PDA

View Full Version : some math help

jrahhali
10-11-2005, 03:22 PM
1. does not exist
2. indeterminate
3. undefined

what exactly do these terms mean, when do they apply and when is it appropriate to use them (concerning math of course)?

for example, if you had f(x) = 1/x, when x = 0 how would you describe it?

another question.

if f(x) = (x^2 - 2x - 3) / (x-3)
f(3) = 0/0 by direct substitution. however, if you factor the dividend it becomes
f(x) = (x-3)(x+1) / (x-3)
f(3) = 4

can an answer only ever be 0/0?
what would you call (using the three terms above if they apply) a 0/0 answer?

thanks

Thantos
10-11-2005, 03:34 PM
for example, if you had f(x) = 1/x, when x = 0 how would you describe it?
Undefined

4 because you can't divide by 0

what would you call (using the three terms above if they apply) a 0/0 answer?
There are times (especially when evaluating limits) that substuting in a particular value will give you 0/0. This is one of the indeterminate forms. There are a couple of methods that will allow you to evaluate it and get an answer, but depending on the exact situation that answer might change. So its not possible to determine 1 answer for 0/0.

the limit as x-> 0 of 1/x is undefined because if you approach from the + you get one answer that is different then the limit as you approach from the -. Since they are different they do not meet all the requirements for a limit and as such the limit does not exist.

Rashakil Fol
10-11-2005, 03:56 PM
1. does not exist
2. indeterminate
3. undefined

what exactly do these terms mean, when do they apply and when is it appropriate to use them (concerning math of course)?

for example, if you had f(x) = 1/x, when x = 0 how would you describe it?

Hi. Try to read this deliberately and carefully.

A 'function' is simply a set of ordered pairs. A cocktail party explanation would be that a function is a 'curve'. For example, the function f, where f(x) = 1/x, contains the points (1,1), (2, 1/2), (-10, -0.1), along with many other ordered pairs.

Each 'function' has a domain. Remember that a function is a set of ordered pairs, of the form (x,y). The domain is a set of numbers. The 'domain' of a function f is a set containing only all the numbers x for which there is an ordered pair of the form (x,y) in f, i.e. an ordered pair whose first element is x. One restriction of functions is that for each value 'x' in the domain, there is exactly one ordered pair with x as the first element. This is sometimes called "passing the vertical line test."

For example, let R denote the set of real numbers. We can let f be a function with domain of R, where f(x) = x*x. Then f contains points such as (1,1), (2,4), (-2,4), (0.5,0.25), and so on. The codomain of the function is R (because that's how we defined it).

In the above example, if we were to write 'f("apple")', that value would be undefined (we've never defined what it means) because we said that f's domain is R, the set of real numbers, and "apple" is not a real number.

Let's consider the function described as g(x) = 1/x. This function implies a domain consisting of all non-zero real numbers. That is, 0 is not in the domain of g. As such, the notation 'g(0)' is undefined. Just like 'g("apple")'. Remember that a function is merely a set of ordered pairs. The set g contains the ordered pairs (1,1), (2, 0.5), (-3/10, -10/3), and such. Our function, g, does not contain any ordered pairs whose first element is 0.

Hopefully this makes sense now that you have an adequate definition of 'function'. (I'll repeat it yet again: a function is a set of ordered pairs.)

[edited to remove irrelevant references to 'codomain']

[edited again to add the section below]

A wrong-headed way to look at function is to look at them as algorithms. Don't think of f, where f(x) = 1/x, as an algorithm that takes a number and divides it into 1. Think of it as a curve.

Well, it's okay to think of a function as an algorithm, as long as you simultaneously think of it as a set of ordered pairs. It's always good to have multiple ways of thinking of the same thing, especially in mathematics and programming.

Thantos
10-11-2005, 04:07 PM
Hopefully this makes sense now that you have an adequate definition of 'function'. (I'll repeat it yet again: a function is a set of ordered pairs.)
I disagree. If you have a function f(x,y,z,w) you are going to have more then a "pair"

http://mathworld.wolfram.com/Function.html
A function is a relation that uniquely associates members of one set with members of another set.

Rashakil Fol
10-11-2005, 04:13 PM
I disagree. If you have a function f(x,y,z,w) you are going to have more then a "pair"

On the contrary, you have pairs of the form ((x,y,z,w), f(x,y,z,w)). E.g. if f(x,y) = x + y, you have f containing pairs such as ((1, 3), 4) and ((2, 9), 11).

About your Mathworld definition: Look up the word 'relation'. http://mathworld.wolfram.com/Relation.html -- my post's definition of a function is the same as theirs.

Rashakil Fol
10-11-2005, 04:21 PM
another question.

if f(x) = (x^2 - 2x - 3) / (x-3)
f(3) = 0/0 by direct substitution. however, if you factor the dividend it becomes
f(x) = (x-3)(x+1) / (x-3)
f(3) = 4

can an answer only ever be 0/0?
what would you call (using the three terms above if they apply) a 0/0 answer?

thanks

If f(x) = (x^2 - 2x - 3) / (x - 3), this implies that 3 is not in the domain of f. Hence f(3) is undefined. The notation "f(3) = 0/0" doesn't mean anything.

