# this thread is for Rpenner

Discussion in 'Physics & Math' started by Jason.Marshall, Dec 8, 2014.

1. ### Jason.MarshallBannedBanned

Messages:
654
Or anyone else that would like a go at it. Here we go ok check out this link man claims pi = 3.125 if you can prove him wrong he will give you 50,000 crowns http://correctpi.com/ don't get this confused with the Babylonians 357.29.../3.125 everyone keeps saying he is a crank I looked at his work I cant find any holes I would love to get the money but I guess am going to have to say he is correct unless someone can prove otherwise. Please I need to understand why this man is wrong doesn't anyone want an easy 50,000 crowns that's about 6500.00 dollars us. In truth I don't really want the money I just want the truth.
He had this out for about 15 years and I cant seem to find anything that supports he is wrong other than he uses one of the first ratios ever used in pi back in the days of the Babylonians but they did not use it in the way he did please someone anyone show me the light.

Last edited: Dec 8, 2014

to hide all adverts.
3. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

Messages:
6,357
Maybe I could divide circumference by diameter.

to hide all adverts.
5. ### exchemistValued Senior Member

Messages:
11,152
Well, he obviously IS a crank, since if everybody was wrong about the value of pi then nothing in our world of technology would work! Anyone who has devoted 35 years to getting a wrong answer for pi is either mad or lying about it. So the sole question is where the errors are in his maths. The fact they may be difficult to spot does not make him right.

Beer w/Straw likes this.

to hide all adverts.
7. ### James RJust this guy, you know?Staff Member

Messages:
36,948
It's difficult to tell exactly what he is claiming, because his book and the rest of his website is so bogged down in arithmetic.

If he is claiming that the ratio of a circle's circumference to its diameter is 3.125, then he is wrong.

Oh good. You've looked at the whole thing thoroughly then, I assume.

So, can you give me the quick summary of what exactly he is claiming? Thanks.

Fine. Let's work together. You summarise his argument for me and I'll tell you where he went wrong. Then we'll claim the money and I can have it, since you don't want it. You'll have your answer and I'll have the money. Deal?

to hide all adverts.
9. ### Captain KremmenAll aboard, me Hearties!Valued Senior Member

Messages:
12,738
Let me break it to you gently James.
You ain't getting any money no matter what happens.

10. ### exchemistValued Senior Member

Messages:
11,152
It seems this fellow Mohammed-Reza Mehdinia is sufficiently well-known to have made it into the "Wackos' Gallery" : http://wackos.gallery

He's in some great company there - some of these people will make you laugh out loud. But I was interested that the writer seems to think Mehdinia's approach is to misunderstand the method ofArchimedes (the earliest method, based on approximating with polygons inside and outside the circle). He's done well to deduce that. When I attempted to navigate round Mehdinia's website, it was so badly laid out and poorly argued that I could not make head or tail of what he is actually trying to do. He seemed to think Archimedes' method involved triangles or squares, so far as I could tell -but I admit I couldn't be arsed to spend a lot of time on it.

The joke is that it seems the sole reason nobody has claimed the prize is that his stuff is so badly done that nobody can work out what he is on about! What a berk. And why does he pick Swedish Krone, when he is an Iranian? Trying to ape the Nobel Prize? How modest.

(By the way, he is using the classic "prove me wrong" gambit of the crank. Actually the onus is on HIM, as the one who would overturn conventional wisdom, to prove HE is RIGHT.)

