# Find $g'(x)$ at $x=0$

The question is:

Question. Let $$f(x-y)=f(x)\cdot g(y)-f(y)\cdot g(x)$$ and $$g(x-y)=g(x)\cdot g(y)+f(x)\cdot f(y)$$ for all $x,y \in \mathbb{R}$.

If right hand derivative at $x=0$ exists for $f(x)$, then find derivative of $g(x)$ at $x=0$.

My try:

By some simple substitutions I figured out that $f(0)=0$ and $g(0)=1$. If in the second equation, we put $x=y$, it will give $g(0)=(g(x))^2+(f(x))^2$. If $g(0)=0$, sum of the two squares becomes $0$ which implies the squares themselves are zero, I neglected $g(x)=f(x)=0$ as a trivial solution and hence took $g(0)=1$. But how do I proceed after this?

• WLOG, can we write $$f(x)=k\cdot\sin x,g(x)=k\cdot\cos x ?$$ – lab bhattacharjee Jan 16 '14 at 17:35
• @labbhattacharjee, why so? Also $g(0)=1$ which implies according to you, $k=1$. – Apurv Jan 16 '14 at 17:38
• Clearly $\;f(0)=0\;$ from the first equation, yet the second one yields $$g(0)=g(0)^2\implies g(0)=0\;\;OR\;\;g(0)=1$$ Whay did you choose $\;g(0)=1\;$ ? Also, what do you mean by "the right hand derivative at $\;x=0\;$ exists for $\;f(x)\;$"? Did you mean the right derivative $\;f'_+(0)\;$ ? – DonAntonio Jan 16 '14 at 17:42
• @DonAntonio, because if in the second equation, we put x=y, it will give $g(0)=(g(x))^2+(f(x))^2$. If $g(0)=0$, sum of the two squares becomes 0 which implies the squares themselves are zero, I neglected $g(x)=f(x)=0$ as a trivial solution and hence took $g(0)=1$. And, right hand derivative means $f'_+(0)$ – – Apurv Jan 16 '14 at 17:53
• @labbhattacharjee See the computation of all the solutions of those functional equations (math.stackexchange.com/a/614228/26489). There are many more solutions than $\sin(kx), \cos(kx)$. – OR. Jan 16 '14 at 18:30

1. From the first equation, putting $x=y=0$, we get $f(0)=0$.

2. From the second equation, putting $x=y=0$, we get $g(0)=g^2(0)$. So, either $g(0)=1$ or $g(0)=0$.

3. If $g(0)=0$ then from the first equation, putting $y=0$, we get $f(x)=0$, for all $x$, and then from the second equation, putting $y=0$, we get $g(x)=0$ for all $x$. From where you can compute that the derivative equals to zero.

Let us assume for the rest that $g(0)=1$. From the first equation, putting $x=0$, we get $f(-y)=-f(y)$. And from the second, putting $x=0$, we get $g(-y)=g(y)$.

So the equations are equivalent to

\begin{align}f(x+y)&=f(x)g(y)+f(y)g(x)\\g(x+y)&=g(x)g(y)-f(x)f(y)\end{align}

Since $g$ is even it is enough to compute the derivative from the right.

\begin{align}\lim_{y\rightarrow0^+}\frac{g(0+y)-g(0)}{y}&=\lim_{y\rightarrow0^+}\frac{g(0+y)-g(0)}{y}\\&=\lim_{y\rightarrow0^+}\frac{-2f(y/2)f(y/2)}{y}\\&=-\lim_{y\rightarrow0^+}\frac{f^2(y/2)}{y/2}\\&=-f(0^+)f'_{+}(0)\\&=0\end{align}

In the second equality we used the formula:

\begin{align}g(x)-g(y)&=-2f(\tfrac{x+y}{2})\,f(\tfrac{x-y}{2})\end{align}

To deduce it we use:

$$g(x)=g(\tfrac{x+y}{2}+\tfrac{x-y}{2})=g(\tfrac{x+y}{2})g(\tfrac{x-y}{2})-f(\tfrac{x+y}{2})f(\tfrac{x-y}{2})$$

$$g(y)=g(\tfrac{x+y}{2}-\tfrac{x-y}{2})=g(\tfrac{x+y}{2})g(\tfrac{x-y}{2})+f(\tfrac{x+y}{2})f(\tfrac{x-y}{2})$$

Subtracting the two equations we get

$$g(x)-g(y)=-2f(\tfrac{x+y}{2})f(\tfrac{x-y}{2})$$

• that was neat! Thanks! Sometimes ideas just don't click... – Apurv Jan 16 '14 at 18:14

Those look like the functional definitions of sin and cos respectively, see Trig functions.

$\sin(x - y) = \sin(x)\cos(y) - \cos(x)\sin(y)$

$\cos(x - y) = \cos(x)\cos(y) - \sin(x)\sin(y)$

You would need a constant $c$ to be able to meet the entire range of functions.

But, $g(0) = c \cos(0) = c = 1$

So you would simply have $f(x) = \sin(x)$ and $g(x) = \cos(x)$ and it would be simple to find $g'(x)$.

Let me know if this helps.

• Why did you take the sin and cos functions only? Some other functions could also satisfy the functional equations.. – Apurv Jan 16 '14 at 17:50
• ABC just posted what I worked out on paper, so I won't duplicate work. I do find it interesting that you can prove g'(0) = 0, and sin(0) = 0. – kleineg Jan 16 '14 at 18:23