $r = \frac{4}{5}$ so it is no surprise that you spout further contradictions yourself when you deny 2 and π are different numbers. Actually $\sqrt{2} \lt \sqrt{3.125} \lt \sqrt{\pi}$.
It's actually worse than "weak" or "circular". It's completely wrong, based on false premises, self-contradictory and advertises the incompetence of the author to discuss geometry and algebra.

Last edited: Jan 12, 2015

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654
16. ### Jason.MarshallBannedBanned

Messages:
654
72 and God are synonymous just like 9 and 90 is also synonymous with God they are factors of each other separated by H

17. ### rpennerFully WiredStaff Member

Messages:
4,833
The rule of 72 is the observation that an effective interest rate (expressed in percentage points) near 7.84687 takes about 72/r periods to double the investment.

In fact $4 \lt r \lt 100 \Rightarrow \left| \frac{ \log 2 }{\log ( 1 + \frac{r}{100} ) } - \frac{72}{r} \right| \lt \frac{1}{3}$ so it's generally good for investments that are expected to double in a human time-frame.

18. ### Jason.MarshallBannedBanned

Messages:
654

So what do you think of my proof format obviously I have to work it all out in detail this is just the short form am basically saying if the sqrt {pi}= sqrt{3.125} =c... and 3.125 is rational number then pi must also converge to a rational number after it is defined with the infinite set described by sin(x) and the zeta function. and then you can derive area of a circle and then transform that area into a square equivalent a^2+b^2=
"c^2 " with is the solution for the quadrature of a circle? What I said about 72 is not part of my proof just speculation which I may prove at another time.

19. ### Jason.MarshallBannedBanned

Messages:
654
"[Y]ou are responsible for what you believe" and you are likewise responsible for what you communicate and assert." Rpenner said

I agree

"so it is no surprise that you spout further contradictions yourself when you deny 2 and π are different numbers. Actually $\sqrt{2} \lt \sqrt{3.125} \lt \sqrt{\pi}$.
It's actually worse than "weak" or "circular". It's completely wrong, based on false premises, self-contradictory and advertises the incompetence of the author to discuss geometry and algebra." Rpenner said

Here I never explained myself correctly, what I was trying to express is that quantities are related in the same family and are the same but only differ by means of translation they either get smaller or bigger but have an identical structure this is demonstrated in my proof.

20. ### LiddzRegistered Member

Messages:
2
Creating a circle and a square with equal areas involving 100% accuracy solved.

From: Liddz.

It is almost impossible to create a circle and a square with equal areas of measure with 100% accuracy due to the transcendental nature of the modern traditional version of the ratio Pi 3.141592653589793. Remember that there are also 2 other modern values of Pi in addition to the traditional most common value of Pi that is 3.141592653589793. The 2 other values of Pi are: 3.146446609406726 and also 3.14460551102969. 3.146446609406726 is the result of 14 subtract square root of 2 = 1.414213562373095 = 12.585786437626905 and then 12.585786437626905 is divided by 4. 3.14460551102969 is the result of 4 divided by the square root of the Golden ratio = 1.272019649514069. 1.272019649514069 squared is the Golden ratio of 1.618033988749895. Please remember that the Golden ratio can be obtained in Trigonometry through the formula Cosine (36) multiplied by 2. All modern Pi values must be approximated meaning the results that are given to us by the 3 modern Pi values must be reduced to 4 or 5 decimal places resulting in multiple approximations for the 3 main Pi values. There is a debate among scholars regarding which version of the 3 values of Pi are better to use when dealing with circles and geometrical figures. The view of this author is that all 3 of the values of Pi are important and all the 3 values of Pi must be studied as much as possible to achieve accuracy. The 3 values of Pi 3.141592653589793 and 3.146446609406726 and 3.144605511029693 must be compared to each other so the geometrician can determine which value is best suited for the desired task.

I think that I have solved the problem of creating a circle and a square with equal areas involving 100% accuracy:

Requirements for creating a circle and a square with equal areas involving 100% include:

1. The surface area of the circle that is to be equal to the surface area of a square can be derived from the surface area of 2 squares with rational widths of measure but does not always have to be.

2. The square that has a surface area equal to the surface area of the circle can have a width of an irrational measure or a width of rational measure.

3. The Pythagorean theorem is required to construct the square that has an irrational width of measure and a surface area equal to the surface area of the circle. The Pythagorean theorem is also required to test the accuracy of the construction of a circle plus a square that has an irrational width of measure with equal areas involving 100% accuracy. The Pythagorean theorem is based upon a scalene triangle and says that the amount of squares on the hypotenuse is equal to the sum of all the squares that are located on both the adjacent edge of the scalene triangle and also the opposite edge of the scalene triangle. The width of the square with a surface area that is equal in measure to the surface area of the circle must be equal in measure to length of the scalene triangle’s hypotenuse.

http://www.jainmathemagics.com/

http://www.iosrjournals.org/iosr-jm/papers/Vol10-issue1/Version-4/C010141415.pdf

My question is it possible to create a circle with a surface area of 106 equal units because if we can create a circle with a surface area of 106 equal units of measure then we can also create a square with a surface area of 106 equal units of measure by creating a scalene triangle with the longest edge length as 9 equal units of measure taken from the diameter of the circle that has a surface area of 106 equal units of measure, while the shortest length of the scalene triangle has 5 equal units of measure. The hypotenuse of a scalene triangle with the second longest length as 9 equal units of measure and the shortest length of the scalene triangle as 5 equal units of measure is equal in measure to the width of a square that has a surface area of 106 equal units of measure. We can use the theorem of Pythagoras: https://en.wikipedia.org/wiki/Pythagorean_theorem to prove that the square with a width equal to the longest length of the scalene triangle also called the hypotenuse also has 106 equal units of measure.

Area of circle = 106.

Diameter of circle = 11.62.

Circumference of circle = 36.5.

9 squared = 81.

5 squared = 25.

81 + 25 = 106.

Most values of Pi can confirm that if a circle has a diameter of 11.62 equal units of measure then the surface area of the circle with a diameter of 11.62 equal units of measure is 106 equal units of measure.

Also is it possible to create a circle geometrically on paper with a surface area of 117 square units because if we can then we can create a square with the same 117 units by simply constructing a scalene triangle with the second longest length of the scalene triangle being equal to 9 units of measure from the circle’s diameter with an area of 117 units of equal measure. The shortest length of the scalene triangle must have 6 equal units of measure that are also taken from the circle’s diameter with an area of 117 equal square units. We can use the theorem of Pythagoras: https://en.wikipedia.org/wiki/Pythagorean_theorem to prove that the square with a width equal to the longest length of the scalene triangle also called the hypotenuse also has a surface area of 117 equal units of measure.

