Re: New variant for the Romberg Integration Method Message #8 Posted by Valentin Albillo on 18 Apr 2012, 9:45 a.m., in response to message #6 by Namir
Quote:
Right you are! And I learned a new function name. Alpha Worlfram recognized the sinc(x) funcion!!
:=)
I'm glad you did, I also learn new things each and every day.
About the sinc(x) function, it has many interesting properties and quirks but the one
I find most uncanny is this: a little computation or theoretical work will quickly stablish the following results:
 I_{1} = Integral( 0, Infinity, sinc(x) dx) = Pi/2
 I_{2} = Integral( 0, Infinity, sinc(x)*sinc(x/3) dx) = Pi/2
 I_{3} = Integral( 0, Infinity, sinc(x)*sinc(x/3)*sinc(x/5) dx) = Pi/2
 I_{4} = Integral( 0, Infinity, sinc(x)*sinc(x/3)*sinc(x/5)*sinc(x/7) dx) = Pi/2
 I_{5} = Integral( 0, Infinity, sinc(x)*sinc(x/3)*sinc(x/5)*sinc(x/7)*sinc(x/9) = Pi/2
 I_{6} = Integral( 0, Infinity, sinc(x)*sinc(x/3)*sinc(x/5)*sinc(x/7)*sinc(x/9)*sinc(x/11) = Pi/2
 I_{7} = Integral( 0, Infinity, sinc(x)*sinc(x/3)*sinc(x/5)*sinc(x/7)*sinc(x/9)*sinc(x/11)*sinc(x/13) = Pi/2
but lo and behold, we unexpectedly find that
 I_{8} = Integral( 0, Infinity, sinc(x)*sinc(x/3)*sinc(x/5)*sinc(x/7)*sinc(x/9)*sinc(x/11)*sinc(x/13)*sinc(x/15) = Pi/2.0000000000294+ !!
You might want to check this amazing fact by trying and computing said integrals I1, I2, ..., I8 using the 34S' extreme precision capabilities, it would be a fine test for any numerical integration procedure such as yours ! ... XD
Best regards from V.
