Find this unique real number!
is an odd function ( that is ) and is twice continuously differentiable in . . We can prove that there is a unique real number such that for any function that satisfies those conditions there exists such that ( can depend on ) what is the value of ?
The answer is 1.
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1 solution
First note that f ( x ) = x satisfies all the conditions and f ′ ′ ( x ) + f ′ ( x ) = 1 for all x . This shows the uniquness of a and a can only be 1 . Then we are left to prove that for any f with stated conditions there exsists 0 < b < 1 such that f ′ ′ ( b ) + f ′ ( b ) = 1 .
Define g ( x ) = f ( x ) − x . Since f ( x ) is odd and f ( 1 ) = 1 , we have g ( 0 ) = 0 , g ( 1 ) = 0 .
From Rolle's Theorem, there exists 0 < c < 1 such that g ′ ( c ) = 0 . This gives us f ′ ( c ) = 1 .
Then is the tricky part....
Define h ( x ) = e x ( f ′ ( x ) − 1 ) . h ′ ( x ) = e x ( f ′ ′ ( x ) + f ′ ( x ) − 1 ) .
If there does not exist 0 < b < 1 such that f ′ ′ ( b ) + f ′ ( b ) = 1 , then, because of the continuity of h ′ ( x ) , for any 0 < x < 1 , h ′ ( x ) > 0 , or for any 0 < x < 1 , h ′ ( x ) < 0 .
Assume h ′ ( x ) > 0 for all 0 < x < 1 then h is strictly increasing in 0 < x < 1 , so h ( 0 ) = h ( 0 − ) ≤ h ( c / 2 ) < h ( c ) = 0 . That is f ′ ( 0 ) < 1 .
Note that when f ( x ) is odd and f is twice differentiable, f ′ ′ ( x ) is odd. Thus f ′ ′ ( 0 ) = 0 . So h ′ ( 0 ) = f ′ ′ ( 0 ) + f ′ ( 0 ) − 1 < 0 .
But h ′ ( x ) > 0 for all 0 < x < 1 and h ′ ( x ) is continuous, so we get h ′ ( 0 ) ≥ 0 . Contradiction!
Similarly if we assume h ′ ( x ) < 0 for all 0 < x < 1 we will also get contradiction.
Done.