Value of zeta(2)

Here shows that

using the sine factorization formula. The session file can be obtained from here.

Maxima 5.20.0 http://maxima.sourceforge.net
using Lisp CMU Common Lisp 19f (19F)
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) block(load(("/Users/yasube/Programming/test-imaxima/cmucl/5.20.0/share/maxima/5.20.0/emacs/imaxima.lisp")), linenum:0)$

(%i1) powerdisp:true;


(%i2) sin(%pi*x)=%pi*x*product(1-x^2/n^2,n,1,inf);


(%i3) sin(%pi*x)=taylor(sin(%pi*x),x,0,10);


(%i4) coeff(rhs(%),x,3);

We see how the first n terms of the infinite product is expanded with n=2,3,4,5,6, and 10 and seet the coincidence of coeffcients of x^3 to the sum of first n terms of the infinite sum definition of zeta function.

(%i5) %pi*x*'product(1-x^2/n^2,n,1,k)=expand(%pi*x*product(1-x^2/n^2,n,1,k)),k=2;


(%i6) 'sum(1/n^2,n,1,k)=sum(1/n^2,n,1,k),k=2;


(%i7) %pi*x*'product(1-x^2/n^2,n,1,k)=expand(%pi*x*product(1-x^2/n^2,n,1,k)),k=3;


(%i8) 'sum(1/n^2,n,1,k)=sum(1/n^2,n,1,k),k=3;


(%i9) %pi*x*'product(1-x^2/n^2,n,1,k)=expand(%pi*x*product(1-x^2/n^2,n,1,k)),k=4;


(%i10) 'sum(1/n^2,n,1,k)=sum(1/n^2,n,1,k),k=4;


(%i11) %pi*x*'product(1-x^2/n^2,n,1,k)=expand(%pi*x*product(1-x^2/n^2,n,1,k)),k=5;


(%i12) 'sum(1/n^2,n,1,k)=sum(1/n^2,n,1,k),k=5;


(%i13) %pi*x*'product(1-x^2/n^2,n,1,k)=expand(%pi*x*product(1-x^2/n^2,n,1,k)),k=6;


(%i14) 'sum(1/n^2,n,1,k)=sum(1/n^2,n,1,k),k=6;


(%i15) %pi*x*'product(1-x^2/n^2,n,1,k)=expand(%pi*x*product(1-x^2/n^2,n,1,k)),k=10;


(%i16) 'sum(1/n^2,n,1,k)=sum(1/n^2,n,1,k),k=10;

Thus the following equation holds:

(%i17) -sum(1/n^2,n,1,inf)*%pi=-%pi^3/6;


(%i18) %/(-%pi);


(%i19)

Euler's Cotangent and Sine factorization formulae

The following session illustrates how to obtain Euler's cotangent formula and sine factorization formula:

They play important role relating to zeta function, which will be treated in other posts in the future.

The file EulerCot.mac can be obtained from here.

Maxima 5.20.0 http://maxima.sourceforge.net
using Lisp CMU Common Lisp 19f (19F)
Distributed under the GNU Public License. See the file COPYING.
Dedicated to the memory of William Schelter.
The function bug_report() provides bug reporting information.
(%i1) block(load(("/Users/yasube/Programming/test-imaxima/cmucl/5.20.0/share/maxima/5.20.0/emacs/imaxima.lisp")), linenum:0)$

(%i1) kill(all);


(%i1) load(fourie)$

(%i2) "In this file, the following two equations are proved.
They are originated by Euler."$

(%i3) %pi*cot(%pi*t)=2*t*sum(1/(t^2-n^2),n,1,inf)+1/t;


(%i4) sinfact:sin(%pi*t)=%pi*t*product(1-t^2/n^2,n,1,inf);


(%i5) assume(t>0);


(%i6) assume(t-n>0);


(%i7) declare(n,integer);


(%i8) an:fourier(cos(t*x),x,%pi);





(%i11) an0:rhs(ev(first(an)));


(%i12) an2:factor(trigexpand(rhs(ev(second(an)))));

As the Fourier coefficients are obtained for cos(t x), we obtain the following formula.

(%i13) fo:cos(t*x)=an0+sum(an2*cos(n*x),n,1,inf);


(%i14) fo2:fo,x=%pi;


(%i15) fo3:fo2*%pi/sin(%pi*t);


(%i16) eucot:trigreduce(expand(fo3));



(%i18) "With this, we can show the factor of sine function:"$

(%i19) sinfact;


(%i20) "Taking the differential of log of left hand side:"$

(%i21) trigreduce(diff(log(lhs(sinfact)),t));


(%i22) "Taking the differential of log of right hand side:"$

(%i23) expand(factor(diff(log(rhs(sinfact)),t)));


(%i24) "According to the Euler's cotangent formula, the previous
two formula are eaual. Hence,"$

(%i25) sf1:sin(%pi*t)=C*t*product(1-t^2/n^2,n,1,inf);


(%i26) sf2:limit(sf1/t,t,0);

Maxima cannot compute this limit. We replace inf with 100 and how the limit is computed.

(%i27) sf3:subst(100,inf,sf2);


(%i28) sf3,nouns;


(%i29) "Hence we obtain:"$

(%i30) sf1, C=%pi;

Let's take a look at the graph of left and right hand side of the equation to make sure this holds.

(%i31) wxdraw2d(explicit(lhs(sinfact),t,-5,5), explicit(ev(subst(100,inf,rhs(sinfact)),simpproduct),t,-5,5));



(%i32)