1. ## derivative problem

If $\displaystyle f(x-y)=f(x)g(y)-g(x)f(y)$ and $\displaystyle g(x-y)=g(x)g(y)+f(x)f(y)$ for all $\displaystyle x,y \in \mathbb{R}$ and the right hand derivative of $\displaystyle f(x)$ at zero exists. Find $\displaystyle g'(x)$ at $\displaystyle x=0$

2. $\displaystyle g(-x) = g(0-x) = g(x)$

Assuming $\displaystyle g$ is differentiable at 0, the limit

$\displaystyle \lim_{d \rightarrow 0} \frac{g(d)-g(-d)}{2d}$ converges and is equal to $\displaystyle g'(0)$; which yields $\displaystyle g'(0)=0$.

3. Originally Posted by Bruno J.
$\displaystyle g(-x) = g(0-x) = g(x)$

Assuming $\displaystyle g$ is differentiable at 0, the limit

$\displaystyle \lim_{d \rightarrow 0} \frac{g(d)-g(-d)}{2d}$ converges and is equal to $\displaystyle g'(0)$; which yields $\displaystyle g'(0)=0$.
You need to provide the explanation of why g is symmetric

CB

$\displaystyle g(x-0) = g(x)g(0)+f(x)f(0)$
$\displaystyle g(0-x) = g(0)g(x)+f(0)f(x)$

5. ## query

Its not given that g is differentiable

6. ## query

Its not given that g is differentiable, so u cant assume g' exists

7. Originally Posted by Chandru1
Its not given that g is differentiable, so u cant assume g' exists
Here we go then!

You have $\displaystyle g(x) = g(x)g(0)+f(x)f(0) = g(x)g(0)$. Hence either $\displaystyle g(0) = 1$ or $\displaystyle g(x)$ is identically zero which instantly yields $\displaystyle g'(0)=0$. Hence suppose $\displaystyle g(0)=1$.

$\displaystyle f(0) = 0$ is easy; set x=y.
$\displaystyle f(x)=-f(-x)$ is also easy.

$\displaystyle f(x+y)=f(x)g(-y)-g(x)f(-y) = f(x)g(y)+g(x)f(y)$ by the above.

$\displaystyle g(x+y)=g(x)g(-y)+f(x)f(-y) = g(x)g(y)-f(x)f(y)$

Now consider the complex numbers

$\displaystyle m(x) = g(x)+if(x)$.

We have $\displaystyle m(x)m(y) = g(x+y)+if(x+y) = m(x+y)$.

In particular $\displaystyle m(0)=1$, and $\displaystyle m(x)m(-x)=1$.

Hence the function $\displaystyle m : \mathbb{R} \rightarrow \mathbb{C}$ is a morphism of groups $\displaystyle m : \mathbb{R}^+ \rightarrow \mathbb{C}^\times$. Hence we have $\displaystyle m(x)=\exp(tx)$ for some $\displaystyle t=a+bi \in \mathbb{C}$. Clearly $\displaystyle a=0$ since $\displaystyle m(0)=1$. Hence $\displaystyle m(x)=\exp(xbi)$

Hence $\displaystyle g(x) = \Re \exp(xbi) = \cos(bx)$,
so that $\displaystyle g'(0) = 0$.

8. Just a little extra :
In case you are wondering why expressing m as an exponential map is possible, consider

$\displaystyle k=m(1)$

Then for all integers n:
$\displaystyle m(n)=m(1+...+1)=k^n$
$\displaystyle k=m(1)=m(1/n+...+1/n)=m(1/n)^n$
so that $\displaystyle m(1/n) = k^{1/n}$

from this it follows easily that for all rational x
$\displaystyle m(x) = k^{x}$.

Hence if $\displaystyle m$ is continuous on any interval, it coincides everywhere on that interval with $\displaystyle k^{x}$ (irrationals included).
But $\displaystyle f$ is certainly continuous at 0 (it's differentiable there!). To see that g is continuous at 0, we have

$\displaystyle g(x/2 + x/2) = g(x/2)^2-f(x/2)^2$

i.e.
$\displaystyle g(x/2)^2-g(x) = f(x/2)^2$

and from this it is easy to show (with sequences for example) that $\displaystyle g$ is continuous at 0 (using the fact that $\displaystyle f(x/2)^2$ is continuous at 0 and is equal to 0)