Area of circle = 117.

Diameter of circle = 12.22

Circumference of circle is 38.39.

Traditional Pi is approximated to: 3.141571194762684.

9 squared = 81.

6 squared = 36.

81 + 36 = 117.

Most values of Pi can confirm that if a circle has a diameter of 12.22 equal units of measure then the surface area of the circle with a diameter of 12.22 equal units of measure is 117 equal units of measure.

Also is it possible to create a circle geometrically on paper with a surface area of 153 square units because if we can then we can create a square with the same 153 units by simply constructing a scalene triangle with the second longest length of the scalene triangle being equal to 12 units of measure from the circle’s diameter with an area of 153 units of equal measure. The shortest length of the scalene triangle must have 3 equal units of measure that are also taken from the circle’s diameter with an area of 153 equal square units. We can use the theorem of Pythagoras: https://en.wikipedia.org/wiki/Pythagorean_theorem to prove that the square with a width equal to the longest length of the scalene triangle also called the hypotenuse also has a surface area of 153 equal units of measure.

Area of circle = 153.

Diameter of circle = 13.96

Circumference of circle is 43.85.

Traditional Pi is approximated to: 3.141117478510029.

12 squared = 144.

3 squared = 9.

144 + 9 = 153.

Most values of Pi can confirm that if a circle has a diameter of 13.96 equal units of measure then the surface area of the circle with a diameter of 13.96 equal units of measure is 153 equal units of measure.

Also is it possible to create a circle geometrically on paper with a surface area of 365 square units because if we can then we can create a square with the same 365 units by simply constructing a scalene triangle with the second longest length of the scalene triangle being equal to 14 units of measure from the circle’s diameter with an area of 365 units of equal measure. The shortest length of the scalene triangle must have 14 equal units of measure that are also taken from the circle’s diameter with an area of 365 equal square units. We can use the theorem of Pythagoras: https://en.wikipedia.org/wiki/Pythagorean_theorem to prove that the square with a width equal to the longest length of the scalene triangle also called the hypotenuse also has a surface area of 365 equal units of measure.

Area of circle = 365.

Diameter of circle = 21.56

Circumference of circle is 67.73.

Traditional Pi is approximated to: 3.141465677179963.

14 squared = 196.

13 squared = 169.

196 + 169 = 365.

Most values of Pi can confirm that if a circle has a diameter of 21.56 equal units of measure then the surface area of the circle with a diameter of 21.56 equal units of measure is 365 equal units of measure.

21. ### LiddzRegistered Member

Messages:
2
Creating a circle and a square with equal areas involving 100% accuracy solved.

From: Liddz.

It is almost impossible to create a circle and a square with equal areas of measure with 100% accuracy due to the transcendental nature of the modern traditional version of the ratio Pi 3.141592653589793. Remember that there are also 2 other modern values of Pi in addition to the traditional most common value of Pi that is 3.141592653589793. The 2 other values of Pi are: 3.146446609406726 and also 3.14460551102969. 3.146446609406726 is the result of 14 subtract square root of 2 = 1.414213562373095 = 12.585786437626905 and then 12.585786437626905 is divided by 4. 3.14460551102969 is the result of 4 divided by the square root of the Golden ratio = 1.272019649514069. 1.272019649514069 squared is the Golden ratio of 1.618033988749895. Please remember that the Golden ratio can be obtained in Trigonometry through the formula Cosine (36) multiplied by 2. All modern Pi values must be approximated meaning the results that are given to us by the 3 modern Pi values must be reduced to 4 or 5 decimal places resulting in multiple approximations for the 3 main Pi values. There is a debate among scholars regarding which version of the 3 values of Pi are better to use when dealing with circles and geometrical figures. The view of this author is that all 3 of the values of Pi are important and all the 3 values of Pi must be studied as much as possible to achieve accuracy. The 3 values of Pi 3.141592653589793 and 3.146446609406726 and 3.144605511029693 must be compared to each other so the geometrician can determine which value is best suited for the desired task.

I think that I have solved the problem of creating a circle and a square with equal areas involving 100% accuracy:

Requirements for creating a circle and a square with equal areas involving 100% include:

1. The surface area of the circle that is to be equal to the surface area of a square can be derived from the surface area of 2 squares with rational widths of measure but does not always have to be.

2. The square that has a surface area equal to the surface area of the circle can have a width of an irrational measure or a width of rational measure.

3. The Pythagorean theorem is required to construct the square that has an irrational width of measure and a surface area equal to the surface area of the circle. The Pythagorean theorem is also required to test the accuracy of the construction of a circle plus a square that has an irrational width of measure with equal areas involving 100% accuracy. The Pythagorean theorem is based upon a scalene triangle and says that the amount of squares on the hypotenuse is equal to the sum of all the squares that are located on both the adjacent edge of the scalene triangle and also the opposite edge of the scalene triangle. The width of the square with a surface area that is equal in measure to the surface area of the circle must be equal in measure to length of the scalene triangle’s hypotenuse.

http://www.jainmathemagics.com/

http://www.iosrjournals.org/iosr-jm/papers/Vol10-issue1/Version-4/C010141415.pdf

My question is it possible to create a circle with a surface area of 106 equal units because if we can create a circle with a surface area of 106 equal units of measure then we can also create a square with a surface area of 106 equal units of measure by creating a scalene triangle with the longest edge length as 9 equal units of measure taken from the diameter of the circle that has a surface area of 106 equal units of measure, while the shortest length of the scalene triangle has 5 equal units of measure. The hypotenuse of a scalene triangle with the second longest length as 9 equal units of measure and the shortest length of the scalene triangle as 5 equal units of measure is equal in measure to the width of a square that has a surface area of 106 equal units of measure. We can use the theorem of Pythagoras: https://en.wikipedia.org/wiki/Pythagorean_theorem to prove that the square with a width equal to the longest length of the scalene triangle also called the hypotenuse also has 106 equal units of measure.

Area of circle = 106.

Diameter of circle = 11.62.

Circumference of circle = 36.5.

9 squared = 81.

5 squared = 25.

81 + 25 = 106.

Most values of Pi can confirm that if a circle has a diameter of 11.62 equal units of measure then the surface area of the circle with a diameter of 11.62 equal units of measure is 106 equal units of measure.

Also is it possible to create a circle geometrically on paper with a surface area of 117 square units because if we can then we can create a square with the same 117 units by simply constructing a scalene triangle with the second longest length of the scalene triangle being equal to 9 units of measure from the circle’s diameter with an area of 117 units of equal measure. The shortest length of the scalene triangle must have 6 equal units of measure that are also taken from the circle’s diameter with an area of 117 equal square units. We can use the theorem of Pythagoras:https://en.wikipedia.org/wiki/Pythagorean_theorem to prove that the square with a width equal to the longest length of the scalene triangle also called the hypotenuse also has a surface area of 117 equal units of measure.

