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# [Math] Classes of functions with specific product integration property

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Does anyone of you know if there are any known sets of real-valued continous functions that for any two memebers g(x) and h(x) satisfy the following? That is, I'm looking for different sets of functions which, when integrated in product, yield the same result as when integrated separately and then multiplied. I'm curious since in one of my color algorithms that are under development I have found that if I could approximate my current functions with functions of such a class the calculations would be simplified tenfold. What are your thoughts on this?

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If the functions g and h are normalized, then replacing the = with <= gives you a special case of Holder's inequality with p = q = 1. Before moving onto this it might be helpful to look at the p = q = 2 case, known as the Cauchy-Schwartz inequality, which is easier to think about because it is just a continuous version of the familiar inequality involving the dot product |a.b| <= |a||b|.

The Cauchy-Schwartz inequality is an equality whenever a and b are parallel, or linearly dependent, and this can be extended to a similar concept for continuous functions.

For the p = q = 1 case, something similar holds except using a different norm, and unfortunately I believe it's only an equality when g and h are constant.

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Looks remarkably similar to Fubini's theorem but possibly different...

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@sQuid
Thanks for your reply. I'm not sure if this will help me but I will certainly look into it. I've only had a crach course in L2 spaces and I'm not familiar with the general properties of Lp spaces. I'll look into this before I get back.

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@etothex
Yes, I've seen the quite elegant solution to the Gaussian integral a few times but I didn't know about this Fubini's theorem. On the other hand I always thought the part where the integral is squared was a bit lofty but now I see that it is well founded. [smile]

So basically it states that the product of two integrals is equal to the product of the integrands integrated with respect to the product of their measures. This under the assumption that the product integral is absolutely convergent. Does this always hold? Thanks for the reply.

[Edited by - staaf on March 3, 2006 2:21:19 AM]

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Another approach:
All continuous functions on a closed interval can be approximated by uniform-sized step functions under the L^n (0 < n < infinity) norms.

Integrals of uniform step functions look a hell of alot like sums.

If a_n are your steps, and C the width of your steps, then
int f =~ C sum a_n

so
int fg = C^2 sum (a_nb_n)
while
int f int g = C^2 sum a_n sum b_n
= C^2 sum a_n b_n + 2 C^2 sum a_j b_i with j!=i
= int fg + 2 C^2 sum a_j b_i with j!=i
(note: this looks a hell of alot like varience from probability.)
or even better:
= 2 int fg - int fg + 2 C^2 sum a_j b_i with j!=i
= - int fg + 2 C^2 sum a_j b_i

so int f int g - int fg =~
- 2 int fg + 2 C^2 sum a_j b_i

Lets look at
2 C^2 sum a_j b_i
= 2 C sum (a_j C sum b_i)
C sum b_i =~ int g
= 2 C sum (a_j int g)
= 2 int g C sum a_j
=~ 2 int g int f

so int f int g - int fg =~
- 2 int fg + 2 int g int f
= 2 int g int f - 2 int fg
=> int f int g - int fg = 0
=> int f int g = int fg

strange. Didn't expect that. Probably made a mistake.

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@NotAYakk
What exactly are you doing in the first part of your derivation?
This is what I arrive at:

int f int g = C2(sum an)(sum bn) = C2 sum aibj

Compare to:

(a1 + a2 + a3)(b1 + b2 + b3) = a1b1 + a1b2 + a1b3 + a2b1 + a2b2 + a2b3 + a3b1 + a3b2 + a3b3

You have a 2 sneaking in there that I can't see where it's coming from.

I hope you are aware that if this would hold then every continous function defined on a closed interval that can be approximated by such step functions would satisfy the equation at the top. [smile] This is very strong and my intuition tells me it can't be so since it is easy to find functions of the given type which does not satisfy the equation.

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@staaf
I believe you meant int{g(x)h(x)dx} = int{g(x)dx}int{h(x)dx}?

Let f(x) and g(x) be polynomials (you can always use Taylor's). In this case rhs doesn't have x^1 term while lhs has (assuming that first term is x^0). Obviously we could set on the coeffs corresponding to x^0 term to zero on both f and g, but this would cause the same situation to propagate to the next term. Therefore I think it's probable that there does not exist such functions f and g.

This reasoning has atleast one loophole: a polynomial having terms from -inf to +inf (this would restrict our domain to [0,inf)). I didn't bother to think this further, but it is easy to get rid of the integrals by expressing the pol. as sum_n=-inf^inf{c_n*x^n} and integrate. From there it is easy to form the relations between the coefficients.

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Quote:
 Original post by staaf@NotAYakkWhat exactly are you doing in the first part of your derivation?This is what I arrive at:int f int g = C2(sum an)(sum bn) = C2 sum aibj

What I was doing was utter crap. =) Did a square instead of a multiply. Which also explains why it looks like varience. ~_~

Quote:
 I hope you are aware that if this would hold then every continous function defined on a closed interval that can be approximated by such step functions would satisfy the equation at the top. [smile] This is very strong and my intuition tells me it can't be so since it is easy to find functions of the given type which does not satisfy the equation.

Going to go and fix the math.

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Quote:
 Original post by WinogradI believe you meant int{g(x)h(x)dx} = int{g(x)dx}int{h(x)dx}?

Yes. My mistake. I didn't think about that it wouldn't be the same thing so I used y to distinguish the two integrals.

Anyway, you are right about polynomials not being eligible as g(x) or h(x), but there are lots of other kinds of continous functions that behave differently when integrated. I think it is wrong to rule out the existence of such functions just based on the result from polynomials. Probably the functions that actually has this property are quite few in number, but nevertheless I'm curious about them.

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Quote:
Original post by staaf
Quote:
 Original post by WinogradI believe you meant int{g(x)h(x)dx} = int{g(x)dx}int{h(x)dx}?

Yes. My mistake. I didn't think about that it wouldn't be the same thing so I used y to distinguish the two integrals.

Anyway, you are right about polynomials not being eligible as g(x) or h(x), but there are lots of other kinds of continous functions that behave differently when integrated. I think it is wrong to rule out the existence of such functions just based on the result from polynomials. Probably the functions that actually has this property are quite few in number, but nevertheless I'm curious about them.

Then you should restrict your search to such functions that do not satisfy the Taylor's theorems conditions. They rule out all continuous functions with infinitely many continuous derivatives (derivatives may vanish of course).

Analysis may get quite involved...

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Quote:
 Original post by WinogradThen you should restrict your search to such functions that do not satisfy the Taylor's theorems conditions. They rule out all continuous functions with infinitely many continuous derivatives (derivatives may vanish of course).

Except for g(x) = h(x) = 0

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Quote:
 Original post by WinogradThen you should restrict your search to such functions that do not satisfy the Taylor's theorems conditions. They rule out all continuous functions with infinitely many continuous derivatives (derivatives may vanish of course).
You're right. I didn't consider that. Thanks.

Quote:
 Original post by Anonymous PosterExcept for g(x) = h(x) = 0
Obviously, but it'd be hard to approximate any other function with that.

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Quote:
Original post by staaf
Quote:
 Original post by Anonymous PosterExcept for g(x) = h(x) = 0
Obviously, but it'd be hard to approximate any other function with that.

You're just not trying hard enough then [smile]

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let fk(x) = k for all x.

Then
int {fk fj} = k * j * [x-x0]
int {fk} int {fj} = k * j * [x-x0]

In other words, it works with constant functions. =)

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No, it doesn't work with constant functions.

j*k*x != j*x * k*x

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If I didn't do any mistakes then your functions should satisfy also following equation:

int{h(x)dx}/h(x) + int{g(x)dx}/g(x) = 1

You can arrive to the result yourself by taking derivatives on both sides:

h(x)g(x) = h(x)*int{g(x)dx} + g(x)*int{h(x)dx}

and first separating h to lhs and g to rhs.

I'm not sure if this is useful at all, but atleast gives some insight to the problem, provides a necessary identity and IMHO it looks beatiful ;).

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EDIT: There was an error in some equations. Lhs was ln(h(x), while it should have been ln(int{h(x)dx}).

OK I have a single set of solutions (there is perhaps more), from the previous formulation with a little work we can get:

h(x) / int{ h(x)dx } = g(x) / ( g(x) - int{ g(x)dx } )

Integrating this from the both sides gives:

ln(int{h(x)dx}) + C0 = int{ g(x) / (g(x) - int{ g(x)dx } )dx } (1)

OK, if I want an easy solution for the rhs I would demand that

g(x) = diff{ g(x) - int{g(x)dx} }

If this would be satisfied then the rule applied for the lhs could be used for rhs too.

So we get that g(x) = g'(x) - g(x) => g'(x) = 2g(x)
I came up only one function that satisfies this: g(x) = exp(2*x) (2)

Now substituting (2) to (1) we get

ln(int{h(x)dx}) + C0 = ln(exp(2*x)) + C1 => ln(int{h(x)dx}) = 2*x + C => int{h(x)dx} = A*exp(2*x), where A is some arbitrary constant.

Thus g(x) = h(x) = exp(2*x) will do. Also h(x) = -2 * exp(2*x) and g(x) = 2*exp(2*x) will do and so on...

Unfortunately this is not very useful for approximation and even more unfortunately this pretty much voids my Taylor polynomial argument :(

Generally h(x) = diff(exp( int{ g(x) / (g(x) - int{ g(x)dx } + C)dx } ),x), so the trick is to be able to calculate the outer integral. Sometimes it is easy when C is 0, but this doesn't always yield a correct solution. Note, that i'm not sure if the C is only constant that has to be introduced. They appear from the integrations.

[Edited by - Winograd on March 6, 2006 2:49:48 AM]

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Quote:
 Original post by WinogradIf I didn't do any mistakes then your functions should satisfy also following equation: int{h(x)dx}/h(x) + int{g(x)dx}/g(x) = 1 ...I'm not sure if this is useful at all, but atleast gives some insight to the problem, provides a necessary identity and IMHO it looks beatiful ;).

I believe it is very useful. It states a reasonable condition to satisfy the original equation.
Quote:
 Original post by WinogradThus g(x) = h(x) = exp(2*x) will do. Also h(x) = -2 * exp(2*x) and g(x) = 2*exp(2*x) will do and so on...

Yes, and this can be generalized to all functions

g(x) = A*exp(p*x)
h(x) = B*exp(q*x)

as long as

1/p + 1/q = 1 [1] and (A, B, p, q != 0)

since for g(x)

int{g(x)dx}/g(x) = int{ A*exp(p*x) dx } / A*exp(p*x) = (A/p)*exp(p*x) / A*exp(p*x) = 1/p

and the same goes for h(x).

Moreover for the original equation:
If we use the same g(x) and h(x) as above we have at the left hand:

int{g(x)h(x)dx} = int{A*B*exp((p+q)*x)dx} = (A*B)/(p+q) * exp((p+q)*x)

and at the right hand:

int{g(x)dx}int{h(x)dx} = (A/p * exp(p*x))(B/q * exp(q*x)) = (A*B)/(p*q) * exp((p+q)*x)

Now if we rearrange [1] we get that p+q = p*q, and by substituting this into the left or right hand side, the equation is satisfied.

Still, as you say, a single A*exp(p*x) is not very useful for approximation. A linear combination of such functions would do better. Hmm...

