Change the value of the built-in constant Pi ?

Posted 8 years ago
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 Is it possible to change the value of Pi in Mathematica ? I found a mathematician who claims the value of Pi is wrong. He claims it is 3.14460551102969314427823434337183571809248823135089http://www.jainmathemagics.com/page/10/default.aspHe offers a mathematical proof.
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Posted 8 years ago
 Rather than change the built-in symbol Pi, it would be better to create a new symbol, for example,  jainPi = 3.14460551102969314427823434337183571807248823135089; Changing Pi in Mathematica would mess up a lot of internal calculations. It is fine if other replies show how to modify built-in symbols. Please do not flame at Mr. Robinson about the other web site.
Posted 8 years ago
 Thank you. I kind of figured it would be something like that. Most CAD programs use the same value of Pi as Mathematica. I have several goals with this question; I mainly use Mathematica to calculate a very complex polynomial algorithm and it never donned on me that Pi could have an alternate value. This could explain why physical models of the complex CAD part fails. It would be nice to run Jain's mathematical proof through Mathematica and see if he is right or not. It would also be nice to calculate jainPi to the same resolution as he has published. A programmable way to 'modify' built in Pi and a way to 'revert' back to the original Pi would be nice. ===============As to 'Changing Pi in Mathematica would mess up a lot of internal calculations.'This is my point exactly. I ran a simple test on Pi * D = C vs the jainPi * D = C and the difference was significant.'of course D = Diameter and C = Circumference ' I use C / 360 for a constant in my algorithm. And if the Pi constant is wrong then my constant is wrong.FYI Jain claims NASA uses a different value of Pi. But I had another inside source that verified that NASA always had to make course corrections for their space vehicles and could not explain why. Victor Schumberg also had an explanation in that planetary orbits were egg shaped and not truly elliptic.ps I hope somebody flames me for referring to Jain. Because he is basically challenging the world on the true value of Pi. He is also challenging Mathematica :)
Posted 8 years ago
 Mathematica rules JainPi to 60 decimal places 3.14460551102969314427823434337183571809248823135089295065961======================JainPhi = N[(1 + Sqrt[5])/2, 60] JainPi = N[4/Sqrt[JainPhi], 60]
Posted 8 years ago
 Maybe this explains why the Brooklyn Bridge is up for sale. It was constructed using the wrong value of Pi.
Posted 8 years ago
 I won't offer a judgement on the validity of your source, but if for some reason you really really want to redefine the value of Pi, you can do so: Unprotect[Pi]; Pi = 3; and then evaluating Pi will return 3.
Posted 8 years ago
 Please do not change the value in any way that will alter Pi Day, as it is already on the calendars for 2015. Pies will typically be divided into eight slices, and that is not subject to change. You are more than welcome to make adjustments that might have the effect of increasing the area or volume of the slices served on that date. Any increase to the calorie count, however, will be viewed with the utmost of displeasure.
Posted 8 years ago
Posted 8 years ago
 $\pi$ as irrational (Lambert, 1761) and transcendent (Lindemann, 1882) number can be described neither as a rational number nor as a zero of a polynomial of finite degree with rational coefficients nor one can deliver all it's digits. So it needs some definition for $\pi$. One definition is: $\pi$ equals to the ratio of a circle's circumference to its diameter in plane euclidean geometry. When is a plane (in general: a two-dimensional manifold) euclidean? If and only if it's Riemannian curvature tensor vanishes.How can the value of $\pi$ be computed following that? There is the well known construction of inscribed resp. circumventing regular n-polygons, giving In[26]:= (* circumventing regular n-polygon *) 2 Limit[n Tan[180 °/n], n -> \[Infinity]] Out[26]= 360 ° In[27]:= (* inscribed regular n-polygon *) 2 Limit[n Sin[180 °/n], n -> \[Infinity]] Out[27]= 360 ° where $\tan(180°/n)$ resp. $\sin(180°/n)$ is half the length of an edge of the considered regular n-polygon. (If one computes not in euclidean geometry, these lengthes were wrong, of course.) You might consider this tautological, because the Degree of Mathematica contains the $\pi$ in it's definition. It is not, because $\sin$ and $\tan$ do not contain $\pi$ in their series definition. But you can convince yourself that Mathematica uses the right value of $\pi$ with the following consideration: Taking only 2^n regular polygons into consideration, the