Quadrature of a circle solved
- Thread starter Jason.Marshall
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I think you are a great person I both admire and respect those that have lived a life of many experiences with a story to tell you seem to be a person as such I did not mean to come off so aggressive I believe I have grown wise myself for I too have lived many years on this planet so a word of advice next time you address me please objectively criticise my work with an honest effort with out jumping to conclusions or using sarcasm there is a time and place for that but it is not here and it is not now. I truly am working on something important to me and I have my reasons although they may not seem obvious to anyone else.Geeze Jason, what a waste of time.
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Jason.Marshall:
Your diagram above has a label "perfect circle created with a protractor all points equidistant from the centroid". The formula given for the area of that circle is:
$$\frac{360}{115.2} r^2 = 3.125 r^2$$
The area of a circle of radius r is actually $$\pi r^2 = 3.14159... r^2$$.
My question to you is this:
Do you believe that circles created using a protractor, with all points equidistant from the centre, are different from "perfect Euclidean circles"? If so, please explain why they are different, and why the area is different. I would like to know why the area of your protractor circle does not involve pi.
On a side point: are you using the word "centroid" to mean "centre"? That is, do these two words mean the same thing to you?
Your diagram above has a label "perfect circle created with a protractor all points equidistant from the centroid". The formula given for the area of that circle is:
$$\frac{360}{115.2} r^2 = 3.125 r^2$$
The area of a circle of radius r is actually $$\pi r^2 = 3.14159... r^2$$.
My question to you is this:
Do you believe that circles created using a protractor, with all points equidistant from the centre, are different from "perfect Euclidean circles"? If so, please explain why they are different, and why the area is different. I would like to know why the area of your protractor circle does not involve pi.
On a side point: are you using the word "centroid" to mean "centre"? That is, do these two words mean the same thing to you?
Yes centroid means center
In a sense the two circles are exactly the same one is just magnified because it could be defined rationally my formula proves this by showing the diameters are cofactors of the ratio 1.0027 and all perfect circles with different diameters are the same they are just scaled up or scaled down versions of each other, if that was not true then my formulas would not work out in its obvious perfection.
$$\frac{360}{115.2} = \frac{3600}{1152} = 2 \times \frac{5^2}{4^2} = \frac{25}{8} = 3.125 \approx \pi - \frac{2821}{170015} \approx \pi - \frac{1}{60} \approx \pi$$
$$H^2 = \frac{\pi}{2}$$ and $$ה = \frac{5}{4} $$ mean $$\frac{H^2}{\pi ה} = \frac{2}{5}$$ and $$\left( \frac{H}{ה} \cdot \frac{360}{\pi} \cdot \frac{H}{ה} \right) \cdot \frac{ה}{1} = \frac{2}{5} \times 360 = 144$$.
$$ \frac{H}{ה} = \sqrt{ \frac{8 \pi}{25} } $$
$$\sqrt{ \frac{8 \pi}{25} } - 1 \approx \frac{6683429}{2520802687} \approx {0.00265131} \approx \frac{6}{2263} \approx \frac{1}{337}$$
$$X = \frac{360}{\pi} \times \left( 1 - \sqrt{ \frac{8 \pi}{25} }\right), \quad Y = - \sqrt{ \frac{8 \pi}{25} } X$$
Is that the entire content of the assertions up to now?
$$H^2 = \frac{\pi}{2}$$ and $$ה = \frac{5}{4} $$ mean $$\frac{H^2}{\pi ה} = \frac{2}{5}$$ and $$\left( \frac{H}{ה} \cdot \frac{360}{\pi} \cdot \frac{H}{ה} \right) \cdot \frac{ה}{1} = \frac{2}{5} \times 360 = 144$$.
$$ \frac{H}{ה} = \sqrt{ \frac{8 \pi}{25} } $$
$$\sqrt{ \frac{8 \pi}{25} } - 1 \approx \frac{6683429}{2520802687} \approx {0.00265131} \approx \frac{6}{2263} \approx \frac{1}{337}$$
$$X = \frac{360}{\pi} \times \left( 1 - \sqrt{ \frac{8 \pi}{25} }\right), \quad Y = - \sqrt{ \frac{8 \pi}{25} } X$$
Is that the entire content of the assertions up to now?
