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A function $f$ is Carathéodory differentiable at $a$ if there exists a function $\phi$ which is continuous at a such that

$$f(x)-f(a)=\phi(x)(x-a).$$

For $f(x) = x^n$, $\phi(x) = x^{n-1} + ax^{n-2} + ... + a^{n-1}$. We can see that $f'(a) = \phi(a) = na^{n-1}$. We get the derivative of $x^n$ directly from this definition.

For $\sin(x)$, can we do the same thing to get $\sin'(x) = \cos(x)$ without using limit?

My question is raised from the booklet Calculus for Mathematicians by D.J.Bernstein, in which he defined derivative this way before introducing the concept of limits.

ahala
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  • If I recalled correctly, we usually use the Caratheodory formulation to show differentiability most of the time; but the process is essentially the same as finding the derivative by first principle. – Chee Han May 29 '15 at 01:43

3 Answers3

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Because of definition of $\sin(x)$, it is very unlikely that we will find a nice representation of $\phi$ by elementary functions. This is why it isn't as easy as for $x^n$:
First, we recall that $\sin(x)$ is analytically defined as infinite series, $\sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1} $. Then, $$ \sin(x) - \sin(a) = \sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n+1)!}(x^{2n+1}-a^{2n+1}) = (x-a) \cdot \sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n+1)!}(x^{2n} + ax^{2n-1} + ... + a^{2n}) = (x-a) \cdot \phi(x).$$ Also, we have $\phi(a) = \sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n+1)!}((2n+1)a^{2n}) = \sum\limits_{i=0}^\infty \frac{(-1)^n}{(2n)!}a^{2n} = \cos(a). $ Notice that, unlike $x^n$-example, here (even if we overlook convergence problems) $\phi$ doesn't have a nice representation by elementary functions, so that's why it isn't as easy to do the same thing as for $x^n$ and get a Caratheodory derivative.

Filip
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The Weierstrass-Caratheodory formulation asserts that $f\colon E\rightarrow\mathbb{R}$ is differentiable at a point $x_0\in E$ if and only if there exists a function $\phi$ continuous at $x_0$ such that $$f(x)=f(x_0)+\phi(x)(x-x_0)$$ and the derivative is given by $f^{\prime}(x_0)=\phi(x_0)$.

Take $f(x)=\sin(x)$, then suppose we have the following equation: $$\sin(x)=\sin(x_0)+\phi(x)(x-x_0)$$ Rearranging gives us: $$\phi(x)=\frac{\sin(x)-\sin(x_0)}{x-x_0}$$ As you can see, if you can show that this quotient has a limit as $x\rightarrow x_0$, then you are done. But this is exactly the same as finding the derivative of $\sin(x)$ from first principle.

I could go on and derive it, but it is an easy exercise to show that it has a limit by using double-angle formula :)

Chee Han
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  • In the booklet I quoted, there is no concept of limit yet when we meet this problem. In that case, $\phi(x)$ in your answer is not well defined at $x_0$ yet. – ahala May 29 '15 at 01:58
  • Ahhh that makes thing a little bit more tricky then, I suspect that it wouldn't be possible to do it then. Proving that the quotient is continuous at $x_0$ doesn't really tell you anything about the derivative itself, it only tells you $\sin(x)$ is differentiable, and because you haven't learn anything about limits yet, it's not possible to define $\phi(x_0)$. – Chee Han May 29 '15 at 02:03
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I am more comfortable with this version, find a continous function q_x such that:

f(x+h)-f(x) = q_x(x+h)*h

Then we have:

f'(x) = q_x(x)

If you try it with f(x)=sin(x), you get this here:

sin(x+h)-sin(x) = 2 cos(h/2 + x) sin(h/2)

                = cos(((x+h)+x)/2) sin(((x+h)-x)/2)/(((x+h)-x)/2) h

                = q_x(x+h) h

The only continuous function q_x(x+h) that does the job, is the one where lim h->0 sin(((x+h)-x)/2)/(((x+h)-x)/2) = 1 is used. So that we have:

f'(x) = q_x(x) = cos(x) 1 = cos(x)

According to Stephen Kuhn in his intro this lim connection seems to be inevitable. It might be evitable if f(x) is a simple polynomial, where we might regress to some reminder theorem.