11. ### rpennerFully WiredValued Senior Member

Messages:
4,833
Note:
$\sin \pi = 0$
Because
$n \geq 1 \Rightarrow 1024 n^2+1280 n > 241$
We have
$n \geq 1 \Rightarrow \frac{5^{8n+2} \, (1024 n^2+1280 n-241) }{2^{12n+9} \, (4n + 3)!} > 0$
Therefore
$m \geq 1 \Rightarrow \sum_{n=m}^{\infty} \frac{5^{8n+2} \, (1024 n^2+1280 n-241) }{2^{12n+9} \, (4n + 3)!} > 0$
so we have
$\sin 3.125 = \sin \frac{25}{8} = \sum_{n=0}^{\infty} \frac{(-1)^n \, 25^{2n+1}}{8^{2n+1} \, ( 2n +1 )! } = \sum_{n=0}^{5} \frac{(-1)^n \, 25^{2n+1}}{8^{2n+1} \, ( 2n +1 )! } + \sum_{n=6}^{\infty} \frac{(-1)^n \, 25^{2n+1}}{8^{2n+1} \, ( 2n +1 )! } \\ = 221860830602575/13715308044877824 + \sum_{n=3}^{\infty} \left( \frac{ 25^{4n+1}}{8^{4n+1} \, ( 4n +1 )! } - \frac{ 25^{4n+3}}{8^{4n+3} \, ( 4n +3 )! } \right) \\ = 221860830602575/13715308044877824 + \sum_{n=3}^{\infty} \frac{5^{8n+2} \, (1024 n^2+1280 n-241) }{2^{12n+9} \, (4n + 3)!} > 0$
Since $\sin 3.125 \neq \sin \pi$ it follows that $\pi \neq 3.125$, Q.E.D.

12. ### Jason.MarshallBannedBanned

Messages:
654
Hala-kin Proof

like I said I don't care about the money, this Is what I got out of it I was able to present his work in a different form.
He is claiming to have used non Euclidean geometry so here we go he is saying that pi equals 360 degrees/115.2 degrees
he claims that lambert did not prove pi was irrational but it proved that a polygon can never transform into a circle and remain a polygon because a circle is not a polygon so infinitely smaller sides will only give approximations to pi. basically the perimeter of the polygons will be less than a circles circumference or more than a circle circumference but never equal so the true radian will always be incorrect. Just think of this a simple polygon like a square is not equidistance from all points from the center it has normal side lengths as well as hypotenuse length every polygon till infinity must share similar qualities because their corners cannot just magical disappear in Euclidean geometry but in non-Euclidean geometry a straight edge can transform into a curve by co factor Hala-kin ratio 72 * 1.25 to get curve side 90 degrees or 72/ 1.25 to straight side true radian of 57.6 ok I have to will be back to continue this discussion. let me know what you guys think put in the numbers for yourself I will explain in more detail when I get back

Last edited: Dec 8, 2014
13. ### Jason.MarshallBannedBanned

Messages:
654
If the definition of a circle is all points are equidistance from the centroid or center point then how can Archimedes ever expect a polygon to transform into a circle, when all polygons mathematically observed closely cannot have all equidistance points from its center origin? by definition alone this method is invalid and does not seem like a good axiom to start from in the first place. So common sense would tell you that you can only get a perimeter that is either lager or smaller than a natural curve but never equal in Euclidean space. Think about it logically you always have to account for the corners in a polygon they cant just magically disappear? If we didn't have non Euclidean geometry then Einstein's work may have not moved along so smoothly.

14. ### Beer w/StrawTranscendental Ignorance!Valued Senior Member

Messages:
6,357
Archimedes was before calculus.

Are you attacking someone 2500 years old?

15. ### Jason.MarshallBannedBanned

Messages:
654
Just his postulate, when I visual the transformations for his series or anyone else's infinite series for calculating pi I never seen a circle but what I did see was more and more complex polygons that the perimeter they had eventually got bigger than a natural circles circumference because it never equaled the mark just shot past it then kept calculating the ratio of more infinitely complex changing polygons to changing diameters ever so slightly . you see? , the problem in the method has to do with the corners of the polygons will scue the measure of a true curve it will not be equidistant therefore making the perimeter larger that the circumference try and visual an infinite series of pi for yourself and see it for yourself...

Last edited: Dec 8, 2014
16. ### Russ_WattersNot a Trump supporter...Valued Senior Member

Messages:
5,051
Do you think mathematicians are that stupid that they don't know that using polygons finds an approximation, not an exact value? Do you think they haven't figured out what level of precision a certain number of polygons gives?

