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Let $R_f,R_g$ be the two radii of convergence. First take $|x|<t<R_f$, with $t>0$. Absolute convergence at $t$ bounds the terms: $|a_n{t}^n|\le M$ for all $n$. ThusAt $x=0$ the right-hand series is finite. Otherwise it converges by the ratio test, because its limiting ratio is $|x|/t<1$. So the differentiated power series converges absolutely at every interior point of $f$, proving $R_g\ge R_f$. This proof also covers $R_f=\infty$ by choosing a finite $t$ above any given $|x|$; if $R_f=0$, that inequality is automatic.

Conversely, if $0<|x|<R_g$, absolute convergence givesThe constant term does not affect convergence, and both series converge at zero. Hence $R_f\ge R_g$, including the zero and infinite cases. Therefore $\boxed{R_f=R_g}$. This proof uses coefficient estimates, rather than assuming the termwise differentiation of a power series result whose radius assertion is being established.

The radius of convergence of a power series is the extended real number $R\in [0,\infty ]$ characterized by absolute convergence for $|x|<R$ and divergence for $|x|>R$. Equivalently,For $R=0$, only the central point is guaranteed; for $R=\infty$, every real point is inside. At finite endpoints $x=\pm R$, convergence must be investigated separately and is not determined by $R$.

Since the common radius of convergence is infinite, the termwise differentiation of a power series theorem applies on the entire real line. For the even series, differentiation and the index change $m=n-1$ giveThe odd series similarly gives $g^{\prime }(x)={\sum }_{n=0}^{\infty }(-1{)}^n{x}^{2n}/(2n)!=f(x)$. Thus $\boxed{f^{\prime }=-g,g^{\prime }=f}$ everywhere. By the product rule,The mean value theorem implies that a differentiable function with zero derivative is constant on the real line. At zero, the series give $f(0)=1$ and $g(0)=0$. Consequently $\boxed{f(x{)}^2+g(x{)}^2=1\text{ for every }x\in \mathbb{R}}$. This derives the identity from the power series without assuming a trigonometric identity in advance.

For any fixed real $x\ne 0$, the ratios between consecutive nonzero terms in the absolute-value series areandThe ratio test proves absolute convergence of both power series at every such $x$; convergence at zero is immediate. Both radii of convergence are therefore $\boxed{\infty }$. Applying the test to the nonzero terms avoids the undefined coefficient ratios caused by alternate zero coefficients when these are viewed as full power series in $x$.

The bipartite graph property means that a walk starting in $V_0$ ends there precisely when its length is even. Thus the even-length walk generating function on a bipartite graph is $Z_G^0(x)={\sum }_{k\ge 0}{\sigma }_{2k}{x}^{2k}$. Deleting the last edge of an odd-length self-avoiding walk leaves an even-length one; each such prefix has at most $\Delta$ possible last-edge choices. Hence ${\sigma }_{2k+1}\le \Delta {\sigma }_{2k}$, and for $x\ge 0$,These inequalities also hold with infinite values. Their finite positive prefactor shows that the two power series have the same radius of convergence $1/\mu$; alternatively apply the root formula to the even subsequence.

The equal rooted counts imply that the self-avoiding walk generating function is $Z_G(x)={\sum }_{n\ge 0}{\sigma }_n{x}^n$. The Cauchy-Hadamard theorem identifies the radius of a power series $\sum a_n{x}^n$ as the reciprocal of $\mathrm{limsup}|a_n{|}^{1/n}$. ConsequentlyEquivalently, the root test gives convergence when $x\mu <1$ and divergence when $x\mu >1$. At the positive boundary $x={\mu }^{-1}$, the preceding infimum formula gives ${\sigma }_n{\mu }^{-n}\ge 1$, so this particular series diverges there as well.