Funny math problem to calculate the Pin Number for a credit card.

08112021, 03:40 PM
(This post was last modified: 08122021 10:45 AM by Albert Chan.)
Post: #4




RE: Funny math problem to calculate the Pin Number for a credit card.
(08102021 05:41 AM)Steve Simpkin Wrote: Here is the solution using a Casio fx991EX. FYI, integrand singularity (at x=1) is removable. (3x^3x^2+2x4) / √(x^23x+2) = (x1)*(3*x^2+2*x+4) / √((x1)*(x2)) = (1x)*(3*x^2+2*x+4) / √((1x)*(2x)) // integral limits 0 to 1, √(1x), √(2x) both real = (3*x^2+2*x+4) * √(11/(2x)) XCAS> ∫((3*x^2+2*x+4) * √(11/(2x)), x=0..1) (202*sqrt(2) + 135*ln(2*sqrt(2)+3))/16 // ≈ 2.98126694401  Proof: assume integral have the form f*√g (f*√g)' = f'*√g + f/(2√g)*g' = (f'*g + f*g'/2) / √g XCAS> f := a*x^2+b*x+c // integrand numerator is cubic, thus quadratic f XCAS> g := x^2  3x + 2 // integrand denominator = √g XCAS> coefs := e2r(f'*g + f*g'/2) [3*a, (15*a+4*b)/2, (8*a9*b+2*c)/2, (4*b3*c)/2] Ignore constant term for now, we match 3 coefs with 3 unknown. XCAS> a, b, c := 1, 13/4, 101/8 XCAS> coefs [3, 1, 2, 199/16] // = [3, 1, 2, 4]  [0, 0, 0, 135/16] I1 = preval(f*√g, 0, 1) = subst(f*√g, x=0) = 101/8*√(2) I2 = ∫(1/√(x^23x+2), x=0..1) // let y = 2*(x3/2)=2x3, dy = 2 dx = ∫(1/√(y^21), y=3 .. 1) // ∫(1/√(y^21), y) = ln(abs(y+sqrt(y^21))) = ln(2*√(2)+3) I = I1 + 135/16 * I2 = 101/8*√(2) + 135/16*ln(2*√(2)+3) 

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