text:
# Axis Functions

Scaling the X and Y axes of a plot box

### Description

The function $$AxisX$$ defines the range of the X-scale of a plot box.

The function $$AxisY$$ defines the range of the Y-scale of a plot box.

The scaling of the axes of a drawing plane in the plot box is performed automatically. The range of scales is calculated so that the axes slightly exceed the positions of the contained points. Mostly it is rounded up to the next whole value.

For example, if you draw a curve whose Y values are between 0.8 and 4.3, the Y axis is likely to be set from 0 to 5.

For a curve with values from 0 to 4.3, the Y-axis is likely to automatically scale from -1 to 5. To obtain an axis from 0 to 5, the scale can be scaled manually by calling the function AxisY (0,5).

### Syntax

AxisX (min, max)

AxisX (min, max, id)

AxisY (min, max)

AxisY (min, max, id)

### Parameters

min

Sets the minimum value of the scale

max

Sets the maximum value of the scale

### Optional Parameters

id

Id number of the plot box, if more than one box is open on the worksheet.

### Show section of a curve

The range of the scale can also be defined smaller than the range of contained data. This will only show a specific area of a curve.

### Example

The following figure shows the complete range of a BesselK function.

The next figure shows a section of the curve by limiting the X and Y axes.


text:
Point Curry Disco

Visual Search Engine!

# Thue - Morse - Prouhet Sequence

Posted on Sat 29 October 2016 in misc

I have a confession of sorts to make. As a little kid, when I wasn't chasing dragonflies or playing carroms or scribbling my name everywhere, I used to be engrossed in this rather weird game.

### Story Time

I would start by doing something trivial using one of my hands, say my left hand. It could be flicking, tapping, touching something, whatever, something very trivial. I would call this a. I would then do something similar with my right hand and call it b.

So as to compensate the imbalance caused in the universe due to that ghastly act a. There now, the world is at peace with itself. Is it though?

## a b

Balanced but not fair. Discernible folks would've realized by now that it isn't fair on b though right? I mean shouldn't it get a chance to lead the sequence too?

I would therefore, repeat b and then a. We have thus arrived at

## a b b a

Justice seems to have prevailed and balance in the universe is as good as ever. Maybe not, after all. I mean b is kinda relieved with the court order but it still sees some injustice that has escaped the jury's attention. 'Justice of first order schmirst order.' b sneers at you, seething with discontent. Onlookers whisper among themselves 'Utter drivel. Token justice. Pandering to the powerful. Isn't b worthy enough to lead a sequence of length four?'

I would therefore proceed ahead with baab , baab and abba and thus complete

## abba baab baab abba

Normally this is where it ended. gasp Slow Clap time. b is happy now guys. We did it. Look at the smile plastered on it's face. Some of you champions of the oppressed might say b isn't exactly beaming with joy but let's not get greedy and talk about justice of third order. [ abbabaabbaababba baababbaabbabaab baababbaabbabaab abbabaabbaababba ] That would be unfair on me, the attorney who took this up pro bono.

The nomenclature part came later of course. When it started, it was all about balancing it out and being fair. By fair I mean, why does *a have to be chosen first before b ? So ba gets appended to ab* and so on.

I can assure you it didn't look as bad as it sounds. I had perfected this game to an extant that If you had ever seen me doing it, you'd have assumed that I was just tapping my fingers to some rhythm and never would've guessed that I was going OCD there. ( unless of course one of a or b was 'picking my nose' or 'licking my lips' )

# OCDCDODOC

Anyway, this wasn't something that consumed me or interfered with my everyday life, like how it plays out for people with debilitating OCD. And over the course of time, it skipped my mind completely as I developed new avocations like

• Summing the digits on number plates to test divisibility by 9
• Rearranging sentences based on syllable and character counts.
• Replacing soft consonants ( Saralas ) with hard consonants (Parushas) and vice versa. Kattappa --- Gaddabba
• Piecewise semantic conjugation of phrases and sentences. He is interesting --- She was boring
• Forcefully fitting a phrase/word into 5 syllables so that I can count it off my fingers. He yers my Hoo li badge -- Heers my Hoo li badge

I remembered the abba game again during my brief stint at Chennai Mathematical Institute and explored the possibility of the set of these sequences being a group under piece wise addition. Googling abbabaabbaababba to find if someone has worked on this before led me to this sequence, something that was discovered over one fifty years back and the astounding fact that this sequence has a history of being discovered over and over again by many people, including a chess grandmaster!

The Thue–Morse sequence was first studied by Eugène Prouhet in 1851, who applied it to number theory. However, Prouhet did not mention the sequence explicitly; this was left to Axel Thue in 1906, who used it to found the study of combinatorics on words. The sequence was only brought to worldwide attention with the work of Marston Morse in 1921, when he applied it to differential geometry. The sequence has been discovered independently many times, not always by professional research mathematicians; for example, Max Euwe, a chess grandmaster, who held the world championship title from 1935 to 1937, and mathematics teacher, discovered it in 1929 in an application to chess: by using its cube-free property (see above), he showed how to circumvent a rule aimed at preventing infinitely protracted games by declaring repetition of moves a draw.

I know! It made me feel kinda special. It felt as if the forces of evolution chose me to ~~kill John Lennon~~ unravel deep structures in the universe or something to that effect. Thanks to the internet however, I understood there are a good number of people who seem to have experienced this sequence playing in their heads during their childhoods. It's almost as if some people are hardwired to discover this pattern. I'm tempted to attribute it to some kind of OCD gene/mutation whatever but my lack of expertise on that subject prevents me from doing so. Whatever be the case, it is incredibly fascinating to see it manifest like this.

# Equitable Distributions

This sequence, called Thue - Morse sequence or should I say Thue - Morse Morse - Thue sequence has many interesting properties.

It can be used as a quick and dirty, not necessarily optimal, algorithm to divide resources equitably among two people. Consider the case of two captains having to pick players for their respective teams from a pool of players. The teams need to be as evenly balanced as possible in terms of the cumulative strengths of the players. Both the captains have an idea of how good each player from the lot is.

The classical way of each captain choosing alternately, doesn't exactly bode well for the captain who gets to choose second. As you can see, the captain who gets to choose first will end up with the $${1}^{st}, {3}^{rd},{5}^{th}, {7}^{th}....$$ best players of the lot whereas the other captain has to settle with the $${2}^{nd}, {4}^{th},{6}^{th}, {8}^{th}....$$ best of the lot. If the captains agree to choose the players in Thue Morse pattern, in other words, if they take turns in taking turns in taking turns and so on, the distribution will be fairer. In this case the first captain gets the $${1}^{st}, {4}^{th},{6}^{th}, {7}^{th},{10}^{th},....$$ best of the lot and the second one gets the $${2}^{nd}, {3}^{rd},{5}^{th}, {8}^{th},{9}^{th},....$$ best of the lot.

Obviously, this doesn't work well if there are extreme outliers in the player pool. Elementary arithmetic shows that if the strengths of these players vary linearly, Thue - Morse distribution works to perfection.

Say we have a lot of $$8$$ players, whose strengths decrease linearly from $$8$$ to $$1$$ when sorted in decreasing order. It can easily be seen that when the teams are split using Thue - Morse sequence, cumulative strengths of both the teams equal $$18$$

Note that we kept it simple by ignoring the existence of say bonding coefficients between players, you know the kind of thing that makes the sum of parts lesser/greater than the collective whole. We would then be looking at the problem of equi-partitioning a set into two groups with minimal difference between their quantified collective capabilities that are expressed as functions of bonding coefficients.

Other example I could think of

• Two parties taking turns for a titular post, if the magnitude of benefits that can be reaped using power decreases with the passage of time or benefits beget more benefits ( interest on money for example )

## Koch Curve

The Thue - Morse sequence is also connected to the ubiquitous Koch curve ( snowflake curve ). The fun never ends I tell you. You can generate it using the following simple algorithm.

For every element in the Thue - Morse sequence, if the element is $$a$$, move forward by one unit and turn anticlockwise by $$60^{\circ}$$ if the element is $$b$$. As simple as that.

## Generating Thue - Morse Sequences

Time for formalization.

There are many ways of generating the Thue - Morse sequence.

### Mirror method

Let $$f(a,b,n)$$ represent the sequence in my head of $${n}^{th}$$ order. I generated it using the equation $$f(a,b,n) = f(a,b,n-1) + f(b,a,n-1) + f(b,a,n-1) + f(a,b,n-1)$$

where + represents concatenation and $$f(a,b,1) = abba$$

The Thue - Morse sequence, on the other hand, starts with $$a, ab$$ and consists of $$abbabaab$$ between $$abba$$ and $$abbabaabbaababba$$. My sequence turns out to be a sub sequence of Thue - Morse sequence. To be more specific, it comprises all the odd terms starting from $$abba$$.

The right recursive relation that describes Thue - Morse sequence of $${n}^{th}$$ order is thus $$f(a,b,n) = f(a,b,n-1) + f(b,a,n-1)$$

It can be generated as follows in Python.

def f(a,b,n):
if n == 0:
return a
else:
return f(a,b,n-1) + f(b,a,n-1) if n > 1 else a + b

Please note that, in literature, there is no such thing as a Thue - Morse sequence of $${n}^{th}$$ order. There is only one Thue - Morse sequence and is equal to $$\lim_{n\rightarrow \infty }f(a,b,n)$$

### Sum of binary digits

Let $$s(n)$$ denote the sum of digits of binary representation of $$n$$

$$t_n$$, the $${n}^{th}$$ term of $$\lim_{n\rightarrow \infty }f(0,1,n)$$ is then equal to $$s(n)\hspace{1 mm} mod\hspace{1 mm} 2$$

##### Proof by induction

$$f(0,1,0)\hspace{1 mm}=s(0)\hspace{1 mm} mod\hspace{1 mm} 2\hspace{1 mm}=0$$

So the statement is true for $$n\hspace{1 mm}=\hspace{1 mm}0$$

Let's assume it is true for all $$m\hspace{1 mm}<\hspace{1 mm}n$$

Let $$2^k \hspace{1 mm}\leq n\hspace{1 mm}<\hspace{1 mm} 2^{k+1}$$

$$\rightarrow \hspace{5 mm}t_n$$ is the $$n^{th}$$ term of $$f(a,b,k+1)$$

$$\rightarrow \hspace{5 mm}t_n$$ is the $$n^{th}$$ term of $$f(a,b,k)\hspace{1 mm}+\hspace{1 mm}f(b,a,k)$$

$$\rightarrow \hspace{5 mm}t_n$$ is the $${(n-2^{k})}^{th}$$ term of $$f(b,a,k)$$

Since $$f(b,a,k)$$ is the mirror image of $$f(a,b,k)$$, we can conclude

$$t_n \hspace{1 mm}=\hspace{1 mm}(t_{n-2^{k}} + 1)\hspace{1 mm}mod\hspace{1 mm}2$$

By induction we have $$t_{n-2^{k}}\hspace{1 mm}=\hspace{1 mm}s(n-2^{k})\hspace{1 mm}mod\hspace{1 mm}2$$

Also, since $$2^k \hspace{1 mm}\leq n\hspace{1 mm}<\hspace{1 mm} 2^{k+1}$$, we have $$s(n)\hspace{1 mm}=\hspace{1 mm}s(n-2^{k})\hspace{1 mm}+\hspace{1 mm}1$$

If the last step isn't obvious, see how it is true for numbers represented in decimal system.

$$\rightarrow \hspace{5 mm}t_n\hspace{1 mm}=\hspace{1 mm}((s(n)-1)\hspace{1 mm}mod\hspace{1 mm}2\hspace{1 mm}+\hspace{1 mm}1)\hspace{1 mm}mod\hspace{1 mm}2$$

$$\rightarrow \hspace{5 mm}t_n\hspace{1 mm}=\hspace{1 mm}s(n)\hspace{1 mm}mod\hspace{1 mm}2$$

The following Python function returns the first $$n$$ characters of the Thue - Morse sequence using the above method. This is a better function for generating the sequence since it doesn't compute $$f(a,b,0), f(a,b,1), f(a,b,2),..., f(a,b,n-1)$$ for generating $$f(a,b,n)$$

def f(a,b,n):
p = ''.join([str(sum_of_digits(number) % 2) for number in range(n)])
return p.replace('0',a).replace('1',b)

def sum_of_digits(number):
return sum([int(numeral) for numeral in str(bin(number))[2:] ])

### Morphism method

Simply put, a morphism is a structure preserving function on a mathematical object under some particular operation.

