Invariably True Integrals

Calculus Level 4

Suppose $f$ and $g$ are non-constant real-valued differentiable functions on $(- \infty, \infty)$ ,

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

$\forall x, y \in \mathbb R$ , and $f'(0) = 0$ . Find $(f(x))^2 + (g(x))^2 , \forall x \in \mathbb R.$

The answer is 1.

2 solutions

Agni Purani
Feb 4, 2019

From the above equation and condition, I figured out that $f(x) = \cos (x)$ and $g(x) = \sin(x)$

$\therefore (f(x))^{2} + (g(x))^{2} = \sin^{2}(x) + \cos^{2} (x) = \boxed{1}$

@Agni Purani , that is your clever observation. But can you show that this is the only solution to the given problem?

Priyanshu Mishra - 2 years, 4 months ago

I will see to that but I believe that that is the only solution. I will come back if I find out. I have other ways to solve too. But to prove that it is the only solution is quite difficult.

Agni Purani - 2 years, 4 months ago

If we put $k = g'(0)$ , then the fact that $f$ and $g$ are differentiable, together with the given conditions, give us that $f(0) = 1$ and $g(0)=0$ , as well as $f'(x) \; = \; -kg(x) \hspace{1cm} g'(x) \; = \; kf(x) \hspace{2cm} x \in \mathbb{R} \hspace{3cm} (\star)$ From this we deduce that $\begin{array}{rclcrcl} f^{(2n)}(0) & = & (-1)^nk^{2n} & \hspace{2cm}& f^{(2n+1)}(0) & = & 0 \\ g^{(2n)}(0) & = & 0 & & g^{(2n+1)}(0) & = & (-1)^k k^{2n+1} \end{array}$ for any $n \ge 0$ . It is clear from $(\star)$ that $f$ and $g$ are infinitely differentiable, and that $\big|f^{(n)}(x)\big| \le |k|^n$ and $\big|g^{(n)}(x)| \le |k|^n$ for all $n \ge 0$ . Taylor's Theorem, with remainder, tells us that, for any $x \in \mathbb{R}$ and $n \in \mathbb{N}$ , we can find $0 < \theta < 1$ such that $f(x) \; = \; \sum_{j=0}^{n-1} \frac{1}{j!}f^{(j)}(0)x^j + \frac{1}{n!}x^n f^{(n)}(\theta x)$ and hence $\left| f(x) - \sum_{j=0}^{n-1} \frac{1}{j!}f^{(j)}(0)x^j\right| \; \le \; \frac{1}{n!} |kx|^n$ from which we deduce that $f(x) \; = \; \sum_{j=0}^\infty \frac{1}{j!}f^{(j)}(0)x^j \; =\; \sum_{j=0}^\infty \frac{(-1)^j}{(2j)!}k^{2j}x^{2j} \; = \; \cos\,kx$ and. similarly, $g(x) = \sin\, kx$ .

Mark Hennings - 2 years, 4 months ago

Interesting.

Agni Purani - 2 years, 4 months ago

Mark Hennings
Feb 3, 2019

Putting $x=y=0$ in the two equations gives $f(0) = f(0)^2 - g(0)^2 \hspace{2cm} g(0) = 2f(0)g(0)$ Since $f(0)(1 - f(0)) = -g(0)^2 \le 0$ , we deduce that either $f(0) \le 0$ or $f(0) \ge 1$ . Thus $f(0) \neq \tfrac12$ , and hence $g(0) = 0$ , and hence $f(0) = 0$ or $1$ . If $f(0) = 0$ then $g(x) = f(x)g(0) + g(x)f(0) = 0$ is constant, which is not possible. Thus we deduce that $f(0) = 1$ .

If we define $h(x) = f(x)^2 + g(x)^2$ , then $h$ is differentiable, and it is easy to show that $h(x+y) = h(x)h(y)$ for all $x,y$ . Since $h(0) = 1$ we see that $h(x)h(-x) = 1$ for all $x$ , and hence $h(x) \neq 0$ for all $x$ . Clearly $h(x) \ge 0$ for all $x$ , and hence it follows that $h(x) > 0$ for all $x$ . Thus $\ln h$ is a differentiable function such that $(\ln h)(x+y) = (\ln h)(x) + (\ln h)(y)$ for all $x,y$ , and hence there exists some $\alpha$ such that $(\ln h)(x) = \alpha x$ , and hence $h(x) = e^{\alpha x}$ for all $x$ . Since $\alpha = h'(0) = 2f(0)f'(0) + 2g(0)g'(0) = 0$ , we deduce that $h(x) = \boxed{1}$ for all $x$ .

Everything in this argument works for general functions $f$ , $g$ up to the point where we use differentiability of $h$ to show that $\alpha =0$ . It might be fun to think of a criterion which is less restrictive than differentiability which would be enough to make this problem work. Requiring both $f$ and $g$ to be bounded functions would do, for example (if $h(x) = e^{\alpha x}$ was a bounded function, we would have to have $\alpha =0$ ).

Invariably True Integrals

The answer is 1.

2 solutions

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