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Spotted the error today. Lhs was missing an integral. I corrected it to my previous post.

Using Maple I found that when g is a first order polynomial h has the following solution:
g(x) = a + b*xh(x) = -2*(a+b*x)*exp((-ln(-2*a-2*b*x+2*a*x+b*x^2)*(b^2+a^2)^(1/2)+2*b*arctanh((-b+a+b*x)/(b^2+a^2)^(1/2)))/(b^2+a^2)^(1/2))/(-2*a-2*b*x+2*a*x+b*x^2)

Note also that the condition or identity I gave earlier isn't accurate. It is satisfied by g and h for which the original equation isn't satisfied. It is a necessary but not sufficient condition.

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Quote:
 Original post by Anonymous PosterNo, it doesn't work with constant functions.j*k*x != j*x * k*x

Ok, it works on constant functions on the interval [0,1].

^_^

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You assume that a laser beam is a single, fine line—as if it would project one tiny, millimeter-wide red dot. But all practical lasers produce divergent beams. You can arbitrarily narrow the beam by shining it out through a telescope, but I'm pretty sure that the narrowest beam we can practically generate still will illuminate many square miles of the lunar ...

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The papers described in the report you've linked to demonstrate examples of sound amplification by spontaneous emission. This is independent of the characteristics of the waves that those source produce. As it happens, for sound (as opposed to light) it is relatively easy to create waves that have high spatial coherence and well-defined wave vectors $\vec k$...

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It is an insightful question, and the answer is "Yes". What you describe is often called "frequency mixing", and can be used to generate practically any wavelength we want. Note, however, that to modulate a laser at 100 THz, you need a source to produce a 100THz signal. And, by the way, you seem to be mixing up wavelengths and frequencies. 630 nm (...

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You will need at least three states for lasing to occur (with reasonable probability) but that does not mean that other states cannot be present. If the pump frequencies are not tuned to reach these other states then they will not be occupied with any appreciable probability and hence will not influence the lasing process.

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The extinction in the atmosphere is a function of atmospheric conditions, dust and aerosol concentrations, the angle to the vertical the laser is fired at, and of course the wavelength. A typical value at 589 mm (sodium laser) in a clean atmosphere and a vertical beam would be about a 15% attenuation. It could easily be a factor of a few worse in more ...

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Well, sorry for the short answer, but "laser flux power density" is just $I_0$, which is the power per unit surface area (which is why it is called flux power density) and the Raman intensity is proportional to $I_R$, so by this equation you can see that when $I_0$ increases, so does $I_R$

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In this link stimulated emission is explained in the third page: Note that the energy of the incoming photon doing the stimulation is $E_2-E_1$ the two energy levels . This means the first photon passes and a second one comes out, which when conditions are appropriate can lead to the lazing phenomenon. This is with pictures and words, but coherence is a ...

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I can rule out momentum transfer from the laser. Even though radiation pressure is a thing, it is far too low to be noticeable. Let's do the math: Radiation pressure p=I/c (c speed of light) is determined by the intensity I of the laser beam. Since pressure in general is p=F/A (F Force and A Area), you get F=I*A/c for the force that results from the ...

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1- 50/50. At the point of inversion both probabilities are the same. 2-I think you are missing some key points here. You always need to include the energy structure. For example in a 4-level system: $E_0$ ground level, $E_1$ upper pump level, $E_2$ upper laser level, $E_3$ lower laser level. The inversion between ground state $E_0$ and upper-pump level $E_1$...

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Given the terms you're using, I'll assume you're asking about a current-pumped diode laser. The key additional information you need is the carrier lifetime due to nonradiative recombination and spontaneous emission events, call it $\tau_R$. Then the current required to maintain a carrier density of $N$ is $$I=\frac{qNV}{\tau_R}$$ where $q$ is the ...

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Q6.    An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

D Divya Prakash Singh

The sides of the triangle are:

$a = 20\ cm,\ b = 50\ cm\ and\ c =50\ cm.$

So, the semi-perimeter of the triangle is given by,

$s = \frac{a+b+c}{2} = \frac{20+50+50}{2} = \frac{120}{2} = 60\ cm.$

Therefore, the area of the triangle can be found by using Heron's formula:

$Area = \sqrt{s(s-a)(s-b)(s-c)}$

$= \sqrt{60(60-20)(60-50)(60-50)}$

$= \sqrt{60(40)(10)(10)} = 200\sqrt{6}\ cm^2$

Now, for the 10 triangular pieces of cloths, the area will be,

$= 10\times200\sqrt6 = 2000\sqrt6\ cm^2$

Hence, the area of cloths of each colour will be:

$= \frac{2000\sqrt6}{2} = 1000\sqrt6\ cm^2.$

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# Proceedings of the nsf/sigplan workshop on object-based concurrent programming

By . In SIGPLAN Notices, vol. 24, no. 4, April 1989, ACM SIGPLAN. 1989.

## BibTeX

@inproceedings{obcp89,
author = "Agha, Gul",
title = "Proceedings of the NSF/SIGPLAN Workshop on Object-Based
Concurrent Programming",
booktitle = "SIGPLAN Notices, vol. 24, no. 4, April 1989, ACM
SIGPLAN",
year = "1989",
}


text:
# To calculate cross elasticity of demand, does percentage change in quantity demanded have to be market demand instead of individual demand? [closed]

If I try to calculate cross elasticity of demand (XED or CED) with individual demand quantity, it seems to break down since even small increase when the base number is small is a big increase in percentage. e.g. 1 to 2 is a 100% change.

## closed as unclear what you're asking by Giskard, ml0105, Wecon, Jamzy, EnergyNumbersJul 19 '16 at 16:02

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

$XED=\frac{\Delta q_i}{q_i}\cdot \frac{p_j}{\Delta p_j}$
It is right that in your case $\frac{\Delta q_i}{q_i}=\frac{2-1}{1}=1$


text:
Göm menyn

Matematiska kollokviet

Organiseras av Anders Björn, Milagros Izquierdo, Vladimir Kozlov och Hans Lundmark.

Du kan få ett dynamiskt uppdaterat schema till din dator eller smartphone från Google Calendar – se Smartphoneinstruktioner.

Onsdag 26 september 2018, Johan Öinert, Blekinge tekniska högskola, Karlskrona

Talare: Johan Öinert, Blekinge tekniska högskola, Karlskrona

Titel: Epsilon-strongly group graded rings, Leavitt path algebras and crossed products by twisted partial actions

Tid och plats: Onsdag 26 september 2018,  Hopningspunkten, 13.15–14.15

Sammanfattning: Epsilon-strongly group graded rings constitute a class of rings which contains all strongly group graded rings and all crossed products associated with unital twisted partial group actions. A result of Năstăsescu, Van den Bergh and Van Oystaeyen (1989) gives a characterization of strongly group graded rings which are separable over their canonical 'degree zero' subrings. A more recent result of Bagio, Lazzarin and Paques (2010) gives a characterization of crossed products, associated with unital twisted partial group actions, which are separable over their coefficient subrings. We are able to simultaneously generalize both of these results by giving a characterization of separable epsilon-strongly group graded rings. We also provide examples of separable epsilon-strongly group graded rings (not strongly graded!) and thereby answer a question of Le Bruyn, Van den Bergh and Van Oystaeyen (1988).

Given an arbitrary group G, we will explain how to equip any Leavitt path algebra over a finite (directed) graph with an epsilon-strong G-gradation.

This talk is based on recent joint work with Patrik Nystedt (University West, Sweden) and Héctor Pinedo (Industrial University of Santander, Colombia).

Onsdag 3 oktober 2018, Håkan Lennerstad, Blekinge tekniska högskola, Karlskrona

Talare: Håkan Lennerstad, Blekinge tekniska högskola, Karlskrona

Titel: Distance-consistent graph labelings, the ampleness of a graph, and graph functionals

Tid och plats: Onsdag 3 oktober 2018,  Hopningspunkten, 13.15–14.15

Sammanfattning: A natural labeling of a simple connected graph $G=(V,E)$ is a labeling $c$ of the nodes with natural numbers $1,2,...,|V|$. Such a labeling induces a labeling distance $c(u,v)=|c(u)-c(v)|$ alongside the usual graph distance $d(u,v)$. A natural labeling that realizes the minimum $$l(G)=\min_{c}\sum_{u,v\in V}(c(u,v)-d(u,v))^2$$ is a distance-consistent labeling, and $l(G)$ is the ampleness of $G$. It trivial that $l(G)=0$ iff $G$ is a path graph, and I'll give the proof that $l(G)≤l(K_{n})$ for all $G$ with $n=|V$|. The normalized ampleness $L(G)=l(G)/(Kn), \, 0 \le L(G) \le1$ is studied for different graph classes such as the bipartite graph $K_{n,n}$, the star graph $S_{n}$, the cycle graph $C_{n}$ and a few other types, particularly for $n \to \infty$.
The quantity $$\min_{c}\sum_{u,v\in V}(c(u,v)-d(u,v))^2$$ is a graph functional; mapping graphs to non-negative integers. It can be thought of as the "inverse listness" of a graph - being zero for lists only (path graphs). The quantity $c(u,v)$ can be replaced by other quantities defining the "inverse cycleness" or "inverse starness" of any graph, in which case the corresponding functional is zero if an only if the graph is $C_{n}$ or $S_{n}$, respectively.

Onsdag 10 oktober 2018, Evgeniy Lokharu, Lunds universitet

Talare: Evgeniy Lokharu, Lunds universitet

Titel: TBA

Tid och plats: Onsdag 10 oktober 2018,  Hopningspunkten, 13.15–14.15

Sammanfattning: TBA

Fredag 12 oktober 2018, Alexandre Karassev, Nipissing University, Canada

Talare: Alexandre Karassev, Nipissing University, Canada

Titel: Dimension and decomposition complexity

Tid och plats: Fredag 12 oktober 2018,  Hopningspunkten, 10.15–11.15

Sammanfattning: In attempts to capture asymptotic properties of finitely generated groups, manifolds, and general metric spaces, various dimension- like properties have been introduced recently, including asymptotic dimension, asymptotic dimension growth, asymptotic property C and asymptotic property D. We prove that if X is a tree-graded space (as introduced by C. Drutu and M. Sapir) and the family of all pieces of X satisfies one of the dimension-like properties, then X satisfies the same property, with explicit control over the parameters used in the property. In particular, the free product of finitely generated groups G*H satisfies a dimension-like property if the property holds for each group G and H. This is a joint with Nikolay Brodskiy

Onsdag 17 oktober 2018, Sergey Vakulenko, St Petersburg, Ryssland

Talare: Sergey Vakulenko, St Petersburg, Ryssland

Titel: TBA

Tid och plats: Onsdag 17 oktober 2018,  Hopningspunkten, 13.15–14.15

Sammanfattning: TBA


text:
## 7.9 change in internal energy

$\Delta U=q+w$

jophan3F
Posts: 5
Joined: Fri Sep 26, 2014 2:02 pm

### 7.9 change in internal energy

7.9 A gas in a cylinder was placed in a heater and gained 550 kJ of energy as heat. If the cylinder increased in volume from 345 mL to 1846 mL against an atmospheric pressure of 750. Torr during this process, what is the change in internal energy of the gas in the cylinder?

