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L'Hopital's Rule, Newton's approximation theorem

1. (L'Hopital's Rule) Show that if f,g:X->R, x0 belonging to X is a limit point of X such that f(x0) = g(x0) = 0, f,g are differentiable at x0, and g'(x0) != 0, then there is some delta > 0 such that g(x) != 0 for all x belonging to (X INTERSECTION (x0 - delta, x0 + delta) and

lim x->x0 [f(x)/g(x)] = f'(x0) / g'(x0) .

Hint: Use Newton's approximation theorem.

Show that the following version of L'Hopital's Rule is not correct. under the above hypothesis then,

lim x->x0 [f(x) / g(x)] = lim x->x0 [f'(x) / g'(x)].

Hint: Consider f(x) = g2(x), and g(x) = x at x0 = 0.

2. Let f: [a,b] -> [a,b], assume there is 0 < c < 1 such that

|f(x) - f(y)| <= c|x-y| for all x,y belonging to [a,b].

(a) Show that f is uniformly continuous on [a,b].

(b) Pick some point y0 belonging to [a,b], and given y_n define inductively y_(n+1) = f(y_n). Show that the sequence {y_n}_(n>=0) is a Cauchy sequence. Show that there is some y belonging to [a,b], such that lim n->infinity y_n = y.

(c) Prove that y is a fixed point, that is, f(y) = y.

(d) Finally prove that given any x belonging to [a,b], then the sequence defined inductively by: x_0 = x, x_(n+1) = f(x_n) converges to y as defined in part (b). (That is the function f has a unique fixed point.)

3. (Jean's favorite problem in multiplicative version) Let f:R->R that satisfies the multiplicative property f(x+y) = f(x)f(y) for all x,y belonging to R. Assume f is not identically equal to zero.

(i) Show that f(0) = 1 and that f(-x) = 1/f(x) for all x belonging to R. Show that f(x) > 0 for all x belonging to R.

(ii) Let a = f(1) (by (i) a > 0). Show that f(n) = a^n for all n belonging to N. Use (i) to show that f(z) = a^z for all z belonging to Z.

(iii) Show that f(r) = a^r for all r belonging to Q.

(iv) Show that if f is continuous at x = 0, then f is continuous at every point in R.

(v) Assume f is continuous at zero, use (iii) and (iv) to conclude that f(x) = a^x for all x belonging to R.

4. (Intermediate Value Theorem for Derivatives or Darboux's Theorem). If f is differentiable on [a,b], and if alpha is a real number in between f'(a) and f'(b) say f'(a) < alpha < f'(b) (or f'(b) < alpha < f'(a)), then there exists a point c belonging to (a,b) such that f'(c) = alpha. (Warning: you can not assume that f' is continuous even if it is defined on all of [a,b], so you can not use the IVT for continuous functions. Consider the example f(x) = x^2 sin(1/x); it is differentiable on [-1,1] but the derivative is not continuous at x = 0.) Hint: to simplify define a new function g(x) = f(x) - alpha x on [a,b]. This function g is differentiable on [a,b] and show that our hypothesis on f imply that g'(a) < 0 < g'(b) (or g'(b) < 0 < g'(a)). Now show that there is a c belonging to (a,b) such that g'(c) = 0. To do the later, show that there exists a point x belonging to (a,b) such that g(a) > g(x), and a point y belonging to (a,b) such that g(b) > g(y). Now finish the proof of Darboux's theorem.

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Solution Summary

Cauchy sequence, Jean's favorite problem in multiplicative version, and Darboux's theorem are clearly contextualized in this case.

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