Copyrighted©PiyushGoel

^{}

3^{2} + 4^{2 }= 5^{2}

9 + 16 = 25

1.**(3/4) ^{2} + (4/4)^{2} = (5/4)^{2}**

0.75^{2} + 1^{2} = 1.25^{2}

0.5625 + 1 = 1.5625

2.**(3/5) ^{2} + (4/5)^{2} = (5/5)^{2}**

0.6^{2} + 0.8^{2} = 1^{2}

0.36 + 0.64 = 1

O.K. Fellas, let’s notice a difference here.

3.**(3/3)^2 +(4/3)^2 = (5/3)^2**

1^2 + **1.3^2** = **1.6**^2

1 + 1.69 = 2.56

Difference => **2.69 – 2.56 = 0.13**

Accuracy =>**0.13/2.69=.048**

4.(3/3)^2 +(4/3)^2 = (5/3)^2

1^2 + **1.33^2** = **1.66^2**

1 + **1.7689** = **2.7556**

Difference => 2.7689 – 2.7556 = 0.133

Accuracy => 0.133/2.7689 = 0.0048

5.(3/3)^2 +(4/3)^2 = (5/3)^2

1^2 + **1.333^2** = **1.666^2**

1 + **1.776889** = **2.775556**

Difference => 2.776889 – 2.775556 = 0.001333

Accuracy => 0.001333/2.776889 = 0.00048

6.(3/3)^2 +(4/3)^2 = (5/3)^2

1^2 + 1.3333^2 = 1.6666^2

1 + 1.77768889 = 2.77755556

Difference => 2.77768889 – 2.77755556 = 0.00013333

Accuracy => 0.0001333/2.77768889 = 0.000048

7.(3/3)^2 +(4/3)^2 = (5/3)^2

1^2 + 1.33333^2 = 1.66666^2

1 + 1.7777688889 = 2.7777555556

Difference => 2.7777688889 – 2.7777555556 = 0.000013333

Accuracy => 0.000013333/2.7777688889 = 0.0000048

8.(3/3)^2 +(4/3)^2 = (5/3)^2

1^2 + 1.333333^2 = 1.666666^2

1 + 1.777776888889 = 2.777775555556

Difference => 2.77777688889 – 2.777775555556 = 0.0000013333

Accuracy => 0.0000013333/2.777776888889 = 0.00000048

So, as you can see its pretty much visible here that irrespective of the precision there is in division, the presence of digits 4 and 8 in the Accuracy Factor is mesmeric to look. This may come under Number Pattern Theory where the operation eventually leads to same numerical digits though quantitatively they are different.

** ****Author: Piyush Goel**

On scienceforums.net, we've implemented a small LaTeX system to allow you to typeset equations (in other words, cut out all the x^2 stuff and make things easier to read for everyone). The basic principle behind it is this: you have a LaTeX string, and you surround it by [math][/math] tags. I'll come to the syntax of the actual string in a moment.

For those who can already use LaTeX (and indeed, those who can't), a few things to note. In the system we've implemented, a tex file is created, surrounding the string you input with a \begin{display} environment so there is no need for $, $, \[ etc. Also note that we've included the standard AMS files for you; if anyone wants any special characters, I'm sure we can probably accommodate your needs.

The images are clickable, so you can see the code that was used to make them by clicking.

Now that's all out of the way, onto some examples

[math]x^2_1[/math] - Indexes (both subscript and superscript) on variables

[math]f(x) = \sin(x)[/math] - A simple function.

[math]\frac{dy}{dxx} = \frac{1}{1+x^2}[/math] - Example of fractions - you can create small fractions by using \tfrac.

[math]\int_{-\infty}^{\infty} e^{-x^2} = \sqrt{\pi}[/math] - A nice integral.

[math]\mathcal{F}_{x} [\sin(2\pi k_0 x)](k) = \int_{-\infty}^{\infty} e^{-2\pi ikx} \left( \frac{e^{2\pi ik_{0}x} - e^{-2\pi ik_{0}x}}{2i} \right)\, dx[/math] - a Fourier Transformation, which is rather large.

