Continuous Functions

MathsIsFun · 2012-11-07 18:03:19

Let me know if you find any errors, or think of some way to improve it.

Jai Ganesh · 2012-11-07 18:23:03

Hi MathsIsFun,

The Continuous Functions page is very well done! The illustrations complement the page!
I don't think there is any error.

A neat page! Thank you!

noelevans · 2012-11-07 18:56:55

Hi MIF!

Looks like an errorless labor of love to me!

bobbym · 2012-11-07 19:29:00

Hi MIF;

Looks good and reads easily. I like it.

Bob · 2012-11-07 20:44:22

hi MathsIsFun,

Another excellent page.

Just one thing.

Is that really undefined at x = 1 ? I'm not sure about the answer to that.

You can say it isn't defined there in the definition, or you could define it to be 2, but it's unclear.

There are lots of situations where we simplify and then substitute after (eg. calculus). This function obeys the continuous definition requirement at x = 1. I accept the point about 0/0 but this example may nevertheless cause controversy amongst your readers.

Suggestion:

Replace with

Bob

bobbym · 2012-11-07 21:50:53

Hi bob bundy;

I have asked that question too, why can't I simplify? I was told you can not simplify there because when you do you have changed the domain of the function. So the original function and the simplified one are not the same.

MathsIsFun · 2012-11-07 22:19:15

Yes, this gets me every time, too. It is just so natural to simplify, how can it be wrong?

It remind me of the 2=1 proof:

Let a=b
a² = ab
a²-b² = ab-b²
(a-b)(a+b) = b(a-b)
a+b=b
1+1=1
2=1

The error is when we simplify by dividing by (a-b)=0

Be that as it may, I can choose a less controversial example.

Bob · 2012-11-08 00:54:11

hi MathsIsFun and bobbym,

Example: f(x) = (x2-1)/(x-1) for all Real Numbers
The function is undefined when x=1:
(x2-1)/(x-1) = (12-1)/(1-1) = 0/0
So it is not a continuous function

It's all in the definition of the domain.

It's not really the 0/0 problem; you could do it with any function.

eg.

y = 2x for x in {reals} except x = 1 where it is undefined.

I've made a discontinuous function at x = 1.

Sure, it's very contrived but it demonstrates that it's all in the definition.

eg y = (x^2-1)/(x-1) for all x in {reals} except x = 1

y = 2 when x = 1

This function is continuous for all x.

I'm happy that it stays unchanged; but it may provoke this discussion again. Maybe that's a good thing ??

bobbym wrote:

changed the domain of the function

There have been several posts recently that seem to suggest that once you have been told the equation for the function then the domain follows of its own accord. I disagree. I think that the domain is part of the definition and hence it's up to the definer to declare whether the domain is changed.

eg.

y = (x^2-1)/(x-1) for all x in {reals} except x=1

y = x+1 when x = 1

Bob

anonimnystefy · 2012-11-08 01:36:06

bob bundy wrote:

hi MathsIsFun,
Another excellent page.
Just one thing.
Is that really undefined at x = 1 ? I'm not sure about the answer to that.
You can say it isn't defined there in the definition, or you could define it to be 2, but it's unclear.
There are lots of situations where we simplify and then substitute after (eg. calculus). This function obeys the continuous definition requirement at x = 1. I accept the point about 0/0 but this example may nevertheless cause controversy amongst your readers.
Suggestion:
Replace with
Bob

Those functions are not continuous, but the sinvularities they have are removable. Try searching removable singularities. I think there is a Wikipeadia article on it.

Bob · 2012-11-08 01:56:13

There is:

http://en.wikipedia.org/wiki/Removable_singularity

and that covers what I was trying to say.

Thanks

Bob

ShivamS · 2012-11-08 02:25:12

Excellent!

anonimnystefy · 2012-11-08 02:58:45

Hi Bob

You're welcome.

Hi MIF

Awesome page!

Last edited by anonimnystefy (2012-11-08 03:01:12)

Calligar · 2012-11-08 13:37:13

Hmm, I'm honestly a little bit confused about 1...indirect thing. Is 0/0 really undefined? I would have honestly thought it equals 0. I'm a little bit confused on that one part, so anyone care to explain (why 0/0 is undefined)? Though I do understand why it's not a continuous function. Other then that, I'd say you explained things clearly enough to be able to understand everything, so I'd say you did a pretty good job.

As for one of the posts, I don't really understand the 2=1 proof. Everything seems right until you get to a+b=b, nor do I understand how they got to that. I fail to see how this is proving 2=1, or rather, making it arguable I should say.

bobbym · 2012-11-08 14:30:04

Hi Calligar;

In line 4 to 5 of the 2=1 proof there is a division by a-b. But if a=b as stated in the first line then a - b =0 and you can not divide by zero.

