Solution to B-1f.
-----------------

Let M0 be the magnetization just before the excitation.
Let M1 be the magnetization just afterward.

Tips are about y.


M1z = M0z*cos(flip) - M0x*sin(flip)
M0z = M1z*E1 + (1-E1)
	where E1 = exp(-TR/T1)

M0x = 0, since we neglect residual transverse magnetization.

Combine these equations:

	M0z = [M0z*cos(flip) - 0]*E1 + (1-E1)
	M0z [1-E1*cos(flip)] = 1-E1
	M0z = [1-E1] / [1-E1*cos(flip)]


************ Notes ************

1.  It wouldn't be that difficult to keep the M0x term around,
    and not neglect the transverse magnetization.

2.  Multiplication by [0 0 0;0 0 0;0 0 1] just before the excitation
    in B-1d would "force" the transverse magnetization to be zero
    in the matrix calculation, so you should get the same answer
    in B-1d as you just got here in B-1f.