Begin2.DVI

Observe that the sum of the angles of the triangle ABC with top angle Aand base angles π 2 −αand π 2 −βis πradians so that one c ...

At a minimum value the derivative must equal zero and so equation (12.17) must be set equal to zero and xmust be solved for. Thi ...

The form of the equations (12.20) and (12.21) can be changed by multiplying equa- tion (12.20) by cos βand multiplying equation ...

Figure 12-4. White light splits into colors. Recall that white light gets split into the colors Red, Orange, Yellow, Green, Blue ...

one equation, two unknowns Any equation of the form F(x, y ) = 0 (12 .28) implicitly defines y as one or more functions of x. Tr ...

If yis held constant, then dy is zero and equation (12.35) yields the result ( dz dx ) y =− ∂F ∂x ∂F ∂z (12 .36) Here the symbol ...

Differentiate equation (12.40) with respect to zand show ∂F ∂z + ∂F ∂w ∂w ∂z = 0 or ∂w ∂z =− ∂F ∂z ∂F ∂w (12 .43) provided ∂F∂w ...

The determinants in the equations (12.48) are called Jacobian determinants of F and Gand are often expressed using the shorthand ...

In a similar fashion one can differentiate the equations (12.51) with respect to the variable yand obtain ∂F ∂y + ∂F ∂u ∂u ∂y + ...

If U =U(x, y ) is converted to polar coordinates to become U =U(r, θ) one can treat r=r(x, y )and θ=θ(x, y )to calculate the fol ...

It is now possible to differentiate the first derivatives given by equations (12.62) and (12.64) to obtain the second derivative ...

three equations, five unknowns A set of equations having the form F(x, y, u, v, w ) =0 G(x, y, u, v, w ) =0 H(x, y, u, v, w ) =0 ...

DIfferentiate the equations (12.69) with respect to yto obtain Fy+Fuuy+Fvvy+Fwwy=0 Gy+Guuy+Gvvy+Gwwy=0 Hy+Huuy+Hvvy+Hwwy=0 (12 . ...

derivatives of the functions in equation (12.76) with respect to say xj, 1 ≤j≤n, one finds ∂F 1 ∂x j+ ∂F 1 ∂y 1 ∂y 1 ∂x j+···+ ∂ ...

Replacing xby −x, the equation (12.82) is sometimes represented Γ(1 −x) = −xΓ(−x) (12 .83) Using the recurrence relation (12.81) ...

Example 12-4. Show that Γ( 1 2 ) = √ π. Solution Substitute x= 1/ 2 into equation (12.79) and show Γ(^1 2 ) = ∫∞ 0 ξ−^1 /^2 e−ξd ...

Product of odd and even integers One can now apply the previous results Γ(n) = (n−1)Γ(n−1), Γ(1) = 1, Γ(^1 2 ) = √ π (12 .88) to ...

If nis odd, say n= 2m− 1 , then equation (12.94) eventually becomes S 2 m− 1 = ( 2 m− 2 2 m− 1 )( 2 m− 4 2 m− 3 ) ··· ( 4 5 )( 2 ...

Various representations for the Gamma function The integral representation of the Gamma function Γ(x) = ∫∞ 0 ξx−^1 e−ξdξ x > ...

Sometime around 1729 Euler defined the Gamma function in the form Γ(x) = limn→∞ (n−1)!nx x(1 + x)(2 + x)(3 + x)···(n−1 + x)= lim ...