Function Mystery

Algebra Level 3

f ( f ( x ) ) = 1 − 1 x 4 + 2 x 2 + 2 f(f(x)) = 1 - \dfrac{1}{x^4 + 2x^2 +2}

Considering the composite function above, what is the value of f ( 0 ) f(0) ?


The answer is 1.

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Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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1 solution

Ravneet Singh
Nov 25, 2016

f ( f ( x ) ) = 1 − 1 x 4 + 2 x 2 + 2 f(f(x)) = 1 - \dfrac{1}{x^4 + 2x^2 +2}

f ( f ( x ) ) = 1 − 1 ( x 2 + 1 ) 2 + 1 f(f(x)) = 1 - \dfrac{1}{(x^2+1)^2 + 1}

f ( f ( x ) ) = ( x 2 + 1 ) 2 ( x 2 + 1 ) 2 + 1 f(f(x)) = \dfrac{(x^2+1)^2}{(x^2+1)^2 + 1} Dividing up and down by ( x 2 + 1 ) 2 (x^2+1)^2

f ( f ( x ) ) = 1 ( 1 x 2 + 1 ) 2 + 1 f(f(x)) = \dfrac{1}{\left(\dfrac{1}{x^2+1}\right)^2+ 1}

⟹ f ( x ) = 1 x 2 + 1 \Longrightarrow f(x) = \dfrac{1}{x^2+1}

⟹ f ( 0 ) = 1 \Longrightarrow f(0) = 1

6 Helpful 0 Interesting 0 Brilliant 0 Confused

I do not see how the last implication is valid without further argument. The function f given is a possibility, but is it unique; and if others exist do they all satisfy f(0)=1?

Will Heierman - 4 years, 6 months ago

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I am not able to see any other possibility. Can you suggest any other function satisfy the above said property. thanks

Ravneet Singh - 4 years, 6 months ago

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I am aware of constructive methods (e. g. different series expansions with undetermined coefficients) that might work, but lack the time to try. I believe if the identity is to hold over an interval of positive length, then any solution must be algebraic (combo of roots, fractions, polys). The fun part of the problem was finding your solution!

Will Heierman - 4 years, 6 months ago

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