In our case, the operation is concatenation denoted by $$+$$. So if $$m(x)$$ is such a structure preserving function under concatenation over strings, we have $$m(x+y)\hspace{1 mm}=\hspace{1 mm}m(x)\hspace{1 mm}+\hspace{1 mm}m(y)\hspace{1 mm}=\hspace{1 mm}m(x)m(y)$$

Define a Thue - Morse morphism $$m$$ such that $$m(a) = a+b=ab, m(b) = b+a=ba$$

$$\rightarrow \hspace{5 mm} m^{2}(a) = m(m(a)) = m(ab) = m(a) + m(b) = ab + ba = abba$$

$$m^{3}(a) = m(m(m(a))) = m(m(ab)) = m(abba) = m(a) + m(b) + m(b) + m(a) = abbabaab$$

and so on. You can see that it is the same as $$f(a,b,3)$$ defined earlier.

Similarly, it can be seen that $$(m^{2}(b)$$ and $$m^{3}(b)$$ are equal to $$f(a,b,2)$$ and $$f(a,b,3)$$ respectively.

We can prove that is indeed the case till death does them apart using induction.

We've just proved that the statement is true for $$n=1, 2,3$$

Assume it is true for $$n$$.

$$m^{n+1}(a)\hspace{1 mm}=\hspace{1 mm}m^{n}(m(a))$$ $$=\hspace{1 mm}m^{n}(ab)$$ $$=\hspace{1 mm}m^{n}(a)+m^{n}(b)$$ $$=\hspace{1 mm}f(a,b,n)+f(b,a,n)$$ $$=\hspace{1 mm}f(a,b,n+1)$$

Similarly, it can be proved that $$m^{n+1}(b) = f(b,a,n+1)$$

Below is the Python code for generating Thue - Morse sequence using this method.

def m(x, a, b):
if len(x) == 1:
if x == a:
return str(a) + str(b)
if x == b:
return str(b) + str(a)
if len(x) > 1:
return ''.join([m(element, a, b) for element in x])

def m_n(x,a,b,n):
m_n = a
for element in range(n):
m_n = m(m_n,a,b)
return m_n

### Recursion

The Thue - Morse sequence can also be generated using the following recurrence relation.

$$t_0\hspace{1 mm}=\hspace{1 mm}0$$

$$t_{2n}\hspace{1 mm}=\hspace{1 mm}t_n$$

$$t_{2n+1}\hspace{1 mm}=\hspace{1 mm}1-t_{2n}$$

Using the recursive relations mentioned above, we can generate the sequence in Python as follows

def tn(n):
if n == 0:
return 0
return tn(n/2) if n % 2 == 0 else 1 - tn(n-1)

def f(a,b,n):
return ''.join([str(element).replace('0',a).replace('1',b)
for element in [tn(index) for index in range(n)]])

### Infinite Product

The $$n^{th}$$ term of Thue - Morse sequence can also be determined by calculating the sign of coefficient of $$x^n$$ in the following infinite product.

### Miscellaneous

Found these formulations on a Facebook group dedicated to Atomic functions, Self-differential equations and Morse -Thue sequences.

Apparently, the first formulation involving Hyper Geometric functions simplifies to the second.

## N - variable Thue - Morse sequence

How about a generalized Thue - Morse sequence comprising $$n$$ variables? I haven't found any literature on the internet concerning this but I suppose a man can dream!

Let's try it out for $$3$$ variables - $$a,b,c$$.

My intuition tells me cyclical permutations might just do the trick. Let's explore that.

###### abc-bca-cab bca-cab-abc cab-abc-bca

How do I know if this is structurally similar to the two variable Thue - Morse sequence apart from the fact that my gut feeling says so?

My sophomoric knowledge of mathematics not withstanding, I can't think of a way to tackle this problem properly. A simple hack however, might just do the job ( sort of ). Remember the fair distribution property of Thue - Morse sequence and how it worked like a charm when the players' competency function was linear? Let's consider the case of picking three balanced teams out of $$9$$ players whose strengths vary from $$9$$ to $$1$$ linearly and see how this sequence fares.

Team 1 comprises $$1^{st}, 6^{th}, 8^{th}$$ best players, Team 2 comprises $$2^{nd}, 4^{th}, 9^{th}$$ best players and Team 3 comprises $$3^{rd}, 5^{th}, 7^{th}$$ best players. So the respective strengths of the three teams are $$9+4+2$$, $$8+6+1$$ and $$7+5+3$$

Voila! There you have it. I can now speak with some confidence that this is indeed a Thue - Morse like sequence of three variables and postulate a general formula for n - variables with I suppose a little less confidence.

Extending the logic, morphism in this case would be $$m(a)\hspace{1 mm}=\hspace{1 mm}abc, m(b)\hspace{1 mm}=\hspace{1 mm}bca, m(c)\hspace{1 mm}=\hspace{1 mm}cab$$

A Thue - Morse like morphism of $$n$$ variables will thus be $$m(a_1)\hspace{1 mm}=\hspace{1 mm}a_1a_2a_3....a_n, m(a_2)\hspace{1 mm}=\hspace{1 mm}a_2a_3....a_na_1,....., m(a_n)\hspace{1 mm}=\hspace{1 mm}a_na_1....a_{n-1}$$

##### Update

It appears that an even more generalized form of Thue - Morse sequence can be generated using a generalized version of the Binary digits' sum method.

The $$n^{th}$$ term of the generalized Thue - Morse sequence $${t_{p,q}}^n$$ can be arrived at as follows

$${t_{p,q}}^n\hspace{1 mm}=\hspace{1 mm}s_{p}(n)\hspace{1 mm}mod\hspace{1 mm}q$$ for $$p ≥ 2$$ and $$q ≥ 1$$, where $$s_p(n)$$ is the sum of of digits of $$n$$ represented in base $$p$$

## References

The ubiquitous Prouhet Thue Morse sequence

Wikipedia

Wolfram Mathworld

An overview of the Thue Morse sequence

The ubiquitous Thue Morse sequence - Talk by Jeffrey Shallit

When Thue Morse meets Koch

Atomic functions, Self-differential equations and Morse -Thue sequences

format: plain


text:
# On the use of Green’s formula for the solution of the Navier-Stokes equations

## Abstract

A general iterative procedure for solving Navier-Stokes system in stream function-vorticity variables  is considered. Using the fundamental solution for Laplace operator we find a boundary condition for vorticity. We define an application such that the vorticity is its fixed point. In order to construct this application we solve a Dirichlet problem for Laplace equation and a boundary value problem for vorticity equation in divergence form. A FEM and a classical BEM are used to accomplish this task. Some numerical experiments are carried out.

## Authors

C.I. Gheorghiu
Tiberiu Popoviciu Institute of Numerical Analysis

## Keywords

Navier-Stokes system; stream function; vorticity; Laplace operator; fundamental solution; iterative solver; FEM; BEM;

### References

See the expanding block below.

## Cite this paper as

C.I. Gheorghiu, On the use of Green’s formula for the solution of the Navier-Stokes equations, Adv. Water Resources, 14 (1991) 113-117
doi: 10.1016/0309-1708(91)90002-6

?

Elsevier

0309-1708

1573-1634

?

?


text:
# Sum of four squares modulo a prime

With $p$ a prime integer, I am interested in showing the existence of four non-negative integers $\alpha_0,\alpha_1,\alpha_2,\alpha_3$, not all zero, such that $$\alpha_0^2+\alpha_1^2+\alpha_2^2+\alpha_3^2\equiv 0\mod p.$$ I am, of course, aware of the Lagrange four square theorem of which this is a special case. However, the fact that I am concerned with only prime $p$, and also that I have the looser modulo condition, leads me to hope that there is a simpler, more direct way of proving this.

My thoughts so far have been directed towards trying to tie quadratic residues into some sort of pigeonhole principle argument. That hasn't panned out. This question is motivated by a (double-star) problem in Herstein's Topics in Algebra. Ideally I am looking for hints rather than anything fleshed out. Thanks in advance.

-
Just a quick idea: If $p \equiv 1 \pmod{4}$, let $i$ be the element such that $i^2 \equiv - 1\pmod{p}$. Then, you can take $\alpha_0 = \alpha_1 = 1$ and $\alpha_2 = \alpha_3 = i$. – JavaMan Nov 21 '12 at 5:31

If $p$ is an odd prime, then there are numbers $x,y$ such that $$1 + x^2 + y^2 = m p,$$ where $0 < m < p.$
Note that Theorem 86 on the same page says you may take $y = 0$ if $p = 4k+1.$
I would characterize the proof, roughly, as counting the squares $\pmod p,$ but care is needed.
Sorry, my first iteration of that comment was worded really clumsily. Your post pointed me in exactly the right direction. The problem in Topics in Algebra is 3.4.13b in my edition, showing quaternions over $\mathbb{Z}/p\mathbb{Z}$ are not a division ring. – Jonathan Nov 21 '12 at 6:59


text:
# American Institute of Mathematical Sciences

• Previous Article
Asymptotic behavior of non-autonomous random Ginzburg-Landau equation driven by colored noise
• DCDS-B Home
• This Issue
• Next Article
Threshold dynamics of a general delayed within-host viral infection model with humoral immunity and two modes of virus transmission

## On existence and uniqueness properties for solutions of stochastic fixed point equations

 1 ETH Zurich, Department of Mathematics, Rämistrasse 101, 8092 Zürich, Switzerland 2 University of St. Gallen, Faculty of Mathematics and Statistics, Dufourstrasse 50, 9000 St. Gallen, Switzerland 3 LMU Munich, Department of Mathematics, Theresienstraße 39, 80333 München, Germany 4 University of Duisburg-Essen, Faculty of Mathematics, Thea-Leymann-Straße 9, 45127 Essen, Germany 5 University of Münster, Faculty of Mathematics and Computer Science, Einsteinstraße 62, 48149 Münster, Germany

Received  April 2020 Published  November 2020

The Feynman–Kac formula implies that every suitable classical solution of a semilinear Kolmogorov partial differential equation (PDE) is also a solution of a certain stochastic fixed point equation (SFPE). In this article we study such and related SFPEs. In particular, the main result of this work proves existence of unique solutions of certain SFPEs in a general setting. As an application of this main result we establish the existence of unique solutions of SFPEs associated with semilinear Kolmogorov PDEs with Lipschitz continuous nonlinearities even in the case where the associated semilinear Kolmogorov PDE does not possess a classical solution.

