New variant for the Romberg Integration Method

[Namir]

Hi All,

I just posted, on my web site, the following article for a new variant for the Romberg method. The article actually looks at several variants and selects the best one.

Enjoy!!

Namir

Edited: 18 Apr 2012, 12:49 a.m.

[Matt Agajanian]

Thanks. This should be intriguing

[Nick_S]

Using both the HP-15c and Wolfram Alpha I get values that differ from yours for these examples:

ln(x)/x integrate 1 to 100 = 10.60378

x in radians

sin(x) integrate 1e-10 to pi/4 = 0.2928932

Nick

Edited: 18 Apr 2012, 5:27 a.m.

[Namir]

Nick,

Thanks for the corrections. I n the case of the sin(x), I meant sin(x)/x. I posted the article with the corrected results.

Namir

Edited: 18 Apr 2012, 7:29 a.m. after one or more responses were posted

[Valentin Albillo]

Quote:

---

Nick,

Thanks for the corrections. I n the case of the sin(x), I meant sin(x)/x. I should posted a corrected article very shortly.

Namir

---

The standard name for the sin(x)/x function is sinc(x), an abbreviation of sinus cardinalis (i.e.: cardinal sine).

Regards from V.

[Namir]

Right you are! And I learned a new function name. Alpha Worlfram recognized the sinc(x) funcion!!

:=)

[Paul Dale]

So does the 34S :-)

- Pauli

[Valentin Albillo]

Quote:

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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:

• I₁ = Integral( 0, Infinity, sinc(x) dx) = Pi/2

• I₂ = Integral( 0, Infinity, sinc(x)*sinc(x/3) dx) = Pi/2

• I₃ = Integral( 0, Infinity, sinc(x)*sinc(x/3)*sinc(x/5) dx) = Pi/2

• I₄ = Integral( 0, Infinity, sinc(x)*sinc(x/3)*sinc(x/5)*sinc(x/7) dx) = Pi/2

• I₅ = Integral( 0, Infinity, sinc(x)*sinc(x/3)*sinc(x/5)*sinc(x/7)*sinc(x/9) = Pi/2

• I₆ = Integral( 0, Infinity, sinc(x)*sinc(x/3)*sinc(x/5)*sinc(x/7)*sinc(x/9)*sinc(x/11) = Pi/2

• I₇ = 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

• I₈ = 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.

[Nick_S]

This brings back memories as the sinc function was one of the first things I plotted as a teenager on my newly acquired Sinclair ZX81 computer.

Nick