Calculus. Prove this

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piercegirl

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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 for all x.

Im supposed to prove that F(x)=G(x) when Q(x)=0
 
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piercegirl said:
Suppose that F, G, and Q are polynomials and F(x)/G(x)=G(x)/Q(x) for all x except when Q(x)=0. Prove for all x
Click to expand...

"Suppose (this) and (that). Prove [???] for all x."

What is [???] that you're supposed to prove?

Thank you! :wink:
 
piercegirl said:
Suppose that F, G, and Q are polynomials and F(x)/G(x)=G(x)/Q(x) for all x except when Q(x)=0. Prove for all x.

Im supposed to prove that F(x)=G(x) when Q(x)=0
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Do it in two parts:
1) \(\displaystyle G(x)\ne 0\) when Q(x)= 0.
2) \(\displaystyle G(x)= 0\) when Q(x)= 0.

In either case, since F(x)/G(x)= G(x)/Q(x) for all x except\(\displaystyle x_0\), where \(\displaystyle Q(x_0)= 0\), take the limit of both sides as x goes to \(\displaystyle x_0\).
 
HallsofIvy said:
Do it in two parts:
1) \(\displaystyle G(x)\ne 0\) when Q(x)= 0.
2) \(\displaystyle G(x)= 0\) when Q(x)= 0.

In either case, since F(x)/G(x)= G(x)/Q(x) for all x except\(\displaystyle x_0\), where \(\displaystyle Q(x_0)= 0\), take the limit of both sides as x goes to \(\displaystyle x_0\).
Click to expand...


you mean,

Lim as xapproaches 0 of F(x)/G(x)=Lim as x approches 0 of G(x)/Q(x)??
 
No, I meant what I said. There is no reason to think that x=0 has anything to do with this problem and no reason to take the limit as x goes to 0. Take the limit as x goes to whatever value makes Q equal to 0.
 
HallsofIvy said:
No, I meant what I said. There is no reason to think that x=0 has anything to do with this problem and no reason to take the limit as x goes to 0. Take the limit as x goes to whatever value makes Q equal to 0.
Click to expand...

Sorry, I still dont understand this is what I got originally
F(x)/Q(x)=G(x)/Q(x)
F(x)/[G(x)*Q(x)]=Q(x)
Then substitute in
F(x)/[F(x)/[G(x)*Q(x)]}=G(x)/[F(x)/[G(x)*Q(x)]]
SIMPLIFY
G(x)*Q(x) = [(G(x))^2*Q(x)]/F(x)
G(x)^2*Q(x)=G(x)*F(x)*Q(x)
THIS IS ONLY TRUE IF
G(x)^2=G(x)*F(x)
Thereforr
G(x)=F(x)
 
HallsofIvy said:
Do it in two parts:
1) \(\displaystyle G(x)\ne 0\) when Q(x)= 0.
2) \(\displaystyle G(x)= 0\) when Q(x)= 0.
Click to expand...
I find this hint somewhat confusing because, unless I am mistaken, G(x) must equal 0 when Q(x) = 0.

To do the problem at all, it must be assumed that it is false that Q(x) = 0 for every real number x because otherwise F(x) and G(x) are insufficiently specified. Furthermore, it must also be assumed that there exists at least one real value of x for which Q(x) = 0 because otherwise the problem makes no sense.

\(\displaystyle \text{Let a be an arbitrary real number such that } Q(a) = 0.\)

\(\displaystyle \text{Consider a neighborhood of a such that }Q(x) \ne 0 \text{ if } x \ne a.\)

\(\displaystyle \text{By hypothesis, in that neighborhood }\dfrac{F(x)}{G(x)} = \dfrac{G(x)}{Q(x)} \implies F(x) * Q(x) = G(x) * G(x).\)

\(\displaystyle \text{But F(x), G(x), Q(x), F(x) * Q(x), and G(x) * G(x) are all polynomials and so are continuous at every x, including a} \implies\)

\(\displaystyle \displaystyle \lim_{x \rightarrow a}F(x) = F(a),\ \lim_{x \rightarrow a}G(x) = G(a),\ and\ \lim_{x \rightarrow a}Q(x) = Q(a) = 0.\)

\(\displaystyle \text{So } \displaystyle \lim_{x \rightarrow a}[G(x) * G(x)] = \lim_{x \rightarrow a}[F(x) * Q(x)] = \lim_{x \rightarrow a}F(x) * \lim_{x \rightarrow a}Q(x) = F(a) * 0 = 0 \implies\)

\(\displaystyle \displaystyle \lim_{x \rightarrow a}G(x) * \lim_{x \rightarrow a}G(x) = 0 \implies G(a) * G(a) = 0 \implies G(a) = 0.\)
 
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piercegirl said:
Sorry, I still dont understand this is what I got originally
F(x)/Q(x)=G(x)/Q(x) But you said the problem was F(x)/G(x) = G(x)/Q(x), not F(x)/Q(x) = G(x)/Q(x).
F(x)/[G(x)*Q(x)]=Q(x)
Then substitute in
F(x)/[F(x)/[G(x)*Q(x)]}=G(x)/[F(x)/[G(x)*Q(x)]]
SIMPLIFY
G(x)*Q(x) = [(G(x))^2*Q(x)]/F(x)
G(x)^2*Q(x)=G(x)*F(x)*Q(x)
THIS IS ONLY TRUE IF
G(x)^2=G(x)*F(x)
Thereforr
G(x)=F(x)
Click to expand...
This does not make ANY sense to me. For one thing, you have changed the problem. For another, you have not worried about dividing by zero, which is a large part of the purpose of the problem.
 
piercegirl said:
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 for all x.

Im supposed to prove that F(x)=G(x) when Q(x)=0
Click to expand...
I think everyone is working too hard.

Suppose, then, that a is such that Q(a) = 0. Then

If F(x)/Q(x)=G(x)/Q(x) for all x except when Q(x)=0, then
F(x)=G(x) for all x except when Q(x)=0.
and then lim[x -> a] F(x) = lim[x -> a] G(x)
But you said F and G are polynomials, which are CONTINUOUS. (apologies for the caps)

So then lim[x -> a] F(x) = F(a) = lim[x -> a] G(x) = G(a)

I think that does it. F and G are surely equal when x /= a, and also when x = a.
 
Everyone was working hard because WRONG problem was posted originally - later corrected.....
 
SAMUELK said:
I think everyone is working too hard.

Suppose, then, that a is such that Q(a) = 0. Then

If F(x)/Q(x)=G(x)/Q(x) for all x except when Q(x)=0, then
F(x)=G(x) for all x except when Q(x)=0.
and then lim[x -> a] F(x) = lim[x -> a] G(x)
But you said F and G are polynomials, which are CONTINUOUS. (apologies for the caps)

So then lim[x -> a] F(x) = F(a) = lim[x -> a] G(x) = G(a)

I think that does it. F and G are surely equal when x /= a, and also when x = a.
Click to expand...
Yes, that does it for THAT problem. Unfortunately, THAT is not the problem originally posed.
 
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