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janhardo

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How to prove this sum is equal to produ...

janhardo 695 Maple
product summation
January 06 2022
2 8

Try to prove those two expressions : a sum and product 
First to know if they are correct defined for some values ?

Eulers productformule in Maple

 
> 

 

Eerst de Euler identiteit in Maple opschrijven
via info over series

 

p-Series

sum(1/n^p, n = 1 .. infinity)

Absolute convergence for p > 1

Diverges for p <= 1 

 

 

> 

restart;

sum(1/n^p, n = 1 .. infinity) = product(1/(1-p^(-s)), i = k .. p)

> 

product(f,i=k..n);

> 

product(1/(1 - p^(-s)), i = 1 .. p);

(1-p^(-s))*(-1/(-1+p^(-s)))^(p+1)

(1.1)
> 

sum(1/n^p, n = 1 .. infinity);

sum(1/n^p, n = 1 .. infinity)

(1.2)
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sum(1/n^p, n = 1 .. 4);

1+1/2^p+1/3^p+1/4^p

(1.3)
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product(1/(1 - p^(-s)), i = 1 .. 4);
            

1/(1-p^(-s))^4

(1.4)
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Product(1/(1 - p^(-s)), i = 1 .. 4)= product(1/(1 - p^(-s)), i = 1 .. 4);

Product(1/(1-p^(-s)), i = 1 .. 4) = 1/(1-p^(-s))^4

(1.5)
> 

sum(1/n^p, n = 1 .. infinity) = product(1/(1 - p^(-s)), i = k .. p);

sum(1/n^p, n = 1 .. infinity) = (-1/(-1+p^(-s)))^(p+1)/(-1/(-1+p^(-s)))^k

(1.6)
> 

 

 How to prove this sum(1/n^p, n = 1 .. infinity) = product(1/(1-p^(-s)), i = k .. p)

For the sum i got some terms , but for the product i don't see factors as i use the commands  

Download the_golden_key_in_Maple.mw

Riemanns formidable formula gives differ...

janhardo 695 Maple
roundoff numeric evalf
January 05 2022
2 5

Some small differences is unavoidable ?

> 

 

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)
> 

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)
> 

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)
> 

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)
> 

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

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

(1.7)
> 

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

0.1e-9+0.1e-9*I

(1.8)
> 

 

Download post_naar_primes_over_triemanns_geweldige_formule.mw

Can i search on persons who answered my ...

janhardo 695 MaplePrimes
January 03 2022
1 3

Best wishes for all for the New Year 2022!

Is it possible to search for my questions who answered this on Maple Primes ?

How to get better 3d plot for Im(logz)...

janhardo 695 Maple
complex special-function 3d
January 01 2022
1 11

Colored axes and showing 5 colored branches (Riemann surfaces) and ln(z)  as a spiral

There is a plotcompare(log(z), expression_plot, 3); branches(ln);

FunctionAdvisor(branch_cuts, log(z));
                         [ln(z), z < 0]

The principal branch is the red colored one : Pi..-Pi (difficult to see here )


                                       pic 1

Compare this pic 1  with the Imaginaire plot log(z) in plotcompare command.

How to get a wanted expression?...

janhardo 695 Maple
multiplication inert duplicate-question
December 30 2021
1 14

As maple 1d input:

expr:=(1/2)*a*(b/c)/d;     

As 2 d input : 

This rightside of the equation i like to get in Maple , but how?

answer

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