Is f f a additive function?

Algebra Level 5

Let f : [ 0 , 1 ] → R f:[0,1]\to \mathbb R be a continuous function such that

{ f ( x 1 ) + f ( x 2 ) + f ( x 3 ) + ⋯ + f ( x 1151 ) = 1 x 1 + x 2 + x 3 + ⋯ + x 1151 = 1 \begin{cases} f(x_1)+f(x_2)+ f(x_3) + \cdots +f(x_{1151})=1 \\ x_1+x_2+x_3 + \cdots + x_{1151} = 1 \end{cases}

for all x 1 , x 2 , x 3 , ⋯ x 1151 ∈ [ 0 , 1 ] x_1, x_2, x_3, \cdots x_{1151} \in [0,1] .

Given that f ( 0 ) = 7 f(0)=7 , find f ( 1 2014 ) f \left(\frac 1{2014}\right) .


The answer is 3.

Souryajit Roy by Souryajit Roy , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Souryajit Roy
Jul 12, 2014

At first, let x 1 = x 2 = ( x + y ) 2 x_{1}=x_{2}=\frac{(x+y)}{2} , x 3 = 1 − x − y x_{3}=1-x-y and x 4 = . . . x 1151 = 0 x_{4}=...x_{1151}=0 .

Hence, 2 f ( x + y 2 ) + f ( 1 − x − y ) + 1148 f ( 0 ) = 1 2f(\frac{x+y}{2})+f(1-x-y)+1148f(0)=1

Next, put x 1 = x , x 2 = y , x 3 = 1 − x − y x_{1}=x,x_{2}=y,x_{3}=1-x-y and x 4 = . . . = x 1151 = 0 x_{4}=...=x_{1151}=0

Hence, f ( x ) + f ( y ) + f ( 1 − x − y ) + 1148 f ( 0 ) = 0 f(x)+f(y)+f(1-x-y)+1148f(0)=0

Therefore, 2 f ( x + y 2 ) = f ( x ) + f ( y ) 2f(\frac{x+y}{2})=f(x)+f(y) .This is Jensen's functional equation with the solution f ( x ) = c x + d f(x)=cx+d for some constants c and d

Put x 1 = 1 x_{1}=1 and remaining all x i = 0 x_{i}=0 .

So, c + d + 1150 f ( 0 ) = 1 c+d+1150f(0)=1 .

Again, put x 1 = x 2 = 1 2 x_{1}=x_{2}=\frac{1}{2} and all other x i = 0 x_{i}=0 .

So, c + 2 d + 1149 f ( 0 ) = 1 c+2d+1149f(0)=1

Solving the two equations, d = f ( 0 ) = 7 d=f(0)=7 and c = 1 − 1151 f ( 0 ) = − 8056 c=1-1151f(0)=-8056

Hence, f ( 1 2014 ) = ( − 8056 ) 1 2014 + 7 = − 4 + 7 = 3 f(\frac{1}{2014})=(-8056)\frac{1}{2014}+7=-4+7=3

2 Helpful 0 Interesting 0 Brilliant 0 Confused

we can find f(0.5) by putting two of the variables equal to 0.5 and the rest equal to 0. then put 1007 of the variables equal to 1/2014. one equal to 1/2 and the rest equal to 0. done.

hemang sarkar - 6 years, 11 months ago

Log in to reply

yup,that is right.....i just wanted to show a generalization

Souryajit Roy - 6 years, 10 months ago

Log in to reply

I like your solution. :)

hemang sarkar - 6 years, 10 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...