Just look carefully!

Calculus Level 5

Let f f be a twice differentiable function defined on x > 0 x>0 satisfying x f ( y ) + y f ( x ) = f ( x y ) x f(y) + y f(x) = f(xy) for all x , y > 0 x,y> 0 . Given that f ( 1 ) = 1 f'(1) = 1 , find 4 x = 1 f ( x ) f ( x + 2 ) \displaystyle 4\sum_{x=1}^\infty f''(x) f''(x+2) .


The answer is 3.

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1 solution

Abhi Kumbale
May 21, 2016

Fun problem!

My approach is similar to Abhi's:

Differentiating with respect to y y gives x f ( y ) + f ( x ) = x f ( x y ) xf'(y)+f(x)=xf'(xy) . Now evaluate at y = 1 y=1 to find x + f ( x ) = x f ( x ) x+f(x)=xf'(x) . Next differentiate with respect to x x to get 1 + f ( x ) = f ( x ) + x f ( x ) 1+f'(x)=f'(x)+xf''(x) or f ( x ) = 1 x f''(x)=\frac{1}{x} . My "end game" is the same as Abhi's.

Otto Bretscher - 5 years ago

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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! ;)

@Otto Bretscher

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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.

Otto Bretscher - 5 years ago

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@Otto Bretscher Yeah. When you drop continuity, you get a function whose graph is dense on R 2 \mathbb{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

Abhi Kumbale - 5 years ago

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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

Abhi Kumbale - 5 years ago

Is this your own problem ? Great problem !! This is so much fun !

nishchith s - 4 years, 8 months ago

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