Citation: Christian Beck, Lukas Gonon, Martin Hutzenthaler, Arnulf Jentzen. On existence and uniqueness properties for solutions of stochastic fixed point equations. Discrete & Continuous Dynamical Systems - B, doi: 10.3934/dcdsb.2020320
##### References:
 [1] C. Beck, F. Hornung, M. Hutzenthaler, A. Jentzen and T. Kruse, Overcoming the curse of dimensionality in the numerical approximation of Allen–Cahn partial differential equations via truncated full-history recursive multilevel Picard approximations, J. Numer. Math., 28 (2020), 197-222. doi: 10.1515/jnma-2019-0074.  Google Scholar [2] C. Beck, M. Hutzenthaler and A. Jentzen, On nonlinear Feynman–Kac formulas for viscosity solutions of semilinear parabolic partial differential equations, preprint, 54 pages, arXiv: 2004.03389. Google Scholar [3] C. Bender, N. Schweizer and J. Zhuo, A primal-dual algorithm for BSDEs, Math. Finance, 27 (2017), 866-901.  doi: 10.1111/mafi.12100.  Google Scholar [4] C. Burgard and M. Kjaer, Partial differential equation representations of derivatives with bilateral counterparty risk and funding costs, The Journal of Credit Risk, 7 (2011), 1-19.  doi: 10.21314/JCR.2011.131.  Google Scholar [5] S. Crépey, R. Gerboud, Z. Grbac and N. Ngor, Counterparty risk and funding: the four wings of the TVA, Int. J. Theor. Appl. Finance, 16 (2013), 31 pages. doi: 10.1142/S0219024913500064.  Google Scholar [6] D. Duffie, M. Schroder and C. Skiadas, Recursive valuation of defaultable securities and the timing of resolution of uncertainty, Ann. Appl. Probab., 6 (1996), 1075-1090.  doi: 10.1214/aoap/1035463324.  Google Scholar [7] W. E, M. Hutzenthaler, A. Jentzen and T. Kruse, Multilevel Picard iterations for solving smooth semilinear parabolic heat equations, preprint, 19 pages, arXiv: 1607.03295. Google Scholar [8] W. E, M. Hutzenthaler, A. Jentzen and T. Kruse, On multilevel Picard numerical approximations for high-dimensional nonlinear parabolic partial differential equations and high-dimensional nonlinear backward stochastic differential equations, J. Sci. Comput., 79 (2019), 1534-1571.  doi: 10.1007/s10915-018-00903-0.  Google Scholar [9] P. Grohs, F. Hornung, A. Jentzen and P. von Wurstemberger, A proof that artificial neural networks overcome the curse of dimensionality in the numerical approximation of Black-Scholes partial differential equations, preprint, 124 pages, arXiv: 1809.02362. Google Scholar [10] I. Gyöngy and N. Krylov, Existence of strong solutions for Itô's stochastic equations via approximations, Probab. Theory Related Fields, 105 (1996), 143-158.  doi: 10.1007/BF01203833.  Google Scholar [11] M. Hairer, M. Hutzenthaler and A. Jentzen, Loss of regularity for Kolmogorov equations, Ann. Probab., 43 (2015), 468-527.  doi: 10.1214/13-AOP838.  Google Scholar [12] P. Henry-Labordère, Counterparty risk valuation: A marked branching diffusion approach, preprint, 17 pages, arXiv: 1203.2369. Google Scholar [13] M. Hutzenthaler, A. Jentzen, T. Kruse, T. A. Nguyen and P. von Wurstemberger, Overcoming the curse of dimensionality in the numerical approximation of semilinear parabolic partial differential equations, Accepted by Proc. Roy. Soc. London A, 30 pages, arXiv: 1807.01212. Google Scholar [14] M. Hutzenthaler, A. Jentzen and P. von Wurstemberger, Overcoming the curse of dimensionality in the approximative pricing of financial derivatives with default risks, Electronic Journal of Probability, 25 (2020), 73 pages, https://doi.org/10.1214/20-EJP423. doi: 10.1007/s13253-019-00378-y.  Google Scholar [15] M. Hutzenthaler and T. Kruse, Multilevel Picard approximations of high-dimensional semilinear parabolic differential equations with gradient-dependent nonlinearities, SIAM J. Numer. Anal., 58 (2020), 929-961.  doi: 10.1137/17M1157015.  Google Scholar [16] A. Kalinin, Markovian integral equations, Ann. Inst. Henri Poincaré Probab. Stat. 56, 1 (2020), 155–174. doi: 10.1214/19-AIHP958.  Google Scholar [17] O. Kallenberg, Foundations of Modern Probability, 2$^nd$ edition, Springer-Verlag, New York, 2002. doi: 10.1007/978-1-4757-4015-8.  Google Scholar [18] I. Karatzas and S. E. Shreve, Brownian Motion and Stochastic Calculus, 2$^nd$ edition, Springer-Verlag, New York, 1991. doi: 10.1007/978-1-4612-0949-2.  Google Scholar [19] S. Lang, Fundamentals of Differential Geometry, Springer-Verlag, New York, 1999. doi: 10.1007/978-1-4612-0541-8.  Google Scholar [20] W. Liu and M. Röckner, Stochastic Partial Differential Equations: An Introduction, Springer, Cham, 2015. doi: 10.1007/978-3-319-22354-4.  Google Scholar [21] A. Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations, Springer-Verlag, New York, 1983. doi: 10.1007/978-1-4612-5561-1.  Google Scholar [22] W. Rudin, Real and Complex Analysis, 3$^rd$ edition, McGraw-Hill Book Co., New York, 1987.  Google Scholar [23] I. Segal, Non-linear semi-groups, Ann. of Math., 78 (1963), 339-364.  doi: 10.2307/1970347.  Google Scholar [24] D. W. Stroock, Lectures on Topics in Stochastic Differential Equations, vol. 68 of Tata Institute of Fundamental Research Lectures on Mathematics and Physics, Tata Institute of Fundamental Research, Bombay; by Springer-Verlag, Berlin-New York, 1982, with notes by Satyajit Karmakar.  Google Scholar [25] F. B. Weissler, Semilinear evolution equations in Banach spaces, J. Functional Analysis, 32 (1979), 277-296.  doi: 10.1016/0022-1236(79)90040-5.  Google Scholar

show all references

##### References:
 [1] C. Beck, F. Hornung, M. Hutzenthaler, A. Jentzen and T. Kruse, Overcoming the curse of dimensionality in the numerical approximation of Allen–Cahn partial differential equations via truncated full-history recursive multilevel Picard approximations, J. Numer. Math., 28 (2020), 197-222. doi: 10.1515/jnma-2019-0074.  Google Scholar [2] C. Beck, M. Hutzenthaler and A. Jentzen, On nonlinear Feynman–Kac formulas for viscosity solutions of semilinear parabolic partial differential equations, preprint, 54 pages, arXiv: 2004.03389. Google Scholar [3] C. Bender, N. Schweizer and J. Zhuo, A primal-dual algorithm for BSDEs, Math. Finance, 27 (2017), 866-901.  doi: 10.1111/mafi.12100.  Google Scholar [4] C. Burgard and M. Kjaer, Partial differential equation representations of derivatives with bilateral counterparty risk and funding costs, The Journal of Credit Risk, 7 (2011), 1-19.  doi: 10.21314/JCR.2011.131.  Google Scholar [5] S. Crépey, R. Gerboud, Z. Grbac and N. Ngor, Counterparty risk and funding: the four wings of the TVA, Int. J. Theor. Appl. Finance, 16 (2013), 31 pages. doi: 10.1142/S0219024913500064.  Google Scholar [6] D. Duffie, M. Schroder and C. Skiadas, Recursive valuation of defaultable securities and the timing of resolution of uncertainty, Ann. Appl. Probab., 6 (1996), 1075-1090.  doi: 10.1214/aoap/1035463324.  Google Scholar [7] W. E, M. Hutzenthaler, A. Jentzen and T. Kruse, Multilevel Picard iterations for solving smooth semilinear parabolic heat equations, preprint, 19 pages, arXiv: 1607.03295. Google Scholar [8] W. E, M. Hutzenthaler, A. Jentzen and T. Kruse, On multilevel Picard numerical approximations for high-dimensional nonlinear parabolic partial differential equations and high-dimensional nonlinear backward stochastic differential equations, J. Sci. Comput., 79 (2019), 1534-1571.  doi: 10.1007/s10915-018-00903-0.  Google Scholar [9] P. Grohs, F. Hornung, A. Jentzen and P. von Wurstemberger, A proof that artificial neural networks overcome the curse of dimensionality in the numerical approximation of Black-Scholes partial differential equations, preprint, 124 pages, arXiv: 1809.02362. Google Scholar [10] I. Gyöngy and N. Krylov, Existence of strong solutions for Itô's stochastic equations via approximations, Probab. Theory Related Fields, 105 (1996), 143-158.  doi: 10.1007/BF01203833.  Google Scholar [11] M. Hairer, M. Hutzenthaler and A. Jentzen, Loss of regularity for Kolmogorov equations, Ann. Probab., 43 (2015), 468-527.  doi: 10.1214/13-AOP838.  Google Scholar [12] P. Henry-Labordère, Counterparty risk valuation: A marked branching diffusion approach, preprint, 17 pages, arXiv: 1203.2369. Google Scholar [13] M. Hutzenthaler, A. Jentzen, T. Kruse, T. A. Nguyen and P. von Wurstemberger, Overcoming the curse of dimensionality in the numerical approximation of semilinear parabolic partial differential equations, Accepted by Proc. Roy. Soc. London A, 30 pages, arXiv: 1807.01212. Google Scholar [14] M. Hutzenthaler, A. Jentzen and P. von Wurstemberger, Overcoming the curse of dimensionality in the approximative pricing of financial derivatives with default risks, Electronic Journal of Probability, 25 (2020), 73 pages, https://doi.org/10.1214/20-EJP423. doi: 10.1007/s13253-019-00378-y.  Google Scholar [15] M. Hutzenthaler and T. Kruse, Multilevel Picard approximations of high-dimensional semilinear parabolic differential equations with gradient-dependent nonlinearities, SIAM J. Numer. Anal., 58 (2020), 929-961.  doi: 10.1137/17M1157015.  Google Scholar [16] A. Kalinin, Markovian integral equations, Ann. Inst. Henri Poincaré Probab. Stat. 56, 1 (2020), 155–174. doi: 10.1214/19-AIHP958.  Google Scholar [17] O. Kallenberg, Foundations of Modern Probability, 2$^nd$ edition, Springer-Verlag, New York, 2002. doi: 10.1007/978-1-4757-4015-8.  Google Scholar [18] I. Karatzas and S. E. Shreve, Brownian Motion and Stochastic Calculus, 2$^nd$ edition, Springer-Verlag, New York, 1991. doi: 10.1007/978-1-4612-0949-2.  Google Scholar [19] S. Lang, Fundamentals of Differential Geometry, Springer-Verlag, New York, 1999. doi: 10.1007/978-1-4612-0541-8.  Google Scholar [20] W. Liu and M. Röckner, Stochastic Partial Differential Equations: An Introduction, Springer, Cham, 2015. doi: 10.1007/978-3-319-22354-4.  Google Scholar [21] A. Pazy, Semigroups of Linear Operators and Applications to Partial Differential Equations, Springer-Verlag, New York, 1983. doi: 10.1007/978-1-4612-5561-1.  Google Scholar [22] W. Rudin, Real and Complex Analysis, 3$^rd$ edition, McGraw-Hill Book Co., New York, 1987.  Google Scholar [23] I. Segal, Non-linear semi-groups, Ann. of Math., 78 (1963), 339-364.  doi: 10.2307/1970347.  Google Scholar [24] D. W. Stroock, Lectures on Topics in Stochastic Differential Equations, vol. 68 of Tata Institute of Fundamental Research Lectures on Mathematics and Physics, Tata Institute of Fundamental Research, Bombay; by Springer-Verlag, Berlin-New York, 1982, with notes by Satyajit Karmakar.  Google Scholar [25] F. B. Weissler, Semilinear evolution equations in Banach spaces, J. Functional Analysis, 32 (1979), 277-296.  doi: 10.1016/0022-1236(79)90040-5.  Google Scholar
 [1] Lorenzo Zambotti. A brief and personal history of stochastic partial differential equations. Discrete & Continuous Dynamical Systems - A, 2021, 41 (1) : 471-487. doi: 10.3934/dcds.2020264 [2] Yueyang Zheng, Jingtao Shi. A stackelberg game of backward stochastic differential equations with partial information. Mathematical Control & Related Fields, 2020  doi: 10.3934/mcrf.2020047 [3] Bixiang Wang. Mean-square random invariant manifolds for stochastic differential equations. Discrete & Continuous Dynamical Systems - A, 2021, 41 (3) : 1449-1468. doi: 10.3934/dcds.2020324 [4] Xuhui Peng, Rangrang Zhang. Approximations of stochastic 3D tamed Navier-Stokes equations. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5337-5365. doi: 10.3934/cpaa.2020241 [5] Fathalla A. Rihan, Hebatallah J. Alsakaji. Stochastic delay differential equations of three-species prey-predator system with cooperation among prey species. Discrete & Continuous Dynamical Systems - S, 2020  doi: 10.3934/dcdss.2020468 [6] Guangjun Shen, Xueying Wu, Xiuwei Yin. Stabilization of stochastic differential equations driven by G-Lévy process with discrete-time feedback control. Discrete & Continuous Dynamical Systems - B, 2021, 26 (2) : 755-774. doi: 10.3934/dcdsb.2020133 [7] Tetsuya Ishiwata, Young Chol Yang. Numerical and mathematical analysis of blow-up problems for a stochastic differential equation. Discrete & Continuous Dynamical Systems - S, 2021, 14 (3) : 909-918. doi: 10.3934/dcdss.2020391 [8] Zhenzhen Wang, Tianshou Zhou. Asymptotic behaviors and stochastic traveling waves in stochastic Fisher-KPP equations. Discrete & Continuous Dynamical Systems - B, 2020  doi: 10.3934/dcdsb.2020323 [9] Wenqiang Zhao, Yijin Zhang. High-order Wong-Zakai approximations for non-autonomous stochastic $p$-Laplacian equations on $\mathbb{R}^N$. Communications on Pure & Applied Analysis, 2021, 20 (1) : 243-280. doi: 10.3934/cpaa.2020265 [10] Siyang Cai, Yongmei Cai, Xuerong Mao. A stochastic differential equation SIS epidemic model with regime switching. Discrete & Continuous Dynamical Systems - B, 2020  doi: 10.3934/dcdsb.2020317 [11] Fabio Camilli, Giulia Cavagnari, Raul De Maio, Benedetto Piccoli. Superposition principle and schemes for measure differential equations. Kinetic & Related Models, 2021, 14 (1) : 89-113. doi: 10.3934/krm.2020050 [12] Yangrong Li, Shuang Yang, Qiangheng Zhang. Odd random attractors for stochastic non-autonomous Kuramoto-Sivashinsky equations without dissipation. Electronic Research Archive, 2020, 28 (4) : 1529-1544. doi: 10.3934/era.2020080 [13] Pengyu Chen. Non-autonomous stochastic evolution equations with nonlinear noise and nonlocal conditions governed by noncompact evolution families. Discrete & Continuous Dynamical Systems - A, 2020  doi: 10.3934/dcds.2020383 [14] Lin Shi, Xuemin Wang, Dingshi Li. Limiting behavior of non-autonomous stochastic reaction-diffusion equations with colored noise on unbounded thin domains. Communications on Pure & Applied Analysis, 2020, 19 (12) : 5367-5386. doi: 10.3934/cpaa.2020242 [15] Nicolas Dirr, Hubertus Grillmeier, Günther Grün. On stochastic porous-medium equations with critical-growth conservative multiplicative noise. Discrete & Continuous Dynamical Systems - A, 2020  doi: 10.3934/dcds.2020388 [16] Chungang Shi, Wei Wang, Dafeng Chen. Weak time discretization for slow-fast stochastic reaction-diffusion equations. Discrete & Continuous Dynamical Systems - B, 2021  doi: 10.3934/dcdsb.2021019 [17] Pengyu Chen, Yongxiang Li, Xuping Zhang. Cauchy problem for stochastic non-autonomous evolution equations governed by noncompact evolution families. Discrete & Continuous Dynamical Systems - B, 2021, 26 (3) : 1531-1547. doi: 10.3934/dcdsb.2020171 [18] Qingfeng Zhu, Yufeng Shi. Nonzero-sum differential game of backward doubly stochastic systems with delay and applications. Mathematical Control & Related Fields, 2021, 11 (1) : 73-94. doi: 10.3934/mcrf.2020028 [19] Stefan Ruschel, Serhiy Yanchuk. The spectrum of delay differential equations with multiple hierarchical large delays. Discrete & Continuous Dynamical Systems - S, 2021, 14 (1) : 151-175. doi: 10.3934/dcdss.2020321 [20] John Mallet-Paret, Roger D. Nussbaum. Asymptotic homogenization for delay-differential equations and a question of analyticity. Discrete & Continuous Dynamical Systems - A, 2020, 40 (6) : 3789-3812. doi: 10.3934/dcds.2020044

