Let f be a twice differentiable function defined on x > 0 satisfying x f ( y ) + y f ( x ) = f ( x y ) for all x , y > 0 . Given that f ′ ( 1 ) = 1 , find 4 x = 1 ∑ ∞ f ′ ′ ( x ) f ′ ′ ( x + 2 ) .
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Fun problem!
My approach is similar to Abhi's:
Differentiating with respect to y gives x f ′ ( y ) + f ( x ) = x f ′ ( x y ) . Now evaluate at y = 1 to find x + f ( x ) = x f ′ ( x ) . Next differentiate with respect to x to get 1 + f ′ ( x ) = f ′ ( x ) + x f ′ ′ ( x ) or f ′ ′ ( x ) = x 1 . My "end game" is the same as Abhi's.
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It would've been more fun if it was only given that f is continuous and we had to prove that f was infinitely differentiable! ;)
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This is an exercise often assigned in calculus classes; essentially Cauchy's functional equation. It becomes (a lot) more interesting when you drop continuity.
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@Otto Bretscher – Yeah. When you drop continuity, you get a function whose graph is dense on R 2 right?
I have added two more problems which are "more fun". Check them out,
1.https://brilliant.org/problems/just-look-carefully-3/?ref_id=1202148
2.https://brilliant.org/problems/just-look-carefully-2-2/?ref_id=1202134
Try the first one first.
@Otto Bretscher, @Deeparaj Bhat
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@Abhi Kumbale – You can try these too,
https://brilliant.org/problems/convergency-and-divergency/?ref_id=1198750
https://brilliant.org/problems/generalization-of-a-problem-unique-symmetry/?ref_id=1198647
https://brilliant.org/problems/a-perfect-set-for-jee/?ref_id=1197855
They are all part of my set,
https://brilliant.org/profile/abhi-pwu19k/sets/my-creations-check-them-out/?ref_id=1198669
Is this your own problem ? Great problem !! This is so much fun !
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