Area of circle = 117.

Diameter of circle = 12.22

Circumference of circle is 38.39.

Traditional Pi is approximated to: 3.141571194762684.

9 squared = 81.

6 squared = 36.

81 + 36 = 117.

Most values of Pi can confirm that if a circle has a diameter of 12.22 equal units of measure then the surface area of the circle with a diameter of 12.22 equal units of measure is 117 equal units of measure.

Also is it possible to create a circle geometrically on paper with a surface area of 153 square units because if we can then we can create a square with the same 153 units by simply constructing a scalene triangle with the second longest length of the scalene triangle being equal to 12 units of measure from the circle’s diameter with an area of 153 units of equal measure. The shortest length of the scalene triangle must have 3 equal units of measure that are also taken from the circle’s diameter with an area of 153 equal square units. We can use the theorem of Pythagoras:https://en.wikipedia.org/wiki/Pythagorean_theorem to prove that the square with a width equal to the longest length of the scalene triangle also called the hypotenuse also has a surface area of 153 equal units of measure.

Area of circle = 153.

Diameter of circle = 13.96

Circumference of circle is 43.85.

Traditional Pi is approximated to: 3.141117478510029.

12 squared = 144.

3 squared = 9.

144 + 9 = 153.

Most values of Pi can confirm that if a circle has a diameter of 13.96 equal units of measure then the surface area of the circle with a diameter of 13.96 equal units of measure is 153 equal units of measure.

Also is it possible to create a circle geometrically on paper with a surface area of 365 square units because if we can then we can create a square with the same 365 units by simply constructing a scalene triangle with the second longest length of the scalene triangle being equal to 14 units of measure from the circle’s diameter with an area of 365 units of equal measure. The shortest length of the scalene triangle must have 14 equal units of measure that are also taken from the circle’s diameter with an area of 365 equal square units. We can use the theorem of Pythagoras:https://en.wikipedia.org/wiki/Pythagorean_theorem to prove that the square with a width equal to the longest length of the scalene triangle also called the hypotenuse also has a surface area of 365 equal units of measure.

Area of circle = 365.

Diameter of circle = 21.56

Circumference of circle is 67.73.

Traditional Pi is approximated to: 3.141465677179963.

14 squared = 196.

13 squared = 169.

196 + 169 = 365.

Most values of Pi can confirm that if a circle has a diameter of 21.56 equal units of measure then the surface area of the circle with a diameter of 21.56 equal units of measure is 365 equal units of measure.

22. ### originTrump is the best argument against a democracy.Valued Senior Member

Messages:
9,453
You say it the 2 values are 100% accurate, but you are using an approximation of pi. If you are using and approximation of pi doesn't that necesarily mean that the are of the circle is an approximation?

23. ### Write4UValued Senior Member

Messages:
3,862
The OP speaks of "squares". Question:
Wouldn't using triangulation (pentagons or hexagons) yield an instant result of total surface area of a globe?Fractals can be used to measure all irregular or variable measurements of all natural shapes, down to the geometry of the fabric of spacetime itself..

Last edited: Apr 21, 2017


text:
# $\mathbb{Z}_2$ Symmetry in Water

I have learned that the critical exponents for phase transitions is independent of the microscopic structure of the substance and is dependent on the symmetry. For instance the phase transition for a ferromagnetic at the Curie temperature and the phase transition for water at the critical point have the same critical exponents. The ferromagnet has $$\mathbb{Z}_2$$ symmetry is related to spin up ground state and the spin down ground state. What will be the $$\mathbb{Z}_2$$ symmetry in water?


text:
# how to get $n$ equidistributed points on the unit sphere

We can get $n$ equidistributed points in the unit circle using CirclePoints. But how do you get $n$ equidistributed points on the unit sphere(surface of a ball)? The preliminary idea is to suppose that the points are some electric charges with same electric quantity on a sphere. And they have the same effective working area. I think their position is what I want. This is my current solution:

point = RandomPoint[Sphere[], 10^6];
Graphics3D[{Sphere[], Blue, PointSize[.02],
Point[Union[point,SameTest -> (EuclideanDistance[#1, #2] < 0.4 &)]]},
Boxed -> False]

But this solution cannot draw the specified number of points, and the space is not very equidistributed in some places.

• The vertices of the Platonic and Archimedean solids are in a way the most optimal way to place points on a sphere. If the number of points you need is not the same as the vertex count of any of those solids, you might consider looking up spherical codes or spherical designs, or look at this article… my point is, this is a math problem and not a Mathematica problem. Jun 25, 2016 at 10:53
• What do you mean by equally spaced? (Points on a circle naturally divide the circle into segments, and we can ask that all the segments to be equal; however, points do not divide a sphere.) Jun 25, 2016 at 12:45
• Perhaps the question and your criteria need to be updated? (Something like you want a stable configuration that minimizes the potential energy of an inverse-square law, such as that for electron on a sphere. The choice of a specific objective function to minimize, as opposed to elementary geometric criteria or another function, seems an important omission. Also requiring stability rules out interesting configurations, although I don't know if that means there is always a unique solution up to rotation/reflection.) Jun 25, 2016 at 14:54
• In any event: equidistributed would be the more appropriate term than equispaced in this context. Jun 25, 2016 at 15:45
• Ifaik the problem is not solved for arbitrary number of points. Electrostatic approach gives some answers, but I don't remember a reference. Mar 18, 2018 at 9:07

Aha~ I suppose this question is created while solving this. Am I correct @yode :P

So here's an easy solution, simple, elegant, and may I say even quite fast after some optimization?

pt = With[{p =
Table[{x[i], y[i], z[i]}, {i, 80(*number of charges*)}]},
p /. Last@
NMinimize[
Total[1/Norm[Normalize[#1] - Normalize[#2]] & @@@
Subsets[p, {2}]], Flatten[p, 1]]];
Graphics3D[{Opacity@.3, Darker@Green, Sphere[], Opacity@1,
PointSize@Large, Darker@Blue, Point@*Normalize /@ pt}]

The result is quite good:

the setting of the minimization variable is crucial, or the point will not be on surface. But fortunately, our 'kindergarten physics' taught us that when charges are freely scattering in a sphere, they'll always be on surface evenly! Thus this must be some sort of 'most even' form of scattering as it follows physical laws.

