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

sao123

Lifer
May 27, 2002
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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?
 

DrPizza

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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...
 

DrPizza

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Originally posted by: TuxDave
Originally posted by: DrPizza
How about y = Abs(x) / x
and y = -x/Abs(x)

Isn't x*y still discontinuous at 0?

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.
 

TuxDave

Lifer
Oct 8, 2002
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Originally posted by: DrPizza
Originally posted by: TuxDave
Originally posted by: DrPizza
How about y = Abs(x) / x
and y = -x/Abs(x)

Isn't x*y still discontinuous at 0?

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.

Ditto... and I'm at work too. How am I supposed to concentrate now? :|
 

TuxDave

Lifer
Oct 8, 2002
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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

???
 

DrPizza

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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?

 

TuxDave

Lifer
Oct 8, 2002
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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?


lol... we think alike.
 

DrPizza

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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.
 

jer0608

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Sep 24, 2004
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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..







 

jer0608

Member
Sep 24, 2004
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Originally posted by: sao123
how exactly is it that
f(0) = 1 + round(0-0.5) is undefined?

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

 

sao123

Lifer
May 27, 2002
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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.
 

sao123

Lifer
May 27, 2002
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106
have no fear though, I have submitted the question to doctor math.
perhaps he will have the true correct answer.
 

jer0608

Member
Sep 24, 2004
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0
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.
 

DrPizza

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I had already proposed simple solutions with rounding, truncating, and stepping functions...
Anyone have any ideas on others?