SOLUTION TO PROBLEM 6

Consider the auxiliary function f(x)=e^x-x-1. We want to prove that $\min f(x)=f(0)=0$, the minimum being over the set $\mathbb{R}$of all reals. This would imply $f(x)\geq f(0)$, for all real x, which is one implication in the equivalence to be proved. Let's study f as in Calculus I, i.e. let's calculate its first derivative $f^\prime (x)=e^x-1$. Clearly e^x-1 equals 0 if and only if x=0, so 0 is the only critical point, and it's very easy to see that $f^\prime$ is negative on the interval $(-\infty,0)$ and positive on $(0,\infty )$. This means that f decreases on $(-\infty,0)$ and increases on $(0,\infty )$, which proves that $f(0)=\min f(x)$ as desired.

We must prove the converse implication now, i.e., we must prove that if a is a fixed positive constant with the property

$$a^x\geq x+1\qquad \forall x\in \mathbb{R}$$

then a must equal e. Observe that for each positive x the property above can be written as

$$\frac{a^x-a^0}{x-0}\geq 1\quad \forall x>0.$$

Let x go to 0⁺ in the inequality above and observe that the conclusion is that the derivative at 0 of the function a^x must be greater than or equal 1. Recall that $(a^x)^\prime =a^x\ln a$, so we obtained that $\ln a\geq 1$. We would like to prove now that $\ln a\leq 1$, which would imply $\ln a=1$ hence a=e, ending the proof. Let's use the same technique, i.e. rewrite the inequality

$$a^x\geq x+1\qquad \forall x\in \mathbb{R}$$

for x<0. Since division by a negative number changes inequalities, one obtains

$$\frac{a^x-a^0}{x-0}\leq 1\quad \forall x<0.$$

and letting x tend to 0^- one gets as above $\ln a\leq 1$. The proof is over.

Questions and/or comments should be directed to Judy Downey or Griff Elder
 

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Last modified:   Sun Feb 11 16:32:48 CST 2001