pouët.net

Math stupid question on continuous curve

category: general [glöplog]
im so shit at this stuff :(
[url]http://en.wikipedia.org/wiki/Discontinuity_(mathematics)[/]
added on the 2010-02-19 00:52:29 by bdk bdk
Quote:
How a function with 1/x can "look" continuous for 0 ?


Not to muddle up the good explanation that's already been provided, but I do have a small addition. What you are witnessing (informally called a "hole", as ryg pointed out) occurs any time the numerator and denominator of a rational function are both equal to zero for some value of x (e.g. f(x) = x(x + 1) / x). This is the only circumstance under which this occurs (afaik).

More math threads!
added on the 2010-02-22 04:43:20 by orby orby
sin(x)/x is not a rational function.
added on the 2010-02-22 11:07:02 by imbusy imbusy
try to represent this function f where f(x) = 1 if x is a rationnal number, and 0 if not.
Lotsa holes ! \o/
added on the 2010-02-22 11:13:15 by faraday faraday
Quote:
sin(x)/x is not a rational function.


Good call, do you know of a more accurate term for a function of the form f(x)/g(x) that includes trig functions?
added on the 2010-02-25 03:38:47 by orby orby
The de L'Hôpital rule: when both the numerator and denominator have the same limit, the limit of the ratio is the ratio of the derivatives.

(sin(4x))'=4cos(x)
x'=1

Therefore sin(4x)/x -> 4cos(x) = 4*1 =4 when x tends to +-0.

added on the 2010-02-25 13:24:52 by ulrick ulrick
Quote:
try to represent this function f where f(x) = 1 if x is a rationnal number, and 0 if not.
Lotsa holes ! \o/

That's plain discontinuities though, not singularities, and the function is well-defined over all reals.

Quote:
Good call, do you know of a more accurate term for a function of the form f(x)/g(x) that includes trig functions?

The next larger family with a well-known name would be meromorphic functions, which are quotients of holomorphic (=complex differentiable almost everywhere) functions.
added on the 2010-02-25 14:12:59 by ryg ryg
Quote:
The de L'Hôpital rule: when both the numerator and denominator have the same limit, the limit of the ratio is the ratio of the derivatives.

(sin(4x))'=4cos(x)
x'=1

Therefore sin(4x)/x -> 4cos(x) = 4*1 =4 when x tends to +-0.


Not exactly.

1. L'Hopital's rule states that when the limit of f(x) = 0 or +-inf and the limit of g(x) = 0 or +-inf, then the limit of f(x)/g(x) is equal to the limit of the ratio of the derivatives.

2. d/dx (sin(4x)) = 4cos(4x)

3. (If we're evaluating the limit as x approaches infinity) The limit of sin / cos is undefined as x approaches infinity, so L'Hopital's rule isn't much help here.
added on the 2010-02-25 15:48:47 by orby orby
Quote:
The next larger family with a well-known name would be meromorphic functions, which are quotients of holomorphic (=complex differentiable almost everywhere) functions.


Cool, thank you. I'll be looking into that.
added on the 2010-02-25 15:53:32 by orby orby
Quote:
The next larger family with a well-known name would be meromorphic functions, which are quotients of holomorphic (=complex differentiable almost everywhere) functions.


Cool, thank you. I'll be looking into that.
added on the 2010-02-25 16:02:16 by orby orby
orbitaldecay: thanks for pointing the error sin(4x)' is not 4cos(x) but 4cos(4x).

But note that krabob raised the case where x tends to 0, not towards infinity.

I explicated the de l'Hôpital following suggestion by iq.
added on the 2010-02-25 16:37:17 by ulrick ulrick
You don't need l'Hôpital to get the limit for x->+-inf in this case.

Since |sin(x)|<=1 for all x, lim x->+inf |sin(x)/x| <= lim x->+inf 1/|x| = lim x->inf 1/x = 0. So the limit must be 0.
added on the 2010-02-25 18:03:27 by ryg ryg
The two limits are supposed to be as x->+-inf (doesn't matter which).
added on the 2010-02-25 18:04:02 by ryg ryg

login