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Consider the Legendre equation $$(1-z^2)u''-2zu'+v(v+1)u=0.$$ Find the roots of the indicial polynomial if we apply the Frobenius method about $z=1$.

My attempt:

Let \begin{align}u=\sum_{k=0}^{\infty} A_k(z-1)^{k+r}&\implies u'=\sum_{k=0}^{\infty} (k+r) A_k(z-1)^{k+r-1}\\ &\implies u''=(k+r)(k+r-1)\sum_{k=0}^{\infty} A_k(z-1)^{k+r-2} \end{align} Substituting this into the Legendre equation, I get: $$\sum_{k=-2}^{\infty} (k+r+1)(k+r+2)A_{k+2}(z-1)^{k+r}+\sum_{k=0}^{\infty} \left((v^2+v)-(k+r)(k+r-1)-2(k+r)\right)A_k(z-1)^{k+r}=0$$

Now what do I do?

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I have found the roots of the indicial polynomial are $r=0, 1$, which contradicts the answer provided, which states that $0$ is a repeated root. Any advice would be greatly appreciated.

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Using $z=1+x$ the equation reads $$ -x(x+2)u''-2(x+1)u'+ν(ν+1)u=0 $$ so that in inserting the power series for $u$ around $x=0$ the coefficient equation for power $k+r$ is $$ -(k+r)(k+r-1)A_k-2(k+r+1)(k+r)A_{k+1}-2(k+r)A_k-2(k+r+1)A_{k+1}+ν(ν+1)A_k=0 \\\iff\\ -2(k+r+1)^2A_{k+1}+[-(k+r)(k+r+1)+ν(ν+1)]A_k=0 $$ so that at $k=-1$ with formally $A_{-1}=0$ you get the indicial equation $$-2r^2=0,$$ so that indeed one gets a double root at $r=0$.


You did something wrong when applying the polynomial factors to the derivatives.

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$$(1-z^2)u''-2zu'+v(v+1)u=0$$ $$u''-\dfrac{2z}{1-z^2}u'+\dfrac{v(v+1)}{1-z^2}u=0$$ with $p(z)=-\dfrac{2z}{1-z^2}$ and $q(z)=\dfrac{v(v+1)}{1-z^2}$, since two points $z=\pm1$ are singularity, so one may find a solution about $z=0$ in $|z|<1$. Hence about $z=1$ $$p_0=\lim_{z\to1}(z-1)p(z)=1 ~\text{ and }~ q_0=\lim_{z\to1}(z-1)^2q(z)=0$$ and indicial equation is $$m^2=0$$ from $m(m-1)+p_0m+q_0=0$.

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