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Change the value of the built-in constant Pi ?

Posted 10 years ago
POSTED BY: Dale Robinson
11 Replies
Posted 10 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....

Addendum

Jain 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....

POSTED BY: Dale Robinson

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 BY: Szabolcs Horvát
7};

In[2]:= g[x_] = x^(n/nx[)6;0

In[2]:= g[x_] = x^(n/nx[)6;0]:= -1M/a2x EIxmt[rNa[Pr
isio n   =N I1n0t0e;grate[(g
 - t)]W)i/t(hS[i{nn[ \=[ P2i3]0 }t,]
, { tN,[ N-eIsntf[i(n2i t#y  /I.,  Sq
, { tN,[ N-eIsntf[i(n2i t#y  /I.,  Sqrt[2] ->   S q r tI[n2f i+n iStqyr tI[}2,] ]W)o r&k,i n8g PSrqercti[s2i o-n  S-q>r t6[02]],] ,2 0n]]], 
 n]
Out][
]= O{u0t.[16817]8=5 966.42284361280563701721072955,8 604.71689728552986647264665250960701527062853}94
875
021I1n6[441]9:4=9 8g8[9x1_8]4 6=1 5x6^3(2n8/\x)

  / 2   I m [ N [1N2I5n7t2e4g1r7a9t9e7[2(5g6[0(619 6+5 0t6)8]4)2/3(4S1i3n5[9\6[4P2i9]6 1t7]3)0,2 6{5t6,4 6-1I3n2f9i4n1i8t7y6 8I9,2 19
164 4 6 3 3 9I\n
nit y        I}, WorkingPrecision  -> 66407]4,9 82105]4]46
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4495O7u4t1[554]1=4 7{806.61589758352926540244863220763761928072851,8 504.513847085579964442045692504667\12
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4495O7u4t1[554]1=4 7{806.61589758352926540244863220763761928072851,8 504.513847085579964442045692504667\12025}

  I n [ 6 ] : =  4g2[1x2_2]9 6=6 1x8^0(5n9/2x9)7;00
7916/444 4I4m2[9N7[0N3I5n7t5e6g4r4a1t5e1[9(0g7[2(319 6+7 4t0)5]6 0-7 4(6g4[9(712 1-3 8t9)3]8)8)9/1(6S\i
\[Pi ]   t ] ) ,   { t1,2 1-3I2n7f0i2n5i5t1y2 3I6,9 20
879 1 9 5 7 6I3n9f3i7n1i1t0y3 9I3}5,8 6W9o8r1k3i9n5g4P7r5e8c1i7s7i3o4n0 9-1>6 56901]5,8 72908]8]37
\
Ou t [ 7 ] =   { 0 .61580738655916648204368290763781026012154,9 609.61088778751916644274765220463781626002055}92
69
89
01
240933525983438\
         366816684508234992463156512294881674612

In[62]:= %61 - N[2 \[Pi], 2 230]
Out[62]= -5.4243834712291862055610789989090382678868805972199128006229870913562404\
          9106372304510215955879291233601848394031005731664413132322487697144048\
          1465451196926136175952959761950159181765162349426015224934226755084498\
          9281529039632922999947944845331120541465222963942167213299797993200467\
          81755555025563023112305372480126725015*10^-140
POSTED BY: Udo Krause

$\pi$ 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 description

Schmiechen

from another book: Michael Schmiechen, Newton's Principia and related 'principles' revisited.

POSTED BY: Udo Krause

enter image description here

POSTED BY: David Reiss

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 BY: Daniel Lichtblau

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 BY: Frank Kampas
Posted 10 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 BY: Dale Robinson
Posted 10 years ago
POSTED BY: Dale Robinson

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 BY: Moderation Team

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