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'[OT] When is a function not continuous?'
2005\12\19@142605
by
Bill Kuncicky

In order to study up on my electronics, I've had to take a detour and
brush up on calculus (there seems to be an infinite number of rabbit
trails to have to chase down on the way to "success.") But I'm
determined, so I bought a used book called QUICK CALCULUS, Second
Edition, by Daniel Kleppner and Norman Ramsey. Found it for a couple of
bucks in the local used book shop. It has a lot of notes in the
margins, but that's OK with me.
I've gotten the preliminary stuff out of the way (functions, graphs,
trig, exponentials and logs) and started on the differential calculus
part. The first section there is "limits," and I've hit a snag in my
thinking. Given a function f(x), such that f(x) = 1 for x .GE. 0, and
f{x} = 0 for x .LT. 0, is that a continuous function or a discontinuous
function?
Since for every possible real number x there is a number f(x) which is
either zero or one, it seems to be that it would be continuous. The
book says it is discontinuous, though. Since they also say "A more
picturesque description of a continuous function is that it is a
function you can graph without lifting your pencil from the paper in the
region of interest," though, I can see where it might possibly be
discontinuous, but it is against intuition. I think that possibly they
meant to write "f(x) = 1 for x .GT 0" rather than using the .GE.
Any thoughts would be appreciated. (I'm going to read over this email
carefully before sending it, to make sure I don't sound like English is
my 27th language. :)
Thanks,
Bill
2005\12\19@144934
by
Mike Hord

> book says it is discontinuous, though. Since they also say "A more
> picturesque description of a continuous function is that it is a
> function you can graph without lifting your pencil from the paper in the
> region of interest," though, I can see where it might possibly be
> discontinuous, but it is against intuition. I think that possibly they
> meant to write "f(x) = 1 for x .GT 0" rather than using the .GE.
That was going to be my definition of it. Dictionary.com says:
"a function which for certain values or between certain values of
the variable does not vary continuously as the variable increases.
The discontinuity may, for example, consist of an abrupt change in
the value of the function, or an abrupt change in its law of variation,
or the function may become imaginary."
Perhaps "piecewise" is a more appropriate term for this function.
It certainly is more intuitive, if we consider the second example,
an abrupt change in its law of variation. There is NO equation to
describe this function; we must resort to the empirical "If >, thus,
and if <=, thus" method.
Mike H.
2005\12\19@145214
by
Spehro Pefhany

