Pi Approximation Day

[Gerson W. Barbosa]

(07-24-2019 10:30 AM)BartDB Wrote:  

(07-23-2019 06:18 PM)Gerson W. Barbosa Wrote:  That is,

π = 22/7 - ∫(0,1,X^4*(1-X)^4/(1+X^2),X)

Similar equalities can be automatically obtained on the HP-50g with help of a small User-RPL program. The Egyptian Fractions part --

  { }
  WHILE SWAP DUP -5. ALOG SQ >
  REPEAT DUP INV CEIL DUP UNROT INV - UNROT +
  END 6. ALOG SQ * SWAP

-- is borrowed code from forumer 3298 here.

For example,

22 ENTER 7

\<< / \->NUM DUP IP R\->I DUP UNROT - { }
  WHILE SWAP DUP -5. ALOG SQ >
  REPEAT DUP INV CEIL DUP UNROT INV - UNROT +
  END 6. ALOG SQ * SWAP NIP DUP SIZE NOT NOT
  { 1 - X SWAP ^ 0 + \GSLIST + } { DROP } IFTE
  4 X 2 ^ 1 + / - COLLECT
\>>

EVAL

--> '(X^8+X^6+3*X^2-1)/(X^2+1)'

Indeed,

'∫(0,1,(X^8+X^6+3*X^2-1)/(X^2+1),X)'

EVAL DISTRIB

--> '-π+22/7'

That is,

π = 22/7 - ∫(0,1,(X^8+X^6+3*X^2-1)/(X^2+1),X)

Notice this is a different integrand polynomial. The original one is more elaborate so that the difference area is continuous, not distributed between both sides of the x-axis.

Likewise,

π = 377/120 - ∫(0,1,(X^61+X^59+X^9+X^7+3*X^2-1)/(X^2+1),X)

and

π = 3 + ∫(0,1,(-3*X^2+1)/(X^2+1),X)