In the solutions manual, it shows the change in internal energy is (5500 kJ - 0.150 kJ = 5500 kJ). Is the number the same because it was not a significantly large difference?

Kayla Denton 1A
Posts: 106
Joined: Fri Sep 26, 2014 2:02 pm

### Re: 7.9 change in internal energy

Yes, you're right. The number is the same due to sig fig rules: when you add or subtract two numbers, you round your answer to the least number of decimal places present in the numbers you are adding or subtracting. Since 5500 has 0 decimal places, you round your answer to the nearest whole number.

Jennifer Aguayo 1I
Posts: 26
Joined: Fri Sep 26, 2014 2:02 pm

### Re: 7.9 change in internal energy

For the pressure, how do we convert torr to atm?
In the solutions manual they divided 750torr by 760 torr x atm^-1 But how do we know it is 760, is that given to us or do we just have to know it? I cannot find it in the equation sheet.

Niharika Reddy 1D
Posts: 127
Joined: Fri Sep 26, 2014 2:02 pm

### Re: 7.9 change in internal energy

The conversion factor (760 torr=1 atm) would be given to us.

Although the back of the periodic table doesn't explicitly state that conversion factor, it gives the gas constant in L·atm/K·mol (.08206) and in L·torr/K·mol (62.364). Since we want the conversion factor of torr to atm, we can get that by dividing the second equivalent of the gas constant listed above by the first. (62.364/0.08206=759.98~760 torr/atm, or 760 torr=1 atm)


text:
# What was the first bit of mathematics that made you realize that math is beautiful? (For children's book)

I'm a children's book writer and illustrator, and I want to to create a book for young readers that exposes the beauty of mathematics. I recently read Paul Lockhart's essay "The Mathematician's Lament," and found that I, too, lament the uninspiring quality of my elementary math education.

I want to make a book that discredits the notion that math is merely a series of calculations, and inspires a sense of awe and genuine curiosity in young readers.

However, I myself am mathematically unsophisticated.

What was the first bit of mathematics that made you realize that math is beautiful?

For the purposes of this children's book, accessible answers would be appreciated.

-
For me Euclid's proof of the infinitude of primes was the first thing that made me realize the beauty of mathematics. –  Manjil P. Saikia Mar 7 '13 at 7:02
Wow. Just last night I had a fierce argument with one of the bartenders of my usual watering hole who is a mechanical engineering student. He insisted that he has a better idea than me of what is mathematics. I am so going to print him a copy of Lockhart's text. Thank you for that link! –  Asaf Karagila Mar 7 '13 at 7:59
I can’t remember a time when I didn’t think that mathematics was beautiful and fascinating. –  Brian M. Scott Mar 7 '13 at 15:06
Although I don't know if it's what you are looking for, try looking up "vihart" on youtube--Even if it's not helpful, I guarantee you will appreciate it. –  Bill K Mar 8 '13 at 2:57
I think it's a shame that this question was voted closed... –  Will Mar 10 '13 at 19:50

There is a series of math children books in Russian by
Владимир Артурович Лёвшин. To list some:
Магистр рассеянных наук (translates roughly as Master (as in M.Sci) of the absent-minded
sciences, though google translates it as Master scattered Sciences),
Новые рассказы Рассеянного Магистра (New stories of the absent-minded Master),
Путешествие по Карликании и Аль-Джебре (google transtales it as Travel Karlikanii and Al Gebre),
Черная маска из Аль-Джебры (The Black Mask from al-ğabr(=al-gebra)).
More of them at http://www.koob.ru/levshin/ (in Russian).

I am a Bulgarian (presently working in New York), and as a child (could have been anything between 6 yr old and 9 yr old) read the Bulgarian translation of Путешествие по Карликании и Аль-Джебре (or it might have been one of the other books listed above).

I was fascinated. At hindsight mathematically the book is fairly simple or even routine (goes on to set and solve an equation, must have been a quadratic one, though it might have even been linear), so once you know how to solve such equations it might appear boring. But amazingly it does it in a way that unfailingly keeps the readers attention. It is written like a detective story (the $X$ with the black mask was enchanted and was to be freed by the Master, and its assistant the Нуличка, i.e. the Naught or the Null), with characters to relate to, number system and operations introduced and, thus, developing in parallel, the necessary math background and an intriguing story to follow and enjoy. I motivated myself to understand the math details (it might have been that we had not yet covered that material in school), so I could keep reading. I was also interested in logic-puzzle books at about the same time. I cannot single out a particular math piece that is exciting in this book (generally it is about real numbers or perhaps integers, setting and solving equations), it is the whole process of taking an ignorant (but intrigued) child and making him willing to follow the story, and to eventually learn algebra (at the level of quadratic equations) on their own, and make them feel great about it (and at the same time to not know at all what exactly feat they have done, that is, there was no feeling at all that I was being educated in "accurate reckoning for inquiring into things, and the knowledge of all things, mysteries...all secrets", to borrow from Ahmes, even if there is no direct relation).

I do not know if an English translation is available of this series (google doesn't seem to know about it, and google knows everything, unless I do not know how to find it). I think this is a great book and would recommend it to anyone who could read it (or, well, would certainly recommend it to children, since adults would be spoiled with what they already know, and might not enjoy it). In my opinion this book has the spirit of adventure, and it might make an interesting reading (it reminds me of another Russian (or Soviet) well-known "adventure" book, The Twelve Chairs, with sequel The Little Golden Calf, though both format and subject are very different, but perhaps one could feel that both represent Russian culture ... don't know what the author(s) would have thought of this alleged affinity though).

I also remember that as a child I had a problem understanding infinity (безкрайност in Bulgarian, literally endlessness), and kept thinking about it. It is not clear if I understood it (it looked like a winding road that kept going no matter what), but at some point I stopped thinking about it. Nowadays I presumably know that there are different related notions: limitless/ boundless / infinite, and different parts of math might have use of one or the other (e.g. manifold without a boundary like the circle, vs the real line which extends both ways, or transfinite numbers which go just one way but could be used for counting). I could not quite accept that infinity exists (and strictly speaking nobody could prove that it does, but anyway it is an accepted convention), my point is that I forced myself to imagine that winding road never-ending, to try to get an idea of infinity, but I don't think I ever convinced myself. I could not see the whole thing, even if any time I approached the end, it kept extending (as I would just generate one more piece of it in my mind and put it there for the sake of the argument), so for me infinity was something that I cannot exhaust by way of observation, but cannot comprehend either. I could not rule out its existence, so I live with that, but I never saw it, so I can't vouch for it. (Much later at university I had a dream, almost a nightmare when I was supposed to pick a rifle from what seemed an endless field of identical rifles, and I couldn't make a choice. Eventually I picked one, its virtue was that is was exactly the same as all the others, but fulfilled the task of picking one. Somehow I tend to relate this to the Axiom of Choice, though strictly speaking you do not need AC to pick just one element of just one set. And I have no idea why it was rifles, and not, say, apples. Also, that was a multiset, not a set, so I don't know what it had to do with AC, I guess I had to come up with something familiar when I woke up, and we had already studied AC. Or perhaps someone could indeed relate this to AC in a meaningful interpretation.)

And of course, I do appreciate things like Euclid's proof that there are infinitely many primes, or that (Pythagoras or Hippasus) $\sqrt{2}$ is irrational (these were some of the first things I enjoyed introducing to my students last semester in a History of Math class), but for me these came "later" when I was already a converted mathematician (or I thought of myself so). I can't tell when this conversion happened, but it might have been in early school. I was good at math (so my teachers were happy and my schoolmates sought my help, and for that matter everyone would keep telling me that my grandmother, whom I newer saw, was a famous (or infamous, because she would uncompromisingly fail bad pupils) math teacher in my home-town), but it is not just about being good at it (as I realized, there were better students than me, once I got into university, in particular some of those coming out from so-called matematicheska gimnazia - a high-school emphasizing math, science and languages), so it is not so much about being good at it, as about being attracted by math, willingness to keep working/discovering, and the adventure and meaning that comes with it.

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In the age four or five i knew: $$2*5=5*2$$
I hope that you understand how this result is wonderful for me on this age, because yet i didn't use to commutativity of multiplication on $\mathbb N$!
In addition i didn't generalize this fact to another numbers!

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Adding to LaceySnr’s answer, I’d like to mention fractals in general. While fractals will probably count as a higher application of maths, they are very often very visually beautiful. So you could easily show a picture of a fractal and explain that there is just a simple formula behind it all.

Some more examples:

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holy crap the last 3.... –  im so confused Mar 7 '13 at 15:43
All the fractals shown in this and other answers are too complex for little kids, but the basic concept is easy for them to understand. Start with a tree with each branch being another tree. It doubles as a introduction to recursion for those more inclined towards programming than pure math. –  Izkata Mar 7 '13 at 17:12
@Izkata But they are visually appealing and that will spark the interest in it. As a kid I wouldn’t care how to prove that 2 is irrational, probably because I wouldn’t know what irrational is to begin with, or that there is some equality with some weird constants I don’t even know ($e^{i\pi} = -1$). It’s things like fractals, or other natural things like Fibonacci flowers that makes math interesting beyond just numbers. –  poke Mar 7 '13 at 17:57
This is fantastic. –  josh Mar 8 '13 at 0:58
@Izkata Perhaps fractal theory is too complicated for kids, but using fractal generation software might be within their reach. And if they start playing with that, they'll be learning a whole ton of math in a way that is incredibly fun. –  Kevin Mar 10 '13 at 4:42

This is rather recent (Less than a year ago), but, since I am 14, I suppose it should still apply. I remember that I was bored in some class, and that I took out my calculator and started playing with it, writing "hello" with numbers upside down. Then I saw this button (this was a scientific calculator) that said "log," and so I pressed it. At first I received "error" for log(0) = -infinity (well, close enough), but then I tried other numbers, 1,2 10. Then I saw that at 10 it would blurt out 1, and at 100 2. I then realized that what log did was find the exponent of a number from a base number (of course, I didn't know that terminology then) but it was still pretty amazing. (I also learned later on that all calculators are log base 10)

Edit: is there something wrong with this answer? Why was it down voted?

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The first thing for me is the working of an equation. it is, to me, like a stanza of a poem that tells us many things in minimum words. No one would have ever thought of describing a geometrical figure. Every one used to draw it before math's entry in the real world. It's awesome for a mathematician to say that write me a circle, ellipse etc.

In order to tell people that math is not only concerned to problem-solving, I have produced my own quote.

" Practice is entirely hollow without understanding ". - Sufyan Sheikh

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The most wonderful thing I've recently seen is this (sorry it's in French) form of the Sieve of Eratosthenes and of course your question too.

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The commutative law doesn't hold for some series. I think this is an amazing fact to teach.

http://www.math.tamu.edu/~tvogel/gallery/node10.html

The example in the link amazed me.

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A few things come to mind:

Here's a beautiful JavaScript demo of these graphs being generated: http://www.datapointed.net/visualizations/math/factorization/animated-diagrams/

• Even as an adult, I think continued fractions and generalized continued fractions are amazing. One of the simplest is the golden ratio: $$\varphi = 1 + \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{1+\ddots}}}$$ And this identity is downright incredible:

$$\frac{\pi}{2} = 1+\cfrac{1}{1+\cfrac{1}{1/2+\cfrac{1}{1/3+\cfrac{1}{1/4+\ddots}}}}$$

I should stop myself now... But math is really filled with astounding phenomena like I've mentioned above...