• You got the idea of God. :)
– yode
Jun 25, 2016 at 14:01
• but i still cannot get the idea of upvoters~ :P
– Wjx
Jun 25, 2016 at 14:02
• This is precisely the Thomson problem mentioned by kirma. Jun 27, 2016 at 3:57
• The generated set of vectors have no unit length, so the sphere is not a unit sphere. Jul 3, 2016 at 9:36
• @plasmacel er, please Normalize them
– Wjx
Jul 3, 2016 at 9:49

If more ad hoc, inexact approaches are welcome, one way to generate relatively uniform density of points on a sphere is to use Monte Carlo Lloyd's algorithm (modified for the spherical case):

With[{points = 200, samples = 40000, iterations = 20},
Nest[With[{randoms = Join[#, RandomPoint[Sphere[], samples]]},
Table[Normalize@Mean@Extract[randoms, Position[#, {i}]], {i, points}] &@
Nearest[# -> Automatic, randoms,
DistanceFunction -> (1 - Dot[#1, #2] &)]] &,
RandomPoint[Sphere[], points], iterations]] //
Graphics3D[{Sphere[{0, 0, 0}, 0.999], Red, Point@#}] &

EDIT:

The above can be written in more concise and much more efficient form as:

With[{points = 200, samples = 40000, iterations = 20},
Nest[
With[{randoms = Join[#, RandomPoint[Sphere[], samples]]},
Normalize@Mean@randoms[[#]] & /@
Values@PositionIndex@Nearest[#, randoms]] &,
RandomPoint[Sphere[], points], iterations]] //
Graphics3D[{Sphere[{0, 0, 0}, 0.999], Red, Point@#}] &

• P.S. Nearest[# -> Range@Length@#, randoms] is more compactly done as Nearest[# -> Automatic, randoms] Jun 27, 2016 at 7:38
• @J.M. I have a strange tendency to miss these Automatic variations... I'll improve the answer. Jun 27, 2016 at 7:52
• I like the idea, but when I tried to use it in high dimensions it got quite slow. I ended up with this approach: Repeatedly find the nearest neighbour for each point; orthogonalize by subtracting the shared component; Reapeat. This converges pretty quickly even for 1000 dimensions. Dec 18, 2020 at 20:01

For an approximately even distribution of points on any surface with cylindrical symmetry, we can use the Golden Angle, the same way that the sunflower uses it on the plane.

To place N points on the surface of a sphere, define an axis. Divide the surface into N equal area strips perpendicular to the axis. For k in 0 to N-1, on the kth strip, place a point at an angle of k*ga, in the centre of the its width. ga is the golden angle, 1/(phi+1) of a circle, about 137.5 degrees / 2.34 rads.

This construction can be generalised to the surface of any volume of revolution, for instance a vase or turned table leg, by keeping the area of each strip constant.

Obviously what is being done here is that as each strip is equal area, the construction automatically makes each point 'serve' the same amount of space. Use of the 'most irrational fraction' then does a reasonable job of spreading the points round the axis without any long range structure developing.

Edit by J. M.

As I noted in a comment to this answer, the phyllotactic arrangement of points on a sphere has been previously featured on the Wolfram Blog. The code there is more general than what is needed here, so I took the liberty to simplify the code a bit for the spherical case, and also used the fact that GoldenAngle is now a built-in constant:

With[{n = Floor[4 π 100]},
Graphics3D[{Sphere[Table[{2 Sqrt[(1 - i/n) i/n] Cos[i GoldenAngle],
2 Sqrt[(1 - i/n) i/n] Sin[i GoldenAngle],
1 - 2 i/n}, {i, n}], 100/n]}, Boxed -> False]]

The 100 in the expression for n controls the point density; increase or decrease as seen fit.

• This is a Wolfram Blog entry that discusses the phyllotaxis strategy. Of note: GoldenAngle is now built-in. Jun 25, 2016 at 22:46
• Thanks for the link, it was a report on the Starshine 3 satellite that first brought this method to my attention, needless to say I couldn't google up the name when I wanted it! Jun 26, 2016 at 6:24
• If you don't mind, I'd like to edit your post later to include some of the code from that link, just to make your post even more prominent (this is a site that expects some amount of code, after all). Jun 27, 2016 at 4:14
• Go for it, I'm all for cooperation. I was surprised to see no mention of phyllotaxis so far in this thread, hence the text-only post from a non-Mathematica user (I'm a python+matplotlib+numpy man). Even given your link I was reluctant to transplant code as I'm not in a position to verify it. I was surprised to see phyllotaxis used for Starshine, as I can't imagine a hard constraint on N. I built a LED sphere recently, and triangular tiled the faces of an icoshedron, a straightforward construction, mathematically, mechanically and electrically. Jun 27, 2016 at 8:43

Correct me if I'm wrong... but I suppose, pedantically speaking, there are only five solutions defined by Platonic solids, and trivial solutions for 0-3 points (extension of CirclePoints).

Thus:

ClearAll[spherePoints];

spherePoints[r_: 1, n_ /; 0 <= n < 4] := {##, 0} & @@@ CirclePoints[r, n];

(spherePoints[r_: 1, PolyhedronData[#, "VertexCount"]] :=
r (Normalize /@ PolyhedronData[#, "VertexCoordinates"])) & /@
PolyhedronData["Platonic"];

Now:

Graphics3D[{Red, PointSize[Large], Point[spherePoints[12]],
Opacity[1/2], White, Sphere[{0, 0, 0}, 0.99]}]

For non-Platonic configurations, function stays unevaluated:

spherePoints[5]

spherePoints[5]

Tammes and Thomson problems can be considered as an extension for non-Platonic cases. For instance, see Math Overflow: Distributing points evenly on a sphere. Sadly, general solutions to these problems are not trivial.

• If I put n electric charges in the sphere,we will get the stable layout?Then we can get the layout.
– yode
Jun 25, 2016 at 8:37
• @yode I think Thompson problem is exactly that. Solving this using numerical optimization on Mma is not that hard, but I don't know how close to optimal solutions achieved this way actually are. Jun 25, 2016 at 8:45
• @yode The following numerically solves the Thomson problem for a ten-point case: With[{points = Array[c, {10, 3}]}, points /. Last@NMinimize[Total[1 / EuclideanDistance @@ # & /@ Subsets[points, {2}]], # \[Element] Sphere[] & /@ points]]. It is not really very fast, but it's reasonably simple... Jun 25, 2016 at 9:00
• And the same for Tammes problem: With[{points = Array[c, {10, 3}]}, points /. Last@NMaximize[Min[EuclideanDistance @@@ Subsets[points, {2}]], # \[Element] Sphere[] & /@ points]]. Jun 25, 2016 at 9:10
• The evaluation can be eased a bit by fixing one of the points to be a "standard" point, such as any of the unit basis vectors, and constraining a second point to lie on a coordinate plane containing the fixed point set earlier. Jun 27, 2016 at 3:47
With[{n = {3, 4, 5, 6, 7, 20}},
Partition[show[points[#]] & /@ n, 3]] // GraphicsGrid

