The proof of the Intermediate Value Theorem does not give a practical method for
ID: 2969256 • Letter: T
Question
The proof of the Intermediate Value Theorem does not give a practical method for finding the point c. This exercise discusses a technique known as the bisection method for approximating the number c. Let f be a continous function defined on the interval [a,b] and let v be any number between f(a) and f(b).
A) Explain why it is sufficient to consider the case in which f(a) and f(b) have opposite signs and v=0.
B) Let m be the midpoint of [a,b]. Unless f(m)=0, show that either f(a)f(m)<0 or f(m)f(b)<0. Thus, the length of the interval in which the point c lies has been shortened by a factor of 2.
C) By continuing the process started in part (B), show that there exists a nested sequence {[a,b]} of closed intervals such that f(an)f(bn)<0 for all n and
(b_{n+1}-a_{n+1}=(b_{n}-a_{n})/2)
for all n. This is where the term bisection comes into play; each interval is bisected and one half of it is used to form the next interval.
D) Let c be the unique point in all of the intervals [ (a_{n}.b_{n}) ] (see the Nested Intervals Theorem). Prove that f(c)=0.
E) Find a formula for (b_{n}-a_{n}) in terms of a,b and n.
Explanation / Answer
(a) if f(a)*f(b) < 0 => f(a) > 0 & f(b) < 0 OR f(a) < 0 & f(b) > 0
without loss of generality let f(a) <0 and f(b) > 0 => f(a) < 0 < f(b) so by choosing our middle point v=0 and applying the Intermediate Mean Value Theorem, we have that=> there exists a c such that f(c) = v = 0
(b) Let f(a) < 0 < f(b); choose m as midpoint of [a,b]
now either f(m) > 0 or <0 (assuming f(m) != 0)
case 1: f(m) > 0 =>
but already we have f(a)<0 => f(a)*f(m) < 0
case 2: f(m) < 0 =>
but already we have f(b) > 0 => f(m)*f(b) < 0
in either case we have shorted the length from [a,b] to either [a,m] or [m,b] which is half the original length
(c) lets design the sequence in this way:
start by choose m = (a+b)/2 => either f(a)*f(m) < 0 or (fm)*f(b) < 0
if f(a)*f(m) < 0 => a1 =a ; b1=m
else if f(m)*f(b) < 0 => a1 = m; b1 = b
^ from part (b) we have b1 - a1 = (b-a)/2
now consider the new interval as [a1,b1]
next choose m2 = (a1+b1)/2 => repeat the same steps as above and get a2,b2
in the same way we have b2-a2 = (b1-a1)/2 ; since [a1,b1] was our new interval and using part (b)
continue this for n steps we have =>
b(n+1) - a(n+1) = (b(n) - a(n))/2
(d) length([a1,b1]) = (b-a)/2
length([a2,b2]) = (b-a)/4 .. and so on
=> length([a(n),b(n)]) = (b-a)/(2^n)
choose n very large,
by continuity, choose epsilon -> 0 => f(an) - f(bn) -> 0 , then delta = (b-a)/(2^n) => f(an) = f(bn)
by f(an)f(bn) <0 =>f(an) = f(bn) = 0 = f(c); since c lies in [an,bn]
(e) as derived in part (d) =>
(bn - an) = (b-a)/(2^n)
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