Question about functions.

[Tom Mattson]

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

 

Which process are you referring to? There is not just one.

[Tom Mattson]

Which process are you referring to? There is not just one.

 

Actually, the real question is: What difference does it make?

 

As I said, we aren't talking about limits when we address the question, "Is f(1)=2?". We would be talking about them if we were addressing the question, "Does the limit of f(z) as z approaches 1 equal 2?"

 

Can you bring the theorems about limits to bear when you take the derivative of sin(x)? Of course, because the derivative is a limit.

 

Can you bring the theorems about limits to bear when you want to know if a certain point is in the domain of that function? No, because the definitions of "domain" and "function" make no reference whatsoever to limits.

[calbiterol]

I guess I was just under the impression that you could simplify something and have it still be equal, even under the condition stated here.

 

By the way, what's the difference between interdeterminate and undefined?

 

And technically, I could bring in limits when talking about dividing by zero. Would it be correct to say that the limit of the above function at f(1) is 2, while the value of the function at f(1) is undefined/interdeterminate?

[Tom Mattson]

I guess I was just under the impression that you could simplify something and have it still be equal' date=' even under the condition stated here.

[/quote']

 

But how can that be? The function [math](z^2-1)/(z-1)[/math] and the function [math]z+1[/math] have two different domains. Of course, the two functions are equal for z not equal to 1. But we were asked about when z=1.

 

By the way, what's the difference between interdeterminate and undefined?

 

Check these out:

 

Indeterminate

Undefined

 

And technically, I could bring in limits when talking about dividing by zero.

 

That would be quite a neat trick, considering that division has nothing to do with taking a limit. The former is an issue of algebra, the latter an issue of analysis.

 

Would it be correct to say that the limit of the above function at f(1) is 2, while the value of the function at f(1) is undefined/interdeterminate?

 

I think that the function is "undefined", not "indeterminate" at z=1, but I would defer to a mathematician. The reason I say this is that in my (limited) experience with pure mathematics, I have not seen the term "indeterminate forms" used except to refer to the result of a limit.

 

But otherwise, yes. The limit is 2 as z approaches 1, and the function is not defined at z=1.

[calbiterol]

But how can that be? The function [math](z^2-1)/(z-1)[/math] and the function [math]z+1[/math] have two different domains. Of course, the two functions are[/b'] equal for z not equal to 1. But we were asked about when z=1.

To be honest, I guess I was so used to simplifying it that I never though of the implications given this particular case. I seem to remember my math teacher saying something about this a while ago, but seeing as I'm out of school, there's no way to ask him, except perhaps emailing him.

 

 

Check these out:

 

Indeterminate

Undefined

 

Can't say the definition of indeterminate is particularly easy to comprehend with the wording they use' date=' but I think I have the idea, and thanks for the links.

 

That would be quite a neat trick, considering that division has nothing to do with taking a limit. The former is an issue of algebra, the latter an issue of analysis.

I guess I meant that AFAIK the concept of division by zero being undefined has something to do with the limit being infinity. I.e., [math]\lim_{x\rightarrow 0}=\frac{k}{x}[/math] when k is any real number, equals infinity, but this is strictly the limit. I'll be the first to admit I don't know the relationship between the function's limit and its interpretation as "undefined." I wasn't referring to division as a whole, only to the divide by zero case.

I think that the function is "undefined"' date=' not "indeterminate" at z=1, but I would defer to a mathematician. The reason I say this is that in my (limited) experience with pure mathematics, I have not seen the term "indeterminate forms" used except to refer to the result of a limit.[/quote']

After finding out the definition of indeterminate (because of your links, I might add, and by the way, thanks) I would have to agree.

[zaphod]

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 ?

 

no. it is not.

 

the limits at z=1 are equal to 2' date=' but that says nothing about the value of f(z) when z=1.

 

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

 

is not the same function as

 

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

 

yes, seem very similar when you graph them out. but one is defined at z=1 and the other is not.

 

stop trying to re-define shit, johnny. you're not going anywhere with this bullshit.

[ydoaPs]

calm down, please.

 

edit: just saw your sig, i guess you've read most of his threads and threads he has hijacked.

[Johnny5]

no. it is not.

 

the limits at z=1 are equal to 2' date=' but that says nothing about the value of f(z) when z=1.

 

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

 

is not the same function as

 

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

 

 

I realize that the limits as z approaches 1 from the right, or the left are each two. And if f(1) =2, then the function is continuous there.

 

Additionally I am not trying to re-define anything, but i'm giving the whole issue some thought, that's all.

 

 

If you stop and think for a moment, the logic presents itself to you, and here it is...

 

The following is a field axiom:

 

Let x denote an arbitrary real number.

