C4 Mathematical Heresy

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Question 6.

Now, I have always done the method in plugging in numbers that makes two other brackets = 0 so you are left with one unknown constant, but seeing the domain in this question leads me to believe, aren't we being blasphemous?

Clearly, not just by the domain, but x =/= -2 or 1/3, due to the 1/0 epidemic, how can we commit such crime without falter?

Reply 1

B_9710

I don't quite follow what you're saying.

Original post by Bobjim12

You can safely ignore that. For the purpose of finding the values of A and B it is irrelevant. I believe. (correct me if im wrong)

Reply 4

Serine Soul

I think it's showing that we can't use -2 or 1/3 in the function, but we most definitely can use it to rearrange the function into a form we like

The domain is the set of values that can be used in the function and when we're expressing a function in partial fractions, we're not really performing the function are we?

I'm probably wrong

(edited 7 years ago)

Reply 5

Zacken

Original post by Bobjim12

You'll want to read this: http://math.stackexchange.com/questions/165118/how-does-partial-fraction-decomposition-avoid-division-by-zero

But essentially, when you're doing partial fractions, you do this:

1. Start off with a bunch of fractions and domains where points are removed because of 1/0.

2. Guess a partial fraction form and multiply so that you have a polynomial equation. Something like Ax + B = C(x+1) + D(x+2).

You're very much right that we can't just substitute in x=-1 and x=-2 since those were not in the domain of the question.

3. What we do is ignore the problem we are solving. We no longer care about partial fractions, we only care about finding C, D such that the equation Ax + B = C(x+1) + D(x+2) is satisfied for all real values of x.

4. Once we have that, we've got a polynomial identity that is satisfied for all real values of x, we then remove the points that were ill-defined in the partial fractions so it's still true for all the other values of x (namely not -1 and -2), and then divide by (x+1)(x+2) to get the partial fraction identity we want. We originally proved it for all x, but we must now remove x=-1, -2 as per the original domain.

tl;dr partial fractions is fractions -> polynomial identity -> forget about partial fractions -> solve polynomial identity -> remember partial fractions -> apply original restriction and get partial fraction.

I'm not sure I explained that very well.

(edited 7 years ago)

Reply 6

Zacken

Original post by Someboady

You can safely ignore that. For the purpose of finding the values of A and B it is irrelevant. I believe. (correct me if im wrong)

Well, yes, you can safely ignore that - but are you not interested in why this is so?

Reply 7

Someboady

Original post by Zacken

Well, yes, you can safely ignore that - but are you not interested in why this is so?

I was just about to message you XD
Is that the explanation above?

(edited 7 years ago)

Reply 8

Zacken

Original post by Someboady

I was just about to message you XD
Why is it?

I've posted above.

Reply 9

Bobjim12

Original post by Zacken

.....

Well, i guess we could just ignore the problem, seems a little too cheeky, but alright then. Thanks for the explanation.

Reply 10

Zacken

Original post by Bobjim12

Well, i guess we could just ignore the problem, seems a little too cheeky, but alright then. Thanks for the explanation.

*shrugs*, if you want...

Consider f(x) / h(x) = g(x) / h(x) for x in some infinite field (say R) satisfying h(x) =/= 0.

This implifies that f(x) = g(x) since p(x) = f(x) - g(x) is some polynomial with infinitely many roots, except the finitely many that have the property h(x) = 0. This implies that p is the zero polynomial since a non-zero polynomial over a field has finitely many roots. So f(x) - g(x) = 0 means f(x) = g(x).

Hence, to solve for variables that appear in g, it is enough to find constants satisfying f(x) = g(x) for all x in your infinite field, even if they contain the roots of h. i.e: you can plug in any x in R.