Deriving $\psi_1(x)=\frac{3\sqrt\pi}{2}\phi_0(x) + \frac{7\sqrt\pi}{4\sqrt2} \phi_1(x) - \frac{\sqrt\pi}{2\sqrt2} \phi_2(x)...$

Question

I tried but I could not get $\psi_1(x)=\frac{3\sqrt\pi}{2}\phi_0(x) + \frac{7\sqrt\pi}{4\sqrt2} \phi_1(x) - \frac{\sqrt\pi}{2\sqrt2} \phi_2(x)...$ from $\psi_1(x) = \cos^3 x + \sin^2 x + \cos x + 1$ using $\phi_0, \phi_1...$

Can any help? Thanks.

Answer 1

To obtain the required decomposition it suffices to sum the left-hand and right-hand sides of the following trigonometric identities:

$$1=\sqrt{\pi}\varphi_0(x),$$ $$\cos x=\frac{\sqrt{\pi}}{\sqrt{2}}\varphi_1(x),$$ $$\sin^2 x =\frac 12(1-\cos 2x)=\frac{\sqrt{\pi}}{2}\varphi_0(x)-\frac {\sqrt{\pi}}{2\sqrt{2}}\varphi_2(x),$$ $$\cos^3 x= \frac 14(\cos 3x+3\cos x)=\frac {\sqrt{\pi}}{4\sqrt{2}}\varphi_3(x)+\frac {3\sqrt{\pi}}{4\sqrt{2}}\varphi_1(x).$$