Level 40

$$\begin{align} j_{40A}(\tau) &= \sqrt{j_{20B}(\tau)} = \sqrt[4]{j_{10A}(2\tau)}\\ j_{40B}(\tau) &= \sqrt{j_{20A}(2\tau)}= \left(\frac{d_4^2\, d_{20}^2}{d_2\,  d_8\, d_{10}\, d_{40}}\right)^2\\ &= j_{40C}(\tau)+\frac1{j_{40C}(\tau)}\\ j_{40C}(\tau) &= \sqrt{j_{20F}(2\tau)}= \left(\frac{d_8\, d_{10}}{d_2\,  d_{40}}\right) \end{align}$$
There is no linear relation between these functions. $j_{40A}$ is one of the few that involves a 4th root.