Wikipedia:Reference desk/Archives/Mathematics/2006 December 11

= December 11 =

continuity question
is the graph of 2*(3root(X)) it should be approching x=0 but never get there, is it called point discontinuity or jump discontinuity thanks --205.222.248.176 12:48, 11 December 2006 (UTC)


 * Do you mean 2 times the cube root of x ? That should include the point (0,0), and I don't think there's any discontinuity. StuRat 13:29, 11 December 2006 (UTC)


 * If y = 2x1/3, for x &ge; 0, then x = (1&frasl;2y)3. If we extend the meaning of cube root for negative values of x to be sign of x times the cube root of the absolute value, then the x equation holds for all y. The formula posted is in error, or the belief in a discontinuity is in error. --KSmrqT 14:14, 11 December 2006 (UTC)

That won't have any discontinuity. However, I'll define them for you. A point discontinuity is a single point on a graph that the value doesn't exist for and such that the lim x -> a- = lim x -> a+. Think of the graph of f(x)=(x+1)²/(x+1). A jump discontinuity is where the graph jumps on the discontinuity such that lim x -> a+ does not equal lim x -> a-. An example would be f(x)= 1/x. --AstoVidatu 00:21, 12 December 2006 (UTC)
 * Actually, in order to count as a jump discontinuity both the right hand and left hand limits have to exist and be finite (see Classification of discontinuities). So 1/x does not have a jump discontinuity at x=0 since the right-hand and left-hand limits aren't finite.  Rather, x=0 would be an essential discontinuity for 1/x.
 * An example of a jump discontinuity would be f(x)=int(x) at x=0. For 0<x<1, f(x)=0, but for -1<x<0, f(x)=-1.  So the right-hand limit is 0 while the left hand limit is -1, making x=0 a jump discontinuity. Dugwiki 16:17, 12 December 2006 (UTC)


 * Once we start talking about continuity, we are probably going to move into calculus. Consider the familiar absolute value function &phi;(x) = |x|; it is continuous everywhere. Now look at the slope of &phi;, or what calculus would call the first derivative, &phi;&prime;. For x &lt; 0 it is −1, for x &gt; 0 it is +1, and at x = 0 it is undefined. Thus &phi;&prime; (a slight variation of the Heaviside step function) is a continuous — in fact, constant — function everywhere but at zero, and there it has a jump discontinuity. Now repeat; take the derivative again, producing &phi;&Prime;. Since &phi;&prime; is constant everywhere except at zero, its slope is zero everywhere except at zero. At zero the slope is undefined, so we have a point discontinuity.
 * It seems a little artificial to treat (x+1)2/(x+1) as having a discontinuity, since we have an irresistible urge to reduce it to simply x+1. So consider instead the "sinc function", which we may define as sinc(x) = sin(x)/x. We cannot simplify, and we get division by zero at x = 0; yet if we let x approach zero from either side, the function approaches 1.
 * These discontinuities are straightforward to explain and easy to picture. Not weird enough? Then define &phi;(x) to be 0 if x is rational, and 1 if x is irrational. Now we have a discontinuity at every point!
 * Let's close with a humdinger, separating discontinuities from derivatives. Define
 * $$\begin{align}

g(x) &{}= \begin{cases} 1+x, & -2 \le x < 0\\ 1-x , & 0 \le x < 2 \\ \text{repeat} , & \text{period 4}\end{cases}\\ f(x) &{}= \sum_{n=1}^{\infty} \frac{g(2^{2^n}x)}{2^n} \end{align}$$
 * In 1953, John McCarthy published an elementary 13-line proof in the American Mathematical Monthly that this is an example of a function that is everywhere continuous but nowhere differentiable. --KSmrqT 18:28, 12 December 2006 (UTC)

Finding a tangent
I need a tangent to a curve. There's an s-t-diagram for a man jogging, and, I don't know the English name for it, but I need the velocity of the jogger at a particular point, specifically after 40 seconds. The travelled distance at that time is 220 meters. How do I calculate the tangent? The answer is 2.3m/s. Sorry I can't explain it properly, hope you guys understand it anyway. Thanks in advance. Jack Daw 21:57, 11 December 2006 (UTC)


 * If I understand your question correctly, what you're looking at is a graph x=f(t) where t is a point in time and x is the total distance travelled. Velocity is a measure of the ratio of change of position x over time, which you can also write as $$\frac{dx}{dt}$$, and this corresponds to the slope of the graph.  To calculate the slope of the graph at a particular time z, you need to find the derivative x'(z). Dugwiki 22:24, 11 December 2006 (UTC)

If you don't have a calculus base you can approximate the tangent by finding a secant line. So find the (x,y) values at 39.999 and 40.001 or something like that (as close together as you can) and then use the slope formula (y1 - y2)/(x1 - x2) = slope. Basically, calculus just makes the difference between x1 and x2, and hence y1 and y2 (under most circumstances) so that it approaches zero. --AstoVidatu 00:27, 12 December 2006 (UTC)


 * I forgot to say I'm looking for the inclination of the tangent. This is the velocity in a certain instant, 40 seconds to be exact, and the answer is 2.3 (meters/second). Now... I don't know how to formulate the variables in an equation, but this is all related to v = (capital delta s)/(capital delta t) where v = velocity in m/s, s = distance and t = time. The instruction for a similar question is, "Draw a tangent to the graph in the point t=42. Calculate the inclination of the tangent..." In that question the tangent is already drawn however, which it isn't in the one I'm trying to solve, and I have no idea how to do it. Should be something similar however. This is junior high physics, so all you mathletes out there ought to be able to solve it (with high school math). Again, how do I find the tangent and its inclination to a point (40;220)? Thank you. Jack Daw 01:03, 13 December 2006 (UTC)


 * Is it the case that the graph is given as black smudges on paper, and not in some more abstract form? --Lambiam Talk  23:29, 15 December 2006 (UTC)