Prove by Induction That All Polynomials Are Continuous
The Limit Rules
For quick reference, these are the limit rules we saw last time.
| | (1) |
| | (2) |
| | (3) |
| | (4) |
| | (5) |
| | (6) |
| | (7) |
Proving More Abstract Theorems
Limit rules are useful not only for torturing calculus students by forcing them to write out highly structure proofs of things they could more easily do by other means. Fundamental theorems are also useful for proving more general theorems with broad applicability.
In this section of the lecture we are going to prove a very useful theorem.
Theorem Let p(x) be any polynomial in x. For all a the limit
exists and equals p(a).
This is a very general statement. In fact, it is so general that we are going to need some more sophisticated tools to solve it. One of those tools is the method of mathematical induction
Mathematical Induction
Suppose you have a statement that you want to prove is true for all positive whole numbers n. Here a method called mathematical induction for proving such general statements.
- Prove the statement for the special case n = 1.
- Prove that if you assume that the statement is true for all n < N then the statement is also true for n = N.
An Example of Mathematical Induction
Theorem For all positive integers n and numbers a
Proof by mathematical induction: The result is true in the special case n = 1 by limit rule (2) above.
Assume the result is true for n = N - 1. By limit rule (6)
and the result holds for n = N.
This result combined with rule (5) above also gives
Proving the Theorem
We are now in a position to prove
Theorem Let p(x) be any nth degree polynomial. For all a the limit
exists and equals p(a).
Proof by mathematical induction on n: The theorem is true for n = 1. The general first degree polynomial p(x) = a1 x + a0 has a limit at a:
To do the induction step, we assume that the theorem is true for all polynomials of degree N-1 and use that assumption to prove that the theorem is true for polynomials of degree N. Any polynomial p(x) of degree N can be written as the sum of a term aN xN and a polynomial g(x) of degree N-1.
By what we have proved already and the induction assumption,
By limit rule (3) we have
Section 2.5: Continuity
The Definition of Continuity
Definition A function f(x) is continuous at a point a if the following three hold:
| | (9) |
| | (10) |
Note that the definition applies to just one point at a time. Most often, we aggregate a large number of points and talk about continuity over intervals. Here are some examples.
Example One
Every polynomial is continuous on the interval 
All polynomials are defined at every point. We proved above that all polynomials have limits at every point, and that
for all a.
Example Two
The function f(x) = 1/x is continuous on the intervals 
f(x) is clearly defined and equal to 1/a for every a except 0. Because
for all a by limit rule (2) above and
by limit rule (7) above, we have that 1/x is continuous for every x but 0. f(x) is undefined at 0, so it is not continous there.
Example Three
The function
is continuous on the interval [-1,1]. At any point in the interior of the interval there is no problem with showing that the function is continuous. The only problem points are the points at either end of the interval. For example, at x = 1 the function is defined and has value 0. However, the limit
does not exist because the function is undefined everywhere to the right of x = 1. The best we can hope for is to say that
exists and equals f(1) = 0. In cases like this we have to fudge and say that the f(x) is continuous from the left at x = 1.
An Important Theorem About Continuous Functions
Intermediate Value Theorem Let f(x) be continuous throughout the closed interval [a,b]. Let m be any number between f(a) and f(b). Then there is at least one c in [a,b] such that f(c) = m.
How to read a theorem
This theorem presents us with a nice opportunity to spend some time thinking about why theorems are put together the way they are. This theorems is a nice example, because on the one hand what it says is fairly obvious while on the other hand if the theorem were not constructed with care it might not be true at all.
The first thing you need to know about theorems is that they all have a common structure:
IF <condition(s)> THEN <consequence(s)>
The <condition(s)> represent one or more specific conditions that have to be fulfilled before we can apply the theorem. For the theorem to apply, all of these conditions have to be met. If one or more of the conditions are not met, the theorem tells us nothing at all: the consequence may or may not be true in that case. The one thing that a theorem does guarantee is that if all of the conditions are satisfied, the consequence(s) must follow.
Whenever you see a new theorem, you should examine the conditions carefully. Why is each particular condition necessary? If we left one of them out, would the theorem continue to be true?
Consider the intermediate value theorem above. That theorem has two conditions. The first is that the function involved must be continuous. The second is that the interval involved must be an open interval. Both of these conditions are necessary: if we leave either one of them out we can quickly concoct examples that satisfy the weakened conditions but not the consequences.
Here is a simple example. Consider the function
f(x) = 1/x
on the interval [-1,1]. f(-1) = -1, while f(1) = 1. Does the function pass through m = 0 anywhere in that interval? No, it does not, even though 
A sample application of the IVT
Here is a problem that demonstrates how one could use the IVT in practice.
Problem Show that the function
passes through zero at least once.
Note that the function is continuous (it is a polynomial). Try a = -100 and b = 100 as endpoints.
Let m = 0. Note that f(a) < m < f(b). Therefore by the IVT, there has to be at least one c between a and b at which f(c) = m = 0.
Note that the theorem merely states that there is such a point c. It tells us nothing about where that point actually is. The IVT is an example of an existence theorem: it states that something exists, but does not help us to find it.
Homework
Section 2.5: 3, 4, 15, 16, 22, 23, 42, 43, 46
Source: http://www2.lawrence.edu/fast/GREGGJ/Math140/022/022Continuity.html
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