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Yes it is this is as far as I have came just the basics... finally you arrive thank you for the simplifications you have helped me a lot in expressing myself. When I get rich I will remember to compensate you handsomely I am not assuming you need money but its just to show my appreciation for taking the genuine effort to critique my work and point out errors if you see them.
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Did you find any errors Rpenner?
Yes. You have your order of operations wrong in your larger "poster" images and you have not solved any problem related to quadrature of a circle.
Can you explain why my proof is not valid?
Your formula's perfection is far from obvious to me.
You seem to be equating 3.125 with pi. They are not the same.
And I agree with rpenner that you haven't addressed the problem of squaring the circle.
Is Rpenner going to give me the same answer?
You have no proof -- you may have a construction which isn't documented well enough to be a proof. If you actually write out the logic of the construction then you have either constructed an approximate quadrature of the circle which fails to be the exact quadrature as claimed or you have cheated by introducing the ratio π in some fashion and ultimately violated the conditions of the proof.
You can't get to √π with straightedge and compass because you can't get to π by any finite sequence of addition, subtraction, multiplication, division and taking square roots. So much of this thread has been nothing but tedious circular reasoning and introduction of weird disguises for π like H or unwanted approximations of π like $$2 \times ה \times ה \quad = \frac{25}{8}$$
Obviously, if you somehow start with two line segments which are in the ratio 1:H, then given any line segment r, you can construct a line of length H r, and then construct an isosceles right triangle with hypotenuse √π r and thus form a square of area π r². The Greeks could do that 2000 years ago, but they knew the difference between constructing ratios like 1:H or 1:√π and using them.
You can't get to √π with straightedge and compass because you can't get to π by any finite sequence of addition, subtraction, multiplication, division and taking square roots. So much of this thread has been nothing but tedious circular reasoning and introduction of weird disguises for π like H or unwanted approximations of π like $$2 \times ה \times ה \quad = \frac{25}{8}$$
Obviously, if you somehow start with two line segments which are in the ratio 1:H, then given any line segment r, you can construct a line of length H r, and then construct an isosceles right triangle with hypotenuse √π r and thus form a square of area π r². The Greeks could do that 2000 years ago, but they knew the difference between constructing ratios like 1:H or 1:√π and using them.
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$$\left( \frac{H}{ה} \cdot \frac{360}{\pi} \cdot \frac{H}{ה} \right) = 360 \times \frac{H^2}{ ה \times ה \times \pi} = 360 \times \frac{1}{ 2 ה \times ה} = \frac{360}{ \quad \frac{25}{8} \quad} = 115.2$$.
The factor of √π that you hid in the two factors of H that you introduced cancels with the factor of π that you also introduced.
The 2 ה² = 25/8 = 3.125 is your approximation to π.
And 360 / 3.125 = 115.2 is the exact same relationship as 360 / 115.2 = 3.125.
It's not a mathematical mystery -- it's a shell game you are playing with π and H and ה and 3.125.
2 H² = π
2 ה² = 3.125
It's not surprising that $$\frac{H^2}{ה^2 \pi} = \frac{1}{3.125}$$ because you introduced those specific definitions for H and ה. It's circular reasoning, not progress.
360 / 3.125 = 115.2 is the exact same relationship as 360 / 115.2 = 3.125. It's circular reasoning, not progress.
oh jeez am really really happy my friend Rpenner you're missing all the fun I wish you could see what I see hope I find the way to show you soon!!
Another question for you Rpenner what would be your method to calculate the square root of 2 using pi??
try this my friend Rpenner you must say this what is your opinion on this please reply
(1/1.25)^2 *3.125 = 2 ....and (1/H)^2*pi=2.00000000000001 approximately so this shows that the Hala-kin construct can calculate the square root of 2 better than pi what have you to say about that my friend???