I could go on and on, but I'll go onto explain some of the basic syntax of LaTeX.**Syntax***Functions & General Syntax*

Basically put, if you want to write a math equation in LaTeX, you just write it. If you wanted f(x) = 3, then bung that between to math tags and you're done, producing [math]f(x)=3[/math]. Don't worry about extra spaces or carriage returns, because in general LaTeX will ignore them. It does get a little more complex than this, but don't worry about that for now. Remember that any letters you type in will be presumed to be some kind of variable and hence will be italicised.

We also have functions to display more complex things like matrices and fractions, and they have the syntax of having a \ before them, usually followed by some kind of argument. For example, \sin will produce the function sin and \frac{num}{denom} will produce a fraction with a specified numerator and denominator. More on these later.

Also remember that LaTeX is case sensitive, so \sigma is NOT the same as \Sigma.*Subscripts and Superscripts*

This is perhaps one of the easiest things to do in LaTeX, and one of the most useful. Let's, for the sake of argument, say you wanted to write x^{2}. Then you'd write x^{2}, producing [math]x^2[/math]. Notice that you don't necessarily need the { and } in cases where you only have 1 thing in the index, for example x^2. But it does care if you want to write something like [math]x^{3x+2}[/math]. Subscripts are done similarly, but you use the _ operator instead of ^. If you want both subscript and superscript, then use the syntax x^{2}_{1} - which is equivalent to x_{1}^{2}.*Fractions and functions*

As I've mentioned, fractions are generated by using the function \frac{num}{denom}. For example:

[math]\frac{1}{3}[/math]

[math]\frac{7}{x^2}[/math]

If you want smaller fractions, you can use \tfrac, to produce things like [math]\tfrac{1}{2}[/math] which will fit into a line nicely without having to seperate it.

LaTeX has some nice in-built functions like \sin, \cos, etc. I'm not going to write them all down here, but I'll point you to a website at the end of the document that contains them. Likewise, you can write symbols (such as infinity by using \infty) and Greek letters (e.g. \phi, \Sigma, \sigma, etc)*Bracketing*

You can get all your usual brackets just by typing them straight in; for instance, (, |, [, etc. However, sometimes they won't be the right size, especially if you want to write something like (1/2)^{n}. You can get around this by using the \left and \right commands, and then placing your favourite brackets after them. For instance, to write (1/2)^{n}, we have:

[math]\left( \frac{1}{2} \right)^{n}[/math]*Integrals, Summations and Limits*

Integrals can be produced by using \int, summations by \sum and limits by \lim. You can put limits on them all in the right places by using the normal subscript/superscript commands. For instance:

[math]\int_a^b x^2 \,dx[/math]

[math]\lim_{n\to\infty} \frac{1}{n} = 0[/math]

[math]\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}.[/math]*Summary*

There's a lot more things you can do with LaTeX, and I'll try to add to this as time goes by. Have a look at:

http://www.maths.tcd.ie/~dwilkins/LaTeXPrimer/'>http://www.maths.tcd...ns/LaTeXPrimer/ - the LaTeX primer

http://omega.albany.edu:8008/Symbols.html'>http://omega.albany....08/Symbols.html - some symbols that you might find useful.

If you have any questions about the system, send me a PM and I'll try to help

Cheers.

The question is: given a binary string that contains 11 digits what is the probability that there is exactly four 1s in the number. I can solve half of it, I know that 11 digits means 2^11 or 2048 total numbers, but I can't figure out how to find out how many contain exactly four 1s through a formula.

]]>I am interested to know in the context of intrinsic curvature but feel I need to get this concept well understood first.

For example must a mathematical "surface" in a 4-D space be 2-dimensional (like a skin) or is it 3-dimensional (like a volume)?

If it is 3-dimensional,what defines it as a surface?

]]>Lot of mathematicians have proved Pythagoras theorem in their own ways. If you google it you will indeed found hundred of ways.

