Recent content by Roberto Bramb

The denominator D factorizes in D=x1*x2*x3*x4, where x1=sqrt(2+i) (first quadrant) x2=-sqrt(2-i) (second quadrant) x3=-sqrt(2+i) (third quadrant) x4=sqrt(2-i) (forth quadrant) The integral is given by the residue theorem int=(2*pi*i)[Res(x1)+Res(x2)] and Res(x1)=-(i/4)*sqrt(2+1)...

Substitute trig. objects with parametric formulae: sinx=2t/(1+t^2), cosx=(1-t^2)/(1+t^2), dx=2dt/(1+t^2) and your trig. integrale becomes the rational integral 2*Integ[-1,1](1/(t^2+2*t+5)dt = pi/4

If 0<x<9, then 0<y<3, so integrate in these limits (x+y) dx dy = 162

Plotting the function (remebering x!=Gamma(x+1)) G(x)=Gamma(x + 1) + Gamma(x - 2) - 16x + 24 you can see that for x>0 there are only four zeros. For x<0 there are infinite solutions near 0,-1,-2,-3...none exacly integer. There Gamma function is singular (has a pole) and change sign.

If x can be a real number (using Gamma function) you can find 4 solutions betweeen 1 and 4.5 (by plotting method or std numeric method). x1=1.1837 x2=1.3134 x3=2.1222 x4=4.4099 There are also 5 negative solution... None is integer.

With the substitution u=y/x equation becomes u''+au=0 which has solution u=c1*cos(kx)+c2*sin(kx) [k=sqrt(a)] so that y=xu=c1*x*cos(kx)+c2*x*sin(kx) So doing y'(0)=c1=0 y(1)=c1*cos(k)+c2*sin(k)=0 i.e. y=0, onlythe trivial solution ?

The series is convergent by Cauchy criterium since C(n+1)/C(n)->0 for n>infinity, as can be easily verified using Striling formula : log(n!) => n*log(n)-n

If M1=Qd is the dipole moment (Q :charge, d:separation as a vector) the potential field is V(r)=1/(4*pi*e) M1.grad(1/r) Having a ground plane, this can be eliminated if you consider an image dipole (symmetrically placed under the ground plane). of moment M2=-M1. Then the interaction energy...

The separability of Laplace and Helmoltz equations in a coordinate system is possible if and only if the cofactor of the Staekel matrix and metric tensor satisfy some appropriate relations. Look at the introductory pages of FIELD THEORY HANDBOOK - P.Moon-D.E.Spencer Springer 1961

It is the Laplace equation in cylindrical coordinates with symmetry about y-axe. You can solve it by variable separation, once given the boundary condition: phi(x,y)=X(x)Y(y) X''+X'/x+cX=0 (in x, 0-order Bessel equation) Y''-cY=0 (in y) c=arbitrary positive/negative real constant

k*n/(m+n)