Unusual Functional Twist
f ( x + y ) = f ( x ) + 2 x y + f ( y )
Let f : Q → R be a function defined on the set of all rational numbers Q satisfying the above functional equation for all x , y ∈ Q and where f ( 1 ) = 2 0 1 5 . Submit the value of f ( 2 0 . 1 5 ) upto three correct places of decimals as your answer.
Bonus : Generalize f ( x ) .
The answer is 40988.122.
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5 solutions
Can you generalize f ( x ) ?
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Functions of the form x 2 + c x (is that what you mean?)
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Yes. Generalize f ( x ) satisfying the above function. And add it in your solution :)
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@Satyajit Mohanty – Is f ( x ) = x 2 + 2 0 1 4 x ?
@Satyajit Mohanty – If that is the generalized function, then why is it stated that the function is defined mapping the rational numbers to the real numbers? Wouldn't that function be valid for all real input?
f ( n x ) = f ( x ) + 2 x 2 ( n − 1 ) + f ( ( n − 1 ) x ) = f ( x ) + 2 x 2 ( n − 1 ) + f ( x ) + 2 x 2 ( n − 2 ) + f ( ( n − 2 ) x ) = n f ( x ) + 2 x 2 ( ( n − 1 ) + ( n − 2 ) + . . . . + 1 ) f ( n x ) = n f ( x ) + 2 x 2 × 2 n ( n − 1 ) = n f ( x ) + x 2 n ( n − 1 ) put x=1, f ( n ) = 2 0 1 5 n + n ( n − 1 ) = n 2 + 2 0 1 4 n f ( 2 0 . 1 5 ) = 2 0 . 1 5 ⋅ 2 0 1 5 + 2 0 . 1 5 ⋅ 1 9 . 1 5 ≈ 4 0 9 8 8 . 1 2 2
f ( x + y ) = f ( x ) + 2 x y + f ( y ) , f ( 1 ) = 2 0 1 5
f ( x + x ) = 2 f ( x ) + 2 x 2
When x = 1, we get value of f(2)
When x = 2, we get value of f(4) and so on.
f ( 1 6 + 4 ) = f ( 2 0 ) = 4 0 6 8 0
f ( 0 . 5 + 0 . 5 ) = f ( 1 ) , f ( 0 . 2 5 + 0 . 2 5 ) = f ( 0 . 5 )
This way we get value of f ( 0 . 2 5 + 0 . 5 ) = f ( 0 . 7 5 )
f ( 5 x ) = f ( x ) + f ( 4 x ) + 8 x 2 = 5 f ( x ) + 2 0 x 2 (Repeat for f(4x) and so on.)
That is how we get value of f(0.15) = 302.1225 (5x = 0.75)
f ( 2 0 + 0 . 1 5 ) = f ( 2 0 . 1 5 ) ≈ 4 0 9 8 8 . 1 2 2
Moderator note:
You should explain that you are using Induction to prove the third statement. And in particular, the induction proof only holds for being an integer. Thus , you haven't proved it for all real values, but only for all integer values thus far.
I'm sorry, but I got lost, could you please explain how you got from your second line to your third line? i don't see where the ( x + n ) ( x + m ) or 2 0 1 5 x terms came from.
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f ( x + 1 ) = f ( x ) + 2 x + 2 0 1 5
Therefore when 1 is added to the value of x, 2x + 2015 is added to the value of f(x). f(x) grows by 2015 every time 1 is added to x so there must be 2015x. f(x) grows by 2x every time 1 is added to x so there must be a 2nd degree polynomial, but we don't know which polynomial it is yet. Using f(0) = 0 and f(1) = 2015, we find that the polynomial is x^2 - x = x(x-1). f(x) = x(x-1) + 2015x (= x(x+2014)) and f(x) satisfies all of the conditions so it must be right.
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You should explain that you are using Induction to prove the third statement. And in particular, the induction proof only holds for x being an integer. Thus , you haven't proved it for all real values, but only for all integer values thus far.
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@Calvin Lin – Is the new solution better?
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@Jesse Nieminen – Yes, this now works as you deal explicitly with the 0 . 1 5 . It would be slightly better to work out what f ( n x ) is in terms of f ( x ) , and use it to calculate f ( 1 ) ⇒ f ( 0 . 0 5 ) ⇒ f ( 0 . 1 5 ) .
Note that such an inductive approach will only work for countable sets. In particular, we would be unable to prove it for all real numbers. In your approach, you can only extend it to all rational x .
F(x) = x(x+1) + 2013x You can prove this by putting x = x-1 and y = 1
f ( x ) = x 2 + 2 0 1 4 x
We may rewrite the given relation as f ( x ) = x f ( 1 ) + x ( x − 1 ) . Therefore, f ( 2 0 . 1 5 ) = 2 0 . 1 5 ⋅ 2 0 1 5 + 2 0 . 1 5 ⋅ 1 9 . 1 5 ≈ 4 0 9 8 8 . 1 2 2 .