[VA] Short & Sweet Math Challenge #25 "San Valentin's Special: Weird Math"

[J-F Garnier]

Now that the challlenge is closed, let me add some comments:

(02-28-2021 02:18 AM)Valentin Albillo Wrote:  My original solution for "Concoction the Third: Weird Integral"



What's so weird about this integral?

Quote:The Sleuthing

As before, I'll describe a typical sleuthing procedure to try and answer the above question. First of all, let's compute the integral's value to full 12-digit accuracy using the HP-71B, right from the command line:

      >DESTROY ALL @ P=(1+SQR(5))/2 [ENDLINE]
      >INTEGRAL(1,P,0,GAMMA(LN(P^2-IVAR))/(GAMMA(LN(IVAR))+GAMMA(LN(P^2-IVAR)))) [ENDLINE]

First, would you have used the Enhanced Math ROM, you could have just typed:

>INTEG(1,P,0,GAMMA(LN(P^2-IX))/(GAMMA(LN(IX))+GAMMA(LN(P^2-IX)))) [ENDLINE]
After all, you are the one who suggested the use of such shortcuts

As I mentioned in my solution, the first aspect I noticed was the speed; I wasn't expecting to get a fast answer on a physical calculator of the 32S class for such complicate-looking equation combining the gamma and log functions.
I also noticed that the speed and the answer were independent of the target accuracy (set by the display mode on the 32S - from FIX 1 to FIX 11). That was really weird.
BTW, using a slow machine sometimes highlights some aspects that can be missed on faster systems such as Free42.

Moving to the HP-71/Emu71, I investigated more:

30 Y=0 @ A=0 @ F=(1+SQR(5))/2
60 DEF FNF(X)
70 A=GAMMA(LN(F*F-X)) @ Y=A/(GAMMA(LN(X))+A)
90 PRINT X;Y
100 FNF=Y
110 END DEF
130 E=.0001
140 DISP INTEGRAL(1,F,E,FNF(IVAR))
150 END

>RUN
1.30901699438 .500000000007
1.09656781074 .169413871978
1.52146617801 .830586128022
1.19554981675 .328227901097
1.422484172 .671772098903
1.02655614795 4.82413106951E-2
1.5914778408 .951758689306

It turned out that only 7 samples were used, whatever was the target accuracy set by the E value.
Having some knowledge of the Romberg algorithm, I recognized the first sample to be the midpoint, then the next two samples used for the 1st estimation and the last four samples used for the 2nd estimation.
It was quickly obvious that the sample pairs (2-3, 4-5, 6-7) gave values that sum to exactly 1.
That gave me the key for the explanation of the speed: the summation for any sample set was the same (and the exact one), so the algorithm reached its termination condition after the first two estimations.
Extrapolating to a single sample at the midpoint, I got the symbolic answer: (φ-1) x 1/2

This is of course a consequence of the symmetry of the function to integrate as Werner and Valentin explained.
It is interesting to note that my observations had also to do with the HP implementation of the Romberg algorithm: although the samples are non-uniform, they are symmetric relative to the midpoint, and that gave the effect I reported.

Thanks for the challenge that was at the intersect of my interests for math puzzles and numerical algorithms (such as used in HP machines).

J-F