Meanwhile I was also sure that maybe one day I could find something new out of this incredible Pythagoras theorem and Recently I got something which I would like to share with you.

To Prove: Deriving the equation of area of trapezium using Arcs

Proof: There is a triangle ABC with sides a b and c as shown in the figure.

Now,

Area of ∆ BCEG = Area of ∆ BDC +Area of ⌂ DCEF + Area of ∆ EFG

c^2=ac/2+ Area of ⌂ DCEF + (c-b) c/2

(2c^2– ac –c^2+ bc )/2=Area of ⌂ DCEF

(c^2– ac+ bc )/2=Area of ⌂ DCEF

c(c– a+ b)/2=Area of ⌂ DCEF

Area of ⌂ DCEF=BC(DE+CF)/2

Copyrighted©PiyushGoel

**√S**≈((S/2)/(e/2)+e)/2

in this S is the number we are trying to find the sqaure of and e is the closest perfect square regardless of weather it is high or low.

For exaple, using 1,863 the number I was initially trying to find the sqaure of, we know that the closest perfect sqaure is 43. So it would look like this

√1,863≈((1863/2)/(43/2)+43)/2

√1,863≈((931.5)/(21.5)+43)/2

√1,863≈(43.325+43)/2

√1,863≈ 86.325/2

√1,863≈ 43.163

Actual square of 1863=43.1624

so the estimate it yielded was very close. I've run a bunch of numbers through it both large and small and so far it appears to give better estimates than the BM (depending on the accuracy of your initial high estimate and number of times you run it through the equation) and without the requirement that the estimate is high, and also we don't have to make an initial estimate ourselves as the starting estimate is fixed for us by the nature of the equation.

What do you guys think? Is there something I've missed or perhaps is there an easier equation like it that I am not familiar with?

Thanks!

Sorry left out a couple words in my haste lol. E is the closest perfect square root.

]]>
You might actually do the experiment and find that the results really did statistically converge to pi as the sample size grew larger or you might find the mathematical solution would indeed result in a probability that is exactly equal to pi. You might notice that there is a cosine of the angle between the needle and the lines on the table involved. You might even be able to construct a circle to describe how the cosine function relates to a circle using trigonometry, but even then you still probably wouldn’t truly have a clear and direct physical understanding of why the problem of randomly scattered needles should be related to circles.

Here, I don’t present the mathematical solution which you can look up online from a number of sources. Instead I present a simple and logical model that explains the problem in the proper physical context which in turn will make it clear why circles are related to randomly distributed needles.

Once again, Once you understand the solution it will seem simple to you as all logical solutions that are properly explained will seem relatively simple compared to the math. Perhaps it will even seem like it should be obvious once you understand it and you may not understand why you didnt think of it before but unless you could physically explain it in fore-site before hearing this solution, then it clearly wasn’t really as obvious in fore-site as it may seem in hind-site.

I present this solution to you not just to show off that I have a gift for solving logic problems, but to provide yet another example that shows why logic really is just as important as math and that logic and math are not the same thing. Neither is logic just an alternative method to mathematics for solving problems that can be used as a substitute for math. It actually performs a completely different function from the math as I’ve said many times before: Logic clarifies our understanding of the problem while math quantifies the numerical results of the properties involved that can then be compared to experimental results. In fact math and logic are actually complementary opposites.

Assume that you have a needle that has a length of *l* and a surface that has parallel lines on it that are all equally spaced at *2l* distance apart.

If you toss a needle in a random manor on that surface such that it can land in any arbitrary position and orientation, then the probability that the needle lands inbetween the lines divided by the number of times that the needle will intersect with a line will be equal to pi (π).

Another words your results will approximate 3.14... etc. with a sufficient sample size and the larger your sample size the better your approximation of π should be.

The mystery of this method is why does it approximate π which we know is a constant that must be somehow related to a circle when there seems to be no circles involved with this method of randomly scattering needles.

There is actually a simple logical explaination for this mystery and to understand it more easily I will provide a probabilistically equivalent scenario to illustrate why.

