help on functions ... Watch

Wrangler
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#21
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#21
(Original post by RichE)
Shouldn't that read

f(x)=f(x)f(0)2?
He seems to have edited the original question. :p:
(Original post by psycho)
oopps ... the question should be ...
find all functions f(R)=R such that f(x+y)=f(x)f(y)f(xy) ...
sorry ..
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RichE
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#22
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#22
Anyway the only differentiable functions which satisfy

f(x+y) = f(x)f(y)f(xy)

are 0, 1, -1.

None of these satisfy f(R)=R (i.e. none are onto) but perhaps you just meant f:R->R?

There may well be loads of pathological functions that satisfy it though. :confused:
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UglyDuckling
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#23
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#23
0,1, -1 ? why is it so ?

what's the difference between f(R)=R and f:R->R ???

f(0) = plus minus 1 is the answer ???? is it that simple ?
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RichE
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#24
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#24
(Original post by psycho)
0,1, -1 ? why is it so ?
Don't have time to type up my proof now.

what's the difference between f(R)=R and f:R->R ???
One says f is onto and the other just says the function maps into R.

f(0) = plus minus 1 is the answer ???? is it that simple ?
what? I am saying there are three differentiable functions namely

f(x)=0 for all x
f(x)=1 for all x
f(x)=-1 for all x

which satisfy your identity.
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BCHL85
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#25
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#25
Oh yes... there're only 3 functions like those
No others :p:
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RichE
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#26
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#26
Because f(x) = f(x)f(0)^2 then if f isn't identically 0 we have f(0) = 1 or -1.

Let's assume f(0)=1.

Rewrite your identity as

[f(x+y) - f(x)]/y = f(x)f(y)[f(xy)-1]/y + f(x)[f(y)-1]/y.

If the function is differentiable, then letting y->0 we get

f'(x) = f(x)xf'(0) + f(x)f'(0).

So f'(x) = af(x)(x+1) where a = f'(0).

Solving this we get f(x) = b exp[1/2 a(x+1)^2]

As f(0) = 1 then b =1.

But f(x) = exp[1/2 a(x+1)^2] satisfies the given identity for all x,y only if a=0.

Hence f = 1.

If f(0) = -1 then -f satisfies the identity and -f(0)=1 and hence -f = 1, so f = -1.
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BCHL85
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#27
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#27
Don't know if it said the function is differentiable or not ... so..

As we can see f(x) = 0 is one solution.
So consider f(x) != 0

We have
f(x+y+z) = f(x).f(y+z).f(xy+xz)= f(x)f(y)f(z).f(xy).f(xz).f(yz).f (x^2.yz)

Similarly we have
f(x+y+z) = f(x)f(y)f(z).f(xy).f(xz).f(yz).f (x.y^2z)

So f(x^2yz) = f(xy^2z)
Let z = 1, so we have : f(x^2y) = f(xy^2)

Let u = x^2y, v = xy^2
If u, v = 0 then f(0) = f(0) = +/-1
Then for u, v are not 0, we can always find x and y in terms of u, v
So we have f(u) = f(v) for all u, v
Therefore f(u) = constant, i.e f(x) = constant.

Hence f(x) = 0, 1 or -1
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RichE
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#28
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#28
(Original post by BCHL85)
Don't know if it said the function is differentiable or not ... so..

As we can see f(x) = 0 is one solution.
So consider f(x) != 0

We have
f(x+y+z) = f(x).f(y+z).f(xy+xz)= f(x)f(y)f(z).f(xy).f(xz).f(yz).f (x^2.yz)

Similarly we have
f(x+y+z) = f(x)f(y)f(z).f(xy).f(xz).f(yz).f (x.y^2z)

So f(x^2yz) = f(xy^2z)
Let z = 1, so we have : f(x^2y) = f(xy^2)

Let u = x^2y, v = xy^2
If u, v = 0 then f(0) = f(0) = +/-1
Then for u, v are not 0, we can always find x and y in terms of u, v
So we have f(u) = f(v) for all u, v
Therefore f(u) = constant, i.e f(x) = constant.

Hence f(x) = 0, 1 or -1
Nice ideas, but aren't you assuming a lot in the middle about where the function is zero and not, with all that cancelling.
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BCHL85
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#29
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#29
well, at first I consider f(x) = 0 for all x in R.
Then I consider f(x) isn't zero all for x, so we can find some x s.t f(x) != 0.

For the f(xy) = 0 then it must occurs for all x, y. It also means f(u) = 0 for all u, so I dont think we need to care about this case.
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RichE
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#30
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#30
(Original post by BCHL85)
well, at first I consider f(x) = 0 for all x in R.
Then I consider f(x) isn't zero all for x, so we can find some x s.t f(x) != 0.

For the f(xy) = 0 then it must occurs for all x, y. It also means f(u) = 0 for all u, so I dont think we need to care about this case.
I just twigged this myself, but I don't get the f(xy) comment :confused:

If f(x) = 0 for some x then by the identity f(x+y) = 0 for all y and so the function is zero.

Nice proof.

EDIT: but generally you can have non-zero functions f,g such that fg=0, so your cancelling at first concerned me.
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