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math guru's again

S

Semidevil

Diamond Member
ok, give me an example of 2 functions who are discontinuous at point C, but when added together, and multiplied together, will be continuous at that point.

I can't think of any that will add and multiply to be continuous. tried 1/x, sqrt(x), some polynomial with the denominator as 1 - x. still cant..

any ideas?
 
Do they have to be one expression or can they be piecewise functions?

Also, after addition and multiplication, do they have to be entirely continuous or JUST continuous at point C?
 
Originally posted by: Legendary
Do they have to be one expression or can they be piecewise functions?

Also, after addition and multiplication, do they have to be entirely continuous or JUST continuous at point C?
Click to expand...

it doesnt say, so I assume it can be anyting you want......

and it doesnt say for the other part too..so maybe, at least point c?
 
I'm finding this hard simply because when you multiply two functions (say f(x) and g(x)) together (h(x)), h(x) can still be factored into f(x)*g(x). Multiplication doesn't get rid of many discontinuities.
 
Originally posted by: Legendary
I'm finding this hard simply because when you multiply two functions (say f(x) and g(x)) together (h(x)), h(x) can still be factored into f(x)*g(x). Multiplication doesn't get rid of many discontinuities.
Click to expand...

exactly what i was thinking. there must be a really easy but non-straightforward answer to this question.
 
an example for C = 0:

f(x) = 0 for x!= 0
f(0) = 2

g(x) = 2 for x!= 0
g(0) = 0

f(x) + g(x) = 2 for all x
f(x)*g(x) = 0 for all x
 
Originally posted by: aux
an example for C = 0:

f(x) = 0 for x!= 0
f(0) = 2

g(x) = 2 for x!= 0
g(0) = 0

f(x) + g(x) = 2 for all x
f(x)*g(x) = 0 for all x
Click to expand...

I thought conditions of functions carried over through multiplication and addition?
 
hm i haven't thought of any non-piecewise functions that would work...

if you can make it piecewise, this is easy. Just have 2 constant functions with "point" discontinuities...have one of them equal 0 everywhere except C and the other equal 0 only at C...and you're done.

Otherwise, the best I've done is:
1/x & (x-1)/x. Their sum removes the discontinuity at 0...but their product makes it worse. bah
 
Originally posted by: Legendary
Originally posted by: aux
an example for C = 0:

f(x) = 0 for x!= 0
f(0) = 2

g(x) = 2 for x!= 0
g(0) = 0

f(x) + g(x) = 2 for all x
f(x)*g(x) = 0 for all x
Click to expand...

I thought conditions of functions carried over through multiplication and addition?
Click to expand...

yes they do, it just happens that in this case (because of the way the functions are chosen) that the values of f+g for x!= 0 and x=0 are the same (you can check this by computing the values in both cases), the same for f*g

e.g. you can write:
f(x) + g(x) = 0 + 2 = 2 for x!=0
f(0) + g(0) = 2 + 0 = 2
and
f(x)* g(x) = 0 * 2 = 0 for x!=0
f(0)*g(0) = 2 * 0 = 0

therefore f+g is always 2 and f*g is always 0
 
Originally posted by: aux
Originally posted by: Legendary
Originally posted by: aux
an example for C = 0:

f(x) = 0 for x!= 0
f(0) = 2

g(x) = 2 for x!= 0
g(0) = 0

f(x) + g(x) = 2 for all x
f(x)*g(x) = 0 for all x
Click to expand...

I thought conditions of functions carried over through multiplication and addition?
Click to expand...

yes they do, it just happens that in this case (because of the way the functions are chosen) that the values of f+g for x!= 0 and x=0 are the same (you can check this by computing the values in both cases), the same for f*g

e.g. you can write:
f(x) + g(x) = 0 + 2 = 2 for x!=0
f(0) + g(0) = 2 + 0 = 2
and
f(x)* g(x) = 0 * 2 = 0 for x!=0
f(0)*g(0) = 2 * 0 = 0

therefore f+g is always 2 and f*g is always 0
Click to expand...

Ah I see, I read your functions a little too fast and missed it.
:beer:
 
Its not an answer, but a different approach to the problem which may lead to a solution (still need brain power from other users)

Given:
F(x) is continuous over R except at point C.
G(x) is continuous over R except at point C.

H(x) = F(x)*G(x)
If H(x) is continuous at C then a reasonable assumption could be
G(x) = could have the form A * Z(x) / F(x)
IE that 1 factor of G(x) is an inverse of F(x)
Multiplying a function and its inverse in this forms would yeild
F(x) * (A * Z(x)) / F(x) = A * Z(x).
A*Z(x) could be a function or a constant or even 1.

So now your task would be to find a function such that


F(x) + A * Z(x) / F(x) is continuous at C
This would take the form

(F(x) F(x) + A * Z(x))
--------------------------
F(x)

F(x) F(x) + A * Z(x) when added togther would then have to be divisibe by F(x) thus removing the disconuity at C

 
After more work I'm starting to believe that such a thing cannot exist.

For F(x) F(x) + A * Z(x) to be divisible by F(x)
A * Z(x) would have to have a factor of F(x) in it.

F(x) * F(x) + A' * Z'(x) * F(x)
F(x) * ( F(x) + A' * Z'(x) )

but then in the original statement
A * Z(x) / F(x) = A' * Z'(x) * F(x) / F(x)
Would be an improper fraction equating to value A' * Z'(x)





 
Originally posted by: Semidevil
ok, give me an example of 2 functions who are discontinuous at point C, but when added together, and multiplied together, will be continuous at that point.

I can't think of any that will add and multiply to be continuous. tried 1/x, sqrt(x), some polynomial with the denominator as 1 - x. still cant..

any ideas?
Click to expand...

tree fiddy
 
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