IVT

The Intermediate Value Theorem

The Intermediate Value Theorem (or IVT) is a theorem of tremendous importance in applied mathematics, as well as in real analysis.

This is because it's an example of an existence theorem for the solution of equations. Solving equations, of course, is an important step in mathematical modeling.

Statement

Thm 4.5.1: The Let $f : [a,b] \to \mathbb R$ be continuous. If $L$ is a real number satisfying $f(a) < L < f(b)$ or $f(a) > L > f(b)$, then there exists a point $c \in (a,b)$ where $f(c) = L$.

Alternative formulation

Suppose that $f:[0,1]\to\mathbb R$ is continuous with $f(0)<0<f(1)$. Then, there is a point $c \in (0,1)$ where $f(c) = 0$.

Two proofs

There are two natural proofs of this theorem that nicely illustrate the distinction between a constructive proof and a pure existence proof.

We're going to take a look at the ideas behind both.

Pure existence

Suppose that $f:[0,1] \to \mathbb R$ satisfies the hypotheses of our alternative formulation. Define

$$S = \{x\in[0,1]: f(x)<0\}.$$

Note that $S$ is clearly non-empty, since $a\in S$. Also, $S$ is bounded above by $1$. Thus $S$ has a supremum, say

$$\sup(S) = \beta.$$

The idea is to show that $f(\beta) = 0$.

Pure existence (cont)

In order to show that $f(\beta) = 0$, you can construct two sequences:

• $(a_n)$ such that $a_n \to \beta^-$ and
• $(b_n)$ such that $b_n \to \beta^+$.

Since $\beta$ is the supremum of all those $x$ with $f(x)<0$, the $a_n$s can be chosen such that $f(a_n)<0$ for every $n$. By continuity, $$f(\beta) = \lim_{n\to\infty}f(a_n).$$ As a result, $f(\beta) \leq 0$.

Similarly, $f(b_n) > 0$ for all $n$ so $f(\beta) \geq 0$. The only choice is then $f(\beta) = 0$.

Comments on pure existence

We call the previous proof a pure existence proof because, while it clearly proves the theorem, it gives us no idea how to find the solution or even estimate it.

A constructive proof, by contrast, yields a technique to find the solution. An example is the bisection method, which can be used to find solutions of equations or estimate those solutions to any desired precision. The bisection method is an algorithm that arises from a proof of the intermediate value theorem.

In practice, when we say that the bisection method finds solutions "to any desired precision", we are using the language of epsilonics.

Proof by bisection

Suppose, again, that $f$ satisfies our alternative formulation of IVT.

Let $a_0 = 0$, $b_0 = 1$, and let $c_1 = (a_0+b_0)/2 = 1/2$. We consider three cases:

• $f(c_1) = 0$ - in which case, we're done,
• $f(c_1) < 0$ - in which case we set $a_1=c_1$ and $b_1=b_0$, and
• $f(c_1) > 0$ - in which case we set $a_1=a_0$ and $b_1=c_1$.

Note that if $f(c_1)\neq 0$, then $f(a_1)<0<f(b_1)$.

Proof by bisection (continued recursively)

The previous process can be continued recursively so that, if we haven't found the solution after $n$ steps, we have

$$f(a_n)<0<f(b_n).$$

We then let $c_{n+1} = (a_n+b_n)/2$ have the same three cases:

• $f(c_{n+1}) = 0$ - in which case, we're done,
• $f(c_{n+1}) < 0$ - in which case we set $a_{n+1}=c_{n+1}$ and $b_{n+1}=b_n$, and
• $f(c_{n+1}) > 0$ - in which case we set $a_{n+1}=a_n$ and $b_{n+1}=c_{n+1}$.

Proof by bisection (finale)

The previous recursive construction yields two possibilites:

• It is possible that $f(c_n) = 0$ exactly for some $n$ - in which case, there's your solution.
• If $f(c_n)\neq0$ for any $n$, then the process yields two bounded, monotone sequences, which must converge to the same value, call it $c$:
  • The sequence of left endpoints $(a_n)$ converges to $c$ from below and satisfies $f(a_n)<0$ for every $n$ and
  • The sequence of right endpoints $(b_n)$ converges to $c$ from above and satisfies $f(b_n)>0$ for every $n$.

As with the previous proof, the only choice for $f(c)$ is zero.

Application

Let's consider the function over the interval . Note that

• and
• .

Furthermore, is continuous, thus, IVT guarantees the existence of such that $f(c)=0$.

Application (cont)

Let's examine how the constructive proof of IVT yields an algorithm to estimate $c$ and apply that algorithm to find an estimate that's accurate to an absolute error tolerance of $\varepsilon = 0.0001$.

Again, we apply this to the function over the interval .

Application (cont)

Note that the midpoint of our initial interval is and that Thus, after one bisection step we find that there's a root in .

The midpoint of that interval is and Thus, our next interval is .

Key question

How many iterates do we need to ensure a result that is accurate to an absolute error tolerance of 0.0001?

Comments

• Here's some computer code using SageMath that carries the process out.
• I hope you see the relationship between this applied problem and the problem of finding $N$ or $\delta$ in terms of $\varepsilon$!