Continuity and Integration by Partial Fractions

• qspeechc
In summary, Stewart is saying that if F(x) and G(x) are not the same then for x close to a, F(x) cannot equal G(x).
qspeechc

Homework Statement

The problem is from Stewart, Appendix G, A58, no.45.

Suppose that F, G, and Q are polynomials, and:
F(x)/Q(x) = G(x)/Q(x)
for all x except when Q(x) = 0. Prove that F(x) = G(x) for all x. [Hint: Use Continuity]

The Attempt at a Solution

I thought the statement was obvious, but ofcourse it isn't if they want you to prove it.

This is under the section "Integration of Rational Functions by Partial Fractions", so I guess the proof must somehow use this technique.

I guess we know it is true for real numbers a, b, c, that if a/c = b/c then a = b. We just haven't proved it for functions, and I think that is what this is about- am I correct? So the proof needs to be based on the fact that it is true for real numbers- is this correct?
That in mind, I decided to take the limit as x approaches a (except if Q(a) = 0) on both sides (sorry about not using Latex- I am in a rush):
lim [F(x)/Q(x)] = lim [G(x)/Q(x)]
lim[F(x)]/lim[Q(x)] = lim[G(x)]/lim[Q(x)]
And because they are all polynomials:
F(a)/Q(a) = G(a)/Q(a)
Now these are all real numbers, and thus F(a) = G(a).

This seems a little dodgy to me, because it does not hold for a when Q(a) = 0, and the questions says prove for all a. Also, we say, when we have:
F(a)/Q(a) = G(a)/Q(a)
that F(a), Q(a), G(a) are all real numbers, but the answer claims they are functions all-of-sudden!
Help is much appreciated.

It does not need to hold for when Q(a)=0, since you have at the very beggining that this must hold true besides for those x-is, this also icludes x=a such that Q(a)=0.

I'm sorry, I don't quite get that?

The trouble with the limits argument applied to the quotients F(x)/Q(x) and G(x)/Q(x) is that those limits may not exist. Argue by contradiction. Assume F(a) is not equal to G(a). Then you can write F(a)=G(a)+c for some non-zero c. Since F and G are continuous there is a value of e>0 such that |F(x)-F(a)|<|c|/2 and |G(x)-G(a)|<|c|/2 for |x-a|<e. But F(x)=G(x) if x is not equal to a. Do you see the contradiction?

Shouldn't we say:
F(x) = G(x) + P(x)? Where P(x) is some polynomial.
I'm not quite sure I have learned the stuff about limits you use in your proof, but I vaguely get it, I think. Stewart relegates this stuff to an appendix (rigorous treatment of limits). I'll go over it and get back to you.
Thank-you for helping.

The point is just that if F(a) and G(a) are different then if x is close enough to a, F(x)=G(x) CAN'T be close to BOTH F(a) and G(a). The rest of the words are just an attempt to express that using the limit definition.

1. What is the concept of continuity in mathematics?

Continuity refers to the smooth and unbroken nature of a mathematical function, where there are no sudden jumps or breaks in the graph. In other words, a function is continuous if its graph can be drawn without lifting a pencil.

2. What is partial fraction decomposition?

Partial fraction decomposition is a method used to break down a rational function into simpler fractions. This is done by representing the rational function as a sum of simpler fractions with denominators that are linear or quadratic expressions.

3. How is partial fraction decomposition used in integration?

Partial fraction decomposition is often used in integration to simplify complex rational functions. By breaking the function into simpler fractions, it becomes easier to integrate each individual term, leading to a solution for the original function.

4. What are the basic steps for integrating using partial fraction decomposition?

The basic steps for integrating using partial fraction decomposition are:
1. Factor the denominator of the rational function into linear and/or quadratic terms.
2. Write the partial fraction decomposition with unknown constants for each term.
3. Equate the original function to the partial fraction decomposition and solve for the unknown constants.
4. Integrate each term separately using basic integration rules.
5. Combine the results to get the final solution.

5. Are there any restrictions or limitations when using partial fraction decomposition?

Yes, there are a few restrictions to keep in mind when using partial fraction decomposition:
- The original function must be a proper rational function, meaning the degree of the denominator must be greater than the degree of the numerator.
- The factors in the denominator must be distinct and irreducible (cannot be further factored).
- In the case of repeated linear factors, different constants must be used for each decomposition term.
- In the case of repeated quadratic factors, different constants and linear terms must be used for each decomposition term.

• Calculus and Beyond Homework Help
Replies
6
Views
188
• Calculus and Beyond Homework Help
Replies
1
Views
426
• Calculus and Beyond Homework Help
Replies
2
Views
949
• Calculus and Beyond Homework Help
Replies
4
Views
805
• Calculus and Beyond Homework Help
Replies
14
Views
328
• Calculus and Beyond Homework Help
Replies
4
Views
73
• Calculus and Beyond Homework Help
Replies
1
Views
451
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
590
• Calculus and Beyond Homework Help
Replies
24
Views
1K