2019 Impact Factor: 1.27

Article outline


text:
# A particle of mass

Question:

A particle of mass $9.1 \times 10^{-31} \mathrm{~kg}$ travels in a medium with a speed of $10^{6} \mathrm{~m} / \mathrm{s}$ and a photon of a radiation of linear momentum $10^{-27} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$ travels in vacuum. The wavelength of photon is times the wavelength of the particle.

Solution:

For photon $\lambda_{1}=\frac{\mathrm{h}}{\mathrm{P}}=\frac{6.6 \times 10^{-34}}{10^{-27}}$

For particle $\lambda_{2}=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.6 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^{6}}$

$\therefore \frac{\lambda_{1}}{\lambda_{2}}=910$

#### Leave a comment

None
Free Study Material


text:
## Multi-directional search with goal switching for robot path planning

• We present a parallel path planning method that is able to automatically handle multiple goal configurations as input. There are two basic approaches, goal switching and bi-directional search, which are combined in the end. Goal switching dynamically selects a fa-vourite goal depending on some distance function. The bi-directional search supports the backward search direction from the goal to the start configuration, which is probably faster. The multi-directional search with goal switching combines the advantages of goal switching and bi-directional search. Altogether, the planning system is enabled to select one of the pref-erable goal configuration by itself. All concepts are experimentally validated for a set of benchmark problems consisting of an industrial robot arm with six degrees of freedom in a 3D environment.

$Rev: 13581$


text:
protroll's blog

By protroll, history, 5 weeks ago, ,

Hi Codeforces Community! I have been struggling with a problem that I find quite hard. Unfortunately, I don't have a link to the English version. I will try to make a short description of it.

You have 2 strings A and B, of length N and M. The strings contain only lowercase Latin letters. For every substring S of B, you have to find the length of the longest common prefix of A and S. Print the sum of these lengths.

1 <= N, M <= 2 000 000
Time Limit: 1.5 seconds
Memory Limit: 32768 kbytes

Examples:

If A = aa, B = abaa, then the answer is 8. If A = aab, B = acaabada, then the answer is 32.

I tried an approach with hashes and binary searching, but I got TLE. Is there any faster way to solve this problem? Thanks in advance.

P.S.: I will explain my approach, if needed.

Hope you'll have a great day, Protroll

• +6

 » 5 weeks ago, # |   0 I think it can be done in linear time using this.
•  » » 5 weeks ago, # ^ |   0 I know the KMP algorithm. I tried an approach with it, but I got TLE as well. Could you give some details?
•  » » » 5 weeks ago, # ^ | ← Rev. 2 →   0 Suppose you have a string B = "abcxxx" and A = "abc"So the match is at index 0. Now starting with index 0 you have 6 substrings and max prefix match is 3.Notice now that after index 2 there won't be any increase in prefix matched. So the answer is (1,2,3,3,3,3) = (3 * 4) / 2 + (3)*3It is easy to see that this is just summing 1...p, where p is the max prefix matched at i and then (n-p-i) which are left, will just have p.So for index i the answer is : (p * (p+1)) / 2 + (n-p-i) * (p)
•  » » » » 5 weeks ago, # ^ |   0 Thanks, But I already found the formula. You need to do this for EVERY INDEX. So, how do you find p, for some i? In order to do that, I used hashes and binary searching. Your formula refers to a fixed i.
•  » » » » » 5 weeks ago, # ^ |   0 Yes do it for every i. p at i can be found using prefix function.
•  » » » » » » 5 weeks ago, # ^ |   0 I did exactly as you said. Here is the code: https://pastebin.com/hSWg2AsR. This gives TLE
 » 5 weeks ago, # |   0 Can you provide the problem link?
•  » » 5 weeks ago, # ^ |   0 As I said in the post, I don't have a link to an English version of it, but here is the original problem link: Http://varena.ro/problema/ultron You can put it into Google Translate.
 » 5 weeks ago, # | ← Rev. 4 →   +1 This problem can be solved with hashing and binary searchThe following is actually the solution to a harder version were for each substring s of B find the longest prefix of s in A. Then print the sum of all these values.This can be solved using suffix array or tree.1)Concatenate the two strings with \$ sign between them.2) find lcp for every two adjacent pair.(This can be found in o(|a| + |b| + 1) for the whole array)3) then you find the answer for each right and do the needed calculations depending on the values found.(you need to use next array)This is an outline of the solution there are some details not mentioned. complexity O(nlogn)
•  » » 5 weeks ago, # ^ |   0 Thanks, but I don't really understand your solution. Maybe you can show those details? Btw, I tried with hashing and binary searching, but I got TLE.
•  » » » 5 weeks ago, # ^ |   0 did you do cumlitive sum on the hash?
•  » » » » 5 weeks ago, # ^ |   0 What is that? I never heard of it
•  » » » » » 5 weeks ago, # ^ |   0 when you do binary search on the prefix you need to calculate the hash of b which you can do easily since you are moving in increasing index. Now, the problem appears when you need to find the hash of the prefix of A.Since you want the hash of a prefix you can just precalculate the hash of every prefix ,and use it to fint he hash of any reange.This explains it. Rolling Hashing
•  » » » » » » 5 weeks ago, # ^ |   0 Oh yeah. I used them, but still got TLE, due to long long operations. My approach is O(NlogN).
•  » » » » » » » 5 weeks ago, # ^ |   +4 Ok I'm stupid.This can be solved in O(n) using z-algorithm.I got a 100, here is the code: https://ideone.com/fQXziaI didn't realize the solution sooner because i kept on thinking on the harder version.
•  » » » » » » » » 5 weeks ago, # ^ |   0 Never thinked of it. Thanks a lot!
 » 5 weeks ago, # |   +4 I think this problem can be solved with Z-algorithm.If we create a string $C$, where $C = A#B$, then run Z-algorithm on this string, which takes $O(n)$ time, we will get for each position $i$ in it, the longest substring of $C$ which is also a prefix of $C$.Here, if we consider the positions after $'#'$, we will get, for each position $j$ of $B$, the longest substring of $B$ which is also a prefix of $A$, let this value be $len_j$.For each substring of $B$, that starts at $j$ and ends at or after $j+len_j-1$, we know the answer, that is $len_j$, so we add $len_j * NumberOfSuchSubstrings$ to the total answer.For each substring of $B$, that starts at $j$ and ends befor $j+len_j-1$, the answer is the length of that substring, so we add $len_j*(len_j-1)/2$ to the total answer.Now, I see a comment which mentioned Z-algorithm to solve this problem before, but I'll let this comment for more details.
•  » » 5 weeks ago, # ^ |   +6 Thanks, but now I know it thanks to Jester.


text:
# How to draw vertical spiral using TiKZ?

I want to draw the following diagram :

I have tried to draw the diagram using the following command :

\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\draw[dotted] (-4,-3) grid (4,7);
\draw[thick] (0,0) ellipse (2cm and 0.5cm);
\draw[thick,->] (0,1) to [in=1, out=1](2,1.5)--(-2,2)--(2,2.5)--(-2,3)--(2,3.5)--(0,4);
\end{tikzpicture}
\end{document}

This yields :
But it does not give the actual diagram which I want. How can I draw the required diagram using TiKZ?

• Search the site for spiral or coil Oct 2, 2014 at 12:14

Only new thing is the use of bending library to have bent arrow head.

