help on functions ... Watch

UglyDuckling
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#1
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#1
Find all functions f(R)=R such that f(x+y)=f(x)f(y)f(xy) ...

thx.
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The D
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#2
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#2
Is it degree year 1 / year 2 level?
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UglyDuckling
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#3
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#3
no .. not at all ..
it's for year 12, A-level or any other qualification ..
it's like a selection test for maths olympiad ..
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Wrangler
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#4
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#4
Set y=0 and so f(x)=f(x)2f(0) for all x...
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UglyDuckling
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#5
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#5
is that the answer ?
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Wrangler
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#6
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#6
Does it look like an answer? No. How about you have a think about the (massive) pointer I've given, and let us know your thoughts. We can guide you from there.
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UglyDuckling
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#7
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#7
ok .. i'll try .. is this question hard ?
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Wrangler
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#8
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#8
(Original post by psycho)
ok .. i'll try .. is this question hard ?
No. The hint I've given should make everything very obvious.
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UglyDuckling
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#9
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#9
ok .. which i'll get f(x)=1/f(0) ... then ?
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KAISER_MOLE
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#10
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#10
F(R) = R you say?
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UglyDuckling
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#11
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#11
well .. the question said so .. f(R)=R ..
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BCHL85
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#12
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#12
f(x) = 0 or f(x) = 1
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UglyDuckling
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#13
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#13
can i trust that ? are u a liar ?
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BCHL85
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#14
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#14
(Original post by psycho)
can i trust that ? are u a liar ?
Yeah I am a liar . Don't believe in me. Hehe
f(x+y)=0=f(x).f(y).f(x+y)=0.0.0
f(x+y)=1=f(x).f(y).f(x+y)=1.1.1
:p:
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UglyDuckling
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#15
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#15
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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UglyDuckling
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#16
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#16
so .. according to wrangler method .. then i'll get:
f(0) = plus minus 1 ... then what ?
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UglyDuckling
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#17
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#17
what should I do next ??
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UglyDuckling
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#18
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#18
how come no body answer ?? is the question too hard ?
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Chewwy
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#19
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#19
(Original post by psycho)
so .. according to wrangler method .. then i'll get:
f(0) = plus minus 1 ... then what ?
you're done.
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RichE
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#20
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#20
(Original post by Wrangler)
Set y=0 and so f(x)=f(x)2f(0) for all x...
Shouldn't that read

f(x)=f(x)f(0)2?
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