At 02:25 PM 12/19/2005 0500, you wrote:
{Quote hidden}>In order to study up on my electronics, I've had to take a detour and
>brush up on calculus (there seems to be an infinite number of rabbit
>trails to have to chase down on the way to "success.") But I'm
>determined, so I bought a used book called QUICK CALCULUS, Second
>Edition, by Daniel Kleppner and Norman Ramsey. Found it for a couple of
>bucks in the local used book shop. It has a lot of notes in the
>margins, but that's OK with me.
>
>I've gotten the preliminary stuff out of the way (functions, graphs,
>trig, exponentials and logs) and started on the differential calculus
>part. The first section there is "limits," and I've hit a snag in my
>thinking. Given a function f(x), such that f(x) = 1 for x .GE. 0, and
>f{x} = 0 for x .LT. 0, is that a continuous function or a discontinuous
>function?
>
>Since for every possible real number x there is a number f(x) which is
>either zero or one, it seems to be that it would be continuous. The
>book says it is discontinuous, though. Since they also say "A more
>picturesque description of a continuous function is that it is a
>function you can graph without lifting your pencil from the paper in the
>region of interest," though, I can see where it might possibly be
>discontinuous, but it is against intuition. I think that possibly they
>meant to write "f(x) = 1 for x .GT 0" rather than using the .GE.
>
>Any thoughts would be appreciated. (I'm going to read over this email
>carefully before sending it, to make sure I don't sound like English is
>my 27th language. :)
BTW, the 'proper' (i.e. rigorous) way to learn the foundations of calculus
is with the deltaepsilon proofs.
Your function f(x) is not continuous (and therefore not differentiable)
at x=0.
You can see this intuitively by seeing that the values of f(x) as x approaches
0 arbitrarily closely from either side are not the same. You can also look up
the formal definition of a 'continuous function'.
>Best regards,
Spehro Pefhany "it's the network..." "The Journey is the reward"
spam_OUTspeffTakeThisOuTinterlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com
>> Inexpensive test equipment & parts http://search.ebay.com/_W0QQsassZspeff
2005\12\19@150018
by
William Couture
On 12/19/05, Bill Kuncicky <.....kuncickyKILLspam@spam@gmail.com> wrote:
> I've gotten the preliminary stuff out of the way (functions, graphs,
> trig, exponentials and logs) and started on the differential calculus
> part. The first section there is "limits," and I've hit a snag in my
> thinking. Given a function f(x), such that f(x) = 1 for x .GE. 0, and
> f{x} = 0 for x .LT. 0, is that a continuous function or a discontinuous
> function?
>
> Since for every possible real number x there is a number f(x) which is
> either zero or one, it seems to be that it would be continuous. The
> book says it is discontinuous, though. Since they also say "A more
> picturesque description of a continuous function is that it is a
> function you can graph without lifting your pencil from the paper in the
> region of interest," though, I can see where it might possibly be
> discontinuous, but it is against intuition.
It is discontinuous.
One of the fundamental ideas of calculus is looking at the change in
a function for some small value, call it epsilon, as epsilon is allowed
to approach 0.
So, look at your function f():
f(0) = 1
and
f(0epsilon) = 0
as epsilon approaches 0, f(0epsilon) does not approach f(0).
Also, again using the idea of epsilon, the derivative of the function f()
at 0 is the limit of
f(0)  f(0  epsilon) / epsilon
as epsilon approaches 0. This is not defined, therefore there is no
derivative of f() at 0, and the function f() is discontinuous at 0.
> I think that possibly they meant to write "f(x) = 1 for x .GT 0"
> rather than using the .GE.
No, if they had defined f() such that f(x) = 1 for x GT 0 and
f(x) = 0 for x LT 0, then f() would not be defined at 0, and nothing
could be said about it's derivative at 0 (and thus, nothing about
the functions discontinuity at 0).
Bill { who seems to be a much bigger geek than he rembered }

Psst... Hey, you... Buddy... Want a kitten? straycatblues.petfinder.org
2005\12\19@150059
by
olin piclist
Bill Kuncicky wrote:
> Given a function f(x), such that f(x) = 1 for x .GE. 0, and
> f{x} = 0 for x .LT. 0, is that a continuous function or a discontinuous
> function?
Discontinuous. This means the function value jumps from one value to
another without ever taking on the values in between. This happens with
your function just below 0.
Another way of looking at it is that the function can't be reversed over its
output range. Your function has an output range from 0 to 1. However there
is nothing you can put into the function to get the value .5 out. This is a
sufficient condition to show discontinuous, although the lack thereof does
not prove continuous.
> Since for every possible real number x there is a number f(x) which is
> either zero or one, it seems to be that it would be continuous.
No, that's not what continuous means. Think of graphing the funtion. If
you have to lift your pencil it's discontinuous.
> The
> book says it is discontinuous, though. Since they also say "A more
> picturesque description of a continuous function is that it is a
> function you can graph without lifting your pencil from the paper in the
> region of interest,"
I see the book says the same thing. I hadn't read this far before writing
the above.
> though, I can see where it might possibly be
> discontinuous, but it is against intuition.
Then you need to adjust your intuitive understanding of continuous. This is
a fine example of a discontinuous function. Can you see how there is a jump
just below 0?
> I think that possibly they
> meant to write "f(x) = 1 for x .GT 0" rather than using the .GE.
That would make it undefined at 0 in addition to being discontinuous.
******************************************************************
Embed Inc, Littleton Massachusetts, (978) 7429014. #1 PIC
consultant in 2004 program year. http://www.embedinc.com/products
2005\12\19@151207
by
olin piclist
Olin Lathrop wrote:
>> Since for every possible real number x there is a number f(x) which is
>> either zero or one, it seems to be that it would be continuous.
>
> No, that's not what continuous means. Think of graphing the funtion.
> If you have to lift your pencil it's discontinuous.
Now that I read this again, I think I see the misconception. Continuous
says something about the function's value, not it's domain. Or to put it in
the context of this example, continuous means abrupt jumps in Y. It is not
about all value so X being "filled in".
******************************************************************
Embed Inc, Littleton Massachusetts, (978) 7429014. #1 PIC
consultant in 2004 program year. http://www.embedinc.com/products
2005\12\19@154409
by
Hector Martin