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations.pathmorphing}
\usetikzlibrary{decorations.markings}
\usetikzlibrary{arrows.meta,bending}
\begin{document}
\begin{tikzpicture}
\draw[dotted] (-4,-3) grid (4,7);
\draw[thick] (0,0) ellipse (2cm and 0.5cm);
\draw[thick,decoration={aspect=0.31, segment length=7mm,
amplitude=2cm,coil},decorate,arrows = {<[bend]-}] (0,4) --(0,1);
\node[draw,fill=white,circle,inner sep=1pt] at (0,1){};
\end{tikzpicture}
\end{document}

An easy way to plot a spiral is to use parametric equations, i.e.,

t = [0..2*n*pi]
x = sin(t)
y = cos(t)
z = t

Very naively speaking, in this form, n indicated the number of spiral rounds and x,y coordinate is actually a circle. By providing some coefficient to the equation one can make conical spirals or generate clockwise/counter-clockwise spirals. I will leave the study of math to you. Here is a basic starting point that you can modify it meet your requirements.

\documentclass{standalone}
\usepackage{pgfplots}
\begin{document}
\begin{tikzpicture}
\begin{axis} [
view={0}{30},
axis lines=none,
ymin=-2,
ymax=5,
xmin=-2,
xmax=2]

\addplot3 [thick, ->, blue, domain=3:7*pi, samples = 100, samples y=0] ({sin(deg(-x))}, {cos(deg(-x))}, {x});
\addplot3 [thick, red , domain=0:2*pi, samples = 100, samples y=0] ({sin(deg(x))}, {cos(deg(x))}, -3);
\addplot3 [thick, only marks, blue, mark=o] ({sin(deg(-3))}, {cos(deg(-3)}, {3});
\end{axis}
\end{tikzpicture}
\end{document}

By playing around the gains, parameters and the axis view, you can generate what you want.


text:
MathSciNet bibliographic data MR751340 35R10 (45K05 47D05) Di Blasio, Gabriella; Kunisch, Karl; Sinestrari, Eugenio \$L\sp{2}\$$L\sp{2}$-regularity for parabolic partial integro-differential equations with delay in the highest-order derivatives. J. Math. Anal. Appl. 102 (1984), no. 1, 38–57. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
# Colección

Música » The Dandy Warhols »

## Not If You Were The Last Junkie On Earth

25 scrobblings | Ir a la página del tema

Temas (25)
Tema Álbum Duración Fecha
Not If You Were The Last Junkie On Earth 3:13 21 Jun 2012, 5:40
Not If You Were The Last Junkie On Earth 3:13 26 Nov 2011, 13:25
Not If You Were The Last Junkie On Earth 3:13 26 Nov 2011, 9:41
Not If You Were The Last Junkie On Earth 3:13 17 Ago 2011, 20:47
Not If You Were The Last Junkie On Earth 3:13 11 Ago 2011, 15:37
Not If You Were The Last Junkie On Earth 3:13 8 Jul 2011, 19:12
Not If You Were The Last Junkie On Earth 3:13 2 Jul 2011, 10:43
Not If You Were The Last Junkie On Earth 3:13 2 Jul 2011, 8:22
Not If You Were The Last Junkie On Earth 3:13 2 Jul 2011, 7:11
Not If You Were The Last Junkie On Earth 3:13 25 Jun 2011, 16:34
Not If You Were The Last Junkie On Earth 3:13 19 Jun 2011, 7:54
Not If You Were The Last Junkie On Earth 3:13 9 Jun 2011, 18:51
Not If You Were The Last Junkie On Earth 3:13 8 Jun 2011, 19:14
Not If You Were The Last Junkie On Earth 3:13 7 May 2011, 9:05
Not If You Were The Last Junkie On Earth 3:13 21 Mar 2011, 20:38
Not If You Were The Last Junkie On Earth 3:13 11 Ene 2011, 6:59
Not If You Were The Last Junkie On Earth 3:13 4 Ene 2011, 6:22
Not If You Were The Last Junkie On Earth 3:13 10 Dic 2010, 13:07
Not If You Were The Last Junkie On Earth 3:13 14 Nov 2010, 11:46
Not If You Were The Last Junkie On Earth 3:13 28 Sep 2010, 17:52
Not If You Were The Last Junkie On Earth 3:13 3 Sep 2010, 17:21
Not If You Were The Last Junkie On Earth 3:13 29 Ago 2010, 6:53
Not If You Were The Last Junkie On Earth 3:13 18 Ago 2010, 18:18
Not If You Were The Last Junkie On Earth 3:13 16 Ago 2010, 4:58
Not If You Were The Last Junkie On Earth 3:13 2 Ago 2010, 9:22


text:
# Find Out

Classical Mechanics Level pending

The GRAVITATIONAL FORCE between two objects is 43.5646516889179172. How much DISTANCE between these objects be DECREASED so that the FORCE between them becomes double?

×

Problem Loading...

Note Loading...

Set Loading...


text:
# $X$ Banach, $u_n \to u$ and $x^*_n \xrightarrow{w^*} x^*$ implie that $\langle x^*_n, u_n\rangle \to \langle x^*,u\rangle$.

Exercise :

Let $$X$$ be a Banach space, $$u_n \to u$$ and $$x^*_n \xrightarrow{w^*} x^*$$. Show that $$\langle x^*_n, u_n\rangle \to \langle x^*,u\rangle$$.

Attempt-Discussion :

I know that a sequence $$x_n^* \in X^*$$ converges to $$x^*$$ provided that $$x_n^*(u) \to x^*(u)$$ for all $$u \in X$$. Knowing that the $$w^*-$$topology coincides the topology of pointwise convergence of linear functionals, this implies that $$x_n^* \xrightarrow{w^*} x^*$$.

Now, I know that the duality brackets given are essentialy the expression of the linear functional, aka : $$u \mapsto \langle x^*, u \rangle := x^*(u)$$. So essentially, since we have haev that the pointwise convergence is coinciding, we essentialy need to prove that $$x^*_n(u_n) \to x^*(u)$$ and I guess since we have the functional convergence, it boils down to $$x^*(u_n) \to x^*(u)$$ ?

How would one work further to prove the convergence asked ?

Alternatively, we would be interested in showing that $$|\langle x^*_n, u_n\rangle - \langle x^*,u\rangle| < \varepsilon$$ but I think I am missing something in re-writting the expression by breaking it up to form an arbitrarily small inequality.

Any hints will be greatly appreciated.

Note that by the Banach-Steinhaus theorem $$\sup_{n}\|x_{n}^{*}\|<\infty$$. We find
$$\lim_{n\rightarrow\infty}|\langle x_{n}^{*},u_{n}\rangle-\langle x^{*},u\rangle|=\lim_{n\rightarrow\infty}|\langle x_{n}^{*},u_{n}-u\rangle+\langle x_{n}^{*},u\rangle-\langle x^{*},u\rangle|$$ $$\leq\lim_{n\rightarrow\infty}|\langle x_{n}^{*},u_{n}-u\rangle|+|\langle x_{n}^{*}-x^{*},u\rangle|\leq \lim_{n\rightarrow\infty}\|x_{n}^{*}\|\|u_{n}-u\|+\lim_{n\rightarrow\infty}|\langle x_{n}^{*}-x^{*},u\rangle|=0$$ as $$x_{n}^{*}\stackrel{w^{*}}{\rightarrow}x^{*}$$, $$u_{n}\rightarrow u$$ and $$(\|x_{n}^{*}\|)$$ is bounded.


text:
# Integration and area under curve

1. Jan 11, 2010

### jmsdg7

I have seen a thread with a similair title but passed up on what i want to know.

I just want somebody to explain to me why definite integration equals the area under between the function and the x axis

Ive just been through indefinite integration, then using the summation formulas in my AP calculus class, and i understand the concept of going from acceleration to velocity to position, but i dont understand how that can equal the area under a curve.

this whole question might be pointless because we havent learned how to evaluate definite integrals yet and maybe my answer lies in that.

Any 2 cents worth is appreciated!!!

2. Jan 11, 2010


text:
Consider the following statements I. The equation 1990x -173y =11 has no solution in integers for x and y .

### Question Asked by a Student from EXXAMM.com Team

Q 1648491303.     Consider the following statements

I. The equation 1990x -173y =11 has no solution in integers for x and y .

II. The equation 3x- 12 y =7 has no solution in integers for x and y.

which of the above statcment(s) is/are correct?
CDS 2015 Paper-1
A

Only I

B

Only II

C

Both I and II

D

Neither I nor II

#### HINT

(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

#### Access free resources including

• 100% free video lectures with detailed notes and examples
• Previous Year Papers
• Mock Tests
• Practices question categorized in topics and 4 levels with detailed solutions
• Syllabus & Pattern Analysis


text:
## anonymous 5 years ago what does the following solve? A=pi r2

1. anonymous

If I am not mistaken this is the solution to finding the area of a circle. $A(area)=2\pi*r$ This is where r is the radius of the circle.

2. anonymous

Wait.... my bad ;) Area is $Area=\pi*r^2$


text:
# Is the Wikipedia article about chordal graphs incorrect?

This is the wiki for chordal graphs. It states that "A perfect elimination ordering in a graph is an ordering of the vertices of the graph such that, for each vertex v, v and the neighbors of v that occur after v in the order form a clique. A graph is chordal if and only if it has a perfect elimination ordering.", but chordal graphs are only defined over the set of cyclic graphs with more than 3 vertices. Should this article be updated to "A graph is chordal if and only if it has a perfect elimination ordering and has more than 3 vertices"? Without the second constraint, we can consider $K_3$ to be chordal, but it is not chordal.

• $K_3$ is in fact chordal. Every cycle of length greater than 3 (none) has a chord (true) – Jernej Apr 10 '16 at 8:03
• Yep, I made a mistake in the logic. I thought it was: IF a graph is chordal, THEN Every cycle of length greater than 3 has a chord AND there exists a cycle of length greater than 3. Didn't realize that the "there exists..." portion wasn't necessary. – acatto Apr 10 '16 at 22:54

Definition. A graph $G = (V,E)$ is said to be chordal if every cycle of length $\geq 4$ has a chord. (Equivalently: it has no induced cycles of length four or more.)
Since $K_3$ does not have any cycle of length $\geq 4$, it is trivially chordal. Similarly, any tree is trivially chordal, for it does not have any cycles at all.
• The definition I've seen is in the first sentence of the wiki article, which I posted as the first sentence of the description. A chord is defined as an edge that is not part of the cycle, but connects two vertices on the cycle. In $K_3$, all edges are part of the cycle, so there can't be a chord. I guess it's a matter of semantics of the term "Chordal Graph"; i figured that chordal graphs must have a chord on them, but this means that $K_3$ is trivially chordal? – acatto Apr 10 '16 at 0:57
• Ah, but the requirement is only for cycles of four or more vertices. Since $K_3$ has no cycles of length $4$ or more, it is trivially chordal. Similarly, any tree is chordal, for it doesn't have any cycles at all (let alone cycles of length $4$ or more). – Josse van Dobben de Bruyn Apr 10 '16 at 1:00


text:
Two previous examples of LU factorizations were given. I found one more to be the key to understanding how to implement this as a matlab algorithm, required for our problem set.

A matrix that contains both pivots and elementary matrix operations is

\label{eqn:luAlgorithm:20}
M=
\begin{bmatrix}
0 & 0 & 2 & 1 \\
0 & 0 & 1 & 1 \\
2 & 0 & 2 & 0 \\
1 & 1 & 1 & 1
\end{bmatrix}

Our objective is to apply a sequence of row permutations or elementary row operations to $$M$$ that put $$M$$ into upper triangular form. At the same time we wish to track the all the inverse operations. When no permutations were required to produce $$U$$, then we end up with a factorization $$M = L’ U$$ where $$L’$$ is lower triangular.

