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hardest maths question, genius whoever solves it?
The function f from the real numbers to the real numbers satisfies f(x+y)= (1+yx+1)f(x)+(1+xy+1)f(y)+x2y+xy+xy2. If f(1)=4, what is the value of f(8)?
1 Answer
- Anonymous9 years agoFavorite Answer
f(x + y) = (1 + xy + 1)f(y) + (x^2)y + xy + x(y^2)
f(x + y) = (xy + 2)f(y) + xy(x + 1 + y)
f(x + y) = xy*f(y) + 2*f(y) + xy(x + y + 1)
f(x + y) = xy(x + y + 1 + f(y)) + 2*f(y)
Let u = x + y, xy = y(u - y)
f(u) = y(u - y)(u + 1 + f(y)) + 2*f(y)
f(1) = 4, so substitute u = 1:
4 = y(1 - y)(2 + f(y)) + 2*f(y)
(y - y^2)(2 + f(y)) + 2*f(y) - 4 = 0
2y - 2y^2 + y*f(y) - (y^2)*f(y) + 2*f(y) - 4 = 0
We need to find a formula for f(y) such that this equation will always hold true for all values for y:
f(y)*(y^2 - y - 2) = (2y - y^2 - 4)
f(y) = (2y - y^2 - 4) / (y^2 - y - 2)
Therefore:
f(u) = y(u - y)(u + 1 + (2y - y^2 - 4) / (y^2 - y - 2)) + 2*(2y - y^2 - 4) / (y^2 - y - 2)
Hence f(8) = y(8 - y)(9 + (2y - y^2 - 4) / (y^2 - y - 2)) + 2*(2y - y^2 - 4) / (y^2 - y - 2))
More information is needed to find a numerical answer.