Bill Kuncicky wrote:
> I've gotten the preliminary stuff out of the way (functions, graphs,
> trig, exponentials and logs) and started on the differential calculus
> part. The first section there is "limits," and I've hit a snag in my
> thinking. Given a function f(x), such that f(x) = 1 for x .GE. 0, and
> f{x} = 0 for x .LT. 0, is that a continuous function or a discontinuous
> function?
>
> Since for every possible real number x there is a number f(x) which is
> either zero or one, it seems to be that it would be continuous. The
> book says it is discontinuous, though. Since they also say "A more
> picturesque description of a continuous function is that it is a
> function you can graph without lifting your pencil from the paper in the
> region of interest," though, I can see where it might possibly be
> discontinuous, but it is against intuition. I think that possibly they
> meant to write "f(x) = 1 for x .GT 0" rather than using the .GE.
You might be thinking in terms of electronics. In a circuit, when say a
digital output swings from 0 to 5V, you _know_ it is at values in
between at some point (the voltage can't make a discrete jump, sinc
everything is analog at least at the macroscopic level. It might be very
fast, but not instantaneous).
In Calculus though, functions can have discrete jumps, just as in a
digital view of things. If we treat that output as digital, it can only
take the value 0 or 1, since those are the two binary digits.
Basically, if the line when you graph the function is "broken" at any
point, it is discontinuous. Note that you do NOT draw a line between the
two states. If, for example, you were to graph the typical ideal clock
pulse the math way, it would look like this:
_____ _____ _____ _____ _____ _____
_____ _____ _____ _____ _____ _____
NOT like:
____ ____ ____ ____ ____ ____
____ ____ ____ ____ ____ ____
Specifically, if you were to graph the function, you would give an
additional piece of information. You would draw small dots at the end of
the lines. If a dot is filled it indicates that the value of the
function AT that point is at the dot's position. An empty circle means
that it isn't. This is because:
_____ _____ _____ _____ _____ _____
_____ _____ _____ _____ _____ _____
/\

At the indicated point, is the value 0 or 1? We're missing that
information. But if we write (assume "o" is a filled circle and "O" is
empty:
o Oo Oo Oo Oo Oo O
Oo Oo Oo Oo Oo Oo
^

At that point, the graph tells us the value is 1 (but, beyond that
point, even if by an infinitesimal amount, it is 0). Essentially that
tells you the GE vs GT difference.
O OO OO OO OO OO O
OO OO OO OO OO OO
^

This graph is not only discontinuous, but it isn't defined at the marked
point. But, anywhere before that point it is 1, and anywhere after it is 0.
At least that is how I believe it works. Note this was written by a Calc
I student and it may be completely wrong :)

Hector Martin (hectorKILLspammarcansoft.com)
Public Key: http://www.marcansoft.com/hector.asc
2005\12\19@155707
by
Danny Sauer