With the code below, with piecewise spring forces (exclusively repulsive) and Cartesian vectors. A different model for force and switch to spherical vectors would probably be wise.

show[points_] :=
Graphics3D[{
Opacity[.5], Sphere[],
Opacity[1], PointSize[Medium], Point[points]},
ImageSize -> Tiny,
Boxed -> False]

points[n_] :=
With[{
steps = 50,
start = RandomPoint[Sphere[], n]},
Nest[move[#, .1] &, start, steps]]

move[points_, dt_] :=
With[{n = Length[points]},
Module[{copy = points},
Do[
copy[[i]] += dt Sum[force[copy[[j]], copy[[i]]],
{j, DeleteCases[Range[n], i]}];
copy[[i]] = copy[[i]]/Norm[copy[[i]]],
{i, n}];
copy]]

force[p1_, p2_] :=
With[{d = VectorAngle[p1, p2], range = Pi},
If[d > range, 0, Normalize[p2 - p1]*(range - d)]]

## Update: Simulated Annealing

Start with n random points on the unit sphere. Then do the main loop: pick a point randomly and make a random move. If the move lowers the energy (electro-static potential), accept the move into a new configuration. Accept the move also, this is crucial, if the energy difference is less then Exp[-difference/T], where T is a temperature-like control parameter. Do K1 * Length[points] of these inner iterations (about 10, so each point is picked about ten times) and K2 outer passes of the inner iterating. Before each outer pass T is set, i.e. gradually lowered from some starting value (few times higher than a typical temperature change) to zero.

Results agree with literature on this, so-called Thomson problem.

start = points[18];
result = Reap[simulation[start, 10, 10]]; // AbsoluteTiming

(* {2.7106, Null} *)

Energy is recorded in the main loop:

ListPlot[result[[2, 1]],
Joined -> True,
ImageSize -> Medium]

Table[Graphics3D[{
PointSize[Medium], Point[xyz /@ p],
Opacity[.75], Sphere[{0, 0, 0}, .99]},
ImageSize -> Small,
Boxed -> False,
SphericalRegion -> True],
{p, {start, result[[1]]}}] // GraphicsRow

Code:

points[n_Integer] :=
With[{x := RandomReal[]},
Table[{2 Pi \[Xi], ArcCos[2 \[Xi] - 1]}, n]]

xyz[{u_, v_}] :=
{Sin[v] Cos[u], Sin[v] Sin[u], Cos[v]}

energy[points_] :=
With[{n = Length[points]},
Sum[1/EuclideanDistance @@ (xyz /@ points[[{i, j}]]),
{i, n - 1}, {j, i + 1, n}]]

(* O(n) instead of energy O(n^2) *)
dE[points_, i_, r_] :=
With[{n = Length[points]},
Module[{indices, before, after},
indices = DeleteCases[Range[n], i];
before = Sum[
1/Norm[xyz@points[[j]] - xyz@points[[i]]], {j, indices}];
after = Sum[
1/Norm[xyz@points[[j]] - xyz@r], {j, indices}];
after - before]]

simulation[start_, K1_, K2_] :=
With[{
n = Length[start],
du = .1 Pi,
dv = .1 Pi,
x := RandomReal[]},
Module[{T, temp = start},
Do[
T = 10 (1 - (k1 - 1)/K1)^5;
temp = Nest[
Module[{i, p, change},
(* monitor energy; must Reap *)
Sow[energy[#]];
i = RandomInteger[{1, n}];
p = #[[i]] + {du (2 x - 1), dv (2 x - 1)};
change = dE[#, i, p];
If[change < 0 || (x < Exp[-change/T]),
ReplacePart[#, i -> p], #]] &, temp, K2 n], {k1, K1}];
temp]]

• Upvote.But it seem the points not very equidistributed such as this angle of view in some place.If you have fixed this,I think I should accept this answer in consideration of the code efficiency
– yode
Jun 25, 2016 at 17:01
• @yode One can tweak the steps, force etc., and see what works best. I remembered another approach which lends to this problem, see update.
– BoLe
Jul 3, 2016 at 18:19

Mathematica 11.1 has a built-in function for this task: SpherePoints. This roughly corresponds CirclePoints for three-dimensional case; thus I wouldn't rely on it to get any sort of randomness on the point placement.

EDIT: An implementation with SpherePoints, with an added random rotation:

Outer[Dot, SpherePoints[200],
RandomVariate@CircularRealMatrixDistribution[3], 1] //
Graphics3D[{Sphere[{0, 0, 0}, 0.999], Red, Point@#}] &

This works with larger number of points (10000), too:

• I think the Wolfram Research guys have seen this question,then go home to build a SpherePoints. :)
– yode
Mar 16, 2017 at 5:18
• And If you want to get a random point,you just rotate a random angle around the centre of sphere
– yode
Mar 16, 2017 at 5:20
• @yode Ah, now I understand your point. Indeed, although set of more or less equidistant points on a sphere are not absolutely random, you can improve the situation by choosing a random direction (random point on a sphere once again), and perform an uniformly distributed random rotation around it. You can alternatively pick a variate from CircularRealMatrixDistribution[3] which provides uniform samples from random 3D rotation matrices - which you can use to rotate the point set. Mar 16, 2017 at 11:17
• Yes,that is what I mean. :)
– yode
Mar 16, 2017 at 15:06

For points uniformly spaced in angular variables, you can use CirclePoints.

spherepoint[m_, n_] := Union@Flatten[Table[Join[{Cos[q]},
Sin[q] #] & /@ CirclePoints[n], {q, 0, Pi, Pi/m}], 1]

ListPointPlot3D[spherepoint[20, 30], BoxRatios -> 1]

For uniformly spaced in Cartesian coordinates, things would be complicated. The best I can suggest is to go for scaled PolyhedronData and try to map them on the sphere. You can find your polyhedron for any given number of face (or vertices, there is a relation I forgot). For example

pts = PolyhedronData["Dodecahedron", "VertexCoordinates"];
PolyhedronData["Dodecahedron", "VertexCount"]
PolyhedronData["Dodecahedron", "FaceCount"]
r = Norm[pts[[1]]];
Show[PolyhedronData["Dodecahedron"],
SphericalPlot3D[r, {\[Theta], 0, Pi}, {\[Phi], 0, 2 Pi},
Mesh -> Full, PlotStyle -> Opacity[0.5]]]

20

12

You have to scale the data with r to bring them on unit circle.

