4
$\begingroup$

I am trying to come up with generalized trigonometric functions using the exponential definition that we use today for the trig functions sine and cosine $$\sin x=\frac{e^{ix}-e^{-ix}}{2i}; \cos x =\frac{e^{ix}+e^{-ix}}{2}$$ I was then thinking, in terms of roots of unity, if we define $$\omega_n^k=e^\frac{2ki\pi}{n}$$ Then $$\cos x =\frac{(\omega_4^0)^0e^{\omega_4^1x}+(\omega_4^2)^0e^{\omega_4^3x}}{2\omega_4^0};\sin x =\frac{(\omega_4^0)^1e^{\omega_4^1x}+(\omega_4^2)^1e^{\omega_4^3x}}{2\omega_4^1}$$ Then I could extend as say $$C_0(x)=\frac{(\omega_6^0)^0e^{\omega_6^1x}+(\omega_6^2)^0e^{\omega_6^3x}+(\omega_6^4)^0e^{\omega_6^5x}}{3\omega_6^0}$$ $$C_1(x)=\frac{(\omega_6^0)^1e^{\omega_6^1x}+(\omega_6^2)^1e^{\omega_6^3x}+(\omega_6^4)^1e^{\omega_6^5x}}{3\omega_6^1}$$ $$C_2(x)=\frac{(\omega_6^0)^2e^{\omega_6^1x}+(\omega_6^2)^2e^{\omega_6^3x}+(\omega_6^4)^2e^{\omega_6^5x}}{3\omega_6^2}$$ I've seen generalizations for trig function using the integral definition of the inverses, and i've seen the defined using lacunary power series. WOuld this be another way to define a trig generalization?

$\endgroup$
  • 1
    $\begingroup$ All the expressions in question are combinations of hyperbolic and trigonometric functions. $\endgroup$ – Lucian Aug 1 '15 at 5:04
  • $\begingroup$ Thank you. I've seen how higher order trig functions have been defined using the hyperbolic and trig function combinations. I was just wondering if there has been a generalized symmetric exponential definition $\endgroup$ – Eleven-Eleven Aug 1 '15 at 13:52
  • $\begingroup$ I recently came across this article which did what you asked. It seems that there are some other applications on those 'higher order trig functions' sciencedirect.com/science/article/pii/S0022314X17302172 $\endgroup$ – Kuhn Dec 7 '18 at 21:34

Your Answer

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

Browse other questions tagged or ask your own question.