"Calculate" doesn't mean use circular algebraic logic to point out that if $$H = \sqrt{\frac{\pi}{2}}$$ then $$\sqrt{2} = \frac{\sqrt{\pi}}{H}$$.
I wouldn't use π in any calculation of √2, I would use rational numbers and iterations of Newton's method for calculating √x , $$a_{n+1} = a_n - \frac{f(a_n)}{f'(a_n)} = a_n - \frac{a_n^2 - x}{2 a_n} = \frac{a_n^2 + x}{2 a_n} = \frac{a_n + \frac{x}{a_n} }{2}$$, for x = 2.
$$\begin{array}{r|cccccc} n & a_0 = 1 & a_0 = \frac{5}{4} & a_0 = \frac{3}{2} & a_0 = \frac{7}{4} & a_0 = 2 & a_0 = \sqrt{2} + \epsilon \\ \hline
\\ 1 & \frac{3}{2} & \frac{ 57}{40} & \frac{ 17}{12} & \frac{ 81}{56} & \frac{ 3}{2} & \sqrt{2} + \frac{\epsilon^2}{2 ( \sqrt{2} + \epsilon )}
\\ 2 & \frac{17}{12} & \frac{ 6449}{4560} & \frac{ 577}{408} & \frac{ 12833}{9072} & \frac{ 17}{12} & \sqrt{2} + \frac{\epsilon^4}{4 ( \sqrt{2} + \epsilon )^3 + 8 ( \sqrt{2} + \epsilon )}
\\ 3 & \frac{577}{408} & \frac{ 83176801}{58814880} & \frac{ 665857}{470832} & \frac{ 329288257}{232841952} & \frac{ 577}{408} & \sqrt{2} + \frac{\epsilon^8}{\left(8 ( \sqrt{2} + \epsilon )^3 + 16 ( \sqrt{2} + \epsilon ) \right) \left( ( \sqrt{2} + \epsilon )^4+12( \sqrt{2} + \epsilon )^2+4 \right) }
\\ 4 & \frac{665857}{470832} & \frac{ 13836760443422401}{9784067139197760} & \frac{ 886731088897}{627013566048} & \frac{ 216861505420438657}{153344241061115328} & \frac{ 665857}{470832}
\\ 5 & \frac{886731088897}{627013566048} & \frac{ 382911879137317758256658486640001}{270759586334881080131743906043520} & \frac{ 1572584048032918633353217}{1111984844349868137938112} & \frac{ 94057825066437841361140592933058817}{66508925928136227558136446852868992} & \frac{ 886731088897}{627013566048}
\\ \vdots & \vdots & \vdots & \vdots & \vdots& \vdots & \vdots
\\ \hline
\\ \infty& \sqrt{2}& \sqrt{2}& \sqrt{2}& \sqrt{2}& \sqrt{2} & \sqrt{2} \end{array} $$
After 5 iterations, even our worst guess has 24 decimal digits of accuracy. Our best has 48, and the number of accurate decimal digits roughly doubles, for every column, on every iteration, forever.
If $$ | \epsilon | << 1$$ and $$a_0 = \sqrt{2} + \epsilon$$ then $$a_n \approx \sqrt{2} + \sqrt{8} \left( \frac{\epsilon}{\sqrt{8}} \right)^{2^n}$$.
That's a calculation that you can bank on.
Since we can never express √2 as a finite decimal or rational number, an iterative procedure that improves our approximation greatly at every step is the best we can do.
In geometry, the ratio 1:√n can be constructed exactly*. Start with any two points, A and B. Draw AB. Extend AB to ABC where AB:BC = 1:n (i.e. by counting). At B construct a line perpendicular to ABC, call a point on that line D. Construct the bisector of AC, call it E. Draw a circle with center E and radius AE. Extend the line BD to intersect the circle with center E, call the point of intersection F.