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I've never seen that one for $\pi$ before! It's much better than the one with integers, which has no pattern at all. –  Ryan Reich Jun 12 '14 at 16:45
I wonder if there exist an intuitive argument explaining the pattern that appears in the continued fraction of $\tfrac\pi2.$ –  Hakim Jun 28 '14 at 23:08

Mine was the discovery of sets in higher order math classes, and how all the lower math classes including physics theories were strictly derived from higher order calculus, and all of the formulas I had ever learned became such simple child's toys.

I don't think those belong in a children's text, however.

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When you realize that taking derivatives is so simple, you look back and realize, "I can't believe people use this as an example of difficult mathematics!" –  Joe Z. Mar 13 '13 at 20:25
I had a similar moment of realization for reducing polynomials, back in middle school when my Sunday School teacher used a really long rational polynomial expression as an example of a "problem that's too hard for you to solve" (it was part of a teaching package). She had to resort to using trigonometry and asking me how I would calculate $\tan 35^\circ$, which I didn't know at the time. The polynomial ended up being something contrived, but it did actually reduce quite a bit. –  Joe Z. Mar 13 '13 at 20:32
Of course, now when I look back at it, I think, that wasn't actually hard! –  Joe Z. Mar 14 '13 at 14:20

I felt like an Einstein and was really interested in mathematics when I myself discovered the truth behind a^0 =1. That is, a^0 = (a)^(1-1) = a^1/a^1 = 1

Yeah, I know this is simple.. But generally it is taught as a formula. Instead this one can be used to change the way of thinking...

Also, multiplication is repeated addition... This used to fascinate me a lot...

2 * 3 = 6 that is, 2 + 2 +2

4 * 3 = 4 + 4 + 4

5 * 8 = 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5

And then in the end you can say that, for very big numbers, you cant sit adding all of them and hence, multiplication is the shortcut to add all of them :)

I am not a writer... But probably you can take some god examples to explain what I am trying to say here... I think this will be really interesting approach to teach multiplication! All the best for your book. Do let us know the name of the book. We will also cherish it... :)

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### Realising why zero is not nothing, and understanding numbers

I first understood the difference between zero and none when thinking about thermometer readings. If you had a ton of thermometers scattered around the world, and you collected their readings periodically and put them in a database, what would you do if any thermometer was broken? If you just put a zero reading, you'll screw up your averages, but if you put a null value, you can handle broken thermometers easily.

That made me realise what a number is.

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I found a book by Isaac Asimov called "Asimov On Numbers" which is a compilation of his essays related to math and numbers.

It was all very fascinating - things like why Roman numerals are inefficient, why zero was such a groundbreaking number to invent, and things like that. You might want to see if you can find that book to get some inspiration.

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One of the things I really like in math is the probability. One of the best examples that I like is on the film 21.

You are in a program show and you have three doors:

One: With the prize, and the other two with monsters;

The presenter tells you to pick up a door. When you finally choose a door, he asks "Are sure about it?

Then for some reason he decides to open one of the wrong doors and asks you: "Are you going to stay with your door or change it?" and he says "But remember I know where the prize is".

So what should have you do?

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Goats aren't monsters! D: –  Joe Z. Mar 7 '13 at 21:33
Of course, switch. The first time I have 1/3 chance of being right. The other two doors 2/3. Now, one wrong door is opened, meaning that the remaining door has the unified 2/3 chance of being right. I am flappergasped how many people do not see this. This is where I demand them to envision this example with 1 million doors, you chose one at random. Then the moderator opens 999.998 other doors leaving you with your first choice, and the one that was not openened. Do you really think not switching would be smart? –  k0pernikus Mar 8 '13 at 13:07
Isn't this called the Monty Hall Problem? –  funkymushroom Mar 8 '13 at 16:28
I remember my teacher telling us this problem, but she had trouble explaining why it was that way. Most of my classmates didn't believe it was better to switch. But I wrote a Basic program on my Commodore 128 running 1000 iterations and found out that it really resulted in 50% chance of winning when switching doors. Several years later I stumbled upon a good explanation of why it is better to switch. –  Anlo Mar 14 '14 at 17:51

The simple and commonly used sum, and divide of apples. I was really bad at math, and using objects instead of numbers really teached me how to love (math, LOL). It's amazing how math can be used on anything.

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Everybody loves fractals. I think this one - The Dragon Curve - is particularly easy to explain, and it is very surprising and aesthetically pleasing:

Here's a video I've seen which explains how it comes about: The Dragon Curve from Numberphile

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I have a friend who is disgusted by pictures of fractals. Might relate to trypophobia. –  NikolajK Mar 7 '13 at 20:31
I seem to recall this fractal making an appearance in Jurassic Park (the book). –  icurays1 Mar 8 '13 at 18:23
I found by myself this fractal by folding and reopening a long strip of paper. The strip was actually the edge (with holes) of the continuous paper used in old pin printers. –  Emanuele Paolini Feb 17 '14 at 14:15

Solving for an unknown. 2x = 4 so x = 2. Beautiful.

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It becomes hideous when forced to solve endless amount of them. –  Phonics The Hedgehog Jun 2 '13 at 2:44

I was pretty good at math from an early age, but what was the clincher for me was the existence of non-Euclidean geometry. In grade 6 my math professor gave me a book on axiomatic Euclidean geometry, and I was totally blown away that the parallel postulate was just that, a postulate, and not an undisputable true fact. If upmto that point I just considered (school) math easy, from that moment I realized is incredibly beautiful. I did not look back since.

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Two instances where I thought math was amazing:

1. In like 4th grade or whenever you learn areas of rectangles, one of the exercises in my book was to estimate the area of some squiggly shape overlaid on a rectangular grid. I thought this was pretty cool, and reasoned that if you could make the grid "smaller" (higher resolution), you could be more accurate. I mentioned this to my mom, who proceeded to tell me that was basically how Calculus 2 worked. :) That was very fun for me.
2. Deriving the quadratic formula in Algebra 1. That was fun--it showed that some totally un-intuitive formula could be easily found using other previously found results.
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Your first sentence is incorrect. :) I had the "grid" epiphany, too! –  columbus8myhw Nov 13 '14 at 1:23

My favourite maths book when I was little was 'Magic House of Numbers' by Irving Adler.

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This was probably the very first mathematic riddle which absolutely got me. It is called Algebrogram in my language, but I couldn't find a reference in English.
I was attending mathematic group after normal school (at age 11-14) and then I made few of my own for my classmates. I loved it ^^

You use characters instead of numbers and you construct some words. You then let others solve it.

F O R T Y
T E N
T E N
---------
S I X T Y

Solution:

2 9 7 8 6
8 5 0
8 5 0
---------
3 1 4 8 6

It was common to construct sentences as well, but it is kind of hard. This is only an example, which is unsolvable ;)

You could specify if there were some other operations or you could let your solvers find it out by themselves.

        O U R
H O U S E
H A S
- T E N
-------------
W I N D O W S

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In English these are sometimes known as cryptarithmetic puzzles. –  Will Mar 7 '13 at 16:41

The first interesting mathematics problem I remember in my limited memory is 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... It never totals to TWO :-)

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Sounds like Zeno's Dichotomy paradox. –  BobStein-VisiBone Mar 7 '13 at 15:28

This was my favourite equation. I was 16 or so, when my father showed it to me. I was amazed, and I programmed an application which drew this:

The interval should be <-6;6> maybe. I made it a looong time ago after all ;)

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My mother repeatedly tells this story about me.

In German television there is a series called Telekolleg (not Kellog you silly, more like in college) which is broadcated for remote learning. One series deals with Math.

I was about 5 or 6 years old, when I sat in front of the TV watching this Telekolleg Mathematik series, turning to my mother and insisting: 'This is a good programme, you have to watch this'.

I don't remember what the exact topic was, perhaps quadratic function graphs.

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When I was young I found a riddle:

• Multiply it by 3
• Multiply it by 3
• Add the number you thought at the beginning
• Tell me the result:)

The number you thought about is your result without the digit 3 at the end, so i.e. if your result is 53, then you thought about 5.

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This kind of joke was very commom im my school ^^ –  Integral Mar 7 '13 at 15:43
@drjimbob: You're missing the "add the number you thought at the beginning" line, which makes the result be $(9x+3) + x = 10x + 3$. –  ShreevatsaR Mar 24 '13 at 11:43

I think the first thing that amazed me in this way was $\pi$. An irrationnal number, which means it has an infinite number of digits, which involves humans can't manage it, we can't know it on the whole, but already the Greeks discovered it. They knew it has something to do in the circumference or the area of a circle, that is, they could manipulate it, and I find this unbelievable.

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1. 17 + 20 = 8
2. 17 − 20 = 26
3. 17 · 20 = 21
4. 17^(−1) = 12 (inverse of 17)

I got really upset when I saw this. The professor explained, to do network communication you will need to understand this.

I found maths awesome after dealing with these. What we are normally learning can not always help (it's real numbers mathematics). But the best things deal with fields. Therefore, the below is the explanation of the above meaningless things.

(i) Addition: 17 + 20 = 8 since 37 mod 29 = 8

(ii) Subtraction: 17 − 20 = 26 since −3 mod 29 = 26

(iii) Multiplication: 17 · 20 = 21 since 340 mod 29 = 21

(iv) Inversion: 17^(−1) = 12 since 17 · 12 mod 29 = 1

The elements of F29 are {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27 28}

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When I was a child, I spent the whole summer at a camping at the coast of Catalonia. There I was always around my grandfather. He himself had no proper education and never finished school. Nevertheless he liked to read books on his own, about many things, grammar, the French language, mechanics, mathematics...

I remember he taught me many things. He was the first to explain me, as I fell asleep in his arms, under the starry night, that the Earth was a ball, and that there were people underneath the ground where we stood, on the other side of the planet, who were standing upside down without falling, because we were all attracted to the center of the ball. I did not understand, at that moment, how was that possible. But I trusted him and knew that there were many things I did not understand about the world.

One particular thing related to mathematics that he told me and that got me thinking, making myself questions and reaching the boundaries of my mind, was that one frog could try to jump her way across a puddle (we also went together to catch frogs), jump first to the half of it, then to the half of the remaining half and so on, and that after an infinite number of jumps she would arrive at the other shore.

This was, I think, one of the first things that made me feel that the world or that reality itself was infinitely bigger, more complex and more beautiful that anything we could understand or even begin to grasp. I guess this sense of real magic is what makes me have a special love for mathematics.

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The Golden ratio

It was presented to me like this: There exists a number that you can square, subtract itself, and you'll get 1. Or, you can inverse the number, add 1, and you'll get the number back. What a beautiful number, I thought. Of course, I later realized the number was just a solution to:

$$x^2 - x - 1 = 0.$$

However, I was really impressed when later I learned this number also shows up in nature in the patterns of plant growth. Wow! Who would have thought?