Calligar · 2012-11-08 17:59:32

If you are talking about the 3'rd line where it says, (a-b)(a+b) = b(a-b), you are multiplying, not dividing, unless I'm missing something...?

Also, I'm not sure I quite understand the reason for simplifying it. Everything seems fine about it too me...

Bob · 2012-11-08 20:17:20

hi Calligar

0 x 63 = 0 x 105

If dividing by zero was ok you'd get

63 = 105 ??

So in basic number theory there is a rule that you shouldn't ever divide by zero

In the fake proof

(a-b)(a+b) = b(a-b) is ok.

The problem arises at the next step because when you cancel a common factor what you are actually doing is dividing both sides of the equation by it.

eg 7 times 3 = 7 times N

divide both sides by 7

3 = N

In the fake proof to get to the next line you have to divide by (a-b), that is the false step as (a-b) = 0

0/0 has an indeterminate value. Why?

Let's go back to what division is.

48 = 6 x 8 so 48/6 = 8

You reverse the multiplication process.

So to work out 0/0 you have to think what would be the number here 0 x ? = 0

As any number will make that work, you cannot say what 0/0 is. (In calculus you have to work with limits to sort out 0/0)

Bob

bobbym · 2012-11-09 03:07:19

Hi Calligar;

Calligar wrote:

If you are talking about the 3'rd line where it says, (a-b)(a+b) = b(a-b), you are multiplying, not dividing, unless I'm missing something...?

Let a=b
a² = ab
a²-b² = ab-b²
(a-b)(a+b) = b(a-b)
a+b=b
1+1=1
2=1

In line three you have a factor of (a-b) on both sides. On line 4 you do not. Both sides were divided by (a-b) which is equal to 0. That is the misstep. Generally, you never divide by a variable unless you are sure that it cannot be 0. Sometimes such divisions are fatal, sometimes they just destroy answers. For instance

Divide both sides by x

Here the decision to divide by x, has cost us the other root which is 0.

MathsIsFun · 2012-11-09 09:01:45

Nice extra example bobby.

And that 0/0 thing ... it is like asking "how many 0s in 0?". Are there no zeros in zero, or perhaps just the one? See: Dividing by Zero

noelevans · 2012-11-09 15:22:45

Hi!

Here's another explanation why division by zero is not possible. Division is a secondary operation,
not a primary operation; that is, division is not given in the field axioms that establish the real
number system. Multiplication is given in the axioms (as well as addition, but not subtraction).
Division is then defined as multiplication by reciprocals.

For example suppose x is a real number whose reciprocal is y. Then given a number z we define
z/x as z*y. So to divide by x, x must have a reciprocal. BUT zero has no reciprocal as stated in
the multiplicative inverse axiom: For each real non-zero number x, there is a real number y such
that xy=1. Of course the usual notation for the reciprocal of x is written 1/x. But this looks like
division, so using 1/x for the reciprocal in the axiom and then later using "/" for division is a bit
confusing and makes the definition via z/x = z*(1/x) look circular.

When all is said and done, the axioms do not allow for a reciprocal of zero, hence division by zero
is a non-issue --- it could never happen since zero has no reciprocal to define division by zero in
terms of.

Having zero in the denominator of a fraction is a similar issue. The set of fractions based on the
set of whole numbers W={0,1,2,3,...} is defined as

F = { p/q | p and q are in W and q is not zero }

If someone asks why we can't have something like 2/0 the answer is simply the definition does
NOT allow zero in the denominator. It has nothing to do with "division by zero."

Analogously the field axiom for the reals do not allow zero to have a reciprocal. Hence to write
an expression one would read as "division by zero" simply violates the field axioms. Trying to
write "a number divided by zero" would be defined as "a number times the reciprocal of zero."
But a "reciprocal of zero" does not exist.

The other explanations of why one can't divide by zero illustrate the problems that would occur
if we did allow zero to have a reciprocal. As such they provide good reasons from disallowing
zero to have a reciprocal in the first place.

A nice physical example of trying to divide by zero can be seen in the operation of the old
mechanical calculators of yesteryear. They operated on division as a "repeated subtraction."
If one tried to divide 2 by zero, the calculator would subtract zero from two, add one to the
quotient, and then check to see if there was enough left to subtract zero from it again. Of course
there was enough left to subtract zero again since 2>0. So it subtracted zero from two again,
added one to the quotient, and checked to see if there was enough left to subtract zero again.
The old mechanical calculators got stuck in an "infinite loop" subtracting zero over and over
again. The quotient register looked like an old gas pump register with the dials spinning as
the quotient grew.

