November 1, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework07
Let f : [a,∞) → ℝ 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.

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.

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.

By way of contradiction, assume that f is a rational polynomial so that f(t) = p(t)∕q(t). Fixing t

- g(t + 1) = f(t + 1) - f(t
_{0}) = f(t) + f(t_{0}) = g(t) so that g is periodic - g(t) = p(t)∕q(t) - f(t
_{0}) = so that g is rational, and - g(t
_{0}) = f(t_{0}) - f(t_{0}) = 0 so that g has a zero at t_{0}.

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.

A rational function f(x) has that

however

Define f : ℝ → ℝ by f(x) = x

which in turn yields

Hence

as desired.

Let f : ℝ → ℝ be smooth with f(0) = 3, f(1) = 2, and f(3) = 8. The Mean Value Theorem yields the existence of c

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 ∈ ℝ.

Assume that the second derivative of f is non-negative for any point of ℝ. 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, f′ is 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.

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

by the MVT so that = v′(c_{1}) which implies v′(c_{1}) > 0 since v(x) > 0. Furthermore there exists an c_{2} ∈ (x,1)
where

so that = v′(c_{2}). Since x < 1 and v(x) > 0, then v′(c_{2}) < 0. Once more, the MVT tells us there exists a c ∈ (c_{1},c_{2})
such that

but since we’ve seen that v′(c_{1}) > 0, v′(c_{2}) < 0, and because c_{2} > c_{1}, 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].

The second derivative of e

Since the previous part of this problem showed e

Thus, letting y = 0, we get e^{x} ≥ x + 1 for all x.

If p(x) = x

We have that p′(x) = 3x^{2} + 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