# help on functions ...Watch

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

thx.
0
12 years ago
#2
Is it degree year 1 / year 2 level?
0
#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 ..
0
12 years ago
#4
Set y=0 and so f(x)=f(x)2f(0) for all x...
0
#5
0
12 years ago
#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.
0
#7
ok .. i'll try .. is this question hard ?
0
12 years ago
#8
(Original post by psycho)
ok .. i'll try .. is this question hard ?
No. The hint I've given should make everything very obvious.
0
#9
ok .. which i'll get f(x)=1/f(0) ... then ?
0
12 years ago
#10
F(R) = R you say?
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#11
well .. the question said so .. f(R)=R ..
0
12 years ago
#12
f(x) = 0 or f(x) = 1
0
#13
can i trust that ? are u a liar ?
0
12 years ago
#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
0
#15
oopps ... the question should be ...
find all functions f(R)=R such that f(x+y)=f(x)f(y)f(xy) ...
sorry ..
0
#16
so .. according to wrangler method .. then i'll get:
f(0) = plus minus 1 ... then what ?
0
#17
what should I do next ??
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#18
how come no body answer ?? is the question too hard ?
0
12 years ago
#19
(Original post by psycho)
so .. according to wrangler method .. then i'll get:
f(0) = plus minus 1 ... then what ?
you're done.
0
12 years ago
#20
(Original post by Wrangler)
Set y=0 and so f(x)=f(x)2f(0) for all x...

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