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The way I first saw it was like this: there is a number that, if you add one to it, becomes its square. Then they showed that you could easily calculate it from just that information with only the quadratic formula (which I had thought was boring and had never suspected of hiding anything beautiful), and that if you kept multiplying it by itself, you'd get a simple expression with fibonacci numbers as coefficients. That was a happy day. –  Michael Shaw Mar 8 '13 at 5:17
Showing that number on a calculator really drives its magic home. Square it, and all the digits after the decimal point stay the same. Or hit the 1/x button, same thing. –  Joubarc Mar 8 '13 at 8:51
I think it's more impressive that if you subtract one from that number, you get its reciprocial. –  Zsbán Ambrus Mar 8 '13 at 11:51
youtube.com/watch?v=ahXIMUkSXX0 Vi Hart explains why the "magic" is actually completely inevitable, and the connection with the Fibonacci numbers. –  Ben Millwood Mar 9 '13 at 23:06

Hilbert's infinite hotel, the realization that $\mathbb{Z}$ is equinumerous with $\mathbb{N}$, and the uncountability of the set of all functions $\mathbb{N} \rightarrow \{0,1\}$. Basically: if it involves infinity, it's interesting.

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The game of Nim and its solution are pretty cool. The proof might be a bit difficult, but I think kids would love to learn a game like that and how to beat their parents at it.

There's a lot of other fun mathematical games like that too. But I think the first thing I learned that turned me towards mathematics was the existence of multiple infinities, and things like Hilbert's infinite hotel.

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+1, Hilbert's hotel could be great for a children's book! See en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hotel. –  Will Mar 7 '13 at 16:25

## protected by Zev ChonolesMar 7 '13 at 22:43

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.


text:
# Domain Name Appraisal

Domain name trading is big business. One of the most useful tools for domain name trading is an automatic appraisal tool, so that you can easily estimate how much a given domain is worth. Unfortunately, many automatic appraisal services require a membership/subscription to use. In this challenge, you will write a simple appraisal tool that can roughly estimate the values of .com domains.

## Input / Output

As input, your program should take a list of domain names, one per line. Each domain name will match the regex ^[a-z0-9][a-z0-9-]*[a-z0-9]$, meaning that it is composed of lowercase letters, digits, and hyphens. Each domain is at least two characters long and neither begins nor ends with a hyphen. The .com is omitted from each domain, since it is implied. As an alternative form of input, you can choose to accept a domain name as an array of integers, instead of a string of characters, as long as you specify your desired character-to-integer conversion. Your program should output a list of integers, one per line, which gives the appraised prices of the corresponding domains. ## Internet and Additional Files Your program may have access to additional files, as long as you provide these files as part of your answer. Your program is also allowed to access a dictionary file (a list of valid words, which you don't have to provide). (Edit) I have decided to expand this challenge to allow your program to access the internet. There are a couple restrictions, being that your program cannot look up the prices (or price histories) of any domains, and that it only uses pre-existing services (the latter to cover up some loopholes). The only limit on total size is the answer size limit imposed by SE. ### Example input These are some recently-sold domains. Disclaimer: Although none of these sites seem malicious, I do not know who controls them and thus advise against visiting them. 6d3 buyspydrones arcader counselar ubme 7483688 buy-bikes learningmusicproduction  ### Example Output These numbers are real. 635 31 2000 1 2001 5 160 1  ## Scoring Scoring will be based on "difference of logarithms." For example, if a domain sold for$300 and your program appraised it at $500, your score for that domain is abs(ln(500)-ln(300)) = 0.5108. No domain will have a price less than$1. Your overall score is your average score for the set of domains, with lower scores better.

To get an idea what scores you should expect, simply guessing a constant 36 for the training data below results in a score of about 1.6883. A successful algorithm has a score less than this.

I chose to use logarithms because the values span several orders of magnitude, and the data will be filled with outliers. The use of absolute difference instead of difference squared will help reduce the effect of outliers in scoring. (Also, note that I am using the natural logarithm, not base 2 or base 10.)

## Data Source

I have skimmed a list of over 1,400 recently sold .com domains from Flippa, a domain auction website. This data will make up the training data set. After the submission period is over, I will wait an additional month to create a test data set, with which the submissions will be scored. I might also choose to collect data from other sources to increase the size of the training/test sets.

The training data is available at the following gist. (Disclaimer: Although I have used some simple filtering to remove some blatantly NSFW domains, several might still be contained in this list. Also, I advise against visiting any domain you don't recognize.) The numbers on the right-hand side are the true prices. https://gist.github.com/PhiNotPi/46ca47247fe85f82767c82c820d730b5

Here is a graph of the price distribution of the training data set. The x-axis the the natural log of price, with y-axis being count. Each bar has a width of 0.5. The spikes on the left correspond to $1 and$6 since the source website requires bids to increment at least $5. The test data may have a slightly different distribution. Here is a link to the same graph with a bar width of 0.2. In that graph you can see spikes at$11 and $16. • Just a heads up, using the most common digraphs to score a domain works abysmally! R² ≅ 0 – user15259 Jul 5 '16 at 17:29 • Someone obviously needs to do a neural network answer for this. – user48538 Jul 6 '16 at 17:31 • Can the program link to the internet (say, query google)? Explicitly not for the purpose of looking up prices of course, but for collecting data to use as a characteristic. – Joe Jul 8 '16 at 19:43 • @Joe Sorry I haven't had a chance to respond, but I have decided to allow internet access. – PhiNotPi Jul 11 '16 at 17:52 ## 1 Answer # Perl, 1.38605 I figured I should go ahead and post my own submission, in the hope that it spurs competition. Its score of 1.38605 means that it is typically off by a factor of 3.999 (that was my stopping point). I didn't use any machine learning libraries, just straight up Perl. It does require access to a dictionary; I used the one from here. Please feel free to use some of the numbers/statistics from my program in your own. use strict; my %dict; my$dictname = "dict.txt";
open(my $dfh, '<',$dictname);
while (my $row = <$dfh>) {
chomp $row;$dict{lc $row} = 1; } my$domain = <>;
chomp($domain); my$guess = 1;

if($domain =~ /^[a-z]*$/){
my @bylength = (200000,20001,401,45,45,41,26,26,26,26,26,24);
if(length($domain) < ~~@bylength+2){$guess *= $bylength[length($domain)-2];
} else {
$guess *= 18; } } elsif ($domain =~ /^[0-9]*$/){ my @bylength = (300000,30001,6000,605,50); if(length($domain) < ~~@bylength+2){
$guess *=$bylength[length($domain)-2]; } else {$guess *= 7;
}
} elsif ($domain =~ /^[a-z0-9]*$/){
my @bylength = (52300,523,28);
if(length($domain) < ~~@bylength+2){$guess *= $bylength[length($domain)-2];
} else {
$guess *= 23; } } else { my @bylength = (50000,500,42,32,32,31); if(length($domain) < ~~@bylength+2){
$guess *=$bylength[length($domain)-2]; } else {$guess *= 12;
}
}

my $wordfact = 1; my$leftword = 0;
for(my $i = 1;$i <= length($domain);$i++){
my $word = substr$domain, 0, $i; if(exists($dict{$word})){$leftword = $i; } }$wordfact *= ($leftword/length($domain))**2 * 0.8 + ($leftword/length($domain)) * -0.1 + 0.9;

if($leftword/length($domain) >= 0.8){
$wordfact *= 2.4; } my$rightword = 0;
for(my $i = 1;$i <= length($domain);$i++){
my $word = substr$domain, length($domain)-$i, $i; if(exists($dict{$word})){$rightword = $i; } }$wordfact *= ($rightword/length($domain))**2 * 0.9 + ($rightword/length($domain)) * -0.2 + 1;

$guess *=$wordfact;

my $charfact = 1; my %charfacts = ( i => 1.12, #500 l => 0.84, s => 1.09, a => 0.94, r => 1.03, o => 0.97, c => 1.22, #400 d => 0.88, u => 1.07, t => 0.95, e => 1.08, m => 0.91, #300 p => 1.08, y => 0.92, g => 0.97, ne => 0.56, #100 n => 1.13, z => 0.67, re => 1.30, es => 0.75, ); while(my ($key,$value) = each %charfacts){ if($domain =~ /$key/){$charfact *= $value; } }$guess *= $charfact;$guess = int($guess + 0.5); if($guess <= 0){
$guess = 1; } print$guess;

Here's a graph made by my scoring program, showing a scatter plot of appraisal over actual price and a histogram of the errors. In the scatter plot .:oO@ mean 10, 20, 30, 40, 50 domains at that point, respectively. In the histogram each O represents 16 domains.

The scale is set at 1 character width = e^(1/3).

There are three main steps to this program. The results from each step are multiplied together.

1. Categorization by character class and length. It determines if the domain is all letters, all numbers, letters and numbers, or if it contains a hyphen. It then gives a numerical value determined by the domain's length. I found that there is a strange dip in value around length 5. I suspect this is due to sampling: shorter domains are valuable due to their length (even if the letters are nonsense), while most longer domains tend to be words/phrases. I order to prevent overfitting, I put a restriction in that domains cannot be penalized for being shorter (so length 5 is at least as good as length 6).

2. Evaluation of word content. I use the dictionary to determine the lengths of the left-hand and right-hand words in a domain name. For example, myawesomesite -> my & site -> 2 & 4. Then, I try to do some fitting based on what proportion of the domain name is made up by those words. Low values typically indicate that the domain does not contain a word, contains a pluralized/modified word not in the dictionary, contains a word surrounded by other characters (internal words are not detected, although I tried that with no improvement), or contains a several-word phrase. High values indicate that it is a single word or a likely a two word phrase.

3. Evaluation of character content. I looked for substrings that were contained in lot of domains and which appeared to affect the values of the domain. I believe this is caused by certain types of words being more popular/more attractive, for various reasons. For example, the letter i appeared in about half of the domains (741 of them), and bumps up domain value by about 12% on average. That's not overfitting; there's something real there, that I don't fully understand. The letter l appears in 514 domains and has a factor of 0.84. Some of the less common letters/digraphs, like ne which appeared 125 times and has a really low factor of 0.56, might be overfitting.

In order to improve upon this program, I would probably need to use machine learning of some kind. Also, I could look for relationships between length, word content, and character content to find better ways of combining those separate results into the overall appraisal value.


text:
# Coterminal & Standard Angles

We define an angle as the union of two non-collinear rays that have a common starting point. The two rays are referred to as the arms of the angle and the common starting point is called its vertex.

We can interpret an angle by rotating a ray from one position to another. When we use this interpretation of an angle, the ray to begin with is called the initial side, and the final position of ray is called the terminal side as shown in the figure.

If we rotate the ray in anti-clockwise direction, the angle formed in this way is termed as positive angle. The angle formed by clockwise rotation of ray is termed as negative angle.

## Angle in Standard Position

''An angle whose vertex lies at the origin of a rectangular coordinate system and whose initial side is directed along the positive direction of x-axis of that coordinate system is said to be in standard position.''

Angles shown in the figures below are in standard position.

## Coterminal Angles

''Angles having the same initial and terminal sides are called coterminal angles.''

Let there be an angle \angleAOB with OA as the initial side and OB as the terminal side as shown in the figures below:

Note that all the three angles \angleAOB in the above figures have the same initial and terminal sides. Hence these are coterminal angles.

That is (\theta+2\pi), (\theta+4\pi) and in general \theta+2(k)\pi are all coterminal angles with \theta where k is an integer.

