Question about functions.

[Johnny5]

Suppose i have the following function of the variable z:

 

[math] f(z) = \frac{z^2-1}{z-1} [/math]

 

Is it true or false that f(1) = 2 ?

[□h=-16πT]

Suppose i have the following function of the variable z:

 

[math] f(z) = \frac{z^2-1}{z-1} [/math]

 

Is it true or false that f(1) = 2 ?

 

Oh yeah, didn't think to simplify it, stupid me. Yeah i get 2.

[Nicoco]

Ah when you simplify the function, it indeed is 2, but then you worked out the fact that the function clearly isnt defined in 1. Actually, as my analysis prof would say, it is absolutely futile to speak of any function without clearly stating on which domain you work.

[Johnny5]

Are you using L' Hopital's rule here, I assume you are?

 

Nope. Just asking whether or not f(1)=2.

[Johnny5]

Ah when you simplify the function, it indeed is 2, but then you worked out the fact that the function clearly isnt defined in 1. Actually, as my analysis prof would say, it is absolutely futile to speak of any function without clearly stating on which domain you work.

 

What do you mean the function isn't defined in 1??

[Tom Mattson]

When you factor the numerator of that function and cancel the common factor, you aren't "simplifying" the function. You are obtaining a new function that agrees with the original function at all but a single point.

 

z=1 is not in the domain of the original function, and so f(1) does not equal 2.

[Johnny5]

When you factor the numerator of that function and cancel the common factor' date=' you aren't "simplifying" the function. You are obtaining a new function that agrees with the original function at all but a single point.

 

z=1 is not in the domain of the original function, and so f(1) does not equal 2.[/quote']

 

Ok, if you don't mind i would like to talk about this more.

 

[math] f(z) \equiv \frac{z^2-1}{z-1} [/math]

 

It is provable using the field axioms that:

 

[math] (z+1)(z-1) = z^2 - 1 [/math]

 

So that the LHS is substitutable for the RHS in any expression, and vice versa.

 

Hence:

 

[math] f(z) \equiv \frac{(z+1)(z-1)}{z-1} [/math]

 

Now, permit z to equal 1, so that we have:

 

[math] f(1) = \frac{(1+1)(1-1)}{1-1} = \frac{0}{0} [/math]

 

Thus, the expression is indeterminate, rather than undefined.

 

But we can cancel (z-1) from the numerator/denominator to obtain:

 

[math] f(z) = (z+1) [/math]

 

Now substituting we have:

 

[math] f(1) = (1+1) = 2 [/math]

 

Now, here is what you said...

 

You said that 1 is not in the domain of the original function.

 

Exactly how do you reach that conclusion. I just want to see the logic.

 

I am inclined to say that the expression was ambiguous, and that cancellation changed that, but that f(1) was defined all the while.

 

But i still want to hear your reasoning.

[Tom Mattson]

You said that 1 is not in the domain of the original function.

 

Exactly how do you reach that conclusion.

 

1 is not in the domain of the original function because the original function does not map 1 onto a real number.

[Johnny5]

1 is not in the domain of the original function because the original function does not map 1 onto a real number.

 

 

Ok follow up question. How are you going about determining what the domain of the original function is exactly?

 

And also, is or isnt the original function equivalent to z+1??

[Tom Mattson]

Ok follow up question. How are you going about determining what the domain of the original function is exactly?

 

I determine what the domain of f(z) is by looking at where it is defined. That is' date=' for which z does f(z) return a real number? I'm assuming of course that z is a real variable and that f is a real valued function.

 

And also, is or isnt the original function equivalent to z+1??

 

I already answered that: No.

 

The two functions agree at all but a single point.

[Johnny5]

I determine what the domain of f(z) is by looking at where it is defined. That is, for which z does f(z) return a real number

 

Well i see that you say that f(z) is not equivalent to z+1, so that i suppose is where to focus.

 

We have transitive property of equality, which doesnt break down.

 

So we have something like this

 

f(z) = X/Y

 

X= AB

 

f(z) = AB/Y

 

Y=B

 

f(z) =AB/B

 

And by axioms of real numbers AB/B=A, hence using transitive property of equality one more time f(z)=A.

 

But i see you very clearly say no, so where is the error in what i've done?

 

To look at things another way, suppose you start off with:

 

f(z) = z+1

 

Then multiply both sides by z-1 to obtain:

 

(z-1) f(z) = (z-1)(z+1)

 

As long as not (z-1)=0 we can divide to obtain:

 

[math] f(z) = \frac{(z-1)(z+1)}{(z-1)} [/math]

 

and then one more step gets us to...

 

[math] f(z) = \frac{(z^2-1)}{(z-1)} [/math]

 

So whatever you are saying is either wrong, or involves something subtle.

[timo]

[...]And by axioms of real numbers AB/B=A [...']

Shame on me but I´ve never heard about the "axioms of real numbers". What are they supposed to be? The definition of a field? If "yes", then your statement isn´t true (see http://mathworld.wolfram.com/FieldAxioms.html)

[Johnny5]

Shame on me but I´ve never heard about the "axioms of real numbers". What are they supposed to be? The definition of a field? If "yes", then your statement isn´t true (see http://mathworld.wolfram.com/FieldAxioms.html[/url'])

 

I meant the field axioms. The list at wolfram is not only superfluous, it is incomplete. But more importantly, what statement of mine are you referring to, which is false?