limit does not change and the circumference of the 2^n regular polygon is expressed by square roots and powers of 2 only: In[52]:= With[{n = 123}, N[Nest[(2 # /. Sqrt[2] -> Sqrt[2 + Sqrt[2]]) &, 8 Sqrt[2 - Sqrt[2]], n], 2 n] ] Out[52]= 6.2831853071795864769252867665590057683943387987502116419498891846156328\ 1257098986162045917883806187127902815967288537806793180869782848721205\ 0555637834194124504452962092149680342499114209242596992338657138220315\ 84148958096646003985015159139416694 In[53]:= %47 - N[2 \[Pi], 123] Out[53]= -1.42813563561047184617226468526801335367923522626*10^-75 What about NASA and their changed value of $\pi$? NASA does not have to do it with euclidean geometry. That's it. They have to get their stuff to the right point in space and time and in order to do so they have to face some nifty disturbations known as the real world. What they in fact use are renormalization techniques: If you have a simple formula, e.g. $W(x,y) = \pi A(x) B(y)$ which does not fit reality, you could go down the model stack considering more complicated models, e.g. $W(x,y)=\pi \int\int A(x')B(y')K(x-x',y-y')dx'dy'$ and so on ... but by chance you sometimes find out, that everything fits perfectly (in the application range) if only you changed $\pi$ a little bit without taking more complicated models into consideration. This is called renormalization and is used in many branches of physics (e.g. the effective mass of the electron in semiconductor physics, dimensional renormalization in quantum field theory, ... ). In our days it has found also the interest of pure mathematics, e.g. search the web for a PDF by Matilde Marcolli called Renormalization for Dummies.Here is some short descriptionfrom another book: Michael Schmiechen, Newton's Principia and related 'principles' revisited.
Posted 8 years ago
 In[60]:= $MaxExtraPrecision = 100; With[{n = 230}, N[Nest[(2 # /. Sqrt[2] -> Sqrt[2 + Sqrt[2]]) &, 8 Sqrt[2 - Sqrt[2]], n], 2 n] ] Out[61]= 6.2831853071795864769252867665590057683943387987502116419498891846156328\ 1257241799725606965068423413596429617302656461329418768921910116446339\ 6474981544670372844957415414786659532250483273698781854534057944059546\ 4212296618059297008796444429703575644151907239674056074649721389388916\ 1213270255123692078791957639371103935869813954758177340916591587988374\ 6503651680389738061149696087711647752438660059276698901240933525983438\ 366816684508234992463156512294881674612 In[62]:= %61 - N[2 \[Pi], 2 230] Out[62]= -5.4243834712291862055610789989090382678868805972199128006229870913562404\ 9106372304510215955879291233601848394031005731664413132322487697144048\ 1465451196926136175952959761950159181765162349426015224934226755084498\ 9281529039632922999947944845331120541465222963942167213299797993200467\ 81755555025563023112305372480126725015*10^-140  Posted 8 years ago  To complement this answer, I'd like to point Mr. Robinson to these articles: These show a detailed description of the calculation Udo Krause mentions with$2^n$sided polygons (or rather,$3\times 2^n$sided ones here), without any references to trigonometric functions such as$\sin$or$\cos$.It seems like Mr. Robinson is not sure whether to believe Jain or the vast majority of scientists who claim otherwise. The good news is that he doesn't need to blindly believe anyone based on their respectability or authority. He can compute an approximation himself. The only practical barrier to calculating upper and lower bounds on the value of$\pi$is the large number of arithmetic operations needed. Other than that, the mathematics is quite basic and only requires Pythagoras's theorem. Fortunately Mathematica is very good at doing arithmetic and can spare us the pen and paper. I strongly recommend that he do the calculation himself, with a bit of help from Mathematica to do the tedious arithmetic.After all the reason why we all need to study mathematics in school is not to memorize boring facts such as "$\pi = 3.1415...\$" but to learn how to do independent deductive reasoning.
Posted 8 years ago
 Cool great food for thought. Especially the lemon merange pi. :) Reading up on this re-normalizing article. Think it is on the right track. The mathematical challenge in this case relates to this physical part.http://www.randcamengine.com/this-is-a-significant-event/Their use of 'complex transcendental formulas' is interesting as one Math Professor criticized Jain for claiming that Pi and Phi are transcendental numbers.NASA geeks ''face some nifty disturbations known as the real world"So now find a value of Pi that gets the job done....AddendumJain describes JainPi as a '4th dimensional equation/polynomial (x^4 + 16x^2  256 = 0)'Jain describes his Pi as the diameter being the square root of 5, whereas Pi as we know it is based on the unit circle of 1....