Let’s express the row operations that we apply to $$M$$ as

\label{eqn:luAlgorithm:40}
U =
L_k^{-1}
L_{k-1}^{-1} \cdots
L_2^{-1}
L_1^{-1}
M,

with

\label{eqn:luAlgorithm:60}
L’ = L_0 L_1 L_2 \cdots L_{k-1} L_k

Here $$L_0 = I$$, the identity matrix, and $$L_i^{-1}$$ is either a permutation matrix interchanging two rows of the identity matrix, or it is an elementary row operation encoding the operation $$r_j \rightarrow r_j – M r_i$$, where $$r_i$$ is the pivot row, and $$r_j, j > i$$ are the rows that we are applying the Gaussian elimination operations to.

For our example matrix, we see that we cannot use the $$M_{1 1}$$ as the pivot element since it is zero. In general, for numeric stability, we wish to use the row with the biggest absolute value in the column that we are operating on. In this case that is row 3. Our first row operation is therefore a $$1,3$$ permutation

\label{eqn:luAlgorithm:80}
L_1^{-1} =
\begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix},

which gives us

\label{eqn:luAlgorithm:100}
M \rightarrow
L_1^{-1}
M
=
\begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
0 & 0 & 2 & 1 \\
0 & 0 & 1 & 1 \\
2 & 0 & 2 & 0 \\
1 & 1 & 1 & 1 \\
\end{bmatrix}
=
\begin{bmatrix}
2 & 0 & 2 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 2 & 1 \\
1 & 1 & 1 & 1 \\
\end{bmatrix}.

Computationally, we do not wish to actually do a matrix multiplication to achieve this permutation. Instead we want to just swap the two rows in question.

The inverse of this operation is the same permutation, so for $$L’$$ we compute

\label{eqn:luAlgorithm:120}
L’ \rightarrow L_0 L_1 = L_1.

As before, we don’t wish to do a matrix operation. Since we have applied the permutation matrix from the right, it results in an exchange of columns $$1,3$$ of our $$L_0$$ matrix (which happens to be identity at this point). We can implement that matrix operation as a column exchange directly using submatrix notation.

We now proceed down the column, doing all the non-zero row elimination operations required. In this case, we have only one

\label{eqn:luAlgorithm:140}
r_4 \rightarrow r_4 – \frac{1}{2} r_1.

This has the matrix form

\label{eqn:luAlgorithm:160}
L_2^{-1} =
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-1/2 & 0 & 0 & 1 \\
\end{bmatrix}.

The next stage of the $$U$$ computation is

\label{eqn:luAlgorithm:180}
M
\rightarrow L_2^{-1} L_1^{-1} M
=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-1/2 & 0 & 0 & 1 \\
\end{bmatrix}
\begin{bmatrix}
2 & 0 & 2 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 2 & 1 \\
1 & 1 & 1 & 1 \\
\end{bmatrix}
=
\begin{bmatrix}
2 & 0 & 2 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 2 & 1 \\
0 & 1 & 0 & 1 \\
\end{bmatrix}.

Again, we do not wish to do this operation as a matrix operation. Instead we act directly on the rows in question with \ref{eqn:luAlgorithm:140}.

Note that the inverse of this matrix operation is very simple. We’ve subtracted $$r_1/2$$ from $$r_4$$, so to invert this we have only to add back $$r_1/2$$. That is

\label{eqn:luAlgorithm:200}
L_2
=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
1/2 & 0 & 0 & 1 \\
\end{bmatrix}.

Observe that when we apply this from the right to $$L_0 L_1 \rightarrow L_0 L_1 L_2$$, the interpretation is a column operation

\label{eqn:luAlgorithm:220}
c_1 \rightarrow c_1 + m c_4,

In general, if we apply the row operation

\label{eqn:luAlgorithm:240}
r_j \rightarrow r_j – m r_i,

to the current state of our matrix $$U$$, then we must apply the operation

\label{eqn:luAlgorithm:260}
r_i \rightarrow r_i + m r_j,

to the current state of our matrix $$L’$$.

We are now ready to move on to reduction of column 2. We will have only a permutation operation

\label{eqn:luAlgorithm:280}
L_3 =
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
\end{bmatrix},

so we apply a $$2,4$$ row interchange to U, and a $$2,4$$ column interchange to $$L’$$. This gives us

\label{eqn:luAlgorithm:300}
M \rightarrow
\begin{bmatrix}
2 & 0 & 2 & 0 \\
0 & 1 & 0 & 1 \\
0 & 0 & 2 & 1 \\
0 & 0 & 1 & 1 \\
\end{bmatrix}.

Our final operation is a regular row operation

\label{eqn:luAlgorithm:320}
r_4 \rightarrow r_4 – \inv{2} r_3,

with matrix
\label{eqn:luAlgorithm:340}
L_4^{-1} =
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & -1/2 & 1 \\
\end{bmatrix}

We can also track all the permutations we have performed, which in this case was

\label{eqn:luAlgorithm:360}
P = L_3 L_1 I.

This should also be computed by performing row interchanges, not matrix multiplication.

Now should we wish to solve the system

\label{eqn:luAlgorithm:380}
M \Bx = L’ U \Bx = \Bb,

we can equivalently solve

\label{eqn:luAlgorithm:400}
P L’ U \Bx = P \Bb,

To do this let $$\By = U \Bx$$, so that we wish to solve

\label{eqn:luAlgorithm:420}
P L’ \By = P \Bb.

The matrix $$L = P L’$$ is lower triangular, as $$P$$ contained all the permutations that we applied along the way (FIXME: this is a statement, not a proof, and not obvious).

We can solve the system

\label{eqn:luAlgorithm:440}
L \By = P \Bb,

using forward substitution. That leaves us to solve the upper triangular system

\label{eqn:luAlgorithm:460}
\By = U \Bx,

which now requires only back substitution.


text:
Felix Hausdorff

Infobox_Scientist
name = Felix Hausdorff

|300px
image_width = 300px
caption =
birth_date = Birth date|1868|11|8
birth_place = Breslau, Germany
death_date = death date and age|1942|1|26|1868|11|8
death_place = Bonn, Germany
residence =
nationality =
field = Mathematician
work_institutions = University of Bonn, University of Greifswald
alma_mater = University of Leipzig
doctoral_students = Karl Bögel

Franz Hallenbach

Gustav Steinbach
known_for = Hausdorff measure
Hausdorff dimension
religion =
footnotes=

Felix Hausdorff (November 8, 1868 – January 26, 1942) was a German mathematician who is considered to be one of the founders of modern topology and who contributed significantly to set theory, descriptive set theory, measure theory, function theory, and functional analysis.

Life

Hausdorff studied at the University of Leipzig, obtaining his Ph.D. in 1891. He taught mathematics in Leipzig until 1910, when he became professor of mathematics at the University of Bonn. He was professor at the University of Greifswald from 1913 to 1921. He then returned to Bonn. When the Nazis came to power, Hausdorff, who was Jewish, felt that as a respected university professor he would be spared from persecution. However, his abstract mathematics was denounced as "Jewish", useless, and "un-German"Fact|date=July 2007 and he lost his position in 1935. Though he could no longer publish in Germany, Hausdorff continued to be an active research mathematician, publishing in the Polish journal Fundamenta Mathematicae. After Kristallnacht in 1938 as persecution of Jews escalated, Hausdorff became more and more isolated. Finally, in 1942 when he could no longer avoid being sent to a concentration camp, Hausdorff committed suicide together with his wife and sister-in-law on the 26th of January.

Work

Hausdorff was the first to state a generalization of Cantor's Continuum Hypothesis; his Aleph Hypothesis, which appears in his 1908 article Grundzüge einer Theorie der geordneten Mengen, and which is equivalent to what is now called the Generalized Continuum Hypothesis.

In 1909, while studying partially ordered sets of real sequences, he stated what is now known as the Hausdorff Maximal Principle; he was the first to apply a maximal principle in algebra.

In his 1914 classic text, Grundzüge der Mengenlehre, he defined and studied partially ordered sets abstractly; using the Axiom of Choice, he proved that every partially ordered set has a maximal linearly ordered subset. In this same book, he axiomatized the topological concept of neighborhood and introduced the topological spaces that are now called Hausdorff spaces.

In 1914 using the Axiom of Choice, he gave a "paradoxical" decomposition of the 2-sphere as the disjoint union of four sets A,B,C, and Q, where Q is countable and the sets A, B, C, and B$cup$C are mutually congruent. This later inspired the Banach-Tarski paradoxical decomposition of the ball in 3-space.

He introduced the concepts now called Hausdorff measure and Hausdorff dimension, which have been useful in the theory of fractals. In analysis, he solved what is now called the Hausdorff moment problem.

Hausdorff also published philosophical and literary works under the pseudonym "Paul Mongré". "Paul Mongre" published a number of books and articles on the philosopher Friedrich Nietzsche, as well as a number of reviews of contemporary literature and drama. Mongre-Hausdorff also published a satirical play which performed in a dozen German cities. In the course of attempts to refute Nietzsche's doctrine of "the eternal return of the same," Hausdorff was led to Cantor's set theory, which set Hausdorff on the road to his set-theoretical discoveries. Hausdorff's Nietzschean philosophical writings appear in volume VII of his collected works.

Important publications

* "Grundzüge der Mengenlehre"
* [Gesamnelte Werke]

ee also

*Baker-Campbell-Hausdorff formula
*Gromov-Hausdorff convergence

*MathGenealogy |id=46991
*MacTutor Biography|id=Hausdorff
* [http://www.aic.uni-wuppertal.de/fb7/hausdorff/ Homepage of the Hausdorff Edition (German)]
* [http://www.ulb.uni-bonn.de/bibliothek/veroeffentlichungen&ausstellungen/veroeffentlichungen/hausdorff.pdf Hausdorff Findbuch]
* [http://www.hausdorff-center.uni-bonn.de/ Hausdorff Center for Mathematics in Bonn]

Persondata
NAME= Hausdorff, Felix
ALTERNATIVE NAMES=
SHORT DESCRIPTION= Mathematician
DATE OF BIRTH= birth date|1868|11|8|mf=y
PLACE OF BIRTH= Breslau, Germany
DATE OF DEATH= death date|1942|1|26|mf=y
PLACE OF DEATH= Bonn, Germany

Wikimedia Foundation. 2010.