Note: If \theta is an angle measured in degrees, then \theta \pm k(360^\circ) , where k is an integer, is an angle coterminal with \theta.

If \theta is measured in radians, then \theta \pm 2(k)\pi is an angle coterminal with \theta.

##### Example 1:

Locate the following angles in standard position on a rectangular coordinate system.

1. 135^\circ
2. -640^\circ
3. 310^\circ

##### Example 2:

Find two +ve and two -ve angles that are coterminal with 60^\circ.

Solution: We are given that \theta = 60^\circ

Two positive angles coterminal with \theta = 60^\circ are:

\theta + 360^\circ = 60^\circ + 360^\circ = 420^\circ and \theta + 2(360^\circ) = 60^\circ + 720^\circ = 780^\circ

Two negative angles coterminal with \theta = 60^\circ are:

\theta - 360^\circ = 60^\circ - 360^\circ = - 300^\circ and \theta - 2(360^\circ) = 60^\circ - 720^\circ = -660^\circ

##### Example 3:

Find one positive and one negative angle coterminal with (2\pi)/3

Solution: One positive angle coterminal with (2\pi)/3 is:

( 2\pi)/3 + 2\pi = (2\pi+6\pi)/3 = (8\pi)/3

One negative angle coterminal with (2\pi)/3 is:

( 2\pi)/3- 2\pi = (2\pi - 6\pi)/3 = ( - 4\pi)/3

#### Become a member today!

Register (it’s Free)


text:
# If a problem says one number is twice as large as another, which number should I choose as my variable?

Lets say that the first number is a $3$, then the second number would be $3 \cdot 2 = 6$ .
If you decided to use variable we would say that $x = 3$ so $x \cdot 2 = 6$.
So your expressions would be $x$ and $2 x$.


text:
Derivation of Aharonov Bohm effect for Quasiparticles

I've noticed the following:

Observation: Central results in the condensed matter physics rely on Aharonov Bohm-type arguments involving quasiparticles with fractional charge.

However, I can't see how these arguments are rigorous. In order for them to work, we would need a partition function of the form

$$Z[x,A]=\int Dx\,\, e^{\frac{i}{\hbar}S[x,A]},~~~~~~S[x,A]=\int_C(qA+\text{terms without A})$$

There are two issues with demonstrating the above: for one, this is describing a quantum-mechanical system, whereas quasiparticles are naturally mentioned in a field-theoretic context. So we'd need to find a way to reduce a field-theoretic partition function to a particle-partition function:

$$Z[\Psi,A]=\int D\Psi\,\, e^{\frac{i}{\hbar}S[\Psi,A]},~~~\to~~~Z[x,A]=\int Dx\,\, e^{\frac{i}{\hbar}S[x,A]}$$

The second issue would lie in finding the field-theoretic Lagrangian for a quasiparticle, but it seems like that is already known (i.e., for the fractionally charged anyons of the FQHE, just write down the Chern-Simons lagrangian). However, the first step is missing, and in a glaring way.

Last Thing:

An even better alternative, if possible, would be to derive everything in the Hamiltonian picture, taking advantage of the adiabatic theorem. The Hamiltonian for the quasiparticles would then take the form $$H=\sum_q E_q \gamma^\dagger_q \gamma_q$$

Does anyone else find it awfully surprising that no one has tried to prove the Aharonov Bohm effect in this context?

The reason that people do not go for such a "second quantized" description is that these $\gamma_q$ operators must be necessarily highly non-local for fractionalized quasiparticles. And there is typically no controlled way to "derive" such a second-quantized Hamiltonian since the system is usually strongly interacting. However, it does not mean that the fractional braiding phase has not been derived using adiabatic theorem. In fact, it was first shown in this way. Take $\nu=1/m$ Laughlin state as an example. First, following Laughlin, we have the (first-quantized) wavefunction for quasiholes, as a function of the complex coordinate of the quasihole. This was proposed by Laughlin. Then one can simply go and calculate the braiding phase of the quasiholes, as an adiabatic Berry phase. Under certain assumption about the plasma screening which has to be checked numerically, one can establish the AB phase as well as the braiding statisitcs. The original derivation was done in http://journals.aps.org/prl/pdf/10.1103/PhysRevLett.53.722.


text:
# Set up LaTeX

## Installing Latex

$$\LaTeX$$ is a great typesetting system that includes a lot of features that allow to produce scientific documents. Many researchers use Latex to produce their papers and presentations, and many journals require authors to hand in their articles in a TeX format.

Latex is free to use. To use the Latex system, a TeX distribution needs to be installed. Detailed instructions for the different platforms are provided below.

### Windows

Download the file install-tl-windows.exe from here and follow the instructions.

### MacOS

You can install MacTeX from the official website or using Homebrew:

brew cask install mactex

### Linux (Ubuntu-based)

Install it from the terminal using:

sudo apt-get install texlive-latex-extra

After following the instructions, check whether everything worked by checking the output of the following command:

tex --version

This should give an output similar to this one, where version numbers and details will change depending on your platform.

TeX 3.14159265 (TeX Live 2019/Debian)
kpathsea version 6.3.1
There is NO warranty.  Redistribution of this software is
covered by the terms of both the TeX copyright and
the Lesser GNU General Public License.
named COPYING and the TeX source.
Primary author of TeX: D.E. Knuth.

Note that additional packages for Tex Live should be installed through the apt package manager as well (using tlmgr leads to problems due to different versions)

## Installing LyX, a $\LaTeX$ Alternative

### Get LyX

LyX is an open source document processor based on Latex. Download LyX.

### Making LyX Available on the Command Prompt

You have just installed LyX and may need to access LyX from the command line.

#### Windows users

For you to be able to use LyX from the command prompt, follow the steps below.

Warning

Making LyX available via the PATH settings on Windows

We need to update our PATH settings; these settings are a set of directories that Windows uses to “look up” software to startup.

• Right-click on Computer.
• Go to “Properties” and select the tab “Advanced System settings”.
• Choose “Environment Variables” and select Path from the list of system variables.
• Choose Edit.
• Environment variable name: LYX_BIN

• Windows 7 and 8 machines: If you chose the default installation directory, copy and paste the following string without spaces at the start or end:

c:\Program Files (x86)\Lyx

• Windows 10 machines:

• Click New and paste the following string:

c:\Program Files (x86)\Lyx

• Click on OK as often as needed.

#### Mac users

For you to be able to use LyX from the command line, you have to add LyX to your environmental variables:

Warning

Making LyX available via the PATH settings on Mac

• Open the Terminal.
• Type nano ~/.bash_profile.
• Add LyX to the environmental variables:
• Add export LYX_BIN=/Applications/LyX.app/Contents/MacOS/lyx to a new line.
• Save by pressing Ctrl + O and exit by pressing Ctrl + X.
• Relaunch the Terminal. Then type source ~/.bash_profile to bring the new .bash_profile into effect.
• Type $LYX_BIN to check availability. Remember to type the $ before LYX_BIN.

\$LYX_BIN

followed by hitting the Return key.


text:
# orth of b onto a

by darthxepher
Tags: orth
 P: 59 1. The problem statement, all variables and given/known data Show that vector orth of b onto a=b-proj of b onto a is orthogonal to a. I totally don't know where to start :( and I don't know hat orth of b onto a means...
Do you remember what the dot product ($\vec a \cdot \vec b$) meant geometrically?


text:
# How do you balance the chemical equation: Fe+H_2O = Fe_3O_4+H_2?

Apr 28, 2016

$3 F e + 4 {H}_{2} O = F {e}_{3} {O}_{4} + 4 {H}_{2}$

#### Explanation:

The $F {e}_{3}$ in $F {e}_{3} {O}_{4}$ on the right implies that there must be some multiple of $3$ $F e$ on the left side.

Similarly the ${O}_{4}$ on the right side implies that there must be some multiple of $4$ ${H}_{2} O$ on the left side.

$\left(3 m\right) F e + \left(4 n\right) {H}_{2} O = \left(p\right) F {e}_{3} {O}_{4} + \left(q\right) {H}_{2}$

Since
$\textcolor{w h i t e}{\text{XXX}} \left(3 m\right) F e \rightarrow \left(m\right) F {e}_{3} {O}_{4}$ (there's no place else for the $F e$ to go)
and
$\textcolor{w h i t e}{\text{XXX}} \left(4 n\right) {H}_{2} O \rightarrow \left(n\right) F {e}_{3} {O}_{4}$ (there's no place else for the $O$ to go)
$\Rightarrow m = n$

Further
$\textcolor{w h i t e}{\text{XXX}} \left(4 n\right) {H}_{2} O \rightarrow \left(q\right) {H}_{2}$
$\Rightarrow q = 4 n$

Therefore we have
$\textcolor{w h i t e}{\text{XXX}} \left(3 n\right) F e + \left(4 n\right) {H}_{2} O = \left(n\right) F {e}_{3} {O}_{4} + \left(4 n\right) {H}_{2}$

Using the simplest version: $n = 1$ gives
$\textcolor{w h i t e}{\text{XXX}} 3 F e + 4 {H}_{2} O = F {e}_{3} {O}_{4} + 4 {H}_{2}$


text:
# Missing } inserted, yet braces are balanced

Regard the following document:

\documentclass[a4paper,reqno]{amsart}

\usepackage{amssymb,amsmath,amsthm,fullpage}

\begin{document}

For each $k$ we denote by $\lambda_k$ the $k$th symmetric polynomial
in the $x_i$s; that is, we define
$\textstyle\lambda_k := \sum_\mathrm{I\subseteq\mathbb{N}_n such that |I|=k}x_I.$

\end{document}

When I try to compile it, I get:

! Missing } inserted.
<inserted text>
}
l.9 ...\mathrm{$I\subseteq\mathbb{N}_n$ such that $|I|=k$}
x_I.\]

Why? The braces are surely balanced.

(Surrounding the call to \mathrm in a set of braces makes no difference.)

-

The \mathrm macro sets its input in a roman font, but it is still in (display) math mode. So the $s inside \mathrm are confusing TeX. It's as if you are trying to embed math mode within display math mode. I think what you want instead is $\textstyle \lambda_k := \sum_{\text{I\subseteq\N_n such that |I|=k}}x_I.$ But that won't work either until you tell TeX what you mean by \N. So if you insert into the preamble or someplace else before the first use: \newcommand{\N}{\mathbb{N}} your document will successfully compile. A few comments: • Why are you using \textstyle? In display math some symbols like the \sum operator are bigger. Try it without; you might like TeX's default choices. • Whitespace is your friend. When I type displayed equations I put the $and$ on their own lines and indent the equation in the middle. That way the source somewhat resembles the final document in that displayed math mode is "displayed." • You might find this discussion we had on colon-equals interesting: How to typseset$:=\$ correctly?


text:
lilypond-user
[Top][All Lists]

## Re: Two simultaneous mark events

 From: Andrew Bernard Subject: Re: Two simultaneous mark events Date: Sun, 19 May 2019 11:08:52 +1000

Why not just use markup for 'fine'? Position it as you wiil.

