England 🏴󠁧󠁢󠁥󠁮󠁧󠁿 Math Olympiad 2009 - Algebra - Find f(x)??

[@MathAPlus]

World 🌎 International Math Olympiad 2019 - Algebra - Find f(x)?! Boost Your Confidence 😆
th-cam.com/video/M8MMf8pm0FM/w-d-xo.html

[@danilmylnikov123]

I think I got good solution that doesn't use any of f(number). Putting x, - x and 2x in pairs instead of x and y you have a great system of equations. And the result is the same: 1 + x and 1 - x

[@shortstick4373]

I got to the f(1)f(-1) = 0 on my own but no matter how much i toyed with it i couldnt get to the solution

[@Zonnymaka]

Interesting. First he proves that f(x) cannot be zero for any x (6:18), then he goes on to claim this 7:34 😇
As far as i can see (frankly speaking i couldn't find any other way to solve the problem), i have to claim that f(x)=f(-x). That leads me to y=-x, hence f(x)=+-sqrt(1-x^2) where f: ]-1,1[ --> ]0,+-1].

[@ayushrudra8600]

No at 6:18 he is proving that f(x) is not zero for all x

[@sbjf]

lol I used truncated taylor series

[@pitreason]

In order to use taylor expansion you need to know that the function is differentiable, and there is no such condition given in the problem

[@jamie31415]

Please don’t call it the England maths olympiad. Britain is much more than just England.

[@MathAPlus]

Indeed! Thanks 🙏

[@comdo777]

answer=xy

[@nasrullahhusnan2289]

From [f(x)][1-f(0)]=0 --> f(0)=1
But not f(x)=0 and it is explained why it is so. I agree that f(x)=0 does not aplly for any x.
Now consider x≠0 and y=0:
[f(x≠0)]f(0)=f(x≠0)
--> [(f(x≠0)][1-f(0)]=0
f(0)=1 or f(x≠0)=0
Therefore f(x)=1 for x=0
=0 otherwise
Kindly give your comment sir.

[@sanidhyavijay2281]

what about when both(x&y) are 0?

[@nasrullahhusnan2289]

@@sanidhyavijay2281: Two solutions you get: f(x)=1+x and f(x)=1-x both give f(0)=1. If x=y=0 LHS=[f(0)][f(0)]=1(1)=1 and RHS=f(0)=1.
The same result will be obtained with
f(x)=1 for x=0
=0 otherwise
as LHS=[f(0)][f(0)]=1(1)=1 and RHS=f(0)=1

[@dandeleanu3648]

You have already proved f(0)=1 now let y = -x
f(x)f(-x) = f(x-x) - x^2 = f(0) - x^2 = 1- x^2 = (1-x)(1+x)
clearly f(x) = 1- x or f(x) = 1+x
q.e.d.

[@Zonnymaka]

Very nice! For some reason, i assumed immediately that it had to be even so in the end i reached [f(x)]^2=1-x^2 and y=-x. That did blind me 😞

[@dandeleanu3648]

@@Zonnymaka thanks

[@Wurfenkopf]

That's a nice way to find some solutions, but you didn't prove that there aren't others

[@RafaelB1717]

8 * 1 = 4 * 2. It doesn't mean 8 = 4 and 1 = 2.

[@dandeleanu3648]

@@RafaelB1717It is possible to add and prove that indeed the solution fulfills the condition imposed