### Look at other dictionaries:

• Felix Hausdorff — (* 8. November 1868 in Breslau; † 26. Januar 1942 in Bonn) war ein deutscher Mathematiker. Er gilt als Mitbegründer der allgemeinen Topologie und lieferte wesentliche Beiträge zur allgemeinen und deskriptiven Mengenlehre, zu …   Deutsch Wikipedia

• Felix Hausdorff — Saltar a navegación, búsqueda Felix Hausdorff Felix Hausdorff (8 de noviembre de 1868, 26 de enero de 1942) fue un matemático alemán que está considerado como uno de los fundadores de la moderna T …   Wikipedia Español

• Felix Hausdorff — (8 novembre 1868 26 janvier 1942) est un mathématicien allemand, considéré comme l un des fondateurs de la topologie moderne. Il contribua aussi significativement à la théorie des …   Wikipédia en Français

• Hausdorff — Felix Hausdorff (Fotografie zwischen 1913 und 1921 entstanden) Felix Hausdorff (* 8. November 1868 in Breslau; † 26. Januar 1942 in Bonn) war ein deutscher Mathematiker. Er gilt als Mitbegründer der allgemeinen Topologie und lieferte wesentliche… …   Deutsch Wikipedia

• Hausdorff — Felix Hausdorff Felix Hausdorff Felix Hausdorff (8 novembre 1868 26 janvier 1942) est un mathématicien allemand, considéré comme l un des fondateurs de la topologie moderne. Il contribua aussi signif …   Wikipédia en Français

• Hausdorff-Trennungsaxiom — Hausdorff Raum (T2) berührt die Spezialgebiete Mathematik Topologie ist Spezialfall von topologischer Raum präregulärer Raum ( …   Deutsch Wikipedia

• HAUSDORFF (F.) — HAUSDORFF FELIX (1868 1942) La renommée du mathématicien allemand Felix Hausdorff repose surtout sur son ouvrage Grundzüge der Mengenlehre (1814), qui en fit le fondateur de la topologie et de la théorie des espaces métriques. Né à Breslau dans… …   Encyclopédie Universelle

• Hausdorff — may refer to:* A Hausdorff space, when used as an adjective, as in the real line is Hausdorff. * Felix Hausdorff, the German mathematician that Hausdorff spaces are named after. * Hausdorff dimension, a measure theoretic concept of dimension. *… …   Wikipedia

• Hausdorff-Besikowitsch-Dimension — Die Hausdorff Dimension wurde von Felix Hausdorff eingeführt und bietet die Möglichkeit, beliebigen metrischen Räumen, wie beispielsweise Fraktalen, eine Dimension zuzuordnen. Für einfache geometrische Objekte wie Strecken, Vielecke, Quader und… …   Deutsch Wikipedia

• Hausdorff-Besikowitsch Dimension — Die Hausdorff Dimension wurde von Felix Hausdorff eingeführt und bietet die Möglichkeit, beliebigen metrischen Räumen, wie beispielsweise Fraktalen, eine Dimension zuzuordnen. Für einfache geometrische Objekte wie Strecken, Vielecke, Quader und… …   Deutsch Wikipedia


text:
# Theoretically, can a solar system evolve around lagrange points

My question is, during the evolution of a solar system is it possible to end up with a massive planet around a star with two small planets around their $L_4$ and $L_5$.
Since we know it is not possible to have any bodies around $L_1, L_2, L_3$ due to their low stability, hence any planet trying to evolve their can be easily nudges out of its position.

• The objects orbiting in the L4 and L5 probably can't be classified as planets due to the impossibility of them clearing their orbits by definition. – A. C. A. C. Aug 4 '17 at 19:19


text:
Digital & Computing

# How Bitcoin works

Updated Wednesday, 25th May 2022
Everything you need to know to get started with Bitcoin.

Bitcoin is an application of public-key cryptography (cyrptographic algorithms which require two separate keys).

Getting started with Bitcoin is relatively simple. You need to download one of many special ‘wallet’ applications available for most computers and smartphones. You must then create a Bitcoin account, your wallet application will then generate a pair of public encryption keys – one private (known only to you) and one public (copies of which will be shared with other Bitcoin users). You are then ready to go.

Assume both of us are Bitcoin users and I want to send some Bitcoins to you.

1. My Bitcoin application performs the transaction by creating a message containing some of my coins and encrypts it using your public key. I can’t unlock the message and neither can any other Bitcoin user. In fact, only your private key can decrypt the message and that’s safe on your computer.
2. My software then signs the message using my private key. This doesn’t make the message any more secure, but it authenticates the message as coming from me.
3. The message is then broadcast to every Bitcoin user. Any Bitcoin user can authenticate the message by checking it with a copy of my public key – if it is genuine, the signature will match, if it doesn’t then the message is false and can be disregarded.
4. If the message is authentic, then your computer uses the single copy of your private key to unlock the encrypted message and place the Bitcoins in your ‘wallet’.

In most cases there is a short delay before a payment can be finalised and money appears in your account. Before you can receive Bitcoins other computers must validate the transaction. This takes on average about ten minutes.

You may have thought of a way to defraud Bitcoin; how can you stop people editing their records to award themselves more money or cancelling transactions? Well this is prevented by having every Bitcoin transaction scrutinised by many computers.

Groups of Bitcoin transactions are bundled into blocks and stored in a list called the block chain that holds a copy of every transaction ever made using Bitcoin. The block chain isn’t stored in a single location, where it would be vulnerable to attack or loss; instead copies of the chain are distributed among all Bitcoin users. Not only is the block chain protected from loss, or attack; but any Bitcoin user can validate any transaction.

Before a transaction can be approved it is added to a new block of pending transactions. The block is then processed by a powerful mathematical technique called hashing  that produces a unique identifier for the block (like a fingerprint) known as the hash. The hash and the block are then added to the block chain. Crucially, the new hash incorporates the hash of the previous block, which, in turn, used the hash of the block before that, and so on and on and on right back to the very first transaction.

If I want to check the validity of a block of transactions, all I need is for my computer to recalculate the hash for that block and compare it to the original hash from the block chain. If any of the data in the block have been changed, even by the smallest amount, the two hashes will be different and it would be obvious the block had been tampered with. Changing the contents of a block far back in the block chain not only creates a conflict with that block’s hash, but all blocks that follow on from it. And it’s not just my computer checking the hashes, millions of Bitcoin users are constantly and independently checking the hashes in their own copies of the block chain and comparing findings. Bitcoin relies on participating computers coming to a consensus about which transactions are valid. A block chain containing tampered records will be quickly identified as containing wrong data and will be ignored by the wider Bitcoin system.


text:
# Reverse Intermediate Value Theorem

What does it mean to say that a real valued function $f : [a, b] \rightarrow \mathbb{R}$ is continuous at $x_0 \in [a, b]$? Assume that $f : [a, b] \rightarrow \mathbb{R}$ is continuous State, without proof, the Intermediate Value Theorem for $f$

Prove that there exist $x_1, x_2 ∈ [a, b]$ such that for all $x \in [a, b]$ , $f(x_1)\leq f(x) \leq f(x_2)$.

[Standard results about sequences of real numbers may be used, without proof, provided they are clearly stated.]

This is a university question from a real analysis course. The way these questions go are they ask you to prove or state a particular theorem in the earlier part and then expect you to use that theorem to solve the problem in the next part. My initial thought to prove the last part was to state and prove the fact that a continuous function on a closed and bounded interval attains its bounds but that is not really in the spirit of the question which clearly wants use of IVT.

So,my attempt:

I split into cases:

1) $f$ was a constant function in which case choosing $x_1 = \dfrac{a+x}{2}$ and $x_2 = \dfrac{x+b}{2}$ would suffice.

2) $f(a)$ or $f(b)$ $\not=f(x)$ in which case you can use IVT on [a,x] or [x,b] one side for one of $x_1$ or $x_2$ and just let the other equal $x$ (I think this must be allowed to let $x=x_1$ $x=x_2$ sometimes otherwise you'd be in trouble with a function which had a maximum point or minimum in [a,b]- You wouldnt be able to find an $x_1$ or $x_2$ both distinct from $x$.I think the question probably means $x_1$ and $x_2$ are both distinct otherwise you could pick $x_1=x=x_2$ which would be stupid.

3) $f(a)$ and $f(b)$ both equal to $f(x)$ . $f$ is not constant so there is a value attained wich is greater than or less than $f(x)$. For the other side just let it equal $x$ again.

Is this fine? I have only used IVT for 2) and I have no idea what to do with the hint in the bracket about sequences of real numbers. If this is faulty or I have misunderstood something or there is an elegant way or any useful feedback I would appreciate it, thanks.

## 1 Answer

Sorry- I was right in first guess- it was just state and prove the fact that a continuous function on a closed and bounded interval attains its bounds. IVT has nothing to do with it. Use of IVT is for a further later part in the question which I have not included here.


text:
## 1047: "Approximations"

This forum is for the individual discussion thread that goes with each new comic.

Moderators: Moderators General, Prelates, Magistrates

rhomboidal
Posts: 801
Joined: Wed Jun 15, 2011 5:25 pm UTC
Contact:

### 1047: "Approximations"

Title Text: Two tips: 1) 8675309 is not just prime, it's a twin prime, and 2) if you ever find yourself raising log(anything)^e or taking the pi-th root of anything, set down the marker and back away from the whiteboard; something has gone horribly wrong.

This would make a cool cheat sheet. And an even cooler tattoo.

bitwiseshiftleft
Posts: 295
Joined: Tue Jan 09, 2007 9:07 am UTC
Location: Stanford
Contact:

### Re: 1047: "Approximations"

Spoiler:
$cos(\pi / 7) + cos(3\pi/7) + cos(5\pi/7) = -\Re(\zeta_7+\zeta_7^2+\zeta_7^3) = \frac12 - \frac12(1 + \zeta_7 + \zeta_7 ^2 + \zeta_7^3 + \zeta_7^4 + \zeta_7^5 + \zeta_7^6) = \frac12$

somebody_else
Posts: 8
Joined: Fri Mar 05, 2010 8:08 am UTC

### Re: 1047: "Approximations"

The error in the first Protip is pretty big though. If only the fractions were added instead of subtracted. I am going to use the radius of the earth one from now on - just to confuse anyone reading my work.

Shidoshi
Posts: 84
Joined: Thu Jul 21, 2011 9:21 am UTC
Location: Brazil - Porto Alegre

### Re: 1047: "Approximations"

The gas constant is expressed in J/gmol.K and not J/kgmol.K which would be the SI unit.

Willl
Posts: 34
Joined: Mon Dec 17, 2007 8:44 pm UTC

### Re: 1047: "Approximations"

The stated accuracy of g is about an order of magnitude off - it's accurate to about 1 in 80,000, not 750,000. I don't know about the others, but I was surprised to learn that an international standard was actually defined to that precision, given that it varies by more than 1 in 200, hence I ended up checking it.

JimsMaher
Posts: 145
Joined: Wed Mar 14, 2012 5:14 pm UTC

### Re: 1047: "Approximations"

Some of the more accurate, yet still elegant ones tempt the idyllic pseudoscientist in me.

G ≈ 1 over {e^(π - 1)^(π + 1)}
... consider {m^3} over {kg*s^2} when off by 1:25000 ... on a galactic scale is still significant. It's the difference between getting a speedy pick-me-up from passing near a black-hole, and winding up with a cyborg's vision of hell set to Disney-music of all things.

bitwiseshiftleft
Posts: 295
Joined: Tue Jan 09, 2007 9:07 am UTC
Location: Stanford
Contact:

### Re: 1047: "Approximations"

Also, e^(pi * sqrt(163)) ~ 640320^3 + 744. But that one is famous. And was found by a method other than brute force.

BlitzGirl
Posts: 9119
Joined: Mon Sep 20, 2010 11:48 am UTC
Location: Out of the basement for Yip 6! Schizoblitz: 115/2672 NP
Contact:

### Re: 1047: "Approximations"

Unfortunately, the only one of these I'm going to remember is the RENT Method.

Present some of the others with heart-wrenching vocals and perhaps I'd retain them.
Knight Temporal of the One True Comic
BlitzGirl the Pink, Mopey Molpy Mome
Spoiler:

<Profile
~.~.FAQ->

hemflit
Posts: 18
Joined: Thu Sep 03, 2009 11:53 pm UTC

### Re: 1047: "Approximations"

𝜋 seconds in a nanocentury, much neater 75^4 in a year.

NeilUK
Posts: 1
Joined: Wed Apr 25, 2012 7:05 am UTC

### Re: 1047: "Approximations"

The number of seconds in a year, to better than 1%, is $pi * 10^7$

MacGeb
Posts: 1
Joined: Wed Apr 25, 2012 7:08 am UTC

### Re: 1047: "Approximations"

hemflit wrote:Pi seconds in a nanocentury, much neater 75^4 in a year.

My favourite as well. What puzzles me is that sometimes I even use it.

scooterboo
Posts: 3
Joined: Wed Jan 26, 2011 5:07 am UTC

### Re: 1047: "Approximations"

My favorite is miles * phi = kilometers
(accurate to .54%)

Posts: 3
Joined: Mon Dec 19, 2011 5:07 pm UTC

### Re: 1047: "Approximations"

Within about 0.5%, there are pi times ten million seconds in a year.

For added fun when trolling, throw in that Earth's orbit being an ellipse instead of a perfect circle is what throws it off from being exactly right.

