Not x 2 − y 2
How many functions
f
:
R
↦
R
are there such that
f
(
x
2
+
y
2
)
=
f
(
x
+
y
)
f
(
x
−
y
)
?
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1 solution
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The only missing step is "We must then determine which possible functions can exist.".
Sir,why f(x) is a constant function.Can u pls explain?
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I am only stating the 3 solutions to the problem. I have provided no justification as yet for why there are only 3 solutions.
Note that I did not say that
f
(
x
)
must be a constant function. In particular, for solution 3, it is not a constant function.
I said "Most people ... think that means it must be constant on the entire domain". From the phrasing, it implies that there is some kind of misconception here.
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oh yes...I am sorry..but can you give me an idea how to solve the problem?? because I am not getting any ideas..so...
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@Rajdeep Brahma – First, show that each point, f ( x ) = 1 or 0.
Second, determine which possible functions can exist. One possibility is to check the 2 2 = 4 cases of values for { f ( 0 ) , f ( 1 ) } = { 0 / 1 , 0 / 1 } , and see conclusions can be drawn.
[This is not a complete solution.]
There are 3 solutions
Most people would get to f ( x ) = 0 , 1 and think that means it must be constant on the entire domain. However, all that we know is at each individual point the value is 1 or 0. We must then determine which possible functions can exist.