You might find this interesting for any arbitrary polygon

Draw an arbitrary convex polyhedron without excess diagonals drawn

• It's seem the spaced isn't equal each other?
– yode
Jun 25, 2016 at 8:06
• They are uniformly spaced in angular variable. Are you looking for uniform spacing in cartesian coordinates? Jun 25, 2016 at 8:20
• Yes,uniform spacing in cartesian coordinates.Can you help me update your solution?
– yode
Jun 25, 2016 at 8:31
• "for any given number of face (or vertices, there is a relation I forgot)." - you might want to look up dual polyhedra. ;) Jun 25, 2016 at 10:59

I posted an implementation of the spherical Lloyd algorithm in Wolfram Community. Applied to this problem, we have the following:

n = 50; (* number of points *)
BlockRandom[SeedRandom[1337, Method -> "MersenneTwister"];
sp = Normalize /@ RandomVariate[NormalDistribution[], {n, 3}]];

With[{maxit = 45,(* maximum iterations *)
tol = 0.001 (* distance tolerance *)},
lp = FixedPoint[Function[pts,
Block[{ch, polys, verts, vor},
ch = ConvexHullMesh[pts];
verts = MeshCoordinates[ch];
polys = First /@ MeshCells[ch, 2];
vor = Normalize[Cross[verts[[#2]] - verts[[#1]],
verts[[#3]] - verts[[#1]]]] &
@@@ polys;
SphericalPolygonCentroid[vor[[#]]] & /@
ch["VertexFaceConnectivity"]]], sp, maxit,
SameTest -> (Max[MapThread[cosDistance, {#1, #2}]] < tol &)]];

Graphics3D[{{Opacity[.75], Sphere[]}, {Green, Sphere[lp, 0.02]}},
Boxed -> False]

To keep this answer self contained, here are the required auxiliary routines:

(* https://mathematica.stackexchange.com/a/97854 *)
vecang[v1_?VectorQ, v2_?VectorQ] := Module[{n1 = Norm[v1], n2 = Norm[v2]},
2 ArcTan[Norm[v1 n2 + n1 v2], Norm[v1 n2 - n1 v2]]]

(* https://mathematica.stackexchange.com/a/167114 *)
vectorRotate[vv1_?VectorQ, vv2_?VectorQ] :=
Module[{v1 = Normalize[vv1], v2 = Normalize[vv2], c, d, d1, d2, t1, t2},
d = v1.v2;
If[TrueQ[Chop[1 + d] == 0],
c = UnitVector[3, First[Ordering[Abs[v1], 1]]];
t1 = c - v1; t2 = c - v2; d1 = t1.t1; d2 = t2.t2;
IdentityMatrix[3] - 2 (Outer[Times, t2, t2]/d2 -
2 t2.t1 Outer[Times, t2, t1]/(d2 d1) + Outer[Times, t1, t1]/d1),

c = Cross[v1, v2];
d IdentityMatrix[3] + Outer[Times, c, c]/(1 + d) - LeviCivitaTensor[3].c]]

(* https://mathematica.stackexchange.com/a/154112 *)
sphereExp[q_?VectorQ, p_?VectorQ] /; Length[q] == Length[p] + 1 :=
With[{n = Norm[p]}, vectorRotate[{0, 0, 1}, q].Append[p Sinc[n], Cos[n]]]

sphereLog[q_?VectorQ, p_?VectorQ] /; Length[q] == Length[p] :=
Most[vectorRotate[q, {0, 0, 1}].p]/Sinc[vecang[p, q]]

SphericalPolygonCentroid[pts_?MatrixQ] := Module[{k = 0, n = Length[pts], cp, h, pr},
cp = Normalize[Total[pts]];
pr = InternalEffectivePrecision[pts];
While[cp = sphereExp[cp, h = Sum[sphereLog[cp, p], {p, pts}]/n];
k++; Norm[h] > 10^(-pr/2) && k <= 30];
cp]

(* https://mathematica.stackexchange.com/a/159465 *)
cosDistance[v1_?VectorQ, v2_?VectorQ] :=
Module[{n1 = Normalize[v1], n2 = Normalize[v2], y},
y = Norm[n1 - n2]^2; 2 y/(Norm[n1 + n2]^2 + y)]

Of course,we have SpherePoints in 11.1,but if we are in before version,we can use IGLayoutSphere to do it,which is in a package IGraphM writed by Szabolcs.As its doucument

IGLayoutSphere[graph] lays out vertices approximately uniformly distributed on a sphere.

It's exactly what I want.

Needs["IGraphM"]
g = IGLayoutSphere[CompleteGraph[50]]

Graphics3D[{Point[GraphEmbedding[g]], Sphere[]}]

Regular equidistribution can be achieved by choosing circles of latitude at constant intervals $$dθ$$ and on these circles points with distance $$dφ$$, such that $$dθ \approx dφ$$ and that $$dθdφ$$ equals the average area per point. This then gives the following algorithm:

(In short, the basic idea is to divide the surface area into small squares (each square effectively contains one point), then find the theta, phi. Convert them into Cartesian coordinates)

lst = {};
num = 100 (*  close to required number of points*);
r = 1.;
(*Discretise theta and phi*)
ar = (4.*Pi*r^2)/num;
len = Sqrt[ar]; ndthe = Round[Pi/len]; len2 = ar/len;
For[m = 0, m <= ndthe - 1, m = m + 1,
theta = (m + 1/2)*(Pi/ndthe);
nphi = Round[2*Pi*Sin[theta]/len2];
For[n = 0, n <= nphi - 1, n = n + 1,
phi = (n)*(2*Pi/nphi);
(* spherical to cart *)
x = r*Sin[theta]*Cos[phi];
y = r*Sin[theta]*Sin[phi];
z = r*Cos[theta];
cor = {x, y, z};
corr = AppendTo[lst, cor]]]
ListPointPlot3D[corr, BoxRatios -> 1]
ListPlot3D[{corr, -corr}, BoxRatios -> 1]

Ref-Short Discription

Ref-Detailed Discription

More precisely, These points are orthogonally equidistant.