TheoGB
Posts: 70
Joined: Fri Aug 22, 2008 1:38 pm UTC

### Re: 1047: "Approximations"

I don't think I've ever seen accuracy used like that in this sort of context. Does someone want to explain what it means? I had a quick look at the seconds in a year but I't not really any clearer to me. In context of a fluid 1 part in 400 would make sense to me but I've been too long away from Uni to begin to remember this stuff.

Smatchmo
Posts: 2
Joined: Mon Jun 13, 2011 4:49 pm UTC

### Re: 1047: "Approximations"

2^42.42

So beautiful. Makes me want to jump in the ol' Heart of Gold and go cruising.

Many Kudos, sir.

Posts: 3
Joined: Mon Dec 19, 2011 5:07 pm UTC

### Re: 1047: "Approximations"

Shidoshi wrote:The gas constant is expressed in J/gmol.K and not J/kgmol.K which would be the SI unit.

I've never seen the abbreviations "gmol" or "kgmol", but the SI units of the gas constant are Joules per mole Kelvin, and moles are defined as if they were based on grams instead of kg. Probably because the chemists threatened insurrection otherwise... grams are a much more convenient laboratory unit.

JimsMaher
Posts: 145
Joined: Wed Mar 14, 2012 5:14 pm UTC

### Re: 1047: "Approximations"

TheoGB wrote:I don't think I've ever seen accuracy used like that in this sort of context. Does someone want to explain what it means?

Approximations. Notice how everything is rounded off with lots of trailing zeroes.

Consider the variation that the given approximation is from the actual standard.
The size of discrepancy is called one part to the actual measure's Whatever-it-gets-rounded-updown-to.

Demonstrating the first item on the list:
One Light-Year (meters) ≈ 99^8 ... "Accurate to Within: One part in 40" Let's say 1:40 ...
99^8 m = 9 227 446 944 279 201 m
1 ly = 9 460 730 472 580 800 m

1 ly - 99^8 = 233 283 528 301 599 = d
d/ly ≈ 0.0246580... that's just less than 2.5% which can be restated concisely (that's the key here) as, "This approximation is accurate to within one part in Forty".
Last edited by JimsMaher on Wed Apr 25, 2012 7:58 am UTC, edited 1 time in total.

Ignitus
Posts: 22
Joined: Fri Jul 02, 2010 8:48 pm UTC

### Re: 1047: "Approximations"

I don't think I've ever seen accuracy used like that in this sort of context. Does someone want to explain what it means? I had a quick look at the seconds in a year but I't not really any clearer to me. In context of a fluid 1 part in 400 would make sense to me but I've been too long away from Uni to begin to remember this stuff.

This took me a while, but it simply means that using this estimation you will be accurate with in one whole number if you don't go beyond this value given

Quick Example
40* 99^8 is a little over 39 light years not 40.

So if you look at the individual estimations they look fairly off since we are used to looking for accuracy up to the 3 or 4th decimal, but in practical terms if your asked does X meters equal a light year X / 99^8th will give a fairly accurate yes or no answer but after 40 light years you start to be off by 1 or 2

Shidoshi
Posts: 84
Joined: Thu Jul 21, 2011 9:21 am UTC
Location: Brazil - Porto Alegre

### Re: 1047: "Approximations"

Shidoshi wrote:The gas constant is expressed in J/gmol.K and not J/kgmol.K which would be the SI unit.

I've never seen the abbreviations "gmol" or "kgmol", but the SI units of the gas constant are Joules per mole Kelvin, and moles are defined as if they were based on grams instead of kg. Probably because the chemists threatened insurrection otherwise... grams are a much more convenient laboratory unit.

I've been recently through ChemEng and now they are starting to use "kilogram moles" and "gram moles", that way everything can be expressed in kg without much problem. Carbon12's molar mass is 12g/gmol or 12kg/kgmol (or even 12lb/lbmol if you swing that way). It's a somewhat new concept yeah, but it sure helps putting everything to SI and making less confusion.

tetsujin
Posts: 426
Joined: Thu Nov 15, 2007 8:34 pm UTC
Location: Massachusetts
Contact:

### Re: 1047: "Approximations"

Is that an Imperial gallon, a US liquid gallon, or one of the various deprecated or antiquated standards for a gallon?
---GEC
I want to create a truly new command-line shell for Unix.
Anybody want to place bets on whether I ever get any code written?

bjornart
Posts: 8
Joined: Thu Mar 04, 2010 12:55 pm UTC

### Re: 1047: "Approximations"

Willl wrote:The stated accuracy of g is about an order of magnitude off - it's accurate to about 1 in 80,000, not 750,000. I don't know about the others, but I was surprised to learn that an international standard was actually defined to that precision, given that it varies by more than 1 in 200, hence I ended up checking it.

You forgot to divide by g.

JimsMaher
Posts: 145
Joined: Wed Mar 14, 2012 5:14 pm UTC

### Re: 1047: "Approximations"

Let:
A = actual standard
B = approximation

Expressed as being accurate to within one part in
{ (A - B)/A } rounded up to nearest greater fraction whose reciprocal decimal form has no more than two significant digits.

Seli
Posts: 17
Joined: Wed Sep 17, 2008 10:01 am UTC

### Re: 1047: "Approximations"

Shidoshi wrote:
Shidoshi wrote:The gas constant is expressed in J/gmol.K and not J/kgmol.K which would be the SI unit.

I've never seen the abbreviations "gmol" or "kgmol", but the SI units of the gas constant are Joules per mole Kelvin, and moles are defined as if they were based on grams instead of kg. Probably because the chemists threatened insurrection otherwise... grams are a much more convenient laboratory unit.

I've been recently through ChemEng and now they are starting to use "kilogram moles" and "gram moles", that way everything can be expressed in kg without much problem. Carbon12's molar mass is 12g/gmol or 12kg/kgmol (or even 12lb/lbmol if you swing that way). It's a somewhat new concept yeah, but it sure helps putting everything to SI and making less confusion.

That just looks, so wrong.

If one would insist on doing this kilomoles would make some sense and proper use of SI, but kilogram moles... sorry no. Weird engineers.

edit: phew, wiki suggests even engineers adhere to the rules at times and modern practice is using kilomole.

Willl
Posts: 34
Joined: Mon Dec 17, 2007 8:44 pm UTC

### Re: 1047: "Approximations"

bjornart wrote:You forgot to divide by g.
How embarrassing. Right you are.

Coderjoe
Posts: 5
Joined: Fri Jan 11, 2008 8:05 pm UTC

### Re: 1047: "Approximations"

according to Wolfram Alpha, the White House switchboard equation comes out to 0.212456141492...

To get ten places after the decimal, you have to round that nine, giving 2124561415, which is NOT the switchboard number.

ETA: Jenny's constant comes out to 867.53090198..., which is acceptable.
Last edited by Coderjoe on Wed Apr 25, 2012 9:19 am UTC, edited 1 time in total.

mrob27
Posts: 1336
Joined: Tue Jun 28, 2011 2:19 am UTC
Location: ]〖
Contact:

### Re: 1047: "Approximations"

Today's RIES date:

(((pi+1/2)^(1/7) * pi)^2 * pi)^2 = 2012.0425...

Robert Munafo - mrob.com/ries

alextrabec
Posts: 8
Joined: Wed Nov 03, 2010 4:08 am UTC

### Re: 1047: "Approximations"

Randall wins again. P.S. You are all nerds fact checking a web comic. Great Job, make your professors proud! Also, your mom plus me makes a "That what she said joke" - Enjoy

Love the comics and nerds, keep it up bitches!

Army1987
Posts: 8
Joined: Fri Aug 12, 2011 2:49 pm UTC

### Re: 1047: "Approximations"

NeilUK wrote:The number of seconds in a year, to better than 1%, is $pi * 10^7$

hthall
Posts: 95
Joined: Wed Oct 10, 2007 3:40 am UTC

### Re: 1047: "Approximations"

Look at me, still talking when there's Science to do.

peaty
Posts: 2
Joined: Thu Nov 24, 2011 6:49 pm UTC
Location: UK

### Re: 1047: "Approximations"

Smatchmo wrote:2^42.42

So beautiful. Makes me want to jump in the ol' Heart of Gold and go cruising.

Many Kudos, sir.

You mean "Much kudos". It's an abstract singular, not a plural.

Red Hal
Magically Delicious
Posts: 1445
Joined: Wed Nov 28, 2007 2:42 pm UTC

### Re: 1047: "Approximations"

alextrabec wrote:You are all nerds fact checking a web comic.
... and you're a nerd commenting on the nerds, and I'm a nerd commenting on a nerd commenting on the nerds. Oh atheia, we're so meta it burns! Loving the "your mom" joke; it makes you look so edgy. Well done!
Lost Greatest Silent Baby X Y Z. "There is no one who loves pain itself, who seeks after it and wants to have it, simply because it is pain..."

scharb
Posts: 32
Joined: Fri Oct 23, 2009 4:09 pm UTC

### Re: 1047: "Approximations"

if you ever find yourself raising log(anything)^e or taking the pi-th root of anything, set down the marker and back away from the whiteboard; something has gone horribly wrong.

I'll say something has gone horribly wrong! YOU'RE DOING MATH

jacksonliam91
Posts: 15
Joined: Thu Oct 14, 2010 8:25 am UTC

### Re: 1047: "Approximations"

9.81 is much easier to remember for g.

mrob27
Posts: 1336
Joined: Tue Jun 28, 2011 2:19 am UTC
Location: ]〖
Contact:

### Re: 1047: "Approximations"

I am Robert Munafo, who made the RIES program. I accidentally re-posted the earlier comment after thinking it was rejected for having a link...

So to comply with the "don't repost, edit instead" rule, and to make this post useful, here are a couple more wacky formulae that didn't make xkcd this time around:

$\sqrt{8+\sqrt{2+\sqrt{0+\sqrt{5}}}} = \pi$

$\sqrt{\sqrt{e^{6}/(x+1)}}$

then use this as the new X, and repeat. It converges on: 3.1415926825119...
Last edited by mrob27 on Wed Apr 25, 2012 3:10 pm UTC, edited 2 times in total.

jutzin
Posts: 2
Joined: Tue Jan 25, 2011 4:08 pm UTC

### Re: 1047: "Approximations"

i believe jenny's constant's last 3 digits are 209, not 902 (http://webcache.googleusercontent.com/search?q=cache:nVZnhecGolcJ:danstheman.com/Jenny.htm).

jutzin
Posts: 2
Joined: Tue Jan 25, 2011 4:08 pm UTC

### Re: 1047: "Approximations"

... and how cool is Robert Munafo's RIES, by the way!

Posts: 6
Joined: Fri Mar 13, 2009 5:20 pm UTC

### Re: 1047: "Approximations"

Even now you're too much of a mathematician to dare to say [imath]\pi \approx \sqrt{10}[/imath]. kudos.

Dobblesworth
Dobblesworth, here's the title you requested over three years ago. -Banana
Posts: 1429
Joined: Wed May 30, 2007 12:06 pm UTC
Contact:

### Re: 1047: "Approximations"

One of the bottom six according to Wolfram is an approximation. The sqrt 5 one. Decimal approximations given by Wolfram for the two terms don't match it seems.
http://www.wolframalpha.com/input/?i=%2 ... 824-4pi%29
http://www.wolframalpha.com/input/?i=sqrt+5

mistapotta
Posts: 2
Joined: Thu Sep 02, 2010 3:40 am UTC

### Re: 1047: "Approximations"

My favorite one is π=355/113, just to irritate my students who insist π is irrational.

355/113 is 3.14159292, so much better than 22/7.

RyanfaeScotland
Posts: 29
Joined: Mon Oct 17, 2011 3:13 pm UTC

### Re: 1047: "Approximations"

I notice Randall uses billions for the world population but millions for the US population. Why is he being dishonest?