Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 13 July 2014

Lecture 18

Derivatives by limits

Let f:[a,b]. Let c[a,b].

The derivative of f at x=c is f(c)= limxc f(x)-f(a)x-c. Alternatively, f(c)= limΔx0 f(c+Δx)-f(x) Δx .

Let f:[a,b] and g:[a,b] and let β,γ. Assume that f(c) and g(c) exist. Then

(a) (βf+γg)(c)= βf(c)+ γg(c),
(b) (fg)(c)= f(c)g(c)+ f(c)g(c),
(c) Assume f:[a,b] is given by f(x)=x. then f(c)=1.
(d) If f(c) exists then f is continuous at x=c.

Define f(c)= (f)(c), f(3)(c)= (f)(c)and f(N)(c)= (f(N-1)) (c).

(Taylor's theorem with Lagrange's remainder). If f:[a,b] and N0 and f(N):[a,b] is continuous and f(N+1):[a,b] exists then there exists c(a,b) such that f(b)=f(a)+f (a)(b-a)+ 12!f(a) (b-a)2++ 1Nf(N) (a)(b-a)N+ 1(N+1)! f(N+1)(c) (b-a)N+1.

Remarks:

(1) The last term in f(b)=f(a)++1(N+1)!f(N+1)(c)(b-a)N+1 is Lagrange's form of the remainder.
(2) The special case N=0 is the Mean Value Theorem. If f:[a,b] and f*:[a,b] is continuous and f:[a,b] exists then there exists c(a,b) such that f(b)=f(a)+f (c)(b-a) i.e. f(c)= f(b)-f(a)b-a a b c x y
(3) The special case N=0 and f(a)=f(b) is Rolle's theorem If f:[a,b] is continuous and f:[a,b] exists then there exists c(a,b) such that f(c)=0 a b x f ( a ) = f ( b ) y
(4) The proof of these theorems uses Intermediate value theorem
(a) If f:[a,b] is continuous and w is between f(a) and f(b) then there exists c(a,b) such that f(c)=w.
(b) If f:[a,b] is continuous then there exist m,M such that f([a,b])=[m,M] a b f ( a ) f ( b ) m M
(5) Let f:[0,2π] be given by f(x)=cosx+ isinx. Then f(0)=f(2π) but f(x) is never 0. Why is not a contradiction to Rolle's theorem?
(6) The first N terms in (all but the remainder term) f(b)=f(a)++ 1N!f(N)(c) (b-a)N+ 1(N+1)!f (c)(b-a)N+1 are the Taylor approximation to f at x=a of order N.
(7) f:[a,b] is differentiable at x=c if f(c) exists.

Approximate 281/3 to 5 decimal places.

Let f(x)=(27+x)1/3. Then (27+x)1/3= a0+a1x+a2 x2+a3x3+ So a0 = (27+0)1/3=3, a1 = (ddx(27+x)1/3) |x=0=1/3 (27+x)-23 |x=0=1/3 132=127, a2 = 12! (d2dx2(27+x)1/3) |x=0=121/3 (-23) (27+x)-53 |x=0= 12·(-2)32 135=-137, a3 = 13! (d3dx3(27+x)1/3) |x=0=12·31/3 (-2)3 (-5)3 (27+x)-83 |x=0= 534·138= 5312, 14!f(4) (c)(28-27)4 = 14!1/3 (-2)3 (-5)3 (-8)3 (27+c)-113 = -24·523·35 (27+c)-113= -2·535 1(27+c)113. So (27+x)1/3=3+127x-137x2+5312x3+ and 281/3 = (27+1)1/3 3+127-137+ 5312 with error equal to 2·535 1(27+c)113 for somec(0,1). So the error is less than 2·535 127113= 2·535·311 =2·5316.

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100419Lect18.pdf and was given on 19 April 2010.

page history