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Question: Riemanns formidable formula gives differences

Question:Riemanns formidable formula gives differences

janhardo 390 Maple
roundoff numeric evalf zeta
January 05 2022
1

Some small differences is unavoidable ?

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Its not symbolic that's why i don't get he same answers ?
Later on i will investigate some values for for this formula  

Riemanns formidabele formule

 
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restart:

Volgens het boekje de riemanns hypothese : de functionaal vergelijking  (zonder bewijs)
Zeta(-z) = ((-2*factorial(z))*(1/(2*Pi)^(z+1)))*sin((1/2)*Pi*z)*Zeta(z+1)

Zeta(-z) = -2*factorial(z)*sin((1/2)*Pi*z)*Zeta(z+1)/(2*Pi)^(z+1)

(1.1)
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Zeta(-z) = -2*z!*sin(Pi*z/2)*Zeta(z + 1)/(2*Pi)^(z + 1);

Zeta(-z) = -2*factorial(z)*sin((1/2)*Pi*z)*Zeta(z+1)/(2*Pi)^(z+1)

(1.2)
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verg:=%;

Zeta(-z) = -2*factorial(z)*sin((1/2)*Pi*z)*Zeta(z+1)/(2*Pi)^(z+1)

(1.3)
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verg2:= subs(z=2+3*I,verg);

Zeta(-2-3*I) = -2*factorial(2+3*I)*sin((1+(3/2)*I)*Pi)*Zeta(3+3*I)/(2*Pi)^(3+3*I)

(1.4)
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verg3:= eval(verg,z=2.+3*I);

.1329711559-.1230533004*I = (2.450819690-44.87297744*I)/(2*Pi)^(3.+3.*I)

(1.5)
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evalf(rhs(verg3));

.1329711558-.1230533005*I

(1.6)
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lhs(verg3)=evalf(rhs(verg3));

.1329711559-.1230533004*I = .1329711558-.1230533005*I

(1.7)
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lhs(verg3)- evalf(rhs(verg3));

0.1e-9+0.1e-9*I

(1.8)
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