Question

rimann hypothesis..
Evaluating the possible values of αThe famous functional equation for η(s) restricted to the domain D is given by(η(s) = ϕ(s)η(1 −s)ϕ(s) = 2(1−2−1+s)(1−2s)πs−1sin pi/2 πs2/Γ (1 −s)(3)Let η(s) = x(s) + iy (s), η (1 −s) = u(s) + iv (s),where x(s)+iy(s) =∞Xn=1(−1)n−1nαcos (βln n), y (s) = −∞Xn=1(−1)n−1nαsin (βln n)u(s) =∞Xn=1(−1)n−1n1−αcos (βln n), v (s) =∞Xn=1(−1)n−1n1−αsin (βln n)(4)The function ϕ(s) can be written as ϕ(s) = ϕ1(s) + iϕ2(s) and there-fore it is given uniquely by the polar form ϕ(s) = ρ(s) exp (iθ (s)) ,where,2 ρ=pϕ21(s) + ϕ22(s)6= 0 and θ(s) = Argϕ (s)∈R.Here, we have ϕ1(s) =ρ(s) cos θ(s) and ϕ2(s) = ρ(s) sin θ(s) since such complex numbers are entirely determined by their modulus and angle. We note that ρ(s)6= 0 sinceϕ(s)6= 0 for all s∈D.The first main result for the proof of the Riemann hypothesis is given asfollow:Theorem 1 The complex number s∈Dis a solution of η(s) = 0 if andonly if x(s) = u(s), y (s) = −v(s)and ϕ(s)6=±1.Proof. If s∈Dis a solution of η(s) = 0,then x(s) = u(s) = 0 andy(s) = −v(s) = 0 since we have from (3) that if x(s) = y(s) = 0,thenu(s) = v(s) = 0 since ϕ(s)6= 0 for all s∈D. If x(s) = u(s), y (s) = −v(s)and ϕ(s)6=±1,then we have η(s) = ¯η(1 −s).Thus, from (3), we haveη(1 −s) = 1ϕ(s)¯η(1 −s) (5)that is, 1−1ϕ(s)u(s)+i/1 + 1ϕ(s)/v(s) = 0.Since ϕ(s)6=±1,i.e., ρ(s)6=1,then u(s) = v(s) = 0,i.e., η(1 −s) = 0, hence η(s) = 0 by using (3) andfinally, s∈Dis a solution of η(s) = 0.Substituting the algebraic forms of η(s) and η(1 −s) in (4) into equation(3), we obtain = x(s)y(s)=A(s)u(s)v(s)A(s) = ϕ1(s)−ϕ2(s)ϕ2(s)ϕ1(s)(6)that is,x(s) = u(s)ϕ1(s)−v(s)ϕ2(s)y(s) = u(s)ϕ2(s) + v(s)ϕ1(s)(7)and the inverse transformation is given by(u(s) = x(s)ρ2(s)ϕ1(s) + y(s)ρ2(s)ϕ2(s)v(s) = y(s)ρ2(s)ϕ1(s)−x(s)ρ2(s)ϕ2(s)(8)The matrix A(s) in (6) is invertible for all s∈Dsince its determinant isρ2(s)6= 0 for all s∈D. If ϕ(s)6=±1,then the only fixed points of the trans-formation defined by this matrix are located in the lines x(s) = 0 and y(s) =3 0.Indeed, x(s)y(s)=A(s)x(s)y(s)implies that (I2−A(s)) x(s)y(s)=00which has one solution (0,0) since the determinant of the matrix(I2−A(s)) is (ϕ1(s)−1)2+ϕ22(s)6= 0 since ϕ(s)6= 1.Here I2is the 2 ×2unit matrix. Clearly, all the images of the roots of η(s) = 0 by the transfor-mation in (7) remain unchanged. The same result holds true for the inversetransformation in (8).?0.Indeed, x(s)y(s)=A(s)x(s)y(s)implies that (I2−A(s)) x(s)y(s)=00which has one solution (0,0) since the determinant of the matrix(I2−A(s)) is (ϕ1(s)−1)2+ϕ22(s)6= 0 since ϕ(s)6= 1.Here I2is the 2 ×2unit matrix. Clearly, all the images of the roots of η(s) = 0 by the transformation in (7) remain unchanged. The same result holds true for the inverse transformation in (8).Assume that x(s) = u(s) and y(s) = −v(s) and looking for the values of s satisfying these equations.Thus from (7) and (8), we obtain(x(s) = u(s)ϕ1(s)−v(s)ϕ2(s) = u(s) = x(s)ρ2(s)ϕ1(s) + y(s)ρ2(s)ϕ2(s)y(s) = u(s)ϕ2(s) + v(s)ϕ1(s) = −v(s) = −y(s)ρ2/(s)ϕ1(s)−x(s)ρ2(s)ϕ2(s)that is(u(s)ϕ1(s)−v(s)ϕ2(s) = x(s)ρ2(s)ϕ1(s) + y(s)ρ2(s)ϕ2(s)u(s)ϕ2(s) + v(s)ϕ1(s) = /−y(s)ρ2(s)ϕ1(s)−x(s)ρ2(s)ϕ2(s)i.e.,−x(s)ϕ1(s)−y(s)ϕ2(s) + u(s)ρ2(s)ϕ1(s)−v(s)ρ2(s)ϕ2(s) = 0−x(s)ϕ2(s) + y(s)ϕ1(s) + u(s)ρ2(s)ϕ2(s) + v(s)ρ2(s)ϕ1(s) = 0(9)by replacing the values of ϕ1(s) = ρ(s) cos θ(s), ϕ2(s) = ρ(s) sin θ(s),and the values of x(s), y (s), u (s) and v(s) from (4), we obtain the following equation∞Xn=1(−1)n−1/nα

Answer 1

Answer:

exactly

Step-by-step explanation:

no, but yes.

Answer 2

hello electro ❤❤❤

sorry to say that but please post the question understand able so that i can answer