Picture a curve, maybe f(x)=x^2. Regardless of how much you zoom in
on any part of the curve, the values form a continous curve. Pick any
point on the curve, and the infinite number of values between whatever
you arbitrarily choose as a minimum and maximum value for x will form
a coninous curve. In the given example, where f(x)=1 when x>= 0 and
f(x)=0 when x<0, if you get close enough, there is a disjunction rught
when you move left of 0. The line has to break. To be a contiguous
line, there would have to be intermediate values of x which would
result in all of the values of y between 0 and 1. There are no such
values in the sample function. You have to have a 1:1 mapping for any
y=f(x) between yMAX and yMIN  referring to a continuous function as a
continuous mapping, as done on
http://mathworld.wolfram.com/ContinuousFunction.html, might help some.
Danny
Bill wrote regarding '[OT] When is a function not continuous?' on Mon, Dec 19 at 13:31:
{Quote hidden}> In order to study up on my electronics, I've had to take a detour and
> brush up on calculus (there seems to be an infinite number of rabbit
> trails to have to chase down on the way to "success.") But I'm
> determined, so I bought a used book called QUICK CALCULUS, Second
> Edition, by Daniel Kleppner and Norman Ramsey. Found it for a couple of
> bucks in the local used book shop. It has a lot of notes in the
> margins, but that's OK with me.
>
> I've gotten the preliminary stuff out of the way (functions, graphs,
> trig, exponentials and logs) and started on the differential calculus
> part. The first section there is "limits," and I've hit a snag in my
> thinking. Given a function f(x), such that f(x) = 1 for x .GE. 0, and
> f{x} = 0 for x .LT. 0, is that a continuous function or a discontinuous
> function?
>
> Since for every possible real number x there is a number f(x) which is
> either zero or one, it seems to be that it would be continuous. The
> book says it is discontinuous, though. Since they also say "A more
> picturesque description of a continuous function is that it is a
> function you can graph without lifting your pencil from the paper in the
> region of interest," though, I can see where it might possibly be
> discontinuous, but it is against intuition. I think that possibly they
> meant to write "f(x) = 1 for x .GT 0" rather than using the .GE.
>
> Any thoughts would be appreciated. (I'm going to read over this email
> carefully before sending it, to make sure I don't sound like English is
> my 27th language. :)
>
> Thanks,
> Bill
> 
2005\12\20@122318
by
Peter
A function is continuous if the left limit at any point equals the right
limit at that point, for all the points where the function is defined.
Iow, if the definition for f(x) is as you said f(x) = { x<0:0, x>=0:1 }
then liml(0)=0 and limr(0)=1, therefore it is not continuous. You can
think of liml as 'approaching x from left (smaller numbers)' and limr as
'approaching x from right (bigger numbers)'. The slope of the function
at that point is infinite, which defies the idea of slope, and thus
that of derivative.
Peter
2005\12\21@002846
by
Mike Singer
The Best mathematical analysis course, in my opinion:
A course of mathematical analysis. In 2 vols.
Nikolsky S.M.
Not that easy reading though.
Mike.
2005\12\21@124925
by
Peter
On Wed, 21 Dec 2005, Mike Singer wrote:
> The Best mathematical analysis course, in my opinion:
>
> A course of mathematical analysis. In 2 vols.
> Nikolsky S.M.
>
> Not that easy reading though.
Especially not in Russian, which I do not speak.
Peter
2005\12\23@041320
by
Mike Singer
Peter wrote:
> > A course of mathematical analysis. In 2 vols.
> > Nikolsky S.M.
> >
> > Not that easy reading though.
>
> Especially not in Russian, which I do not speak.
In your country at least a half of IT folks do :)
Mike
2005\12\23@112932
by
Peter
On Fri, 23 Dec 2005, Mike Singer wrote:
> Peter wrote:
>>> A course of mathematical analysis. In 2 vols.
>>> Nikolsky S.M.
>>>
>>> Not that easy reading though.
>>
>> Especially not in Russian, which I do not speak.
>
> In your country at least a half of IT folks do :)
No.
Peter
2005\12\23@152012
by
Mike Singer
My mistake. ( 10% ?)
Mike.
On 12/23/05, Peter <.....plpKILLspam.....actcom.co.il> wrote:
{Quote hidden}>
> On Fri, 23 Dec 2005, Mike Singer wrote:
>
> > Peter wrote:
> >>> A course of mathematical analysis. In 2 vols.
> >>> Nikolsky S.M.
> >>>
> >>> Not that easy reading though.
> >>
> >> Especially not in Russian, which I do not speak.
> >
> > In your country at least a half of IT folks do :)
>
> No.
>
> Peter
> 
2005\12\24@123716
by
Peter
On Fri, 23 Dec 2005, Mike Singer wrote:
> My mistake. ( 10% ?)
I can't count in russian either, you know...
Peter
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