• You really don't need to use For[]: Block[{r, num, ar, len, ndthe, theta, nphi, phi}, r = 1.; num = 100; ar = (4 Pi r^2)/num; len = Sqrt[ar]; ndthe = Round[Pi/len]; Flatten[Table[theta = (m + 1/2) (Pi/ndthe); nphi = Round[2 Pi Sin[theta]/len]; Table[phi = 2 Pi n/nphi; r {Sin[theta] Cos[phi], Sin[theta] Sin[phi], Cos[theta]}, {n, 0, nphi - 1}], {m, 0, ndthe - 1}], 1]] Mar 7, 2019 at 5:28
• @J.M., Thank you! Mar 7, 2019 at 6:37

Marsaglia's method: uniform on disk (here rejection method), transform by Lambert azimuthal equal-area projection:

uniformDisk[n_] :=
Take[Select[Table[RandomReal[{-1, 1}], {3 n}, {2}], #.# <= 1 &], n];

diskToSphere[p_] := {2 p[[1]] Sqrt[1 - Dot[p, p]],
2 p[[2]] Sqrt[1 - Dot[p, p]],
1 - 2 Dot[p, p]}

Graphics3D[{
Sphere[{0, 0, 0}, .999],
Specularity[Blue, 10],
Sphere[Map[diskToSphere, uniformDisk[1256]], .015]},
Boxed -> False]

• This is more of an answer to this question, but Marsaglia's method is already there. Jun 27, 2016 at 8:49
• Didn't see it. Rereading does the OP mean "well-spaced" then and not " equidistributed"... like some low discrepancy on $\mathbb{S}^2$? Jun 27, 2016 at 9:06
• I would say that the more appropriate term is "uniformly distributed" if you are talking about (quasi)-randomly generated points. Jun 27, 2016 at 9:08

Just in case what you were really asking was how to distribute $n$ points relatively evenly on an $d$-dimensional hypersphere, here's a code snippet that will work... :)

    genSphere[n_, d_] :=
Table[
Normalize[
RandomVariate[
MultinormalDistribution[
ConstantArray[0, d],IdentityMatrix[d]
]
]
], {n}]

• How to understand this?
– yode
Apr 7, 2017 at 20:09
• Oops, had to take out the "Abs[ ]", which was putting all points in the positive quadrant. Use this to generate data to test multi-dimensional geometric algorithms. Apr 7, 2017 at 20:17
• Thanks,and do you test Graphics3D[{Point[genSphere[20, 3]], Sphere[]}]?It's not a good result still.
– yode
Apr 7, 2017 at 20:32
• The points are distributed as per a uniform distribution over the surface (each patch has same probability of getting a point as any other patch of the same size) but that inevitably results in a clumpy appearance. This could be a good seed for the simulated annealing or other iterative algorithm. It also readily generalizes to arbitrary dimension, which I find fascinating, based on its simplicity. Apr 7, 2017 at 21:22
• This is more of an answer to this question. Apr 7, 2017 at 23:33


text:
Under the auspices of the Computational Complexity Foundation (CCF)

REPORTS > KEYWORD > MEMBERSHIP QUERY:
Reports tagged with membership query:
TR08-091 | 10th September 2008
Vitaly Feldman

On The Power of Membership Queries in Agnostic Learning

Revisions: 1

We study the properties of the agnostic learning framework of Haussler (1992)and Kearns, Schapire and Sellie (1992). In particular, we address the question: is there any situation in which membership queries are useful in agnostic learning?

Our results show that the answer is negative for distribution-independent agnostic learning and positive ... more >>>

TR09-005 | 7th December 2008
Emanuele Viola

Bit-Probe Lower Bounds for Succinct Data Structures

We prove lower bounds on the redundancy necessary to
represent a set $S$ of objects using a number of bits
close to the information-theoretic minimum $\log_2 |S|$,
while answering various queries by probing few bits. Our
main results are:

\begin{itemize}
\item To represent $n$ ternary values $t \in \zot^n$ in ... more >>>

ISSN 1433-8092 | Imprint


text:
#### dropzone and laravel, filename is getting numbers prepended to it

I’m using dropzone and laravel in order to upload images to aws and save the filename in the database. The problem is that it saves the filename to aws as a bunch of numbers before the filename of the uploaded file. In other words, if I upload blueSquare.png it comes into aws as 102472349837blueSquare.png

In the controller code that uploads, I dumped the filename to see what’s going on and it shows the filename as php701f.tmp:

if($request->hasFile('file')){$file = $request->file('file'); dd($file->getFileName()); //this dumps php701f.tmp
$extension =$file->getClientOriginalExtension();
$originalName =$file->getClientOriginalName();
$dir='component_images';$path = $dir.DIRECTORY_SEPARATOR.$originalName;
$url = 'https://test.s3.amazonaws.com/'.$path;
}

How can I get it to just upload the blueSquare.png filename to aws?

Source: Laravel StackOverflow


text:
# Search by Topic

#### Resources tagged with Visualising similar to Dicey Operations:

Filter by: Content type:
Stage:
Challenge level:

### Semi-regular Tessellations

##### Stage: 3 Challenge Level:

Semi-regular tessellations combine two or more different regular polygons to fill the plane. Can you find all the semi-regular tessellations?

### On the Edge

##### Stage: 3 Challenge Level:

If you move the tiles around, can you make squares with different coloured edges?

### Square It

##### Stage: 1, 2 and 3 Challenge Level:

Players take it in turns to choose a dot on the grid. The winner is the first to have four dots that can be joined to form a square.

### Square Coordinates

##### Stage: 3 Challenge Level:

A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides?

### Marbles in a Box

##### Stage: 3 and 4 Challenge Level:

How many winning lines can you make in a three-dimensional version of noughts and crosses?

##### Stage: 3 Challenge Level:

How many different symmetrical shapes can you make by shading triangles or squares?

### Route to Infinity

##### Stage: 3 Challenge Level:

Can you describe this route to infinity? Where will the arrows take you next?

### Isosceles Triangles

##### Stage: 3 Challenge Level:

Draw some isosceles triangles with an area of $9$cm$^2$ and a vertex at (20,20). If all the vertices must have whole number coordinates, how many is it possible to draw?

### Christmas Chocolates

##### Stage: 3 Challenge Level:

How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?

### Cuboid Challenge

##### Stage: 3 Challenge Level:

What size square corners should be cut from a square piece of paper to make a box with the largest possible volume?

### Konigsberg Plus

##### Stage: 3 Challenge Level:

Euler discussed whether or not it was possible to stroll around Koenigsberg crossing each of its seven bridges exactly once. Experiment with different numbers of islands and bridges.

### Frogs

##### Stage: 2 and 3 Challenge Level:

How many moves does it take to swap over some red and blue frogs? Do you have a method?

### Squares in Rectangles

##### Stage: 3 Challenge Level:

A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?

### An Unusual Shape

##### Stage: 3 Challenge Level:

Can you maximise the area available to a grazing goat?

### Painted Cube

##### Stage: 3 Challenge Level:

Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?

### Picturing Square Numbers

##### Stage: 3 Challenge Level:

Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?

### Picturing Triangle Numbers

##### Stage: 3 Challenge Level:

Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?

### Eight Hidden Squares

##### Stage: 2 and 3 Challenge Level:

On the graph there are 28 marked points. These points all mark the vertices (corners) of eight hidden squares. Can you find the eight hidden squares?

