2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ WLOG, can we write $$f(x)=k\cdot\sin x,g(x)=k\cdot\cos x ?$$ $\endgroup$ – lab bhattacharjee Jan 16 '14 at 17:35
  • $\begingroup$ @labbhattacharjee, why so? Also $g(0)=1$ which implies according to you, $k=1$. $\endgroup$ – Apurv Jan 16 '14 at 17:38
  • $\begingroup$ 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 $\;f'_+(0)\;$ ? $\endgroup$ – DonAntonio Jan 16 '14 at 17:42
  • $\begingroup$ @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 $f'_+(0)$ – $\endgroup$ – Apurv Jan 16 '14 at 17:53
  • 2
    $\begingroup$ @labbhattacharjee See the computation of all the solutions of those functional equations (math.stackexchange.com/a/614228/26489). There are many more solutions than $\sin(kx), \cos(kx)$. $\endgroup$ – OR. Jan 16 '14 at 18:30
4
$\begingroup$
  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

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

$\endgroup$
  • $\begingroup$ that was neat! Thanks! Sometimes ideas just don't click... $\endgroup$ – Apurv Jan 16 '14 at 18:14
0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Why did you take the sin and cos functions only? Some other functions could also satisfy the functional equations.. $\endgroup$ – Apurv Jan 16 '14 at 17:50
  • $\begingroup$ 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. $\endgroup$ – kleineg Jan 16 '14 at 18:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.