= Solution
At fixed gradient and magnet length, $f\propto p$. If $E$ is total <energy>,
$$
\frac{d\ln p}{d\ln E}=\frac{E^2}{E^2-m_e^2c^4}=\frac1{\beta_0^2},\qquad
\boxed{f(E)=f_0\left(1+\frac\delta{\beta_0^2}\right)+O(\delta^2).}
$$
Here $\beta_0=v_0/c$, not one of the <Courant-Snyder parameters>. For ultrarelativistic electrons this becomes $f\simeq f_0(1+\delta)$, or $1/f\simeq(1-\delta)/f_0$: higher-energy particles are bent and focused less strongly. If the symbol $E$ instead denotes <kinetic energy> $K$, the coefficient is $(K+mc^2)/(K+2mc^2)$, approaching $1/2$ in the nonrelativistic limit. Stating the energy convention avoids applying the ultrarelativistic result to a slow beam.
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