### Conway's Chequerboard Army

##### Stage: 3 Challenge Level:

Here is a solitaire type environment for you to experiment with. Which targets can you reach?

### World of Tan 6 - Junk

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of this junk?

### You Owe Me Five Farthings, Say the Bells of St Martin's

##### Stage: 3 Challenge Level:

Use the interactivity to listen to the bells ringing a pattern. Now it's your turn! Play one of the bells yourself. How do you know when it is your turn to ring?

### World of Tan 28 - Concentrating on Coordinates

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of Little Ming playing the board game?

### World of Tan 25 - Pentominoes

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outlines of these people?

### World of Tan 8 - Sports Car

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of this sports car?

### World of Tan 26 - Old Chestnut

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of this brazier for roasting chestnuts?

### World of Tan 27 - Sharing

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of Little Fung at the table?

### World of Tan 29 - the Telephone

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of this telephone?

### World of Tan 5 - Rocket

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of the rocket?

### World of Tan 13 - A Storm in a Tea Cup

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of these convex shapes?

### World of Tan 12 - All in a Fluff

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of these rabbits?

### Colour Wheels

##### Stage: 2 Challenge Level:

Imagine a wheel with different markings painted on it at regular intervals. Can you predict the colour of the 18th mark? The 100th mark?

### Cubes Within Cubes Revisited

##### Stage: 3 Challenge Level:

Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?

### Instant Insanity

##### Stage: 3, 4 and 5 Challenge Level:

Given the nets of 4 cubes with the faces coloured in 4 colours, build a tower so that on each vertical wall no colour is repeated, that is all 4 colours appear.

### Sprouts

##### Stage: 2, 3, 4 and 5 Challenge Level:

A game for 2 people. Take turns joining two dots, until your opponent is unable to move.

### Coin Cogs

##### Stage: 2 Challenge Level:

Can you work out what is wrong with the cogs on a UK 2 pound coin?

### Domino Numbers

##### Stage: 2 Challenge Level:

Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?

### World of Tan 4 - Monday Morning

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of Wai Ping, Wah Ming and Chi Wing?

### Makeover

##### Stage: 1 and 2 Challenge Level:

Exchange the positions of the two sets of counters in the least possible number of moves

### Fence It

##### Stage: 3 Challenge Level:

If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?

### World of Tan 24 - Clocks

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outlines of these clocks?

### Nine-pin Triangles

##### Stage: 2 Challenge Level:

How many different triangles can you make on a circular pegboard that has nine pegs?

### World of Tan 11 - the Past, Present and Future

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of the telescope and microscope?

### World of Tan 7 - Gat Marn

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outline of this plaque design?

### Turning Cogs

##### Stage: 2 Challenge Level:

What happens when you turn these cogs? Investigate the differences between turning two cogs of different sizes and two cogs which are the same.

### World of Tan 18 - Soup

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outlines of Mai Ling and Chi Wing?

### World of Tan 17 - Weather

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outlines of the watering can and man in a boat?

### World of Tan 16 - Time Flies

##### Stage: 2 Challenge Level:

Can you fit the tangram pieces into the outlines of the candle and sundial?

### Put Yourself in a Box

##### Stage: 2 Challenge Level:

A game for 2 players. Given a board of dots in a grid pattern, players take turns drawing a line by connecting 2 adjacent dots. Your goal is to complete more squares than your opponent.

### Four Triangles Puzzle

##### Stage: 1 and 2 Challenge Level:

Cut four triangles from a square as shown in the picture. How many different shapes can you make by fitting the four triangles back together?

### Muggles Magic

##### Stage: 3 Challenge Level:

You can move the 4 pieces of the jigsaw and fit them into both outlines. Explain what has happened to the missing one unit of area.


text:
# animate button presses in a sequence (tkinter)

I am attempting to animate the series of button presses in the simon game I am creating with tkinter. The animate method is called for each button that needs to be pressed in the series. It currently presses all of the buttons at the same time rather than in a series as it is meant to.

from tkinter import *
from random import choice
root = Tk()
Grid.rowconfigure(root, 0, weight=1)
Grid.columnconfigure(root, 0, weight=1)
frame = Frame(root)
frame.grid(row=0, column=0, sticky=N+E+S+W)

def animate(btn):
btn.config(relief=SUNKEN, state=ACTIVE)
btn.after(200, lambda: btn.config(relief=RAISED, state=NORMAL))

class Game(object):
def __init__(self):
self.colors = ["green", "red", "yellow", "blue"]
self.i = 0
self.round = 1
self.moves = []
self.cpu_moves = []

self.btns = []
for row in range(2):
Grid.rowconfigure(frame, row, weight=1)
for col in range(2):
Grid.columnconfigure(frame, col, weight=1)
btn = Button(frame, width=150, height=150, bg=self.colors[0],
command=lambda b=self.i: self.user_select(b))
btn.grid(row=row, column=col)
self.btns.append(btn)
self.colors.remove(self.colors[0])
self.i += 1

def user_select(self, btn):
self.moves.append(btn)
print(self.moves)
print(self.cpu_moves)
if self.moves == self.cpu_moves:
self.clear_user_moves()
self.round += 1
self.cpu_move()
elif len(self.moves) == len(self.cpu_moves) and self.moves != self.cpu_moves:
self.clear_cpu_moves()

def clear_user_moves(self):
self.moves = []

def clear_cpu_moves(self):
self.cpu_moves = []

def cpu_move(self):
nums = [0, 1, 2, 3]
self.cpu_moves.append(choice(nums))
print(self.cpu_moves)
for i in self.cpu_moves:
root.after(200, animate(self.btns[i]))

class App(object):
def __init__(self, master):
self.master = master
game = Game()
game.cpu_move()

root.geometry("300x300")

app = App(root)
root.resizable(width=False, height=False)
root.mainloop()

Let's look at this code:

for i in self.cpu_moves:
root.after(200, animate(self.btns[i]))

First, you're not using after correctly. You're doing the equivalent of this:

for i in self.cpu_moves:
result = animate(self.btns[i])
root.after(200, result)

You might as well not call root.after at all, since result will be None. If you need to pass parameters to the function you can do so by adding arguments to after itself:

root.after(200, animate, self.btns[i])

Unfortunately, you can only do this with positional parameters, not named parameters. You still need to use lambda the way you are doing it in animate.

Second, this loop is going to end up calling every function at the same time - the first one at "now+200ms", the next one at "now+200ms", the third at "now+200ms" and so on. What you want is the first one to be at "now+200ms", the next at "now+400ms" and so on. A simple solution is to add 200 for each iteration of the loop:

delay=200
for i in self.cpu_moves:
root.after(delay, animate(self.btns[i]))
delay += 200