And, ψ(z + n) diminishing without limit, we have

a

z 		· φ(z+1)

φz 		= a

z+ 		a

(z+1)+ 		a

(z+2)+ 		a

(z+3)+ ...

Let z = ½; and let 4a = -x2. Then

a

z		φ(z+1) is -x2

2		 1 - x2

2·3 		+ x4

2·3·4·5...		 or -x

2		sin x.

Again

φz is 1 - x2

2 		+ x4

2·3·4 		or cos x:

and the continued fraction is

- ¼x2

½+ 		- ¼x2

(3/2)+ 		- ¼x2

(5/2)+ ... 		or -x

2 		x

1+ 		- x2

3+ 		- x2

5+ ...

whence