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Seek a Frobenius solution $y=z^r{\sum }_{n\ge 0}{a}_n{z}^n$ with $a_0\ne 0$. Since $(z+z^3{)}^2=z^2+2z^4+z^6$, its lowest coefficient gives the indicial equationThus the repeated indicial root is $\boxed{r=-1}$. Equating the coefficient of $z^{n+r}$ before setting $r=-1$ givesFor $n\ge 1$ this becomes the linear recurrence relationEvery coefficient is fixed by $a_0$, since the denominator never vanishes for $n\ge 1$. In particular $a_1=0$ and then every odd coefficient is zero. Therefore there is $\boxed{\text{exactly one independent pure Frobenius series}}$, not two merely because the indicial root repeats. If "power series" means an ordinary Taylor series with nonnegative integer powers, the only such solution is zero: its coefficient equation starts with $a_0=0$ and inductively forces all coefficients to vanish.

A second independent local solution is a logarithmic solution from a repeated Frobenius exponent. Indeed, Abel's identity gives $W=C{z}^{-3}[1+O(z^2)]$. For the normalized first solution $y_1=z^{-1}[1+O(z^2)]$, reduction of order givesAfter choosing the harmless multiplicative constant in the numerator, its integrand is $1/s+O(s)$, so $y_2=y_1\log z+z^{-1}\times \text{an analytic series}$. The nonzero logarithmic coefficient proves that this second solution is not another pure Frobenius solution.

Normalize the free leading coefficient to $a_0=1$. The linear recurrence relation from part (b) givesConsequently the first four nonzero coefficients areand the single pure Frobenius solution isFor clarity about the second independent solution, a choice with no added multiple of $y_1$ isThese companion coefficients follow by reduction of order with the normalized Wronskian $W=z^{-3}(1+z^2{)}^{3/2}\exp [3z^2/(2(1+z^2))]$. The logarithm requires a local branch; it does not supply a second ordinary power series solution.

At zero the indicial equation is $\nu (\nu -1)+\nu ={\nu }^2=0$, with repeated indicial root zero. To exhibit the analytic solution rather than only its candidate exponent, clear the denominator in the normal form:Substitute $y_1={\sum }_{n\ge 0}{b}_n{x}^n$ and equate coefficients. With $b_j=0$ for negative indices, the linear recurrence relation isEvery leading coefficient on the left is nonzero. Hence $b_0\ne 0$ is free and all later coefficients are determined; the Frobenius method guarantees local convergence up to the next singularity. In particular,For a second solution, use reduction of order. Writing $y_2=y_{1}v$ gives $W=y_1^2{v}^{\prime }$, so part (c) yieldsThe additive constant in the integral only adds a multiple of $y_1$. As $x\downarrow 0$ within $(0,m)$, its integrand isThereforeThe nonzero Wronskian proves linear independence, and the logarithm is unavoidable for this second local solution.

In the shifted variable $t=y-1$, substitute the Frobenius solution $x={\sum }_{j\ge 0}{a}_j{t}^{j+\sigma }$ into $t(t+2)x^{\prime \prime }+x^{\prime }+x=0$. Equating the coefficient of $t^{j+\sigma }$ givesThus the linear recurrence relation, valid for $j=0,1,\dots$, isFor the indicial root $\sigma =0$, this becomesFor the indicial root $\sigma =1/2$, rename the coefficients $b_j$; thenWith $a_0=b_0=1$, the first terms of the two Frobenius solutions are

The Fibonacci recurrence, with the supplied initial values, is $F_{n+2}=F_{n+1}+F_n$ for $n\ge 0$, with $F_0=0,F_1=1$. Two initial values are needed for this second-order linear recurrence relation.

Let $\varphi =(1+\sqrt{5})/2$, the golden ratio. Since ${\varphi }^2=\varphi +1$, multiplication by ${\varphi }^n$ gives ${\varphi }^{n+2}={\varphi }^{n+1}+{\varphi }^n$. Adding this recurrence relation to the one for $A_n$ yieldsThus the extra exponential term changes the initial values, but not the linear recurrence relation.

Adding the Fibonacci recurrence and the identical recurrence relation for the Lucas numbers givesLinearity preserves this linear recurrence relation; the initial values come from $0+2$ and $1+1$.

A nonzero root $\mu$ of multiplicity $m$ supplies $m$ independent solutions $n^j{\mu }^n$ to a constant-coefficient linear recurrence relation. Substitution $y_n={\mu }^n{z}_n$ reduces its repeated factor to the $m$th forward difference operator. A zero root needs separate initial-data treatment: for characteristic polynomial ${\lambda }^m$, the recurrence is $y_{n+m}=0$, leaving the first $m$ values arbitrary.

A terminating Frobenius series is a Frobenius method solution whose analytic power series factor has only finitely many nonzero coefficients. Its coefficient linear recurrence relation must allow the first omitted coefficient and all following coefficients to be zero. At an integer separation of characteristic exponents at a regular singular point, a resonant coefficient equation can become $0=0$, permitting such a choice rather than forcing a logarithmic term. With exponent zero, termination produces a polynomial solution.

Consider the real matrixThe Cayley-Hamilton theorem gives $A^2=3A-I$. ConsequentlyThusAll four entries come from just two scalar coefficients. If $A^n=a_{n}A+b_{n}I$, multiplying once by $A$ and replacing $A^2$ gives the linear recurrence relationEquivalently $a_{n+2}=3a_{n+1}-a_n$, with $a_0=0,a_1=1$. The appearance of Fibonacci numbers is explained by $A=Q^2$, where $Q=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ advances the Fibonacci recurrence. No eigenvectors need to be computed.

The same identity computes an inverse. Rearrange it as $A(3I-A)=I$, givingThis works because the constant term of this characteristic polynomial is nonzero and the coefficients lie in a field. Generally, when $c_0\ne 0$ over a field, rearranging the Cayley-Hamilton theorem expresses $A^{-1}$ as a polynomial evaluated at a matrix. Over a general commutative ring, a nonzero constant need not be a unit; invertibility requires that extra condition.

A variable-coefficient affine recurrence is a linear recurrence relation with coefficients allowed to depend on the index. If every $a_n$ is nonzero, a discrete integrating factor reduces it to a telescoping series.