Now, as x gets arbitrarily close to 3, you'll find that f(x) gets arbitrarily close to the value 4. But the ordered pair (3, 4) is not contained in f.

jrahhali
10-11-2005, 04:33 PM
but isn't

(x^2-2x-3)/(x-3) = (x-3)(x+1)/(x-3) = x+1?

edit:
just as
just has 16/8 = 8/2 = 4?

Thantos
10-11-2005, 04:44 PM
No, he's right. Even if you factor and simplify you still have to make sure your answers don't violate the orignal function.

I've never really thought about functions as a set of inputs ordered with an output.

Rashakil Fol
10-11-2005, 04:52 PM
Another way of breaking down the problem is, does 0*x/0 = x, in general? Suppose that were the case. And let's suppose that y and z are numbers where y does not equal z. Then

y = (0*y)/0 ; Following the assumption
= (0)/0 ; 0*y = 0
= (0*z)/0 ; 0 = 0*z
= z

We would then conclude that y = z. (This is a common trick used in the 'proofs' that 1 = 0.) So simplifying 0*x/0 = x is not permissible.

Hence (x-3)(x+1)/(x-3) equals x+1 only when x != 3.

I'd say that (x-3)(x+1)/(x-3) is "syntactically incorrect" if x is allowed to equal 3. Fortunately, it isn't; when f(x) = (x-3)(x+1)/(x-3), the number 3 is not in the domain of f.

jrahhali
10-11-2005, 05:18 PM
woh!! lol! coooool!!!
i completely missed the fact that (x-3)(x+1)/(x-3) = 0*(x+1)/0 at x = 3!
holy doo doo crap!
man, i'm excited lol. you just cleared up a weak of function limit confusion right there!

another question. 0/0 is indeterminate. someone explained it like this

0*1 = 0 so 0/0 = 1
0*2 = 0 so 0/0 = 2
...

and so we conclude that 0/0 is indeterminate (rather than undefined)

is this really correct way of thinking about it? i mean, if 0/0 can equal any real number, and shouldn't 0/0 = inf.
I know there was that post on dividing by 0, but i think my question is different.

jrahhali
10-11-2005, 05:30 PM
a function is a set of ordered pairs.

wow.. woweee.
that's awsome! THANK YOU! it makes everything make so much more sense now!

edit:

whew.. oh man...whewy!
you may think this is spam, but it's spam from my heart.
i am so excited right now. you see there is no one in the house at the moment that i can let my emotions out on. MAN!! ORDERED PAIRS!!
if f at x does not have an ORDERED PAIR in the form (x, y) then x is not in the DOMAIN of f !! nice!!!

One restriction of functions is that for each value 'x' in the domain, there is exactly one ordered pair with x as the first element. This is sometimes called "passing the vertical line test."

so if you have a function:

f(x) = 2, x <= 1
3, x >= 1
x at 1 would not be in the domain of f(x)?

Rashakil Fol
10-11-2005, 05:42 PM
I'd say that 0/0 is 'indeterminate' if it's inside a limit. For example, if lim x->c f(x) = 0, and lim x->c g(x) = 0, then lim x->c (f(x)/g(x)) is indeterminate. I.e. we currently don't have enough information to determine the limit's value.

But outside a limit, the cold hard way of describing 0/0 is 'undefined'; it has no value. (Or is that the cold-hearted way? Warm-hearted people like to find value in everything :-))

Since 0 = -0, then if you were to let 0/0 = infinity, then (-0)/0 = inf, then -(0/0) = inf, then -inf = inf.

And -inf = inf is kind of weird :)

BobMcGee123
10-11-2005, 07:29 PM
and if you plug the numbers into the equation, and you get some number over zero, then the limit doesn't exist at that point. If you get zero over some number, the limit is zero.

>>x at 1 would not be in the domain of f(x )?
Well, it is. It doesn't have a limit at that point, and it isn't a function if at x = 1 the function has two values. This is akin to the fact that if you graph a circle using polar coordinates it is a relation but not a function because for certain values of x you have two values of y. You can graph a semi circle and have it be a function, but not a full circle and have it be a function.

Maragato
10-11-2005, 08:12 PM
I'd say that 0/0 is 'indeterminate' if it's inside a limit. For example, if lim x->c f(x) = 0, and lim x->c g(x) = 0, then lim x->c (f(x)/g(x)) is indeterminate.
AFAIK the limit of f'(x)/g'(x) can replace the original limit, this is called rule o L'Hopital. So limit of f(x)/g(x) = lim of f'(x)/g'(x) so your afirmation is wrong.

JaWiB
10-11-2005, 08:21 PM
and if you plug the numbers into the equation, and you get some number over zero, then the limit doesn't exist at that point. If you get zero over some number, the limit is zero.

I thought f(x) at c doesn't have to exist for lim f(x) x->c to exist. Like:
f(x) = (x^2+x-2)/(x-1)

lim f(x) x->1 = 3 right?

Maragato
10-11-2005, 08:24 PM
I thought f(x) at c doesn't have to exist for lim f(x) x->c to exist. Like:
f(x) = (x^2+x-2)/(x-1)

lim f(x) x->1 = 3 right?
Yes you can get it by manipulation or by L'Hopital.
Lim f'(x) when x->1 = lim 2x + 1 = 3.

Thantos
10-11-2005, 08:48 PM
Actually his assertion is correct. The form is indeterminate. L'Hopital can only be used on indeterminate forms.