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Termwise differentiation of the convergent Frobenius series and collection of powers in the Bessel differential equation gives the lowest coefficient $({\sigma }^2-{\nu }^2)a_0$, the next coefficient $[(\sigma +1{)}^2-{\nu }^2]a_1$, and, for $n\ge 2$, the coefficient $[(\sigma +n{)}^2-{\nu }^2]a_n+a_{n-2}$. Since $a_0\ne 0$, the stated choices $\sigma =\pm \nu$, $a_1=0$ and the recurrence make all coefficients vanish.

For $\nu =1/2$, the two resulting even-coefficient series areandThese are linearly independent on $x>0$ and give the Bessel functions of order one half, up to normalization. The specified condition $a_1=0$ is sufficient; in the negative-exponent branch at this particular order the next coefficient equation alone would allow $a_1$ to be free, adding a multiple of the positive-exponent solution.

For the Legendre differential equation, substitution gives the full conditionsThus $\sigma =0$ permits arbitrary $a_1$, while $\sigma =1$ requires $a_1=0$. This Frobenius recurrence for the Legendre equation separates the even and odd series. The nearest singular points of the differential equation are $\pm 1$, so the series about zero has radius at least one, and the generic nonterminating solutions have radius exactly one. Particular cancellations can give a larger radius.

Termination occurs when the degree reaches a nonnegative integer $N$ with $\lambda (\lambda +1)=N(N+1)$. For $\lambda \ge 0$ this givesChoose the even branch for even $N$ or the odd branch for odd $N$ to obtain a Legendre polynomial. The recurrence then terminates and gives a solution for all real $x$, with infinite radius of convergence. A nonzero entire solution at other $\lambda \ge 0$ is impossible: each nonterminating parity has coefficient ratio tending to one in the variable $x^2$, and so radius one.