Pi Approximation Day

[Albert Chan]

Best Pi approximation using decimal digits permutations:

85910 / 27346 = 355 / 113 ≈ 3.141592920, abs error ≈ 2.67e-7

Proof:

Since 355/113 is one of Pi's convergents, and the next convergent is 103993/33102
A better ratio, if exist, must be a semi-convergent: (52163+355k)/(16604+113k), k>=0

(52163+355k) + (16604+113k) (mod 9) ≡ 7 ≠ 0

Thus, better ratio does not exist.

Update: With hex digits permutations, this produce best pi approximation

0xFE86C29B / 0x5104AD73 = 4270244507 / 1359261043 ≈ 3.141592653590, abs error ≈ 2.94e-13

[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)

[Bill Duncan]

Yeah, mine was always 355/113. I used it a lot (with 4-bangers!) before calculators became available with a dedicated Pi key. It was easy to remember and pretty darned accurate.

[Gerson W. Barbosa]

(07-22-2019 03:14 PM)Gerson W. Barbosa Wrote:  Now, time for a little riddle.

The following appears to be a pretty bad approximation. It really is, depending on how we look at it. However, when I change only one digit or, equivalently, when I remove one of its parts, it returns a perfect 10-digit result on my HP-41C, which I am using to evaluate it. BTW, I have used the HP-41C for this one because of its 10-digit display and a seldom used useful built-in function which most of my Voyagers lack. Too many tips, but it doesn't matter :-)

Have fun!

\(\frac{\frac{26}{7}-\frac{6}{11211}}{\left (\frac{4141}{3313}\right ) ^{\frac{3}{4}}}\)

There hasn’t been any response to this puzzle, but it’s my fault. As I said, I forgot to mention something else had to be assumed. I will post the answer tomorrow, but I’ll give you another tip, in case you still want to give it a try. Remember numbers are not always what they look, as in the phrase “There are 10 types of people, those who understand binary and those who don’t”.

[Leviset]

See my latest post ‘Proving the Duffin-Schaeffer conjecture’- if I’d read this one first I’d have used it as a reply instead.

[Albert Chan]

(08-14-2019 09:14 PM)Gerson W. Barbosa Wrote:  \(\frac{\frac{26}{7}-\frac{6}{11211}}{\left (\frac{4141}{3313}\right ) ^{\frac{3}{4}}}\)

Puzzle solved
If assumed all octal numbers, numerator converted back to decimal:

22/7 - 6/4745 = 104348/33215 = 3.141592654 (10 digits, rounded)

This value happened to be one of Pi convergents, from continued fraction terms: [3;7,15,1,292,1]
Thus, all is needed is to "remove" the denominator, by changing exponent to 0/4

[Gerson W. Barbosa]

(08-14-2019 10:46 PM)Albert Chan Wrote:  

(08-14-2019 09:14 PM)Gerson W. Barbosa Wrote:  \(\frac{\frac{26}{7}-\frac{6}{11211}}{\left (\frac{4141}{3313}\right ) ^{\frac{3}{4}}}\)

Puzzle solved
If assumed all octal numbers, numerator converted back to decimal:

22/7 - 6/4745 = 104348/33215 = 3.141592654 (10 digits, rounded)

This value happened to be one of Pi convergents, from continued fraction terms: [3;7,15,1,292,1]
Thus, all is needed is to "remove" the denominator, by changing exponent to 0/4

Exactly, congrats!

Perhaps I should have chosen a higher base to make it a bit more difficult to check :-)

[Albert Chan]

(08-14-2019 11:09 PM)Gerson W. Barbosa Wrote:  Perhaps I should have chosen a higher base to make it a bit more difficult to check :-)

It is even harder if the base goes negative.

For negative base, we do not need the minus sign.
Example, your original puzzle in negaoctal base: (166/7 + 6/172627) / (14241/15473)^3/4

[Gerson W. Barbosa]

(07-26-2019 11:02 PM)Bill Duncan Wrote:  Yeah, mine was always 355/113. I used it a lot (with 4-bangers!) before calculators became available with a dedicated Pi key. It was easy to remember and pretty darned accurate.

That's a really good one. (Made in China)