Definition of continuity in math help

In summary, the homework statement is that w is any bounded 2pi periodic function of one variable. and u(x,y) is a function in cartesian coordinates. The function u is continuous at the origin if w(theta) is differentiable.
  • #1
sara_87
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Homework Statement


given: w is any bounded 2pi periodic function of one variable. and u(x,y) is a function in cartesian coordinates.
show that u(x,y)=r*w(theta) is continuous at the origin.

Homework Equations



u(x,y)=r*w(theta) is equal to v(r,theta) where v is a function in polar coordinates

The Attempt at a Solution



I know that the definition of continuity is:
f(x) is said to be continuous at x=x0 if the limit of f(x) as x tends to x0 is equal to f(x0).

but i don't know how to apply it in this case just that our x0 is the origin but how do i apply this in terms of polar coordinates?
 
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  • #2


You want to show that u(x,y)->0 as (x,y)->(0,0).
Use the fact that r(x,y)->0 as (x,y)->(0,0) and w(theta(x,y)) is bounded.
 
  • #3


A nice thing about polar coordinates is that the distance from a point to the origin is given by r alone. [itex]r w(\theta)[/itex] goes to 0 as r goes to 0 no matter what [itex]\theta[/itex] is so the limit as (x,y) goes to (0,0) is 0, the value of the function.
 
  • #4


ok, so clearly as (x,y) tends to zero, then r also ttends to zero since x=rcos(theta) and y=rsin(theta); but how do we use the fact that w(theta) is bounded?
 
  • #5


because, if w tended to infinity as theta went to a particular value, r w(theta) would not necessarily go to 0 as r goest to 0. Knowing that w has a bound, say M, rw(theta)< rM and goes to 0 as r goes to 0.
 
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  • #6


Thank you, that makes a lot of sense.
The question then says: give an example where u is not continuous outside of the origin.
I take it i can choose any function as long as it's not continuous outside the origin.
I thought tan(x) but tan(x) is not bounded. I need a bounded function.
How about u=tanh(x) ? would this be a good example for u?
 
  • #7


There's another part to the question that says:
give a condition on 'w' such that u is also continuous outside of the origin.
Does anyone have any idea about this condition?
Thank you
 
  • #8


sara_87 said:
Thank you, that makes a lot of sense.
The question then says: give an example where u is not continuous outside of the origin.
I take it i can choose any function as long as it's not continuous outside the origin.
I thought tan(x) but tan(x) is not bounded. I need a bounded function.
How about u=tanh(x) ? would this be a good example for u?

Shouldn't u be of the same form as above, i.e. u=r*w(theta)? What discontinuous functions do you know? Remember that tanh is a nice infinitely differentiable function on R, so this will not work.

There's another part to the question that says:
give a condition on 'w' such that u is also continuous outside of the origin.
Does anyone have any idea about this condition?
Thank you

You want u to be continuous, so what would be the obvious condition for w? Keep in mind that you can always divide u by r if r is not 0, and that the quotient of continuous functions is continuous where the denominator is not 0.
 
  • #9


I think you are right. u should be of the form r*w(theta).
The discontinuos functions i know are:
y=1/x; discontinuous at x=0
y=sin(1/x); discontinuous at x=0
but i need a function where it's discontinuous outside the origin. I can't think of one. How about: y=x (for integer values of x).

For the second part:
can the condition for w be: that it has to be differentiable? (becuase if it is differentiable, then it is continuous).
 
  • #10


sara_87 said:
I think you are right. u should be of the form r*w(theta).
The discontinuos functions i know are:
y=1/x; discontinuous at x=0
y=sin(1/x); discontinuous at x=0
but i need a function where it's discontinuous outside the origin. I can't think of one. How about: y=x (for integer values of x).

