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

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He likes to play smoke and mirrors with symbol letter thingies.

3. rpennerFully WiredValued Senior Member

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That doesn't follow, that only demonstrates that $3.125 \neq \pi$ both of which follow from $0 < 3.125 < \pi$.

Gaddy's pi is generally accepted to be $\sqrt{2} + \sqrt{3}$. This is not the same as 3.146264 because $0 < \frac{1}{2703127} < \sqrt{2} + \sqrt{3} - \frac{3146264}{10^6} < \frac{1}{2703126}$.

Thus Gaddy's (approximation to) pi is the largest root of $x^4-10 x^2+1 = 0$

My geometric construction approximated pi with $\left( \sqrt{\frac{3}{2}} + \sqrt{\frac{3}{10}} \right)^2 = \frac{9 + 3 \sqrt{5}}{5}$ which is the largest root of $25 x^2-90 x+36 =0$.

Since $\left( \frac{9 + 3 \sqrt{5}}{5} \right)^4 - 10 \left( \frac{9 + 3 \sqrt{5}}{5} \right)^2 +1 = - \frac{419 - 108 \sqrt{5})}{625} < 0$ it follows that my construction does not equal Gaddy's approximation.

No, it should not. Approximately pi doesn't equal pi. Approximately 1 does not equal 1.

Because $\frac{1}{215} < \sqrt{2} + \sqrt{3} - \pi < \frac{1}{214}$ it is not too surprising that $\frac{1}{428} < \tan \frac{\sqrt{2} + \sqrt{3}}{4} - 1 < \frac{1}{427}$.

Likewise because $\frac{1}{215} < \frac{3146264}{10^6} - \pi < \frac{1}{214}$ it is not too surprising that $\frac{1}{428} < \tan \frac{3146264}{4 \times 10^6} - 1 < \frac{1}{427}$.

Last edited: Dec 12, 2014

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What is the area of your circle?

7. Jason.MarshallBannedBanned

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Last edited: Dec 12, 2014
8. Jason.MarshallBannedBanned

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[

No, it should not. Approximately pi doesn't equal pi. Approximately 1 does not equal 1.

Because $\frac{1}{215} < \sqrt{2} + \sqrt{3} - \pi < \frac{1}{214}$ it is not too surprising that $\frac{1}{428} < \tan \frac{\sqrt{2} + \sqrt{3}}{4} - 1 < \frac{1}{427}$.

Likewise because $\frac{1}{215} < \frac{3146264}{10^6} - \pi < \frac{1}{214}$ it is not too surprising that $\frac{1}{428} < \tan \frac{3146264}{4 \times 10^6} - 1 < \frac{1}{427}$.[/QUOTE]

Yes you are correct there is no argument here.

9. rpennerFully WiredValued Senior Member

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I pointed out you failed to carry your burden of proof for the claim, a statement of fact which requires you to provide information, argument and clear reasoning, not ask more questions which are beyond your educational background.
Your problem appears to be with Euclid.
Since all circles in the diagram have radius equal to the length AB, the blue square with the red boundary has an area equal to the circle. What is at issue is
• What separates the steps used to make your diagram in post #106 to my diagram?
• Was there any legitimate construction of the ratio 1:√π in your diagram when there is no connection between my green lines and my red lines?
and since you have not answered those questions in the specific way that mathematics says you cannot, then I again assert that your have failed to carry your burden of proof for the claim made with respect to the diagram in post #106, and that your diagram is a fraud and a humbug.

10. Jason.MarshallBannedBanned

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The point no one makes is everyone is stuck in a sqaure universe of euclidean space you cannot merge them accurately without reaching infinity that defines the boundaries of one dimension to a next infinities only desribe boundaries of the given dimension that is percieved it does not describe existence or conscious awareness because that is Gods domain and is described as infinity + 1 these are just conceptual example metaphors I have not yet verified any math with a valid proof as yet to back up all of the claims but some of them are definitely clear to me already just from a stand point of common sense.

Last edited: Dec 12, 2014
11. krash661[MK6] transitioning scifi to realityValued Senior Member

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have you ever heard of god's algorithm or/and 10^10^123

12. rpennerFully WiredValued Senior Member

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Archimedes asserted that a circle's area and perimeter are intermediate between those of a regular polygon inscribed in that circle and a regular polygon circumscribed about it. What, specifically, do you find fault with in that statement?

Then we have for circumference, C, of a circle of radius R, the relation, $\forall n,m \in \mathbb{N}, n, m \geq 3 \rightarrow 2 n R \sin \frac{\pi}{n} \lt C \lt 2 m R \tan \frac{\pi}{m}$ and for Area, A, $\forall n,m \in \mathbb{N}, n, m \geq 3 \rightarrow n R^2 \sin \frac{\pi}{n} \cos \frac{\pi}{n} \lt A \lt m R^2 \tan \frac{\pi}{m}$.

Not only is this compatible with $C = 2 \pi R$ and $A = \pi R^2$, but this requires them to be true.

If you wish to argue for different formulas, then you must use different axioms and do all the heavy lifting yourself, but even though you will be right in your little universe, you won't be talking about the circles of Euclid.

Not part of Archimedes's argument.
$3.125 = \frac{25}{8}$ regardless of what Archimedes might have said.

