How to represent a sum of exponentials as one exponential

Question

Let us say, we have a sum of exponentials, such as: $$ a (e^{iwt}+e^{-iwt})+b (e^{i2wt}-e^{-i2wt})+c (e^{i3wt}+e^{-i3wt}). $$ Is there a way that represents the above sum as a sum of two exponentials with exponents that only differ in sign, i.e., $$ a (e^{iwt}+e^{-iwt})+b (e^{i2wt}-e^{-i2wt})+c (e^{i3wt}+e^{-i3wt})\stackrel{!}{=}A e^{ip(t)}+B e^{-ip(t)}\, ? $$ I know, one can write the pairwise sums in terms of sin and cos, in that case the question would be, if one can find for any combination of oscillations a single trigonometric function that contains the combinations in its argument?

Answer 1

We have that

$$a (e^{iwt}+e^{-iwt})+b (e^{i2wt}-e^{-i2wt})+c (e^{i3wt}+e^{-i3wt})=$$$$=2a \cos (wt)+2ib \sin (2wt)+2c \cos (3wt)=\ldots$$

then by trigonometric identities

• $\cos (3wt)=4\cos^3 (wt)-3\cos (wt)$
• $\sin (2wt)=2\cos (wt)\sin (wt)$

therefore

$$\ldots=2a \cos (wt)+4ib\cos (wt)\sin (wt)+8c\cos^3 (wt)-6c\cos (wt)=$$

$$=(2a-6c)\cos (wt)+8c\cos^3 (wt)+4ib\cos (wt)\sin (wt)=$$

$$=(a-3c)(e^{iwt}+e^{-iwt})+c(e^{iwt}+e^{-iwt})^3+b(e^{iwt}+e^{-iwt})(e^{iwt}-e^{-iwt})$$

that is an expression in term of $e^{iwt}$.