Andrew

On Sun, 19 May 2019 at 03:15, Gianmaria Lari <address@hidden> wrote:
I'm often in this situation:

\version "2.21.0"
fine = {
\once \override Score.RehearsalMark #'break-visibility = #'#(#t #t #f)
\once \override Score.RehearsalMark #'self-alignment-X = #RIGHT
\mark \markup \italic  "Fine"
}

\fixed c' {
c d e f \bar "|." \fine \break
\mark A c c c c
}

So, I would like the text "Fine" at the end of the first measure (and in this case at the end of the line) and the rehearsal mark "A" at the beginning of the next measure (and in this case at the beginning of the second line).

reply via email to


text:
# Preface

This tutorial was made solely for the purpose of education and it was designed for students taking Applied Math 0340. It is primarily for students who have some experience using Mathematica. If you have never used Mathematica before and would like to learn more of the basics for this computer algebra system, it is strongly recommended looking at the APMA 0330 tutorial. As a friendly reminder, don't forget to clear variables in use and/or the kernel. The Mathematica commands in this tutorial are all written in bold black font, while Mathematica output is in regular fonts.

Finally, you can copy and paste all commands into your Mathematica notebook, change the parameters, and run them because the tutorial is under the terms of the GNU General Public License (GPL). You, as the user, are free to use the scripts for your needs to learn the Mathematica program, and have the right to distribute and refer to this tutorial, as long as this tutorial is accredited appropriately. The tutorial accompanies the textbook Applied Differential Equations. The Primary Course by Vladimir Dobrushkin, CRC Press, 2015; http://www.crcpress.com/product/isbn/9781439851043

Introduction to Linear Algebra

# Laplace equation in a wedge

Recall that both the real and imaginary parts of an analytic function satisfy Laplace’s equation in two dimension. Suppose the region of interest is defined by the angular wedge W = 0 ≤ θ ≤ α. Consider the analytic function
$f(z) = z^{\pi /\alpha} = r^{\pi /\alpha} \left( \cos \frac{\pi\theta}{\alpha} + {\bf j}\,\sin \frac{\pi\theta}{\alpha} \right) ,$
of which we consider the principal branch. If α = π/m for some integer m, then f(z) is analytic everywhere. If α is an arbitrary real number, then f(z) may have a branch point at the origin, but we may choose the branch cut so that f(z) is still analytic in our region everywhere except at the origin. In fact the function $$f_n (z) = z^{n\pi /\alpha}$$ for any integer n has the same nice properties. Then its real part $$u(t, \theta ) = \Re f(z) = r^{\pi /\alpha} \cos \frac{\pi\theta}{\alpha}$$ and imaginary part $$v(t, \theta ) = \Im f(z) = r^{\pi /\alpha} \sin \frac{\pi\theta}{\alpha}$$ are both solutions of the Laplace's equation:
$\Delta u =0 \qquad\mbox{and}\qquad \Delta v =0$
in the wedge. Thus, the potential in a wedge-shaped region W with opening angle α and conducting boundaries at potential V0 is described by the complex potential
$\Phi (z) = c + {\bf j}\,V_0 + \sum_n a_n z^{n\pi /\alpha} .$
Its real part
$u(r, \theta ) = c + \sum_{n=-\infty}^{\infty} a_n r^{n\pi /\alpha} \cos \frac{\pi\theta}{\alpha}$
and the imaginary part
$v(r, \theta ) = V_0 + \sum_{n=-\infty}^{\infty} a_n r^{n\pi /\alpha} \sin \frac{\pi\theta}{\alpha}$
are solutions of the Laplace's equation subject to the boundary conditions:
$\left. \frac{\partial u}{\partial r} \right\vert_{\theta =0 \ \mbox{ or } \ \alpha} =0 \qquad\mbox{and} \qquad v(r, \theta = 0 \ \mbox{ or } \ \alpha ) = 0.$
The coefficients an must be chosen to satisfy any remaining boundary conditions in r.

Since the origin is included within our wedge-shaped region W, the sum is over positive n only, so that the potential remains finite. Then the potential near the origin (small r) is dominated by the first (n = 1) term,and the field near the origin has components:

$\Delta \Phi (z) = \left( \frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\, \frac{\partial u}{\partial r} + \frac{1}{r^2}\, \frac{\partial^2 u}{\partial \theta^2} \right) \Phi =0.$
D[r^(n*Pi/a)*Cos[n*Pi/a*t] + I*r^(n*Pi/a)*Sin[n*Pi/a*t], r, r] + D[r^(n*Pi/a)*Cos[n*Pi/a*t] + I*r^(n*Pi/a)*Sin[n*Pi/a*t], r] /r + D[r^(n*Pi/a)*Cos[n*Pi/a*t] + I*r^(n*Pi/a)*Sin[n*Pi/a*t], t, t]/r^2
FullSimplify[%]
0
This is true only when $$-1 + \frac{\pi}{\alpha} > 0$$ Otherwise, π < α, the field is unbounded unless the first coefficient is zero.

As one might expect, the behavior of solution near r = 0 has to be restricted:

$v(r,\theta ) = c_0 + c_1 \ln r + o(1) \qquad\mbox{as} \quad r \to 0 ,$
where c0 and c1 are some constants. The constant c0 cannot be chosen arbitrary. (It is analogous to the so-called blockage coefficient in other potential flows.) The above condition on behavior of harmonic function near the corner point is called the wedge condition.

For instance, if we consider the Neumann boundary conditions on the two sides of the wedge,

$\begin{split} \frac{1}{r}\, \frac{\partial u}{\partial \theta} = 0 \qquad\mbox{when} \quad \theta = \alpha , \quad r> 0, \\ \frac{1}{r}\, \frac{\partial u}{\partial \theta} = f(r) \qquad\mbox{when} \quad \theta = 0 , \quad r> 0, \end{split}$
where f is a specified function. Evidently, the solution of the problem, u, is not unique because we can always add c0 + c1 log r, where c0 and c1 are arbitrary constants.

Example: Consider Laplace's equation in a corner:

$\begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < \alpha , \\ u(r,0) =0 , \quad u(r, \alpha ) =0 , \qquad u(a, \theta ) = f(\theta ) , \\ u(r, \theta ) \ \mbox{ is bounded as }\ r \to 0 \qquad ({\bf wedge\ condition}). \end{split}$
w = Graphics[{Orange, Disk[{0, 0}, 1, {0, 35 Degree}]}];
text1 = Graphics[ Text[Style["$CapitalDelta] u = 0", FontSize -> 16], {0.6, 0.2}]]; text2 = Graphics[Text[Style["u = 0", FontSize -> 16], {0.4, 0.4}]]; text3 = Graphics[Text[Style["u = 0", FontSize -> 16], {0.6, -0.05}]]; text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.18, 0.35}]]; txt0 = Graphics[Text[Style["0", FontSize -> 16], {-0.03, 0.0}]]; txtr = Graphics[Text[Style["r = a", FontSize -> 16], {1.1, 0.0}]]; Show[w, text1, text2, text3, text4, txt0, txtr] To solve the given boundary value problem, we apply separation of variables. So we seek partial nontrivial (not identically zero) solutions of the auxiliary problem \[ \begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < \alpha , \\ u(r,0) =0 , \quad u(r, \alpha ) =0 , \\ u(r, \theta ) \ \mbox{ is bounded as }\ r \to 0 \qquad ({\bf wedge\ condition}), \end{split}$
and represented as a product: u(r,θ) = R(r) ⋅ Θ(&theta). Substitution of this product into the Laplace equation yields
$r^2 \frac{R'' + \frac{1}{r}\,R'}{R} = - \frac{\Theta'' (\theta )}{\Theta (\theta )} = \lambda^2 ,$
where we denote by λ² the constant of separation. From the above equations, we derive two differential equations containing a parameter λ:
$\begin{split} r^2 R'' (r) + r\, R' (r) - \lambda^2 R =0 , \\ \Theta'' (\theta ) + \lambda^2 \Theta (\theta ) =0 , \qquad \Theta (\theta ) = \Theta (\theta + 2 \pi ) . \end{split}$
The latter is the Sturm--Liouville problem that is not hard to solve:
$\Theta_n (\theta ) = \sin \left( \frac{n\pi\theta}{\alpha} \right) \qquad \mbox{for} \quad \lambda_n = \frac{n\pi}{\alpha} , \quad n=1,2,\ldots .$
Note that λ = 0 is not an eigenvalue because the corresponding eigenfunction Θ0 = a+ bθ must be identically zero to satisfy the homogeneous boundary condition.

Therefore, we get a discrete sequence of positive eigenvalues

$\lambda_n = \frac{n\pi}{\alpha} , \quad n=1,2,\ldots ,$
that we will use to solve the Euler differential equation for R(r). If we choose R(r) to be a power function R(r) = rk, we get
$R(r) = r^k , \quad r\, R' (r) = k\, r^k , \quad r^2 R'' (r) = k(k-1)\,r^k .$
So we obtain an algebraic equation for k:
$k(k-1) + k - n^2 =0 \qquad \Longrightarrow \qquad k = \pm n.$
Hence, the general solution for the Euler equation $$r^2 R'' + r\, R' - n^2 R =0$$ becomes
$R_n (r) = a_n r^n + b_n r^{-n} , \qquad n=1,2,3,\ldots .$
Since this function must be bounded at the origin (wedge condition), we are forced to set bn = 0. Adding all nontrivial solution, we obtain
$u(r,\theta ) = \sum_{n\ge 1} R_n (t)\, \Theta_n (\theta ) = \sum_{n\ge 1} a_n r^{n\pi /\alpha} \, \sin \left( \frac{n\pi \theta}{\alpha} \right) .$
To satisfy the boundary condition for r = a, we have to determine the values of coefficients an from the equation
$u(a, \theta ) = f(\theta ) = \sum_{n\ge 1} a_n a^{n\pi /\alpha} \, \sin \left( \frac{n\pi \theta}{\alpha} \right) ,$
which is actually the Fourier sine series. Therefore,
$a_n = a^{-n\pi/\alpha} \,\frac{2}{\alpha} \int_0^{\alpha} f(\theta )\, \sin \left( \frac{n\pi \theta}{\alpha} \right) {\text d}\theta , \qquad n=1,2,\ldots .$

Example: Consider the Dirichlet problem for Laplace's equation in a corner:

$\begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < \alpha , \\ u(r,0) =u_0 , \quad u(r, \alpha ) = u_{\alpha} , \qquad u(a, \theta ) = f(\theta ) , \\ u(r, \theta ) \ \mbox{ is bounded as }\ r \to 0 \qquad ({\bf wedge\ condition}), \end{split}$
where u0 and uα are given numbers.
w = Graphics[{Orange, Disk[{0, 0}, 1, {0, 35 Degree}]}];
text1 = Graphics[ Text[Style["$CapitalDelta] u = 0", FontSize -> 16], {0.6, 0.2}]]; text2 = Graphics[ Text[Style[ "u(r,\[Alpha]) = \!$$\*SubscriptBox[\(u$$, $$\[Alpha]$$]\)", FontSize -> 16], {0.3, 0.4}]]; text3 = Graphics[ Text[Style["u(r,0) = \!$$\*SubscriptBox[\(u$$, $$0$$]\)0", FontSize -> 16], {0.6, -0.05}]]; text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.18, 0.35}]]; txt0 = Graphics[Text[Style["0", FontSize -> 16], {-0.03, 0.0}]]; txtr = Graphics[Text[Style["r = a", FontSize -> 16], {1.1, 0.0}]]; Show[w, text1, text2, text3, text4, txt0, txtr] We reduce the given boundary value problem to the problem considered in the previous example with homogeneous boundary conditions by representing the unknown function u(r,θ) as a sum of two functions: \[ u(r, \theta ) = w(r, \theta ) + v(r, \theta ) , \qquad\mbox{where} \quad w(r, \theta ) = \left( u_{\alpha} - u_0 \right) \frac{\theta}{\alpha} + u_0 .$
Actually, instead of w can be used any function that satisfies the prescribed boundary conditions: $$w(r,0) = u_0$$ and $$w(r,\alpha ) = u_{\alpha} .$$ Then for function v(r,θ) we get the following boundary value problem
$\begin{split} v_{rr} + \frac{1}{r}\,v_r + \frac{1}{r^2}\, v_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < \alpha , \\ v(r,0) =0 , \quad v(r, \alpha ) =0 , \qquad v(a, \theta ) = f(\theta ) - w(a,\theta ), \\ v(r, \theta ) \ \mbox{ is bounded as }\ r \to 0 \qquad (\mbox{\bf wedge condition}). \end{split}$
This is essentially the same problem that was solved in the previous example. ■

Example: .

