=+1.6. Put M(t) = (de15,(w) dw, and show by successive differentiations under the integral that

(1.38)

M(k)(0) = f's(w) do.

Over each dyadic interval of rank n, s,

(c) has a constant value of the form

+1+1+ . .. + 1, and therefore M(t)=2-"Eexpt(+1+1+ . .. +1), where the sum extends over all 2" n-long sequences of + 1's and - 1's. Thus

-

(1.39)

M(1)=

-

= (cosht)".

2