Instead of using needles we can use clear plastic discs that have the needle embedded in the disc such that they perfectly bisect the circumference of the discs. After all it will still represent a random position and orientation just as the needles would. In fact they would probably be more random than the needles themselves since needles are not perfectly symetrical and they may be tossed in such a way that may be biased while the disc surrounding the needle would ensure a more random or unbiased result.

Given in this new context, it should now be clear that the source of pi is linked to the circular shape of the disc.

Put another way, think of the position of the disks and the orientation of the needles as independant properties. It is the probability distribution of the needle’s orientation that has an even disc like distribution about their center of gravity. By taking all the angles accross the entire sample space then the orientations of all the needles would stack up to be a random sample of all angles between 0 and 2π or between 0 and π if the needle is symmetrical. So you can see that on average, the orientations of all the needles should combine to be distributed in the shape of a disk.

If by some extreme long shot, they did not create a reasonable disc like distribution, then you probably would not get a reasonable approximation of pi as your result.

]]>

everything dean has shown was known at the time godel did his proof but no one meantioned any of it

http://gamahucherpress.yellowgum.com/books/philosophy/GODEL5.pdf

look

godel used the 2nd ed of PM he says

“A. Whitehead and B. Russell, Principia Mathematica, 2nd edition, Cambridge 1925. In particular, we also reckonamong the axioms of PM the axiom of infinity (in the form: there exist denumerably many individuals), and the axioms of reducibilityand of choice (for all types)”

note he says he is going to use AR

but

Russell following wittgenstien took it out of the 2nd ed due to it being invalid

godel would have know that

russell and wittgenstien new godel used it but said nothing

ramsey points out AR is invalid before godel did his proof

godel would have know ramseys arguments

ramsey would have known godel used AR but said nothing

Ramsey says

Such an axiom has no place in mathematics, and anything which cannot be

proved without using it cannot be regarded as proved at all.

This axiom there is no reason to suppose true; and if it were true, this

would be a happy accident and not a logical necessity, for it is not a

tautology. (THE FOUNDATIONS OF MATHEMATICS* (1925) by F. P. RAMSEY

every one knew AR was invalid

they all knew godel used it

but nooooooooooooo one said -or has said anything for 76 years untill dean

the theorem is a fraud the way godel presents it in his proof it is crap

]]>Also, is there a general equivalent of the quadratic formula for circles?

I would like to basically have a simple equation that gives me the coordinates of an intersection (x, y, z) for a given value of x, y, and time.

After that, the derivative of an intersecting function is going to be treated as an axis for a change in slope of the other function over time (the slope will 'reverse' over the perpendicular line to the other function's derivative at the point of intersection).

The equations would be for wave equations when interactions between infinitesimally thin waves are involved.

Right now I have written the equations for this type of wave interaction, but only if I can make up a value for the other function's origin for every single intersection I am evaluating. I can't yet use them to determine what each secessive point of intersection will be without having to change the radius and origin of one of the functions for each intersection I calculate.

I am not sure if the term would be 'implicit', but with circular and spheroid functions I think what I am looking for would be called a "Implicit Quadratic Formula", as they involve variables paired with a constant together as a square root, which of course don't have a simple algebraic way of being determined. Rediscovering a proof for those types of intersections would take me weeks/months/years, and I have already tried and failed numerous times.

With graphing algorithms, I could approximate the change in intersections over time to a certain number of decimal places accurately through repeatedly estimating factors and adjusting them based on the range of difference, but this is labor intensive even for a computer. It'd be a lot easier if there was a more general formula like the Quadratic formula; does anyone know if it exists?

I couldn't just find the intersection between one function and the line equation perpendicular to the other's derivative, because this does not tell me what the first point of intersection would be for that perpindicular - there must be an equal increase in each function's radius over time.

]]>

How do we move on from there if we want to build a model with 4 such planes ,all presumably orthogonal to each other in the same way?

