1549901369-Elements_of_Real_Analysis__Denlinger_

7.7 *Elementary Transcendental Functions 433

Proof. Suppose F: JR~ JR has properties (a)-(c). By (b), F(O) = sinO.
So, suppose x-:/:-0. Define H(x) = F(x) -sinx. Then H has derivatives of all
orders at x, and by Taylor's theorem, Vn EN, :3 Cn between 0 and x 3

H(x) = H(O) + H' (0) X + H" (0) x2 + ... + H(n) (0) xn + H(n+1) (en) xn+l (42)
1 2! n! (n+l)!
for some Cn between 0 and x. Now,

H(O) = F(O) -sinO = 0 - 0 = 0

H' (0) = F' (0) -cos 0 = 1 - 1 = 0

H"(O) = F"(O) + sinO = -F(O) + 0 = 0

H'"(O) = F"'(O) + cosO = -F'(O) + 1=-1+1=0

H(^4 l(O) = p(^4 )(0) -sinO = (F")"(O) = (-F)"(O) = -F"(O) = F(O) = 0

H(n)(O) = 0.

Thus, by ( 42), :3 Cn between 0 and x such that

H( )

= H(n+l)(cn) n+l

x (n+l)! x. (43)

Now, we take a closer look at H( n+^1 ) ( Cn), first for even values of n + 1 and

then for odd values. From (a) we get F(^2 nl(x) = (-l)nF(x). We also know that
cos(^2 n) x = (-l)nsinx. Thus,

IH(^2 n) (en) I = 1(-l)n F(cn) - (-l)n sin(cn)I = IF( en) -sin( en) I

= IH(cn)I.

Differentiating H one more time, we see that

Now, both H and H' are differentiable over the closed interval between 0 and
x, hence they are bounded there. Thus, :3 B > 0 3 Vn E N, in the expression
(43) we have
IH(2n+l)(cn)I '.SB.

Putting together (43) and (44), we have

Blxn+ll

Vn EN, IH(x)J '.S (n + l)!.

(44)

Jxn+ll
Now, lim ( )' = 0 by Corollary 2.3.11. Thus, since limits preserve
n-+oo n + 1.
inequalities, IH(x)I '.S 0, so
H(x) = 0.
Therefore, F(x) = sinx. •