Now 1:√n:√(1+n) :: AB:BF:AF :: BF:BC:CF :: AF:CF:AC because we have similar right triangles. So AB:BF:AF:CF are in the ratio 1:√n:√(n+1):√(n²+n)
So the same method gives two easy constructions for √2.
n=1 (B=E) -> AB:BF:AF:CF :: 1:1:√2:√2
n=2 -> AB:BF:AF:CF :: 1:√2:√3:√6
*Obviously, this exactness is an abstraction in the limit of perfect tools, but then so again everything in geometry is.
Yes. $$\left( \frac{4}{5} \right)^2 \times \frac{25}{8} = \frac{16}{8} = 2$$.
No. Because $$H \equiv \sqrt{\frac{\pi}{2}}$$ you can't say approximately. Your over-reliance on calculators has robbed you of a chance to learn elementary algebraic manipulations.
Thus you waste your time with circular reasoning instead of progress.
$$\left( \frac{1}{H} \right) ^2 \times \pi = \frac{1}{H^2} \times \pi = \frac{1}{ \left( \sqrt{\frac{\pi}{2}} \right)^2 } \times \pi = \frac{1}{\; \frac{\pi}{2} \; } \times \pi = \frac{2}{\pi} \times \pi = 2$$, exactly.
None of your post relates to the calculation of √2 or demonstrates that calculations with H can result in √2 or even 2 without also including π.
I believe it is the unanimous evaluation of all moderators and myself that you have an inflated sense of your own mathematical competency and your obsession blinds you to any chance of further learning. The typical prognosis is that this will result in conflict and necessary marginalization and infraction points. Staff and members have seen this phenomenon many times before.
You have prioritized self-gratification over clear communication with your correspondents, which defeats the purpose of a discussion forum.
You have prioritized self-gratification over questioning if you are using the most reliable methodology, which defeats the purpose of a science forum.
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In geometry, the ratio 1:√n can be constructed exactly*. Start with any two points, A and B. Draw AB. Extend AB to ABC where AB:BC = 1:n (i.e. by counting). At B construct a line perpendicular to ABC, call a point on that line D. Construct the bisector of AC, call it E. Draw a circle with center E and radius AE. Extend the line BD to intersect the circle with center E, call the point of intersection F.
Now 1:√n:√(1+n) :: AB:BF:AF :: BF:BC:CF :: AF:CF:AC because we have similar right triangles. So AB:BF:AF:CF are in the ratio 1:√n:√(n+1):√(n²+n)
So the same method gives two easy constructions for √2.
n=1 (B=E) -> AB:BF:AF:CF :: 1:1:√2:√2
n=2 -> AB:BF:AF:CF :: 1:√2:√3:√6
*Obviously, this exactness is an abstraction in the limit of perfect tools, but then so again everything in geometry is.
Yes. $$\left( \frac{4}{5} \right)^2 \times \frac{25}{8} = \frac{16}{8} = 2$$.
No. Because $$H \equiv \sqrt{\frac{\pi}{2}}$$ you can't say approximately. Your over-reliance on calculators has robbed you of a chance to learn elementary algebraic manipulations.
Yes I must say you are right about this the calculator has robbed me of vital skills and made me lazy...Who knows I could have been a great mathematician had I thoroughly developed my skills then I could get my point across with crystal clear certainty following the formal rules of mathematical discipline.
Thus you waste your time with circular reasoning instead of progress.
$$\left( \frac{1}{H} \right) ^2 \times \pi = \frac{1}{H^2} \times \pi = \frac{1}{ \left( \sqrt{\frac{\pi}{2}} \right)^2 } \times \pi = \frac{1}{\; \frac{\pi}{2} \; } \times \pi = \frac{2}{\pi} \times \pi = 2$$, exactly.
None of your post relates to the calculation of √2 or demonstrates that calculations with H can result in √2 or even 2 without also including π.
