Question

The function satisfies the identity

f(x) + f(y) = f(x+y) (1)

for all x and y. Show that 2f(x) = f(2x) and deduce that f '' (0) = 0. By considering the Maclaurin series for f(x),
find the most general function that satisfies (1).
[Do not consider issues of existence or convergence of Maaclaurin series in this question.]

Answer 1

@TuringTest help me! :P

Answer 2

@experimentX

Answer 3

First, it is clear to you that f(2x) = 2f(x), yes?

Answer 4

talk to me, I'll help you, but you need to be here.

Answer 5

yes it is, sorry I wasn't there. But everything after that I am not ok with. How do I deduce that f''(0) = 0 ?

Answer 6

Differentiate the equation
f(2x) = 2f(x)
twice

Answer 7

2f'(2x) = 2f'(x)
4f''(2x) = 2f''(x)
?

Answer 8

Right. Now evaluate that at zero and what do you get?

Answer 9

uhh 4f''(0) = 2f''(0) or you mean when f''(x) = 0 ? Sorry I'm probably being stupid

Answer 10

Yes. Notice that all of the higher derivatives at x = 0 are also zero.

Answer 11

Oh because f''(0)=0 is the only solution to 4f''(0) = 2f''(0) ?

Answer 12

so I can see anyway now that there will only be 2 terms in the maclaurin series, the rest will be zero

Answer 13

Yes, that shows that f''(0) = 0.

Now keep on differentiating the original expression and you can show that
\[ f^{(n)}(0) = 0 \] i.e., the nth derivative of f is also zero when x = 0.

Answer 14

ok

Answer 15

so is it just ax + b?

Answer 16

Those are the only possibilities at this stage. Now put that into the original equation and see if you can say anything else.

Answer 17

2ax + b = 2(ax + b) so it's not possible unless b = 0 ?

Answer 18

Right. Alternatively, ask yourself this: can we say anything about f(0)? Yes, it must be ...

Answer 19

Wow, you're great :)

Answer 20

My mother says so too. But thanks.

Answer 21

Hah :) thank you