17. ### Jason.MarshallBannedBanned

Messages:
654
Its clear that its just an approximation what is not clear to non mathematicians is why waste time trying to get better approximations that boost your false postulates that don't even match up with a correct definition of a circle when you can get the correct answer that does match up with the definition of a circle. why not just say pi is equal to the ratio of the perimeter of a polygon to its diameter now there is no contradiction to logic here but an approximation is not the same as the truth just like almost doesn't count.

18. ### Russ_WattersNot a Trump supporter...Valued Senior Member

Messages:
5,051
Not sure what to say -- pi is a useful relation, so it is useful to have an accurate value for it. That seems like it should be self evident. But it is irrational, so you can only ever make approximations for it. That seems like it should be self-evident too.

Last edited: Dec 8, 2014
19. ### rpennerFully WiredValued Senior Member

Messages:
4,833
pi is defined in Euclid geometry, so "claiming to have used non Euclidean geometry" is a non-starter. The nice thing about Euclidean geometry is that you have have similar circles and triangles which are not congruent.
The ratio between the measure of a circle's circumference and diameter in non-Euclidean geometry is a function of diameter (and for general non-Euclidean geometry, of position as well) but in the limit of very small circles, the ratio is again $\pi$. Circles of different diameters are not similar.

$\sin -x = - \sin x$ so $\sin^{-1} -x = - \sin^{-1} x$.

$\pi = \frac{ 2 \int_{-R}^{R} \sqrt{1 + \left( d\, \frac{\sqrt{R^2 - x^2}}{d \, x} \right)^2} dx }{ R - (-R) } \\ = \frac{1}{R} \int_{-R}^{R} \sqrt{\frac{R^2 - x^2}{R^2 - x^2} + \frac{x^2}{R^2 - x^2}} dx \\ = \int_{-1}^{1} \frac{1}{\sqrt{1 - u^2}} du \\ = 2 \int_{0}^{1} \frac{1}{\sqrt{1 - u^2}} du \\ = 2 \int_{0}^{1} \sum_{n=0}^{\infty} \begin{pmatrix} n - \frac{1}{2} \\ -\frac{1}{2} \end{pmatrix} u^{2n} du \\ = \sum_{n=0}^{\infty} \frac{2}{2n + 1} \begin{pmatrix} n - \frac{1}{2} \\ -\frac{1}{2}\end{pmatrix}$
Which, since all terms are positive, converges (slowly!) from below. Thus if we have enough terms, this number is going to be bigger than $\frac{25}{8}$.
$\sum_{n=0}^{\infty} \frac{2}{2n + 1} \begin{pmatrix} n - \frac{1}{2} \\ -\frac{1}{2} \end{pmatrix} \\ = \lim_{N \to \infty} \sum_{n=0}^{N} \frac{2}{2n + 1} \begin{pmatrix} n - \frac{1}{2} \\ -\frac{1}{2} \end{pmatrix} \\ = \pi - \lim_{N \to \infty} \frac{2}{2N+3} \begin{pmatrix} N + \frac{1}{2} \\ -\frac{1}{2} \end{pmatrix} \, _3 F_2 \left( \begin{matrix} 1, \; N + \frac{3}{2}, \; N + \frac{3}{2} \\ N + 2, \; N + \frac{5}{2} \end{matrix} , 1 \right)$

And for large N, $\frac{2}{2N+3} \begin{pmatrix} N + \frac{1}{2} \\ -\frac{1}{2} \end{pmatrix} \, _3 F_2 \left( \begin{matrix} 1, \; N + \frac{3}{2}, \; N + \frac{3}{2} \\ N + 2, \; N + \frac{5}{2} \end{matrix} , 1 \right) \approx 2 \begin{pmatrix} N +\frac{1}{2} \\ -\frac{1}{2} \end{pmatrix} \approx 2 \sqrt{\frac{1}{\pi N}}$

Based on this, you need to add perfectly about 1160 4625 terms before you can prove that $\frac{25}{8} \lt \pi$

Whew, tedious, and exactly why we have mathematical expressions that converge to pi faster.

That's why I gave the sine method first because it requires many less terms to check.