I wasn't using "H" Rpenner I don't think you notice this?? and I also never used pi. But I was still able to get an eact calculation of 2.
I believe it is the unanimous evaluation of all moderators and myself that you have an inflated sense of your own mathematical competency and your obsession blinds you to any chance of further learning. The typical prognosis is that this will result in conflict and necessary marginalization and infraction points. Staff and members have seen this phenomenon many times before.
You have prioritized self-gratification over clear communication with your correspondents, which defeats the purpose of a discussion forum.
You have prioritized self-gratification over questioning if you are using the most reliable methodology, which defeats the purpose of a science forum.[/QUOTE]
Who knows this statement may or may not be true by ego stroking I will ultimately gain nothing so I seek not this I truly believe I am correct but I must be explaining it in the wrong fashion and no one else will defend the truth of this discovery so since I could see its value I refuse to turn my back on the discovery, I just demonstrated to you the calculation of the number 2 without using pi, I must find my error in communication because it is not an error in my math.
Now 1:√n:√(1+n) :: AB:BF:AF :: BF:BC:CF :: AF:CF:AC because we have similar right triangles. So AB:BF:AF:CF are in the ratio 1:√n:√(n+1):√(n²+n)
So the same method gives two easy constructions for √2.
n=1 (B=E) -> AB:BF:AF:CF :: 1:1:√2:√2
n=2 -> AB:BF:AF:CF :: 1:√2:√3:√6
*Obviously, this exactness is an abstraction in the limit of perfect tools, but then so again everything in geometry is.
Yes. $$\left( \frac{4}{5} \right)^2 \times \frac{25}{8} = \frac{16}{8} = 2$$.
No. Because $$H \equiv \sqrt{\frac{\pi}{2}}$$ you can't say approximately. Your over-reliance on calculators has robbed you of a chance to learn elementary algebraic manipulations.
Yes I must say you are right about this the calculator has robbed me of vital skills and made me lazy...Who knows I could have been a great mathematician had I thoroughly developed my skills then I could get my point across with crystal clear certainty following the formal rules of mathematical discipline.
Thus you waste your time with circular reasoning instead of progress.
$$\left( \frac{1}{H} \right) ^2 \times \pi = \frac{1}{H^2} \times \pi = \frac{1}{ \left( \sqrt{\frac{\pi}{2}} \right)^2 } \times \pi = \frac{1}{\; \frac{\pi}{2} \; } \times \pi = \frac{2}{\pi} \times \pi = 2$$, exactly.
None of your post relates to the calculation of √2 or demonstrates that calculations with H can result in √2 or even 2 without also including π.
I wasn't using "H" Rpenner I don't think you notice this?? and I also never used pi. But I was still able to get an eact calculation of 2.
I believe it is the unanimous evaluation of all moderators and myself that you have an inflated sense of your own mathematical competency and your obsession blinds you to any chance of further learning. The typical prognosis is that this will result in conflict and necessary marginalization and infraction points. Staff and members have seen this phenomenon many times before.
You have prioritized self-gratification over clear communication with your correspondents, which defeats the purpose of a discussion forum.
You have prioritized self-gratification over questioning if you are using the most reliable methodology, which defeats the purpose of a science forum.[/QUOTE]
Who knows this statement may or may not be true by ego stroking I will ultimately gain nothing so I seek not this I truly believe I am correct but I must be explaining it in the wrong fashion and no one else will defend the truth of this discovery so since I could see its value I refuse to turn my back on the discovery, I just demonstrated to you the calculation of the number 2 without using pi, I must find my error in communication because it is not an error in my math.
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"None of your post relates to the calculation of √2 or demonstrates that calculations with H can result in √2 or even 2 without also including π." Rpenner said
Did you not look at the equation 3.125 (1/1.25) ^2 = 2 here I never used pi or H but my calculation was more accurate than if I used pi or H???
Did you not look at the equation 3.125 (1/1.25) ^2 = 2 here I never used pi or H but my calculation was more accurate than if I used pi or H???