Consider the Laplace equation subject to mixed boundary conditions:
$\begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < \alpha , \\ u_{\theta} (r,0) = 0 , \quad u_{\theta} (r, \alpha ) = 0 , \qquad u(a, \theta ) = f(\theta ) , \\ u(r, \theta ) \ \mbox{ is bounded as }\ r \to 0 \qquad ({\bf wedge\ condition}). \end{split}$
w = Graphics[{Orange, Disk[{0, 0}, 1, {0, 35 Degree}]}];
text1 = Graphics[ Text[Style["$CapitalDelta] u = 0", FontSize -> 16], {0.6, 0.15}]]; text2 = Graphics[ Text[Style[ "\!$$\*SubscriptBox[\(u$$, $$\[Theta]$$]\)(r,\[Alpha]) = 0", FontSize -> 16], {0.3, 0.4}]]; text3 = Graphics[ Text[Style["\!$$\*SubscriptBox[\(u$$, $$\[Theta]$$]\)(r,0) = 0", FontSize -> 16], {0.6, -0.05}]]; text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.18, 0.35}]]; txt0 = Graphics[Text[Style["0", FontSize -> 16], {-0.03, 0.0}]]; txtr = Graphics[Text[Style["r = a", FontSize -> 16], {1.1, 0.0}]]; Show[w, text1, text2, text3, text4, txt0, txtr] Example: Consider the boundary value problem for the Laplace equation in circular domain: \[ \begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad a < r < b, \quad \alpha < \theta < \beta , \\ u_{\theta} (r,0) = 0 , \quad u_{\theta} (r, \alpha ) = 0 , \qquad u(a, \theta ) = g(\theta ) , \quad u(b, \theta ) = f(\theta ) . \end{split}$
w = RegionPlot[ 1/2 <= x^2 + y^2 <= 2 && 0 < x && x < y , {x, 0, 2.1}, {y, 0, 2}, Frame -> False, PlotStyle -> LightOrange];
text1 = Graphics[ Text[Style["$CapitalDelta] u = 0", FontSize -> 16, FontWeight -> "Bold"], {0.4, 1.0}]]; text2 = Graphics[ Text[Style["u(r,\[Beta]) = 0", FontSize -> 16], {-0.3, 0.74}]]; text3 = Graphics[ Text[Style["u(r,\[Alpha]) = 0", FontSize -> 16], {1.0, 0.75}]]; text4 = Graphics[ Text[Style["u(b,\[Theta]) = f(\[Theta])", FontSize -> 16], {0.8, 1.4}]]; text5 = Graphics[ Text[Style["u(a,\[Theta]) = g(\[Theta])", FontSize -> 16], {0.3, 0.45}]]; Show[w, text1, text2, text3, text4, text5] Example: Consider the boundary value problem for Laplace's equation \[ \begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < \pi , \\ u_{\theta} (r,0) = 0 , \quad u (r, \pi ) = 0 , \qquad u(a, \theta ) = f(\theta ) , \end{split}$
w = Graphics[{Orange, Disk[{0, 0}, 1, {0, Pi}]}];
text1 = Graphics[ Text[Style["$CapitalDelta] u = 0", FontSize -> 16, FontWeight -> "Bold"], {-0.05, 0.45}]]; text2 = Graphics[ Text[Style["\!$$\*SubscriptBox[\(u$$, $$\[Theta]$$]\)(r,0) = 0", FontSize -> 16], {0.5, -0.15}]]; text3 = Graphics[ Text[Style["u(r,\[Pi]) = 0", FontSize -> 16], {-0.5, -0.1}]]; text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.0, 0.8}]]; p = Graphics[{PointSize[Medium], Point[{0, 0}]}]; rec = Graphics[{Brown, Rectangle[{1, -0.05}, {0, 0}]}]; Show[w, text1, text2, text3, text4, p, rec] Substituting the assumed form u(r,θ) = R(r) ⋅ Θ(&theta) into Laplace's equation, and separating variables, we obtain two differential equations \[ \begin{split} r^2 R;; (r) + r\,R' (r) - \lambda^2 R(r) =0 , \\ \Theta'' (\theta ) + \lambda^2 \Theta (\theta ) =0 . \end{split}$
Adding the boundary conditions at θ = 0 and θ = π, we get the Sturm--Liouville problem:
$\Theta'' (\theta ) + \lambda^2 \Theta (\theta ) =0 , \qquad \Theta' (0) =0, \quad \Theta (\pi ) =0.$
So we derive the eigenvalues and corresponding eigenfunctions:
$\Theta_n (\theta ) = \cos \left( \frac{\theta \left( 1 + 2n \right)}{2} \right) , \qquad \lambda_n = \frac{\left( 1 + 2n \right)}{2} , \qquad n=0,1,2,\ldots .$
Note that λ = 0 is not an eigenvalue. Then the solution to the given boundary value problem is the sum of all partial nontrivial solutions
$u (r,\theta ) = \sum_{n\ge 0} c_n \,r^(n+1/2) \,\cos \left( \frac{\theta \left( 1 + 2n \right)}{2} \right) .$
To satisfy the boundary condition at r = a, we have to choose coefficients cn so that
$u (a,\theta ) = f(\theta ) = \sum_{n\ge 1} c_n \,a^{n+1/2} \,\cos \left( \frac{\theta \left( 1 + 2n \right)}{2} \right) .$
Since this is just Fourier series over orthogonal set of eigenfunctions, we determine its coefficients as
$c_n = a^{-n-1/2} \,\frac{2}{\pi} \int_0^{\pi} f(\theta )\,\cos \left( \frac{\theta \left( 1 + 2n \right)}{2} \right) {\text d}\theta , \quad n=0,1,2, \ldots .$

Example: Consider the boundary value problem for Laplace's equation

$\begin{split} u_{rr} + \frac{1}{r}\,u_r + \frac{1}{r^2}\, u_{\theta\theta} =0 , \qquad 0 < r < a, \quad 0 < \theta < 2\pi , \\ u_{\theta} (r,0+0) = u_0 , \quad u (r, 2\pi -0 ) = u_1 , \qquad u(a, \theta ) = f(\theta ) , \end{split}$

w = Graphics[{Orange, Disk[{0, 0}, 1, {0.01, 2*Pi - 0.01}]}];
p = Graphics[{PointSize[Medium], Point[{0, 0}]}];
p = Graphics[{PointSize[Medium], Point[{0, 0}]}];
text1 = Graphics[ Text[Style["\[CapitalDelta] u = 0", FontSize -> 16, FontWeight -> "Bold"], {-0.05, 0.45}]];
text2 = Graphics[ Text[Style["u(r,0+0) = \!$$\*SubscriptBox[\(u$$, $$0$$]\)(r)", FontSize -> 16], {0.5, 0.1}]];
text3 = Graphics[ Text[Style["u(r,2\[Pi]-0) = \!$$\*SubscriptBox[\(u$$, $$1$$]\)(r)", FontSize -> 16], {0.5, -0.1}]];
text4 = Graphics[ Text[Style["u(a,\[Theta]) = f(\[Theta])", FontSize -> 16], {1.0, 0.8}]];
Show[w, text1, text2, text3, text4, p, rec]

If the boundary conditions on the crack are the same, u0 = u1, the solution can be obtained from our first two examples by taking the limit $$\lim_{\alpha \to 2\pi} u(r, \theta ) .$$ When they are not the same, the problem becomes very hard to solve. ■

# Laplace equation in a corner

1. Peirce, A., Circular domains, 2018.


text:
Mid-Central USA Programming Contest 2018

#### Start

2018-11-03 17:30 UTC

## Mid-Central USA Programming Contest 2018

#### End

2018-11-03 22:30 UTC
The end is near!
Contest is over.
Not yet started.
Contest is starting in -446 days 17:40:30

5:00:00

0:00:00

# Problem FGoat Rope

You have a fence post located at the point $(x, y)$ in the plane, to which a goat is tethered by a rope. You also have a house, which you model as an axis-aligned rectangle with diagonally opposite corners at the points $(x_1, y_1)$ and $(x_2, y_2)$. You want to pick a length of rope that guarantees the goat cannot reach the house.

Determine the minimum distance from the fence post to the house, so that you can make sure to use a shorter rope.

## Input

The input consists of a single line containing six space-separated integers $x$, $y$, $x_1$, $y_1$, $x_2$, and $y_2$, each in the range $[-999, 999]$.

It is guaranteed that $x_1 < x_2$ and $y_1 < y_2$, and that $(x, y)$ is strictly outside the axis-aligned rectangle with corners at $(x_1, y_1)$ and $(x_2, y_2)$.

## Output

Print the minimum distance from the goat’s post to the house, with a relative or absolute error no more than $0.001$.

Sample Input 1 Sample Output 1
7 3 0 0 5 4

2.0

Sample Input 2 Sample Output 2
6 0 0 2 7 6

2.0

Sample Input 3 Sample Output 3
3 -4 -3 -1 -1 2

5.0


text:
# Excluded Point Space is T5

## Theorem

Let $T = \left({S, \tau_{\bar p}}\right)$ be an excluded point space.

Then $T$ is a $T_5$ space.

## Proof

Let $A, B \subseteq S$ such that:

$A^- \cap B = A \cap B^- = \varnothing$

where $A^-$ denotes the closure of $A$.

Every closed set of $T$ contains $\left\{{p}\right\}$.

From Limit Points in Excluded Point Space and the definition of closure:

$A^- = A \cup \left\{{p}\right\}, B^- = B \cup \left\{{p}\right\}$

So if $A^- \cap B = A \cap B^- = \varnothing$ then $p \in A^-$ and $p \in B^-$.

Thus, $p \notin A$ and $p \notin B$ and so by definition $A$ and $B$ are open sets in $T$.

It follows that $A \cap B = \varnothing$.

Hence the result, as we have shown that:

$\forall A, B \subseteq X, A^- \cap B = A \cap B^- = \varnothing: \exists U, V \in \tau_{\bar p}: A \subseteq U, B \subseteq V, U \cap V = \varnothing$

where $U = A$ and $V = B$.

$\blacksquare$