Last edited: Dec 8, 2014
20. ### rpennerFully WiredValued Senior Member

Messages:
4,833
No "if" about it.

On page 15 he attempts to divide squares into three classes, but since he ignores units, his classification system has no value.
Example: if a square has a side of 4 meters, then it's area is 16 meters² and its perimeter is 16 meters. Because the author doesn't understand units, he concludes the perimeter equals the area and puts that square in a certain class. However, if measured in centimeters, the same square has side of 400 centimeters, area 160,000 centimeters² and perimeter 1600 centimeters, which the author would classify the square in a difference class for a choice of units that cannot make a physical difference. The mistake is thinking one can compare apples (units of area) with oranges (units of linear measure). The ratio DIVIDED by perimeter is seen to be 1 meter = 100 centimeters in both cases because this is a sensible operation on these physical magnitudes.

Thus he doesn't have a classification of squares with respect to other squares, but rather squares with respect to a choice of "units of length", abbreviated u.l.

On page 21 he attempts to classify circles by the same valueless system
"When π was used for calculating values of the circle with the diameter 4 u.l. the circumference was larger than the area".
P = 2 π r = π D
A = π r² = π D²/4
If D = 4, then A = 4 π = P, so he is wrong.

What is this magic where D=4 means A = P for both circles and squares? This comes from a study of regular polygons.

Look at an regular polygon of N sides and side of a certain side length L. All the angles and sides are identical. So if you draw the perpendicular bisectors of all the sides, they must meet at a single point in the interior. Let's call this point the center. Let's call the length from the center to the middle of any side, r, and the length from the center to any vertex, R. Because r is perpendicular to a side, we can use the Pythagorean theorem and compute the area of this triangle.
r² = R² – (L/2)² = (4 R²-L²)/4
A₁ = 1/2 r L = 1/2 L √( R² – (L/2)² ) = 1/4 L √(4 R²-L²)

Because the polygon has N such sides, we have the total area, A = N A₁, and the total perimeter, P = NL
So the ratio A/P = A₁/L = r/2 = 1/4 √(4 R²-L²)

This ratio holds for equilateral triangles, squares, regular pentagons, regular hexagons, etc. And since D = 2r for squares , circles, and any regular polygon with an even number of sides, the A/P = D/4 which unsurprisingly equals 1 u.l. when D = 4 u.l.

So how do we go from N-sided polygons to a circle? With the concept of geometric limits.

For a regular polygon, L/2 = R sin (π/N) and r = R cos(π/N) so L = 2 r tan(π/N) and R = r sec(π/N).

So P = NL = 2 N r tan(π/N) = 2 N R sin (π/N) and A = N A₁ = N r² tan(π/N) = N R sin (π/N) √( R² – (R sin (π/N))² ) = N R^2 sin( 2 π /N ) / 2

Thus the behavior of $\lim_{N\to\infty} N \tan \frac{\pi}{N} = \lim_{x\to 0} \frac{\tan (\pi x)}{x} = \pi \left. \frac{d \tan x}{d x} \right|_{x=0} = \pi \sec^2 0 = \pi$ and $\lim_{N\to\infty} \frac{N}{2} \sin \frac{2 \pi}{N} = \lim_{x\to 0} \frac{\sin (\pi x)}{x} = \pi \left. \frac{d \sin x}{d x} \right|_{x=0} = \pi \cos 0 = \pi$ prove the formulas for a circle: $A = \pi r^2 = \pi R^2, \; P = 2 \pi r = 2 \pi R$.

21. ### rpennerFully WiredValued Senior Member

Messages:
4,833
Also the difference $N \tan \frac{\pi}{N} - \pi$ is approximately $\frac{\pi^3}{3 N^2}$ for large N.
$\pi - \frac{N}{2} \sin \frac{2 \pi}{N}$ is approximately $\frac{2 \pi^3}{3 N^2}$ for large N.