To my untrained eye it seems that the new( 4th) plane will have "nowhere to go"

ie ,if it is orthogonal to the x=0 plane then it *feels* like it* should *be identical to the y=0 or the z=o plane which already occupy those areas("area" is a poor choice of word ,but I hope the meaning is clear)

How do we "shoehorn " this extra plane into the model when it feel like there is no space available.

PS I am obviously thinking about the spacetime model but the difficulty I have seems to apply to any 4d model with 4 spatial axes (which is what the spacetime model can be viewed as sine "ct" is a spatial distance)

]]>

Can this be considered a field?

Can this be considered a solution for division by zero?

Can this sufficiently create varying amounts of zero?

Allow that there exists an integer zero element ( -0 ).

0 =/= (-0)

|0| = |-0|

0 |=| (-0)

Where |=| is defined as “Zero Element Equivalency”, where any two unique or similar additive identities are considered equal because they share the same absolute value and cardinality but may or may not possess different multiplicative properties.

Allow that :

0: possess the additive identity property and possess the multiplicative property of zero.

(-0): possess the additive identity property and possess the multiplicative identity property.

The addition of any two additive identities is not expressible as a sum, except with |=|.

0 + 0 =/= 0

0 + 0 |=| 0

0 + ( -0 ) |=| 0

( -0 ) + ( -0 ) |=| 0

Where n =/= 0:

n + 0 = n = 0 + n

n + ( -0 ) = n = ( -0 ) + n

Multiplication of any two additive identities is not expressible as a product, except with |=|.

0 * 0 =/= 0

0 * 0 |=| 0

0 * ( -0 ) |=| 0

( -0 ) * ( -0 ) |=| 0

Where n =/= 0:

n * 0 = 0 = 0 * n

n * ( -0 ) = n = ( -0 ) * n

1 * 0 = 0 = 0 * 1

1 * ( -0 ) = 1 = ( -0 ) * 1

The division of any two zero elements is not expressible as a quotient, except with |=|.

0 / ( -0 ) =/= 0

0 / ( -0 ) |=| 0

( -0 ) / 0 |=| 0

( -0 ) / ( -0 ) |=| 0

Where n =/= 0:

0 / n = 0

( -0 ) / n = ( -0 )

n / 0 = n

n / ( -0 ) = n

Therefore the multiplicative inverse of 1 is defined as ( -0 )

1 * ( -0 ) = 1

1/( -0 ) * ( -0 )/1 ) = 1

0 remains without a multiplicative inverse.

Examples containing the distributive property:

a( b + c) = a * b + a * c

Where: a=1, b= 0, c=0

1( 0 + 0) = 1* 0 + 1* 0

1 * 0 = 1 * 0 + 1 * 0

Where: a=1, b=0, c=( -0 )

1( 0 + ( -0 ) = 1 * 0 + 1 * ( -0 )

0 + 1 = 0 + 1

Where: a=1 b=( -0 ) , c=( -0 )

1( ( -0 ) + ( -0 ) ) = 1 * ( -0 ) + 1 * ( -0 )

1 + 1 = 1 + 1

Therefore, non-zero elements divided by zero elements are defined.

Therefore, the product of non-zero elements multiplied by zero elements is relative to which integer zero element is used in the binary expression of multiplication.

The rules for exponents and logarithms exist without change. It continues that multiplication of any zero elements by any zero elements is not expressible as a product except with |=|.

n^0 = 1

n^(-0) = 1

0^0 = 1

( -0 )^0 = 1

0^( -0 ) = 1

( -0 )^( -0 ) = 1

0^n = 0

( -0 )^n = 1

0^(-n) = 1

( -0 )^(-n) = 1

log_{0} |=| 0

log_{( -0 )} |=| 0

]]>

How about 0^0.

]]>

How do I take the derivative of dln(y)/dln(x)

ln is natural log...

Can someone do like a step by step kind of a thing...