13. Jason.MarshallBannedBanned

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like the minority report and the concept used in limitless crude comparisons to God functionally but compared to a human stand point it will be all you need just like in the movie limitless its easy to control a planet when the majority of the individuals on it are glorifiers of ego this is why elites will always control markets and enslave everyone else to build their singular vision. These are some of the reasons on why I will write the paper "minority report" because sometimes the solutions are not as complex as they may seem but it all starts with the defeat of mankind's master the ego.

14. Jason.MarshallBannedBanned

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Last edited: Dec 13, 2014

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While you're at it, you might try to really master the run-on sentence and the mindless logorrhea that seems to afflict your posts. Just saying...

16. Jason.MarshallBannedBanned

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thank you for your help in pointing out my horrible grammar I am fully aware of that fact.

17. krash661[MK6] transitioning scifi to realityValued Senior Member

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wait, what ? wtf ?
are you serious ?

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No, he's a

, obviously.

19. Jason.MarshallBannedBanned

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I may have missed your point if I did just disregard the message I will admit I assumed I knew what you were talking about without proper investigating but yes I am serious. This is just my opinion you need not take it as fact. and I do believe existence is created by a conscious creator "God" hence your probable calculations 10^ 10^ 123 and I also believe we were not the only creations of intelligent beings that was created by God

Last edited: Dec 13, 2014
20. Beer w/StrawTranscendental Ignorance!Valued Senior Member

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That's some pretty funky stuff!

Thanks.

21. OnlyMeValued Senior Member

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I believe you have gotten your forums mixed up, this thread is Physics & Math, in the Science section... Not religion or theology.

22. Russ_WattersNot a Trump supporter...Valued Senior Member

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The problems here go well beyond grammar and remedial math. The writing isn't disorganized because of poor understanding of grammar, it is disorganized because of disorganized thought processes. That's a problem for a psychiatrist, not a math or English tutor.

23. rpennerFully WiredValued Senior Member

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I never said anything about two polygons meeting, so this didn't address the question. I'm talking about area and perimeter.
I don't think so, but then again you never defined your terms or made an argument, so I'm not obliged to follow your train of thought.
I never said they were not, so this didn't address the question. I'm talking about area and perimeter.
Then stop doing investing hallucinogens and start doing math.
No it wasn't. I never said anything about polygons substituting for circles, so this didn't address the question. I'm talking about area and perimeter.
No one is talking about points, so this didn't address the question. I'm talking about area and perimeter. In geometry, everything that has non-zero measure is made of "infinite points" so you are inventing off-topic stumbling blocks for yourself without thinking about the geometry.
Well, then, you are ignoring the meaning of regular polygon and circle. Those terms have definitions. Nothing in the question requires verification of equidistance because all assumptions of equidistance are built into the definitions used.

If we have a line marked off as AB, and construct a perpendicular BC, extend AB to ABD such that AD = AC, and then construct a perpendicular at D such that it meets AC at E, then we have the following relationships between lengths:

AB² + BC² = AC²
AB:AD = BC: DE = AC:AE

It's pretty clear that a circular arc with center A and radius AD cuts the quadrilateral BCED at C and D. Thus the area of the circular sector has to be less than the area of the triangle ADE while it must be larger than the area of triangle ABC.

Since the area of triangle ABC is in ratio to the area of triangle ADE by AB²:AC² = AB²: (AB² + BC²) it follows that in the limit of BC → 0, the ratio of the areas approaches 1 and the area of the circular sector has to be sandwiched to the common ratio.

So the only question left to determine is how many angles of ∠BAC make up a full rotation or four right angles.

If AB = BC, we have ∠BAC = ∠BCA and so ∠BAC = ½ of a right angle so 8 copies of this diagram can be fit in a circle of radius = AC.

∠BCA = tan⁻¹ ( BC/ AB) so in general we need ( 8 tan⁻¹ 1 )/(tan⁻¹ ( BC/ AB)) copies of this diagram to fill a circle of radius = AC.

Thus the area of a circle of radius R = AC is
sup { ½ ( 8 tan⁻¹ 1 )/(tan⁻¹ ( BC/ AB)) × (AB) × (BC) : BC ∈ ℝ⁺ } = inf { ½ ( 8 tan⁻¹ 1 )/(tan⁻¹ ( BC/ AB)) × (AB² + BC²) × (BC) / (AB) : BC ∈ ℝ⁺ }
sup { ½ ( 8 tan⁻¹ 1 )/(tan⁻¹ ( x )) × R² × x / (1 + x²) : BC ∈ ℝ⁺ } = inf { ½ ( 8 tan⁻¹ 1 )/(tan⁻¹ ( x )) × R² × x : BC ∈ ℝ⁺ }
R² × sup { ½ ( 8 tan⁻¹ 1 )/(tan⁻¹ ( x )) × x /(1 + x²) : x ∈ ℝ⁺ } = R² × inf { ½ ( 8 tan⁻¹ 1 )/(tan⁻¹ ( x )) × x : x ∈ ℝ⁺ }
R² × sup { π (x + x⁻¹)/ tan⁻¹ x : x ∈ ℝ⁺ } = R² × inf { π x / tan⁻¹ x : x ∈ ℝ⁺ }
π R²

http://en.wikipedia.org/wiki/Infimum_and_supremum