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Math problem got me stumped... (not my homework)

sao123

sao123

Lifer
ok, so I started thinking about this math problem in this thread.
I dont give a crap about the guys hw...but now im interested in what the real answer is. I kinda stumped myself, but I dont wanna quit until I find some sort of answer.

Can someone verify my approach and solution is correct or provide an alternative way to approach the problem?
 
shoot. The sum isn't continuous at 0, because it's undefined.
Technically, the product is undefined at 0 as well, unless you simplify the product.

without thinking too long about it, I think you can tie together some abs value function with a cos function...
 
Originally posted by: TuxDave
Originally posted by: DrPizza
How about y = Abs(x) / x
and y = -x/Abs(x)
Click to expand...

Isn't x*y still discontinuous at 0?
Click to expand...

If you simplify after multiplying, it's -1, as are all other values. But, from the previous response to this, I guess not, since x still exists in the denominator, even though it can be cancelled with the x in the numerator.

Thanks, Sao, now this question is bugging ME.
 
Originally posted by: DrPizza
Originally posted by: TuxDave
Originally posted by: DrPizza
How about y = Abs(x) / x
and y = -x/Abs(x)
Click to expand...

Isn't x*y still discontinuous at 0?
Click to expand...

If you simplify after multiplying, it's -1, as are all other values. But, from the previous response to this, I guess not, since x still exists in the denominator, even though it can be cancelled with the x in the numerator.

Thanks, Sao, now this question is bugging ME.
Click to expand...

Ditto... and I'm at work too. How am I supposed to concentrate now? :|
 
Does it have to be a normal math function? Can I make one up like...

isgreaterthanzero(x) = 1 if x >0 and 0 if x <= 0
islessthanzero(x) = 1 if x < 0 and 1 if x >= 0

???
 
Okay... just drew a picture for myself. The problem with my Abs(x) / x types of functions is that it's undefined at x=0

Instead, function A = 1 when x is less than zero, and -1 when x >= 0
Function B = - function A
The product will always be -1
The sum will always be 0

very easy to do with a piecewise defined function. I'm 99% sure it can be done without a piecewise defined function, but I'm having a brain fart at the moment. (edit: because I've come up with a dozen different functions that are 1 for x> 0 and -1 for x < 0 but are all undefined for x=0.

To create such a simple function, am I allowed to use a rounding, truncating, or stepping function?

 
Originally posted by: DrPizza
Okay... just drew a picture for myself. The problem with my Abs(x) / x types of functions is that it's undefined at x=0

Instead, function A = 1 when x is less than zero, and -1 when x >= 0
Function B = - function A
The product will always be -1
The sum will always be 0

very easy to do with a piecewise defined function. I'm 99% sure it can be done without a piecewise defined function, but I'm having a brain fart at the moment.

To create such a simple function, am I allowed to use a rounding, truncating, or stepping function?
Click to expand...


lol... we think alike.
 
Yeah, but darn it, no one else has posed a solution not consisting of a piecewise defined function. Now I've gotta start thinking again today.
 
Stating the problem more mathematically may be instructive:

You have 2 functions, f(x) and g(x) that are discontinuous at x = p

Discontinuity at p means

lim f(x) = a
x->p+

and

lim f(x) = b
x->p-

with a ~= b

also

lim g(x) = c
x->p+

and

lim g(x) = d
x->p-

with c ~= d

1) For continuity of the sum, we require

lim [f(x)+g(x)]
x->p+

=

lim [f(x)+g(x)]
x->p-

or, recalling distributive property of limits:

lim f(x)+ lim g(x)
x->p+

=

lim f(x)+ lim g(x)
x->p-

thus we need

a + c = b + d

along with

f(p)+g(p) =
lim [f(x)+ g(x)] = a + c = b + d
x->p

2) For continuity of the product, we require

lim f(x)g(x)
x->p+

=

lim f(x)g(x)
x->p-

or, recalling the properties of limits again:

lim f(x)*lim g(x)
x->p+

=

lim f(x)*lim g(x)
x->p-

thus we need

ac = bd

along with

f(p)*g(p) =
lim [f(x)* g(x)] = ac = bd
x->p

Functions fitting the two criteria, will satisfy the problem.

Ex: if a = 1, b = 0, c = 0, d = 1

a + c = 1 + 0 = 0 + 1 = b + d

and

ac = 1*0 = 0*1 = bd

if p = 0, the following two step functions satisfy sum and product continuity at p, while being discontinuous there themselves:

f(x) = 1 + round(x-0.5)
g(x) = round(x-0.5)


You can verify this easily in Matlab, MathCAD, Maple, Mathematica, etc..







 
Originally posted by: sao123
how exactly is it that
f(0) = 1 + round(0-0.5) is undefined?
Click to expand...

It most certainly is defined at x = 0. However discontinuity at x = p is not equivalent to undefined at x = p. A function can be discontinuous, yet still defined, at a point.

Or am I misunderstanding your question?

If it clears anything up, for negative numbers, the round function, as I have used it, corresponds to the following: round(x) = sign(x)*round(abs(x))

so f(0) = 1 + round(0-0.5) = 0

 
I was interpreting it to be discontinuous at C such as:

F(x) = 1 / X is discontinuous (undefined) at 0.
F(x) = Tan X is discontinuous (undefined) at pi.
 
have no fear though, I have submitted the question to doctor math.
perhaps he will have the true correct answer.
 
I should also probably clarify that what I have laid out takes care of a jump discontinuity, but does not completely address the possibility of introducing a removable discontinuity with the composite functions. In the latter case, it is possible, I believe, to end up with a situation where the left and right limits are identical at point p, but the composite function at p is different. Care would need to be taken to "match up" the functional values of the discontinuous functions at p so this does not happen. I apologize, but I've yet to work out the formal mathematical statement of the condition this imposes on them 🙂


Edit: OK, I've edited the original post and I think I covered the situation I described.
 
I had already proposed simple solutions with rounding, truncating, and stepping functions...
Anyone have any ideas on others?
 
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