So for a regular 36-gon we compute $\frac{25}{8} \lt \frac{9377}{3000} \lt 18 \sin \frac{\pi}{18} \lt \pi \lt 36 \tan \frac{\pi}{36} \lt \frac{3474}{1103}$ which is a geometric proof that $3.125 \neq \pi$. In practice, we know we can't construct a 36-gon with the classic tools of geometric construction, compass and straightedge. But we can construct a regular 40-gon or 48-gon, though precision in manipulating such small sides makes the effort difficult, and we have using classical Euclidean geometry $\frac{25}{8} \lt \frac{122}{39} \lt 20 \sin \frac{\pi}{20} \lt \frac{72}{23} \lt 24 \sin \frac{\pi}{24} \lt \pi$

22. ### Dr_ToadIt's green!Valued Senior Member

Messages:
2,527
Collect your prize! As if that could happen: You'd have to go to Tehran or someplace just as lovely, and who knows what would be waiting?

23. ### rpennerFully WiredValued Senior Member

Messages:
4,833
The middle part of those should read $\lim_{x\to 0} \frac{\tan (\pi x)}{x} = \left. \frac{d \tan \pi x}{d x} \right|_{x=0} = \pi \sec^2 (0 \pi) = \pi \sec^2 0$ and $\lim_{x\to 0} \frac{\sin (\pi x)}{x} = \left. \frac{d \sin \pi x}{d x} \right|_{x=0} = \pi \cos (0 \pi) = \pi \cos 0$, respectively.

Sorry for the sloppiness.

24. ### rpennerFully WiredValued Senior Member

Messages:
4,833
From page 29, we have the definitions
$Q(s) = \left( \ln \sqrt{ \left( e^{\ln s} \right)^2 \times 2 } / \ln e^{\ln s} \right)^2/2 = \left( \ln \sqrt{ s^2 \times 2 } / \ln s \right)^2/2 = \frac{1}{2} \left( \ln \left( s \sqrt{ 2 } \right) / \ln s \right)^2 = \frac{1}{2} \left( 1 + \ln \sqrt{ 2 } / \ln s \right)^2 = \frac{1}{2} \left( 1 + \frac{1}{2 \log_2 s} \right)^2$
$r(s) = \frac{1}{2Q(s)} \\ b = 1 - Q(s) \\ h = 1 - r(s)$

Which already gives more insight into the formula than found in this text. These formulas rest on apparently nothing and are so much noise.

On page 41, it is asserted for a circle inscribed in a square of side S=4, that the ratio of the circle's area to the square's ratio is M = Q(S) = Q(4) = 25/32
The conventional ratio is π/4 ≈ 355/452 ≈ 0.785398 ≈ (25 + 538/4053 ) / 32.

On page 46, it is asserted that for squares and circles of the same area, the side of the square is 7 and the diameter of the circle is 7/M = 224/25 = 8.96
But the conventional diameter is √(49 × 4 /π ) = 14/√π ≈ 328663/41610 ≈ 7.898654

Generalizing from this claim, AreaCircle(s/M) = AreaSquare(s) = s² or AreaCircle(2r) = 4 M² r² = 625/256 r²
which is much smaller than the conventional π r² ≈ 804.25/256 r²
and incompatible with page 41's AreaCircle(4) = 25/2 from the same document.

Code:
Radius Area(Euclid) Area(page 41) Area(page 46)
1.000  3.141592654  3.125000000  2.4414062500
2.000  12.566370614 12.500000000  9.7656250000
3.500  38.484510007 38.281250000  29.9072265625
3.949  48.991883937 48.733128125  38.0727563477
4.480  63.053021195 62.720000000  49.0000000000

Clearly, this isn't viable as circles by these definitions don't share the properties of area and similarity as defined in geometry, nor does the dependence on units reflect any behavior of areas for physical reasons. Add in that the author contradicts himself, this is an example why you should never blindly trust a self-promoting guy on the Internet with a self-published book.

exchemist likes this.
25. ### Farsight

Messages:
3,492
rpenner: I don't know why you waste your time with stuff like this. One can check that a disk is circular with say a compass, then measure its diameter and circumference. With facile threads like this, I wouldn't be surprised if the person asking the question was pimping his own wares.