You could, for example, take the so called http://en.wikipedia.org/wiki/Everywhere_discontinuous_function" . It is defined as follows: If x is a rational number, then f(x)=1, otherwise (if x is irrational) f(x)=0. This function is nowhere continuous (and difficult to draw)! It is also obviously bounded, which is what you are looking for.
You could also use the fractional part or "floor" function.

For the second part:
can the condition for w be: that it has to be differentiable? (becuase if it is differentiable, then it is continuous).

It is even simpler than that, you only need continuity (do you see why this is sufficient?).
 
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  • #11


Thank you for the dirichlet function. It's first time i hear of it but it's interesting.

For the second part,
what do you mean in this case by sufficient?
can i say that u has a limit at all points??
 
  • #12


sara_87 said:
Thank you for the dirichlet function. It's first time i hear of it but it's interesting.

For the second part,
what do you mean in this case by sufficient?
can i say that u has a limit at all points??

Do you see why, if w is continuous, then u will also be continuous? (Hint: the product of continuous functions is continuous)

One more thing: I am assuming that theta is restricted to some range like [0,2pi). Otherwise you will need to choose a [tex]2\pi[/tex]-periodic function for w, i.e [tex]w(\theta)=w(\theta+2\pi)[/tex]. The Dirichlet function D is not [tex]2\pi[/tex] periodic, but [tex]D(x/(2\pi))[/tex] is.
 
  • #13


ok, so a condition on w is that it is continuous because the product of two continuous functions is continuous (as you said).

theta is between 0 and 2pi; so in that case, can i take the dirichlet function D or must i take D(x/(2pi)) ?
 
  • #14


sara_87 said:
theta is between 0 and 2pi; so in that case, can i take the dirichlet function D or must i take D(x/(2pi)) ?

If theta is in the interval [0,2pi), so 2pi is excluded, then it's no problem. But if theta can be both 0 and 2pi, then you need to choose w so that it gives the same value for 0 and 2pi (because that is the same angle). You probably defined polar coordinates with theta in [0,2pi). If you want to be on the safe side you can take D(x/(2pi)), which works in both cases.
 
  • #15


2pi is included so theta can be 0 and 2pi.
so do i say that w=D(x/(2pi)) or is it u=D(x/(2pi))?
 
  • #16


sara_87 said:
2pi is included so theta can be 0 and 2pi.
so do i say that w=D(x/(2pi)) or is it u=D(x/(2pi))?

We already agreed that u=r*w(theta), remember? So w(theta)=D(theta/(2pi)).
 
  • #17


ok that's good thanks a lot.
so u=r*D(theta/(2pi))right?
 
  • #18


Yes, that is one possible example. Also, remember how D was defined, since the notation is not standard like sin,tanh.
 
  • #19


ok, thank you this helps a lot.
 

What is the definition of continuity in math?

The definition of continuity in math is the idea that a function has no sudden jumps or breaks in its graph. This means that as the input values get closer and closer, the output values also get closer and closer, without any sudden changes.

How can I test for continuity in a function?

To test for continuity in a function, you can use the three-part definition of continuity. This states that for a function to be continuous at a specific point, the limit of the function as x approaches that point must exist, the function must be defined at that point, and the limit and the function value must be equal.

What are the three types of continuity?

The three types of continuity are pointwise continuity, uniform continuity, and global continuity. Pointwise continuity means that a function is continuous at each individual point in its domain. Uniform continuity means that a function is continuous over its entire domain. Global continuity means that a function is continuous at every point in its domain and its range.

How does continuity relate to differentiability?

Continuity is a necessary condition for differentiability. This means that if a function is differentiable at a point, it must also be continuous at that point. However, a function can be continuous without being differentiable, as it may have sharp turns or corners in its graph that prevent it from having a defined derivative at certain points.

Can a function be continuous but not differentiable?

Yes, a function can be continuous but not differentiable. This occurs when a function has sharp turns or corners in its graph, which make it impossible to have a defined derivative at those points. These points are known as points of discontinuity for the derivative, but the function itself is still continuous.

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