Post by thread:[OT] When is a function not continuous?

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

> 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.

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 delta-epsilon 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

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(0-epsilon) = 0 as epsilon approaches 0, f(0-epsilon) 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

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) 742-9014. #1 PIC consultant in 2004 program year. http://www.embedinc.com/products

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) 742-9014. #1 PIC consultant in 2004 program year. http://www.embedinc.com/products

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 O----o O----o O----o O----o O----o O-- O----o O----o O----o O----o O----o O----o ^ | 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 O----O O----O O----O O----O O----O O-- O----O O----O O----O O----O O----O O----O ^ | 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

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 > --