Question

Find f(x) IF:
f(x+y) = f(x) + f(y) - f(x)f(y)
And f(0) is not 1
And f'(0) = -1

Answer 1

@experimentX
The basic fault there in your thing is that,
x and y are INDEPENDENT variables.
Therefore change in x does not change y in any way.
Hence,
While differenciating you can differenciate it w.r.t. x as
f'(x+y) = f'(x) - f'(x)f(y).
Now I put x=0 ,
=> f'(y) + 1 = f(y)
Thats how far I've gotten so far.

Answer 2

dy/dx -y = -1

Answer 3

that's a linear DE.

Answer 4

let me tell u what i did
f'(x) = lim y->0 f(x+y) - f(x) / y
= lim y->0 f(y)(1-f(x)) / y
put x=0
-1 = ......
since the limit exist , there must be common factor that will cancel out
1-f(0) cannot be y.it must be constant
so f(y) = y
f(x) = x

Answer 5

is there any mistake ^^ ??

Answer 6

hmm.

Answer 7

that's pretty cool @hartnn

Answer 8

Okay.
Yeah. I see. Linear D.E.

Answer 9

@hartnn Thats pretty cool but,
I guess,
f(y) = y When y tends to zero only. Otherwise we cannot say anything.

Answer 10

yup, i also realised that f'(0) is not -1 then...so i m incorrect

Answer 11

lim y->0 f(x+y) - f(x) / y

wouldn't you need the limit def. of the partial though?

Answer 12

f'(0) is -1 for the record.

Answer 13

Nice.
Thanks guys.

Answer 14

1-e^(x+y) = (1 - e^x)+(1-e^y) - (1 - e^x)*(1-e^y)
= 2 - (e^x+e^y) - 1 + (e^x+e^y) - e^(x+y)
= 1 - e^(x+y)