 

[math] not(x=0) \Rightarrow \exists x^{-1} x \cdot x^{-1} = 1 [/math]

 

Now, consider the following theorem.

 

[math] \text{Theorem: For any real number x: } 0 \cdot x = 0 [/math]

 

Let a,b denote arbitrary real numbers.

 

One of the field axioms provides us with:

 

0+b = b

 

Now use the following statement:

 

for any real numbers x,y,a: if x=y then ax=ay

 

to obtain

 

a(0+b) = ab

 

The distributive axiom provides us with:

 

a0+ab = ab

 

Now, every real number has an additive inverse, ab has an additive inverse which we will represent by -ab, and the additive inverse of ab is such that ab+(-ab)= 0.

 

Now use use the statement that

 

for any real numbers x,y,b: if x=y then x+b=y+b

 

to obtain

 

(a0+ab) + (-ab) = ab + (-ab)

 

Now use associativity of addition to obtain

 

a0+(ab + (-ab)) = ab + (-ab)

 

From which it follows that:

 

a0+ 0 = 0

 

From which it follows that

 

a0 = 0

 

And by commutativity of multiplication we finally have:

 

0a = 0

 

Thus, zero times a is equal to zero.

 

But the symbol a represented an arbitrary real number throughout the proof, hence the theorem is true for any real number a.

 

QED

 

Suppose that we remove the restriction that division by zero is not allowed, and adopt the following field axiom:

 

Let x denote an arbitrary real number.

 

[math] \exists x^{-1} x \cdot x^{-1} = 1 [/math]

 

Since zero is a real number, we can use universal instantiation to obtain the following true statement:

 

[math] 0 \cdot 0^{-1} = 1 [/math]

 

But by the previous theorem we know that

 

[math] 0 \cdot 0^{-1} = 0 [/math]

 

as a consequence of the field axioms used in that proof.

 

Now, using the transitive property of equality it follows that:

 

[math] 0=1 [/math]

 

Which is false, since one of the field axioms is

 

[math] not(0=1) [/math]

 

We considered taking the following statement as an axiom:

 

[math] \forall x \in \mathbb{R} \exists x^{-1} \in \mathbb{R}: x \cdot x^{-1} = 1 [/math]

 

Since 0 is an element of the real number system, we can use universal instantiation to obtain the following true statement:

 

[math] \exists 0^{-1} \in \mathbb{R}: 0 \cdot 0^{-1} = 1 [/math]

 

Hence:

 

[math] 0 \cdot 0^{-1} = 1 [/math]

 

Whence we can now erroneously conclude that 0=1.

 

So the real field axiom is this:

 

[math] \forall x \in \mathbb{R} [not(x=0) \Rightarrow \exists x^{-1} \in \mathbb{R}: x \cdot x^{-1} = 1 ][/math]

 

Now consider the original function again:

 

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

 

We can write it as:

 

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

 

We can now factor the z^2-1 term and write:

 

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

 

Now comes the question, what is f(1)?

 

Putting 1 in the expression above we have:

 

[math] f(1) = (1+1)(1-1)(1-1)^{-1} [/math]

 

From which it follows that:

 

[math] f(1) = (2)(0)(0)^{-1} [/math]

 

And the parenthesis are superfluous, so that we can write:

 

[math] f(1) = 200^{-1} [/math]

 

From which it follows that:

 

[math] f(1) = 00^{-1} [/math]

 

Which we know leads to the conclusion that 0=1, which is false.

 

I would say that Tom is right. The original function cannot be defined for z=1.

 

So you are right zaph, they cannot be the same function, since f(z) = z+1 is defined for all z.

[matt grime]

Johnny, part of the definition of a function is its domain (and range). Your f is not a proper function: you fail to define the range. It is at best "an expresion".

 

Let us do an example to show why this is important:

Let S be the the integers mod 2, ie 0 and 1 with addition such that 1+1=0. The functions x^2+1 and x^4+1 are obviously different as functions of a real variable, but they are equivalent as functions from S to S.

 

Now, in your case, assuming you were taking about f being a function from C to C (as z usually implies) the function isn't defined at 1, but 1 is a removable singularity, that is there is a fucntion defined on all of C and continuous (analytic actually) that agrees with f on its natural domain.

 

If you recall I tried to explain removable singularities to you in an attempt to expllain why 0^0 does not have a universal meaning but can be taken to equal 1 when doing Taylor series.

[Johnny5]

If you recall I tried to explain removable singularities to you in an attempt to expllain why 0^0 does not have a universal meaning but can be taken to equal 1 when doing Taylor series.

 

I remember that, perhaps we can discuss it again. Let me think about your answer to this first.