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Inspired by this recent question. Starting with the function definitions:

$$C(x) := \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$$ $$S(x) := \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}$$

and the following properties that can be derived from them (in the above-linked thread):

  • $C(0) = 1$, $S(0) = 0$
  • $C(-x) = C(x)$, $S(-x) = -S(x)$
  • $C'(x) = -S'(x)$, $S'(x) = C(x)$
  • $C(x + y) = C(x) C(y) - S(x) S(y)$
  • $S(x + y) = C(x) S(y) + S(x) C(y)$
  • $C(x)^2 + S(x)^2 = 1$

Now, let $h(a, b)$ be the length of the hypotenuse of a right triangle with leg lengths $a$ and $b$ (in standard Euclidean geometry).

How can we use the $C$ and $S$ functions to help prove that $h(a, b) = \sqrt{a^2 + b^2}$?

The identity $C(x)^2 + S(x)^2 = 1$ looks promising, but as presented, it's just an artifact of the observation that $ f(x) = S(x)^2 + C(x)^2$ has the derivative $f'(x) = 2S(x)C(x) + 2C(x)(-S(x)) = 0$, so $f(x)$ has the constant value $S(0)^2 + C(0)^2 = 1$. I haven't connected it with geometry yet.

One observation that I've made is that if $a$, $b$, and $c$ are the side lengths of a right triangle, then multiplying them all by a positive scalar $r$ forms a similar right triangle with side lengths $ra$, $rb$, and $rc$. So:

$$h(ra, rb) = rc = r h(a, b)$$

Let $\epsilon$ be a very small positive number, and let $r = 1 + \frac{\epsilon}{a}$. Then:

$$h((1 + \frac{\epsilon}{a})a, (1 + \frac{\epsilon}{a})b) = (1 + \frac{\epsilon}{a}) h(a, b)$$ $$h(a + \epsilon, b + \frac{b\epsilon}{a}) = h(a, b) + \frac{\epsilon}{a}h(a,b)$$

Let $h_a$ and $h_b$ denote the partial derivatives of $h$. Then:

$$h(a, b) + \epsilon h_a(a, b) + \frac{b\epsilon}{a} h_b(a, b) \approx h(a, b) + \frac{\epsilon}{a}h(a,b)$$ $$\epsilon h_a(a, b) + \frac{b\epsilon}{a} h_b(a, b) \approx \frac{\epsilon}{a}h(a,b)$$ $$a \epsilon h_a(a, b) + b\epsilon h_b(a, b) \approx \epsilon h(a,b)$$

In the limit as $\epsilon \rightarrow 0$, the approximation of using the tangent plane becomes exact. But as long as $\epsilon$ isn't equal to 0, we can divide by it:

$$a h_a(a, b) + b h_b(a, b) = h(a, b)$$

Now, let $\theta \in \mathbb{R}$, and let $a = C(\theta)$ and $b = S(\theta)$.

$$C(\theta) h_a(C(\theta), S(\theta)) + S(\theta) h_b(C(\theta), S(\theta)) = h(C(\theta), S(\theta))$$

I would like to show that $h(C(\theta), S(\theta)) = 1$, but my PDE-solving skills are rusty. Can y'all help?

Eric Wofsey
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Dan
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  • I suppose that in $C(x)$ some of the $n$ should be $(2n)$ – Claude Leibovici May 24 '22 at 06:48
  • There is no PDE solving here. Also you already have $h(C(\theta),S(\theta))^2 = C(\theta)^2 + S(\theta)^2 = 1$. – AHusain May 24 '22 at 13:36
  • If I understand correctly, you would like to prove that $C(x)$ and $S(X)$ are the same as geometrically defined functions $\cos x$ and $\sin x$. But that has already been proved in the linked thread, if I'm not mistaken. Where's the difference? – Intelligenti pauca May 24 '22 at 14:18
  • If $a=c\sum \alpha_n$ and $b=\sum \beta_n$ then $a^2+b^2=c^2((\sum \alpha_i)^2+(\sum \beta_i)^2)$ – Piquito May 24 '22 at 14:30
  • @Intelligentipauca: That question already assumes (based on its area integral and curve parametrization) that the equation of the unit circle is $x^2 + y^2 = 1$, which is related to the PT. Mine does not make the that assumption. – Dan May 24 '22 at 16:30
  • How do you define $h(a,b)$ exactly? It is impossible to prove anything about it without a precise definition. (And I don't see any reason to expect that your $C(x)$ and $S(x)$ functions would be helpful for proving anything about $h$, for any reasonable choice of definition.) – Eric Wofsey Sep 05 '22 at 18:59

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