]]>
$$

\dot{x_1}(t) = \alpha_1 f_1(x_1,t) + \beta_1 u(t) \\

\dot{x_2}(t) = \alpha_2 f_2(x_2,t) + \beta_2 u(t)

$$

They are constrained by an algebraic equation

$$ x_1(t) + x_2(t) = k $$

where $\left( \alpha_1,\alpha_2, \beta_1,\beta_2 , k \right) \in \mathbb{R}$ are known constants (i.e. parameters). $f_1(t)$ and $f_2(t)$ are both unknown.

Starting from a rich set of input-output **noise-free** data available from simulating a complex proxy system, what would be the best procedure to identify (even a subset of repeatable/characteristic properties) the unknown **_possibly time-varying_** functions $f_1(x_1,t)$ and $f_2(x_2,t)?$ I am almost certain that $f_1(x_1,t)$ and $f_2(x_2,t)$ are both linear.

I am looking for a grey-box system-id approach that shall work well to arbitrary excitations in all future simulations. (NOT merely a curve-fitting procedure to match a specific excitation input-output dataset)

]]>

This is my final work on the Prime product problem.

I know it is just x^2 * y^2 = PNP^2

However the terms would just cancel out. Instead I have decided to let x^2 equal a pattern of x and PNP. So I just substituted the equation which is more complex and will not equal the right side of the equation for x^2.

In calculus where you have a complex derivative where you let du/dx equal a portion of the derivative so you can understand and simplify the manipulation of the integral. I am instead taking a more complex pattern and leaving it so x^2 does not cancel x squared. By doing this I hope it solves the pattern.

So if you could solve this polynomial equation you would solve the factorization problem.

If you couldn’t solve the polynomial? Well you could just write an algorithm that plugged in Prime numbers from smallest to largest. And because the polynomial is set up to find PNP you would get a feel for the range x was in. I mean, that this time when you try a number how far away the computed value is from PNP is significant.

So if this works it is faster than using division to factor.

But of course I await any disagreeing opinions. I now this problem gets that. But it was my final attempt before moving on to a different problem to pursue.

(((((x^2*PNP^4 + 2*PNP^2 * x^5) + x^8)/ PNP^4 ) - ((1 - x^2/(2*PNP)))) * ((PNP^2/x^2 ))) == PNP^2 Above is the pattern of x^2 * y^2 = PNP^2 It is not to be simplified yet x put and tested in that place. It is faster than division since the equation approaches PNP as the proper x is used. Smallest to largest Prime numbers are to be used. PNP = 85 x = 5 (((((x^2*PNP^4 + 2*PNP^2 * x^5) + x^8)/ PNP^4 ) - ((1 - x^2/(2*PNP)))) * ((PNP^2/x^2 ))) 85 5 4179323/578 N[4179323/578, 14] Sqrt[7230.66262975778546712802768166089965397924`14.] 85.033303062728 ((((x^2*PNP^4 + 2*PNP^2 * x^5) + x^8)/ PNP^4 ) - ((1 - x^2/(2*PNP)))) 4179323/167042 N[4179323/167042, 14] 25.019593874594 ((PNP^2/x^2)) 289 Questions to ask: Is it unique to factors of PNP or does it just give a decimal to all real numbers? Does it work for other values of PNP and x? Is it faster than factoring (recursion)?. Is it just x = x and as so not a useful pattern? Can the error be programmed? Verify then post. That is what I need to do. But I actually believe there is a pattern here. The question is does it work for all PNP and x values. I have many patterns. Some answer some of the questions. But I haven't found a polynomial I can solve after these questions have been answered. For example if this worked, I would need to solve the given equation. And this proves to be challenging. PNP = 85 x = 3 (((((x^2*PNP^4 + 2*PNP^2 * x^5) + x^8)/ PNP^4 ) - ((1 - x^2/(2*PNP)))) * ((PNP^2/x^2 ))) 85 3 847772947/130050 N[847772947/130050, 14] 6518.8231218762 Sqrt[6518.8231218762] 80.739229138481]]>

On 2018. 01. 27. at 8:44 AM, Strange said:

On the other hand, mathematics shows us that there are an infinite number of real numbers between 0 and 1 and between 1 and 2. These infinities are the same "size" (but larger than the infinite number of integers).