A friend of mine in high school rented a mechanical calculator to do lots of arithmetical homework.
On the front in bold letters it said "DO NOT DIVIDE BY ZERO!" Of course, that was an invitation.
So about half way through his assignment he starts a division by zero. The calculator was run
on electricity, so it just kept spinning the digits in the quotient. Quite fun to watch, but tiresome
after a while. So he punched "clear" and all the other buttons and levers he could find. Nothing
stopped the process. Finally he got the idea to pull the plug, and presto! it stopped.

So he thought that he'd better get back to his homework. He plugged it back in and presto! it
resumed the division calculation. He did the rest of his assignment by hand. It appears that
something had to be reset internally to get it to stop the division. I suspect when he returned
the machine he just set it on the counter and quickly headed for the door!

The old electric mechanical calculators (which were made of metal and weighed about 60 lbs)
demonstrated quite well the "repeated subtraction" algorithm for division. You could watch it
as it proceeded through a calculation as the dials spun and the carriage shifted. Slow by today's
standards, but it got the job done (if not dividing by zero!).

Calligar · 2012-11-11 10:56:39

Okay, I understand the fact that you can not divide x by 0, but i was strictly talking about 0 / 0, not 1, 2, or any other number. I don't understand why you can't divide by 0 in that case, because I do see 0 x 0 = 0 if u changed it to multiplication. Also, I didn't realize that rule applied to 0 as well, that was why I was asking. I already knew about stuff like 1/0 is undefined though and already understand why that is. Also I have read every post, and it still doesn't seem to prove to me that 0 / 0 is undefined, maybe because I'm looking at it the wrong way? Though I also worry about arguing this further, as this might only end up being something just like 0.999...=1. So, I'm just going to assume its a rule right now I don't... personally agree with. If there's better reasoning for it though, I would not mind hearing it.

Calligar · 2012-11-11 13:15:32

Actually, I just realized I was looking at it differently. This in my opinion is another futile argument, I understand why it's a rule now. No need to further explain to me. I also understand the 2 = 1 proof now and see the error, I was having a...similar kind of issue with it. So with all my questions answered, I go back to with what I was originally going to say, but didn't say it quite clearly. I think the continuous functions page is very clear and well done. I didn't have an issue understanding it, only understanding indirectly relating things.

MathsIsFun · 2012-11-11 13:33:25

Thanks Calligar. Questioning is an important part of mathematics.

Calligar · 2012-11-11 14:21:44

I'd agree, because I have high emphasis on myself for understanding things. It really bothers me if I walk away from something without understanding it well. It has also done nothing but help me with...well pretty much everything I know now (though there is still a lot I don't understand). So yeah, I will often question something I don't understand to better understand it.

anonimnystefy · 2012-11-11 21:40:35

Hi MIF

You can put on the same page the definition of a smooth function.

MathsIsFun · 2012-11-12 08:33:35

Smooth function would need a page of its own, I think.

Math Is Fun Forum

#1 2012-11-07 18:03:19

Continuous Functions

#2 2012-11-07 18:23:03

Re: Continuous Functions

#3 2012-11-07 18:56:55

Re: Continuous Functions

#4 2012-11-07 19:29:00

Re: Continuous Functions

#5 2012-11-07 20:44:22

Re: Continuous Functions

#6 2012-11-07 21:50:53

Re: Continuous Functions

#7 2012-11-07 22:19:15

Re: Continuous Functions

#8 2012-11-08 00:54:11

Re: Continuous Functions

#9 2012-11-08 01:36:06

Re: Continuous Functions

#10 2012-11-08 01:56:13

Re: Continuous Functions

#11 2012-11-08 02:25:12

Re: Continuous Functions

#12 2012-11-08 02:58:45

Re: Continuous Functions

#13 2012-11-08 13:37:13

Re: Continuous Functions

#14 2012-11-08 14:30:04

Re: Continuous Functions

#15 2012-11-08 17:59:32

Re: Continuous Functions

#16 2012-11-08 20:17:20

Re: Continuous Functions

#17 2012-11-09 03:07:19

Re: Continuous Functions

#18 2012-11-09 09:01:45

Re: Continuous Functions

#19 2012-11-09 15:22:45

Re: Continuous Functions

#20 2012-11-11 10:56:39

Re: Continuous Functions

#21 2012-11-11 13:15:32

Re: Continuous Functions

#22 2012-11-11 13:33:25

Re: Continuous Functions

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Re: Continuous Functions

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Re: Continuous Functions

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Re: Continuous Functions

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