684 Appendix C • Answers & Hints for Selected Exercises
(b) [(x -t)k+^1 J(k+ll(t) - l J(k+l)(t)(k + l)(x -t)k(-dt)J:
= [(x - t)k+l j(k+^1 l(t)J: + (k + 1) l:(x - t)k j(k+ll(t) dt
= 0. J(k+l)(x) - j(k+l)(a)(x -a)k+l + (kk!l)! lax(x -t)k J(k+l)(t) dt
= -j(k+ll(a)(x - a)k+^1 + (k + l)!Rk(x)
= -J(k+^1 l(a)(x - a)k+l + (k + l)![f(x) -Tk(x)]
= (k + l)! { j(x) - [rk(x) + !~::~;~) (x -a)k+^1 J}
= (k + l)! {f(x) -Tk+i(x)} = (k + l)!Rk+1(x).- 'Vx E [a, b], f(a) s; f(x) s; f(b), so by Thm. 7.5.2 (b), f(a)(b-a) s; l: f s;
f(b)(b - a). Define g on [a, b] by g(x) = j(a)(x - a)+ f(b)(b - x). Then g is
continuous and g(a) = j(b)(b-a), g(b) = f(a)(b-a). Thus, g(a) s; l : f s; g(b).
By the intermediate value theorem (5.3.9), 3c E [a, b] 3 l: f = g(c) = f(c)(c -
a)+ j(b)(b -c).
- By Thm. 5.7.3, there is a bounded, monotone increasing (hence integrable)
function f: JR---+ JR whose set of discontinuities is Q. By Cor. 5.2.19, all of the
discontinuities of fare jump discontinuities. Thus, f has a jump discontinuity
at every rational number.
If a< c < d < b, then f has jump discontinuities on a dense subset of [c, d].
So, by Exercise 6.3.12, f cannot be the derivative of any function on [c, d]. - (a) Let x 0 E (a, b). By Thm. 5.2. 17 , f(x 0 ) and f(xci) exist, and by Ex.
20, f is differentiable from the left and right at xo, and F! ( xo) = f ( x 0 ),
F~(xo) = f(xci). Suppose f is not continuous at xo. Then, by Thm. 5.2. 17 ,
j(x 0 ) < f(xci), so F!(xo) < F~(xo) .. '.Fis not differentiable at xo.
(b) Let a< band A= Qn[a, b]. By Thm. 5.7.3, there is a bounded, nonneg-
ative, monotone increasing f : [a, b] ---+JR having A as its set of discontinuities.
Define F(x) = l: f on [a, b]. Then,- F is continuous on [a, b], by Thm. 7.6.6;
- Fis monotone increasing, by Ex. 20.
- Fis differentiable at every irrational number in [a, b] since f is continuous
there. (See FTC-II) - 'Vx E A, f is discontinuous at x, so by Part (a), F is not differentiable
at x. .'. 'V rational numbers x in [a, b], F is not differentiable at x.
2n 2n+l
23. F~(x) = :L)-l)kp2n-k'l/J~k+^1 )(x) = L (-lr-1p2n-m+1'1/!~m)(x) =
k=O m=l
[f
0(-1r-1p2n-m+l~~ml(x) ]-(-l)p2n+l~~o)(x)+(-l)2n-lp'l/J~2n+l\x)= [-pt, (-l)mp2n-m~~ml(x)] + p2n+l~n(x) = -pFn(x) + p2n+l'l/Jn(x).