Functional analysis.

Algebra Level 5

Let a function f : W → W f : \mathbb W \rightarrow \mathbb W be such that for every m , n ≥ 0 m, n \ge 0

2 f ( m 2 + n 2 ) = ( f ( m ) ) 2 + ( f ( n ) ) 2 \large\ 2f(m^2 + n^2) = (f(m))^2 + (f(n))^2 .

Find f ( 25 ) f(25) .

Notation: W \mathbb W denotes the set of whole numbers.


The answer is 50.

Priyanshu Mishra by Priyanshu Mishra , 20, 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

X X
May 27, 2018

2 f ( 0 + 0 ) = f 2 ( 0 ) + f 2 ( 0 ) → f ( 0 ) = 0 o r 1 2f(0+0)=f^2(0)+f^2(0)\rightarrow f(0)=0\space or\space 1

2 f ( 0 + 1 ) = f 2 ( 0 ) + f 2 ( 1 ) 2f(0+1)=f^2(0)+f^2(1) , if f ( 0 ) = 1 f(0)=1 , then the whole function will be 1 1 , so f ( 0 ) f(0) can't be 2.

Then f ( 0 ) = 0 , f ( 1 ) = 0 o r 2 f(0)=0,f(1)=0\space or\space 2 .

If f ( 1 ) = 0 f(1)=0 , then the whole function will be 0 0 . So f ( 1 ) = 2 f(1)=2

2 f ( 1 + 1 ) = 4 + 4 → f ( 2 ) = 4 2f(1+1)=4+4\rightarrow f(2)=4

2 f ( 4 + 1 ) = 16 + 4 → f ( 5 ) = 10 2f(4+1)=16+4\rightarrow f(5)=10

2 f ( 25 + 0 ) = 100 + 0 → f ( 25 ) = 50 2f(25+0)=100+0\rightarrow f(25)=50

1 Helpful 0 Interesting 0 Brilliant 1 Confused

2f(0+0)=2 f(0)**2 implies f(0)=0 or 1 not 2 or 0

Dan Timsit - 2 years, 7 months ago

Log in to reply

Thanks. I see, I'll update my solution later.

X X - 2 years, 7 months ago

Why can f not be a constant function at 0 or 1? i.e. for all a in 'the whole numbers' (unclear whether you mean natural numbers or integers) we have f(a) = 0, or alternately f(a) = 1.

Daniel Ellesar - 11 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...