Homework 7 <me@tylerlogic.com>
November 1, 2014
http://coursework.tylerlogic.com/courses/upenn/math508/homework07

1 Prove that smooth f : [a,∞) →ℝ with bound first derivative is uniformly continuous.

Let f : [a,) be smooth with M bound the first derivative, i.e. |f(x)|≤ M. Let ε > 0 and set δ = ε M. Then for any x,y [a,), assuming without loss of generality that x < y, the mean value theorem informs us of a c (x,y) such that

Hence if |y - x| < δ, we have

so that f is uniformly continuous.

2

(a) Show that sinx is not a polynomial.

The function sinx is zero at 2πn for all integers n, i.e. it has infinitely many zeros. Polynomials have a finite amount of zeros, and so sinx cannot be a polynomial.

(b) Show that sinx cannot be a rational function.

A rational function p(x)∕q(x) is zero if and only if p(x) is zero. Therefore a rational function is zero at only finitely many points, and just as we saw in the previous part of the problem, this implies sinx cannot be a rational function.

(c) If f(t + 1) = f(t) for all real t, and f is not constant, show that f is not a rational function.

By way of contradiction, assume that f is a rational polynomial so that f(t) = p(t)∕q(t). Fixing t0 and by putting g(t) = f(t) - f(t0) we have
1. g(t + 1) = f(t + 1) - f(t0) = f(t) + f(t0) = g(t) so that g is periodic
2. g(t) = p(t)∕q(t) - f(t0) = so that g is rational, and
3. g(t0) = f(t0) - f(t0) = 0 so that g has a zero at t0.

Putting the above three things together informs us that g is a rational function with infinitely many zeros, but rational functions can only have a finite number of zeros; a contradiction.

(d) Show that ex is not a rational function.

A rational function f(x) has that

however

3 Show that limn→∞(n + 1)- n= 0

Define f : by f(x) = x. Since this is a polynomial, it’s smooth on . For an integer n, the mean value theorem tells us that there is an x (n,n + 1) such that

which in turn yields

Hence

as desired.

4

Let f : be smooth with f(0) = 3, f(1) = 2, and f(3) = 8. The Mean Value Theorem yields the existence of c1 (0,1) and c2 (1,3) such that
and
Because f is smooth, fis continous since f′′ is differentiable, and thus we can appeal to the Mean Value Theorem again to obtain a c (c1,c2) such that
Since c2 > c1 > 0 this implies f′′(c) > 0, as desired. Furthermore, because, more precisely, 3 > c2 > 1 > c1 > 0 we have that 3 > c2 - c1 so that f′′(c) > according to the above equation. So let M = .

5

By “a convex function f” we mean one for which every point of the graph of f lies above all of its tangent points; i.e. one for which

for all x,y .

(a) Show that a smooth function f is convex if f′′(x) ≥ 0 for all x.

Assume that the second derivative of f is non-negative for any point of . For any reals x,y with x < y the Mean Value Theorem (MVT) gives us a z (x,y) such that
 (5.1)

Since f is smooth, fis differentiable on [x,z], and so the MVT gives us a w (x,z) such that

Since the second derivative is non-negative, then so is the left hand side of the above equation. Since z > x this implies f(z) f(x) which in light of Equation 5.1 implies f(x). This yields

as desired for the convexity of f.

(b) Prove that v(x) ≤ 0 for all 0 ≤ x ≤ 1 if v′′(x) > 0 for 0 ≤ x ≤ 1 and v(0) = v(1) = 0

Assume for later contradiction that there is a point x (0,1) with v(x) > 0. Then there exists a c1 (0,x) with

by the MVT so that = v(c1) which implies v(c1) > 0 since v(x) > 0. Furthermore there exists an c2 (x,1) where

so that = v(c2). Since x < 1 and v(x) > 0, then v(c2) < 0. Once more, the MVT tells us there exists a c (c1,c2) such that

but since we’ve seen that v(c1) > 0, v(c2) < 0, and because c2 > c1, the above equation yields v′′(c) 0. This contradicts the fact that v′′(x) for 0 x 1. Hence there is no point x (0,1) with v(x) > 0, and therefore v(x) 0 for all x [0,1].

(c) Prove that ex is convex.

The second derivative of ex is ex, which is always positive. By the first part of this problem we know that ex is convex.

(d) Prove that ex≥ 1 + x for all x

Since the previous part of this problem showed ex is convex, then for any x,y we have

Thus, letting y = 0, we get ex x + 1 for all x.

6

(a) What constraints are on c and d so that p(x) = x3+ cx + d has three distinct real roots?

If p(x) = x3 + cx + d were to have three distinct real roots, then there would exist real x1 < x2 where p(x1) > 0 is a local maximum and p(x2) < 0 is a local minimum. Since limx→-∞p(x) = -∞ and limx→∞p(x) = , then we can find x0,x3 with x0 < x1, x3 > x2, p(x0) < 0, and p(x3) > 0. Thus the intermediate value theorem implies, since x1 < x2, p(x1) > 0, and p(x2) < 0, the existence of c1,c2,c3 where x0 < c1 < x1 < c2 < x2 < c3 < x3 and p(c1) = p(c2) = p(c3) = 0. Thus it is indeed possible for there to exist three distinct real roots.

We have that p(x) = 3x2 + c so that x = ± when p(x) = 0. Hence in order for there to be three real roots, c must be less than zero. Since x = ± are the two local maximum and minimum, then p() < 0 so that

At this point, since we have c < 0, then the right hand side of the above inequality is positive so that
and this is the constraint on d.