Another multivariable calculus question...?

Question

Give an example of a function f: R^2 -> R which is continuous in each variable separately, but is not continuous at (0,0).

I presume that I need a function which is continuous along the x- and y-axes, but not along a line/curve like x = y. Does that sound right?

(I really need to take a class on this stuff so it won't bug me!)

Doug: I believe the function needs to be defined at the origin, just not continuous there.

Prince Ali: neither of your functions are defined at the origin, much less continuous in x and y there.

Doug: on your comment to mathematician:

"f(0,0) = 0*0/(0² + 0²) = 0/0 = undefined (*not* 0 )"

please note that he defined f(0,0) to be 0.

Also, your function:

f(x,y) = (x² + y²)/(x + y)

is not defined at (0,0). The function we're looking for *must* be defined on all of R^2, be continuous in both x and y everywhere, but not continuous at (0,0).

I believe your function, if defined to be 0 at (0,0), meets all the requirements, as it's clearly defined everywhere, continuous everywhere except possibly at the origin, and (I believe) continuous at the origin along the x- and y-axes. Let's see:

Your function in polar coords is f(r, t) = r/(cos t + sin t) for r != 0 and 0 for r = 0.

On the x-axis, we have f(r, t) = -r for x < 0 and f(r, t) = r for x > 0, so yes, f is continuous along the x-axis, and a similar argument holds for the y-axis.

But I believe f is continuous at (0, 0) in all directions, as lim r -> 0 r/(sin t + cos t) = 0.

And I believe it is clear to me that mathematician's answer of f(x,y) = xy/(x^2 + y^2) for (x,y) not (0,0) and 0 for (x,y) = (0,0) satisfies all conditions.

In polar coords, we have f(r, t) = cos t sin t for r != 0, and 0 for r = 0. f is clearly continuous everywhere except possibly the origin, and defined everywhere.

On the x- and y-axes, f(r, t) = 0, so f is continuous at (0,0) in x and y, therefore continous in x and y everywhere.

But through any other path passing through the origin, the limit of f(r, t) as r -> 0 must be non-zero on at least one side of the path, thus f is not continuous at (0,0) anywhere but along the x- and y-axes.

Thanks all for helping me think... I think I aced my prelim yesterday :). Now to get ready for topology and algebra classes.

Answer 1

Let f(x,y)=xy/(x^2+y^2) for (x,y) not the origin and f(0,0)=0.

Answer 2

You're right. You really need to take an upper division analysis course if you're gonna be doing a lot of this

What they want is a g(x(t),y(t)) such that x(t) and y(t) are continuous, but g is not continuous at g(0,0)

How 'bout letting x(t) = t and y(t) = t² and then defining g as

g(x,y) = x(t)/y(t)

Clearly x(t) and y(t) are continuous over all t. But at t=0 g vanishes since it becomes 0/0 at t=0.


Doug

For 'mathematician'
If f(xy) = xy/(x² + y²) then

f(0,0) = 0*0/(0² + 0²) = 0/0 = undefined (*not* 0 )

Doug

For 'celebrated summer'

Sorry. I used g(x,y) instead of the f(x,y) you were probably expecting. My bad

Put then I 'parameterized' the x and y so they were both functions of t. Technically that is a function from R² -> R but maybe it would have been better if I'd said something along the lines of

f(x,y) = (x² + y²)/(x + y)

which is continuous in each variable, but not continuous at x = 0, y = 0 (since it has a 'hole' in it there)

e-mail me at [e-mail] and we can talk.

Doug

Answer 3

If you have y = f(R(x,y)) then an example of singularity would be when R->0 in an equation such as y = 1/R. The function does not exist at the zero points. The same can be said of y = 1/R^2

Interestingly, singularity using the theories of relativity at the atomic level is exactly what caused physicists to come up with string theory. That is, relativity broke down because of the discontinuities.

There are also step-wise functions that go along at one y (dependent variable) value until they hit an independent variable value that causes the y value to jump to a new level. Potential energy wells at the atomic level are examples of such functions.

Answer 4

Actually exceptional colleges divide Calculus into exceptional portions. I cross to UT, or even right here they divide it into 2 exceptional units. You can both take M408K, M408L, and M408M or you'll be able to take M408C, and M408D. They quilt the identical fabric however in exceptional time intervals. Likewise, AP Calculus AB covers Differential and Integral Calculus. AP Calculus BC quilt Differential, Integral, and Sequences and Series Calculus. No AP path covers Multivariable Calculus. So you ask, "is multivariable calculus simply one more identify for calculus III?" Not always, however sure. At UT if you're taking the 3 semester series, sure, but when you're taking the 2 semester series, no. "Is calculus bc one more identify for Calculus II?" Not always, however sure. At UT if you're taking the 3 semester series, style of, but when you're taking the 2 semester series additionally style of. I do not suppose you've gotten time to head over the BC fabric side of the AP scan in the sort of few minutes, the AP exams are speedily drawing near in May.

Answer 5

It sounds to me like you need a parabola that is not continuous with the x/y axis but instead need something that is an independent variable, but it has been about 5 years since i took high school/college calculus. hope this helps. the reason that i say is because the question details not continuous at the origin.
best of luck
sw

Answer 6

What about something like f(x,y) = 1/ (x+y)...simple yet elegant. f(x,y) is discontinuous at the origin.

Or if you really wanted to cheat (which is not really cheating), define your own function and make it piecewise...hehehe...something like

g(x,y) = 1/(x+y) if x=0, y=0
g(x,y) = x+y otherwise