On 2018. 01. 27. at 5:16 AM, OroborosEmber said:

if this is true why we need any numbers beyond 0 and 1? Feels like different scaling of the natural numbers.

What would be the difference between the infinite numbers between 0 and 1 or the infinite numbers between 0 and 1 000

]]>If I roll a fair dice I know I have a one in six chance of getting a "1".

What is my chance of getting a "1" if I roll it twice? (It could be either on the first roll or the second roll or both. I am just concerned with getting a "1" at some point.)

What is my chance of getting a "1" if I roll it ten times? (Same thing: I'm just concerned with getting a "1" (could be more than one "1") at some point.)

I realize this may be too basic of a question for this forum, but I'm hoping someone will have the time to help me out with this really simple question.

Thank you.

]]>

So I'll start off:

**1)** http://linuxfreak87.googlepages.com/

**1)** Covers a lot of stuff, Maths and some physics.

**2)** http://mathworld.wolfram.com/

**2)** Amazing maths resource, lot of advanced stuff.

**3)** Basic and advanced maths here. Good tutorials.

**4)** Again more good tutorias and weekly challanges.

**5)** http://mathforum.org/dr.math/

**5)** LOTS of question solutions here, examples too. This one has helped me a lot in the past and still does

**6)** http://www.ics.uci.edu/~eppstein/junkyard/

**6)** Lots of fun geometry, useful stuff and interesting stuff here.

**7)** http://en.wikipedia.org/wiki/Category:Mathematics

**7)** As always Wikipedia is a great resource for one and all.

**8)** http://www.research.att.com/~njas/sequences/Seis.html

**8)** If your interest is number sequences this is the place to go. Useful for research.

**9)** Equations, equations and yes you guessed - MORE equations. Very useful resorce for reference.

**10)** http://home.att.net/~numericana/

**10)** Lots of interesting stuff and some other useful links too.

**11)** http://www.mcs.surrey.ac.uk/Personal/R.Knott/

**11)** Lots of interesting stuff, the mysteries of the Fibonacci Numbers etc.

**12)** http://integrals.wolfram.com/

**12)** Very useful too, online integral solver!

**13)** Maths in music, what next?

If you have more to add pease share them

Cheers,

Ryan Jones

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I am trying to measure a curved profile of a surface(2D) to determine the surface availability at different rate of testing. I have attached an image for a rough picture. Actually I have a reference geometry and performed two tests to determine the behavior. They two had different profiles at top portion. So I considered an axis for reference geometry and drew radial lines with 15deg angle difference each and marked those points as 1 (15 deg anticlockwise from vertical axis),2(30 deg anticlockwise from vertical axis),3,... and measured length of the radial lines which gives the distance of the top profile at different points(L1 at pt1, L2 at pt2...). I did the same on my first test results and got the lengths L1',L2'...and with second test lengths L1",L2".... Now I estimated the deviation of the profile by calculating the error like this test1: Error = ((L1-L1')/L1)*100% test2: Error = ((L1-L1")/L1)*100% As the number of experiments increases it is a bit tedious to divide lines based on angles and measure the deviation of the top profile and hence would like to look for an alternative. One approach would be to eliminate the axis and split the complete area with square grids and get the coordinates (x,y) and find the error, but this seems not too good. Could any one please suggest an easy method to do find out the difference in the top profile? Thanks in advance |

What's a Piangle? Maybe this will make it clear.

The Piangle is an unraveled circle. Imagine cutting a radius, then draw some inner circles.

Next unroll each outline to the right.

This is a right triangle, so by the Pythagorean theorem the length of the hypotenuse is , which is or .

The Piangle is not distorted, it's just an unrolled circle. It even has the same area as its corresponding circle. Its area is 1/2*b*h = = .

Proof that I discovered this: the hypotenuse = ≈ 6.3622651. Googling that doesn't return anything about the triangle.

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