Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Trigonometric functions are . . . . . . somewhat like exponential functions.

If $f$ is an exponential function, then $\displaystyle \frac{\prod_i f(\theta_i)}{f\left(\sum_i \theta_i\right)} = 1$. So just suppose we had a function $f$ satisfying $\displaystyle \frac{\prod_i f(\theta_i)}{f\left(\sum_i \theta_i\right)} = \text{something well-behaved}$. So we wonder whether $$ \text{something well-behaved} \equiv 1 \pmod{\text{whatever}} $$ where "whatever" is something that can serve as the denominator when forming some kind of quotient structure.

Now let $e_n$ be the $n$th-degree elementary symmetric polynomial in the variables $\tan\theta_i$ for $i\in\text{some at-most-countably-infinite index set}$. As I have observed in some other postings, we have $$ e_0 - e_2 + e_4 - e_6 + \cdots = \frac{\prod_i f(\theta_i)}{f\left(\sum_i \theta_i\right)}\text{ if } f = \sec = \text{the secant function}.\tag{1} $$

So:

  • Is $\sec$ the only (reasonable?) solution to the functional equation $(1)$?
  • What's the quotient structure?

(If I hadn't thought of the trigonometric identity $(1)$, I might never have realized that the series to the left of "$=$" in $(1)$ converges if $\sum_i\theta_i$ converges absolutely. Quite simply: the left side converges because the right side converges.)

share|improve this question

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.