[Tom Mattson]

I meant the field axioms.

 

Well if you mean the field axioms then you should not be using division at all. You should be using multiplicative inverses. Then we would have the following:

 

 

f(z) = XY^-1

 

X= AB

 

f(z) = ABY^-1

 

Y=B

 

f(z) =ABB^-1

 

 

But if B=(z-1) then B^-1 is not defined for z=1.

[timo]

I meant the field axioms. The list at wolfram is not only superfluous, it is incomplete. But more importantly, what statement of mine are you referring to, which is false?

The only one I quoted. And to emphasize the line on Wolfram referring to this:

a*a^-1 = 1 = a^-1 * a if a != 0

[Johnny5]

What do you both say to this...

 

For any real numbers A,B, if not (B=0) then

 

[math] \frac{A}{B} \equiv AB^{-1} [/math]

 

?

 

Here is one of the field axioms:

 

[math] \forall x \in \mathbb{R} \exists (x^{-1}) \in \mathbb{R} \text{such that } [ \neg(x=0) \Rightarrow xx^{-1}=1] [/math]

[timo]

I wouldn´t know of any other (sensible) definition of division othen than multiplication by the inverse, so "yes".

[Tom Mattson]

I say just what I've been saying all along: There's no problem with making that identification as long as B^-1 exists.

[Johnny5]

Well, and this is for either of you...

 

In the case I gave, we had an indeterminate form

 

[math] f(z) = \frac{0}{0} [/math]

 

Or to put things differently

 

[math] f(z) = \frac{0}{0} = 00^{-1} [/math]

 

And you object, and now i at least see why.

 

However, 0/0 is indeterminate, which means that we aren't necessarily dealing with division by zero error, upon performing some operations, we can determine that f(z) is equal to some number.

 

So there seems to be some confusion about this.

 

And then there is my proof that

 

if f(z) = (z+1) then f(z) = (z^2-1)/(z-1)

 

Mmm i am starting to think the whole issue boils down to this now

 

is (z-a)(z-a)^-1 = 1 regardless of a?

 

I think it boils down to that one question.

 

And i know your response now.

 

No, because if z=a, then (z-a)^-1 is undefined.

 

Something is still bugging me though.

[Tom Mattson]

However' date=' 0/0 is indeterminate, which means that we aren't necessarily dealing with division by zero error,

[/quote']

 

Yes, you are. Admitting 0^-1 is nothing other than allowing division by zero.

 

upon performing some operations, we can determine that f(z) is equal to some number.

 

The operations you refer to are defined on the reals. You have no justification for using those operations on objects that are not real numbers, such as 0^-1.

[calbiterol]

Johnny5, there's another way of saying what you just said. Tom, I'm assuming you've been through limits before. Bear with me, I'm new to using [math]LaTeX[/math].

 

So, first off, through the definition of a limit, the limit of a whole number is the whole number.

 

Now...

 

[math]lim_{z\to 1}f(z) = lim_{z\to 1} \frac{z^2 - 1}{z - 1}[/math]

 

When we do limits, we're allowed to simplify - if you couldn't simplify, then you couldn't find the derivative of sin or cos, etc, and you can - so, we'll go about it this way.

 

[math]lim_{z\to 1}f(z) = lim_{z\to 1}\frac{(z - 1)(z + 1)}{z - 1}[/math]

 

Just as before, and so on:

 

[math]lim_{z\to 1}f(z) = lim_{z\to 1} \frac{(1)(z + 1)}{1}[/math]

 

Now, by definition of limits, you can break this apart into the following:

 

[math]lim_{z\to 1}f(z) =\frac{lim_{z\to 1}1 * lim_{z\to 1}(z + 1)}{lim_{z\to 1}1}[/math]

 

Which becomes

 

[math]lim_{z\to 1}f(z) = lim_{z\to 1}(z + 1)[/math]

...

[math]lim_{z\to 1}f(z) = (lim_{z\to 1}z) + 1[/math]

... and by definition of limits, we can just plug in 1 for z...

[math]lim_{z\to 1}f(z) = 1 + 1=2[/math]

 

AFAIK, you can do this.

[Tom Mattson]

Just as before' date=' and so on:

 

[math']lim_{z\to 1}f(z) = lim_{z\to 1} \frac{(1)(z + 1)}{1}[/math]

 

How did "z-1" turn into "1"?

[calbiterol]

Cancelling it out. Hence the "When we do limits, we're allowed to simplify - if you couldn't simplify, then you couldn't find the derivative of sin or cos, etc, and you can - so, we'll go about it this way." part. We're allowed to do this in this case, just like you are when finding the derivative of sin or cos.

[Tom Mattson]

Cancelling it out.

 

Ah' date=' of course.

 

Hence the "When we do limits, we're allowed to simplify

 

Yes, you can do simplification operations on functions when you take limits. But that's only because a theorem was proved that says that, if f(z) and g(z) agree at all but one point, and that point is z=c, then:

 

[math]

\lim_{z\rightarrow c}f(z)=\lim_{z\rightarrow c}g(z)

[/math]

 

But no one asked about the limit of f(z) as z approaches 1. The question "Does the limit of f(z) as z approaches 1 exist?" is completely different from "Does the function f(z) exist at z=1?", and the theorem I just mentioned for limits cannot be brought to bear.

[calbiterol]

Forgive me if I'm misunderstanding you, but then how do explain the process behind finding the derivative of sine?