24x2f(x) +f(1-x)=2x-x4 -------(1)
Replace x→1-x
(1-x)2f(1-x) +f(x)=2(1-x)-(1-x)4 ------(2)
Solving (1) and (2)
[x2(1-x)2-1]f(x)=(1-x)2(2x-x4)-2(1-x)+(1-x)4
f(x)=(1-x)2(2x-x4)-2(1-x)+(1-x)4
[x2(1-x)2-1]
This question is not for the advanced students.. but for those who fear the name Functional equations and have not tried too many....
SOlve this one to get some hold on the topic..
Determine f(x) such that...
x^2f(x)+f(1-x)=2x-x^4
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11 Answers
24
62you have made a small mistake in solving the two equations...
See properly
3eure its not for advanced students :P as u have solved the functional eqn buks by bargave for olympiads.
24hehe.........
I realised that later....but sinceno one posted ths soln for 30 mins I thought i should do that
1708x2f(x) +f(1-x)=2x-x4 -------(1)
here f(x) must be in the form of 2 degree polynomial
and degree of leading coefficient is -1.
bcz max. power of variable x in R.H S is -1
so f(x) = - (x-a)*(x-b)
put x=0 in eq .(1) we get f(1) = 0 means (x-1) is a factor of f(x)
so f(x) = -(x-1)*(x-b).........(2)
again put x = 1 in eq..(1) we get f(1) +f(0) = 1
again put x = 0 in eq....(2) we get f(0) = -b
so f(1) =0 and f(0) =-b
put these value in f(1) +f(0) = 1
we get b = -1
put value of b in eq....(2)
f(x) = -(x-1)(x+1) =1-x2
ans = (1-x2)
and this solution was given by hsbhatt sir.
62The solution is indeed awesome :)
but I did not understand why f(x) should be a polynomial..? ?
could it not be a function where there is one thing like sin(x) - sin (1-x) which gets canceled?
Am i missing something?
1708No No Sir bcz R.H .S is a polynomial so L.H.S must be a polynomial.
62but that is not guaranteed because they could cancel each other... ?? one -ve term
there could be something like |x| ?? Or somethign still more complex.. say in an exponential form?
srry exponential will not be possible..
19even i have the same doubt !!! the way mavarick is assuming f(x) to be a polynomial is a bit doubtful....if this is the case then we can also suitably divide or multiply the lhs and rhs by suitable quantities and then call it some other polynomial.....
what i want to ask is ,,what is the standard form in these cases to determine the degree of polynomial or the polynomial itself ?
62I saw that.. and i was also thinking about it.. but I am still not fully convinced.. (may be I am overlooking something much more fundamental...)