Another Functional Equation!

Algebra Level 5

f ( x + y ) = f ( x ) + f ( y ) + α \large{f(x+y) = f(x) + f(y) + \alpha}

Let f : Q → Q f : \mathbb Q \to \mathbb Q , where Q \mathbb Q is the set of all rational numbers, be such that the above functional equation holds true for all x , y ∈ Q x,y \in \mathbb Q . If f ( β ) = α f(\beta) = \alpha , find the value of f ( α ) f(\alpha) when β = 403 , α = 2015 \beta=403, \alpha=2015 .

Note - α , β ∈ N \alpha, \beta \in \mathbb N .

Bonus - Also generalize the value of f ( α ) f(\alpha) in terms of α \alpha and β \beta .


The answer is 18135.

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Satyajit Mohanty by Satyajit Mohanty , 24, India

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3 solutions

Satyajit Mohanty
Aug 5, 2015

We have f ( x + y ) = f ( x ) + f ( y ) + α f(x+y) = f(x) + f(y) + \alpha .

Let g ( x ) = f ( x ) + α . . . ( 1 ) \large{g(x) = f(x) + \alpha \quad ...(1)} .

⟹ g ( x + y ) = f ( x + y ) + α \Longrightarrow g(x+y) = f(x+y) + \alpha = f ( x ) + f ( y ) + α + α = ( f ( x ) + α ) + ( f ( y ) + α ) = g ( x ) + g ( y ) = f(x) + f(y) + \alpha + \alpha = (f(x) + \alpha) + (f(y) + \alpha)= g(x) + g(y)

Therefore, it's obvious that g ( x ) = λ x g(x) = \lambda x for some λ ∈ Q \lambda \in \mathbb Q .

Now put x = β x=\beta in ( 1 ) (1) to obtain; g ( β ) = f ( β ) + α = α + α = 2 α g(\beta) = f(\beta) + \alpha = \alpha + \alpha = 2\alpha But g ( β ) = λ ⋅ β g(\beta) = \lambda \cdot \beta ⇒ λ ⋅ β = 2 α ⟹ λ = 2 α β ⟹ g ( x ) = 2 α x β \Rightarrow \lambda \cdot \beta = 2\alpha \Longrightarrow \lambda = \dfrac{2\alpha}{\beta}\Longrightarrow g(x) = \dfrac{2\alpha x}{\beta}

Now put x = α x=\alpha in ( 1 ) (1) to obtain; g ( α ) = f ( α ) + α ⟹ 2 α 2 β = f ( α ) + α g(\alpha) = f(\alpha) + \alpha \Longrightarrow \dfrac{2\alpha^2 }{\beta} = f(\alpha) + \alpha ⟹ f ( α ) = 2 α 2 β − α \Longrightarrow f(\alpha) = \dfrac{2\alpha^2 }{\beta} - \alpha

The generalized value of f ( α ) f(\alpha) would be 2 α 2 β − α \dfrac{2\alpha^2}{\beta} - \alpha .

Substituting, α = 2015 , β = 403 \alpha = 2015, \beta = 403 , we obtain f ( 2015 ) = 18135 f(2015) = 18135 .

Note : To check out another generalized solution involving another approach, please visit my another solution in the comments section of @Harsh Shrivastava 's solution.

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Well, @Harsh Shrivastava Good Work! :D My motive behind the problem was not this. I framed the problem a little wrong probably. What if the problem is like this:

f ( x + y ) = f ( x ) + f ( y ) + α \large{f(x+y) = f(x) + f(y) + \alpha}

Let f : Q → Q f : \mathbb Q \to \mathbb Q , where Q \mathbb Q is the set of all rational numbers, be such that the above functional equation holds true for all x , y ∈ Q x,y \in \mathbb Q . If f ( β ) = α f(\beta) = \alpha , find the value of f ( α ) f(\alpha) .

Note : α , β ∈ N \alpha, \beta \in \mathbb N

Can you solve this problem with your approach? For your info, I can certainly say that this problem can be solved by your approach :)

Satyajit Mohanty - 5 years, 10 months ago

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Well your solution to this prob is too good 😀!

BTW I m getting the answer to your problem as 2a^2 - a, is it correct?

Harsh Shrivastava - 5 years, 10 months ago

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@Harsh Shrivastava Look at this. A solution involving the idea of your approach;

I looked for a general pattern of f ( n x ) f(nx) by substituting certain integers n = 2 , 3 , . . n=2,3,.. and came upto something.

First of all, we claim that f ( n x ) = n f ( x ) + α ( n − 1 ) . . . ( 1 ) f(nx) = nf(x) + \alpha(n-1) \quad ...(1) . For n = 1 n=1 , it's trivial. Assume it holds for some integer n n . To prove our claim, let's go for induction. We have: f ( ( n + 1 ) x ) = f ( n x + x ) = f ( n x ) + f ( x ) + α f((n+1)x) = f(nx + x) = f(nx) + f(x) + \alpha ⇒ f ( ( n + 1 ) x ) = n f ( x ) + α ( n − 1 ) + f ( x ) + α = ( n + 1 ) f ( x ) + α ( ( n + 1 ) − 1 ) \Rightarrow f((n+1)x) = nf(x) + \alpha(n-1) + f(x) + \alpha = (n+1)f(x) + \alpha((n+1)-1) hereby, completing the proof of our claim.

Now, in ( 1 ) (1) , put n = α , x = β n=\alpha, x=\beta to obtain: f ( α β ) = α f ( β ) + α ( α − 1 ) f(\alpha \beta) = \alpha f(\beta) + \alpha(\alpha-1)

Again in ( 1 ) (1) , put n = β , x = α n=\beta, x=\alpha to obtain: f ( α β ) = β f ( α ) + α ( β − 1 ) f(\alpha \beta) = \beta f(\alpha) + \alpha(\beta -1)

Equating the above two equations, we get; α f ( β ) + α ( α − 1 ) = β f ( α ) + α ( β − 1 ) \alpha f(\beta) + \alpha(\alpha-1) = \beta f(\alpha) + \alpha(\beta -1)

Since f ( β ) = α f(\beta) = \alpha , substituting it, ⇒ α 2 + α 2 − α = β f ( α ) + α β − α \Rightarrow \alpha^2 + \alpha^2 - \alpha = \beta f(\alpha) + \alpha \beta - \alpha ⇒ β f ( α ) = 2 α 2 − α β ⇒ f ( α ) = 2 α 2 β − α \Rightarrow \beta f(\alpha) = 2\alpha^2 - \alpha \beta \quad \Rightarrow f(\alpha) = \boxed{\dfrac{2 \alpha^2}{\beta} - \alpha}

Satyajit Mohanty - 5 years, 10 months ago

f ( 403 ) = 2015 f ( 806 ) = f ( 403 + 403 ) = f ( 403 ) + f ( 403 ) + 2015 = 6045 f ( 1612 ) = f ( 806 + 806 ) = f ( 806 ) + f ( 806 ) + 2015 = 14105 f ( 2015 ) = f ( 1612 + 403 ) = f ( 1612 ) + f ( 403 ) + 2015 = 18135 \begin{aligned} f(403) & = 2015 \\ f(806) & = f(403+403) = f(403) + f(403) + 2015 = 6045 \\ f(1612) & = f(806+806) = f(806) + f(806) + 2015 = 14105 \\ f(2015) & = f(1612+403) = f(1612) + f(403) + 2015 = \boxed{18135} \end{aligned}

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