
# 3.6: The Method of Frobenius II


In this section we discuss a method for finding two linearly independent Frobenius solutions of a homogeneous linear second order equation near a regular singular point in the case where the indicial equation has a repeated real root. As in the preceding section, we consider equations that can be written as

\label{eq:3.6.1}
x^2(\alpha_0+\alpha_1x+\alpha_2x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y' +(\gamma_0+\gamma_1x+\gamma_2x^2)y=0

where $$\alpha_0\ne0$$. We assume that the indicial equation $$p_0(r)=0$$ has a repeated real root $$r_1$$. In this case Theorem $$(3.5.3)$$ implies that \eqref{eq:3.6.1} has one solution of the form

\begin{eqnarray*}
y_1=x^{r_1}\sum_{n=0}^\infty a_nx^n,
\end{eqnarray*}

but does not provide a second solution $$y_2$$ such that $$\{y_1,y_2\}$$ is a fundamental set of solutions. The following extension of Theorem $$(3.5.2)$$ provides a way to find a second solution.

## Theorem $$\PageIndex{1}$$

Let

\label{eq:3.6.2}
Ly= x^2(\alpha_0+\alpha_1x+\alpha_2x^2)y''+x(\beta_0+\beta_1x+\beta_2x^2)y' +(\gamma_0+\gamma_1x+\gamma_2x^2)y,

where $$\alpha_0\ne0$$ and define

\begin{eqnarray*}
p_0(r)&=&\alpha_0r(r-1)+\beta_0r+\gamma_0,\\
p_1(r)&=&\alpha_1r(r-1)+\beta_1r+\gamma_1,\\
p_2(r)&=&\alpha_2r(r-1)+\beta_2r+\gamma_2.\\
\end{eqnarray*}

Suppose $$r$$ is a real number such that $$p_0(n+r)$$ is nonzero for all positive integers $$n$$, and define

\begin{eqnarray*}
\begin{array}{ccl}
a_0(r)&=&1,\\
a_1(r)&=&-\displaystyle{p_1(r)\over p_0(r+1)},\\
\end{array}
\end{eqnarray*}

Then the Frobenius series

\label{eq:3.6.3}
y(x,r)=x^r\sum_{n=0}^\infty a_n(r)x^n

satisfies

\label{eq:3.6.4}
Ly(x,r)=p_0(r)x^r.

Moreover,

\label{eq:3.6.5}
{\partial y\over \partial r}(x,r)=y(x,r)\ln x+x^r\sum_{n=1}^\infty a_n'(r) x^n,

and

\label{eq:3.6.6}
L\left({\partial y\over \partial r}(x,r)\right)=p'_0(r)x^r+x^rp_0(r)\ln x.

Proof

Theorem $$(3.5.2)$$ implies \eqref{eq:3.6.4}. Differentiating formally with respect to $$r$$ in \eqref{eq:3.6.3} yields

\begin{eqnarray*}
{\partial y\over \partial r}(x,r)&=&\displaystyle{{\partial\over\partial r}(x^r)\sum_{n=0}^\infty a_n(r)x^n +x^r\sum_{n=1}^\infty a_n'(r)x^n}\\
&=&\displaystyle{x^r\ln x\sum_{n=0}^\infty a_n(r)x^n +x^r\sum_{n=1}^\infty a_n'(r)x^n}\\
&=&y(x,r) \ln x + x^r\sum_{n=1}^\infty a_n'(r)x^n,
\end{eqnarray*}

which proves \eqref{eq:3.6.5}.

To prove that $$\partial y(x,r)/\partial r$$ satisfies \eqref{eq:3.6.6}, we view $$y$$ in \eqref{eq:3.6.2} as a function $$y=y(x,r)$$ of two variables, where the prime indicates partial differentiation with respect to $$x$$; thus,

\begin{eqnarray*}
\end{eqnarray*}

With this notation we can use \eqref{eq:3.6.2} to rewrite \eqref{eq:3.6.4} as

\label{eq:3.6.7}
x^2q_0(x){\partial^2 y\over \partial x^2}(x,r)+xq_1(x){\partial y\over \partial x}(x,r)+q_2(x)y(x,r)=p_0(r)x^r,

where

\begin{eqnarray*}
q_0(x)&=&\alpha_0+\alpha_1x+\alpha_2x^2,\\
q_1(x)&=&\beta_0+\beta_1x+\beta_2x^2,\\
q_2(x)&=&\gamma_0+\gamma_1x+\gamma_2x^2.\\
\end{eqnarray*}

Differentiating both sides of \eqref{eq:3.6.7} with respect to $$r$$ yields

\begin{eqnarray*}
x^2q_0(x){\partial^3y\over \partial r\partial x^2}(x,r)+ xq_1(x){\partial^2y\over \partial r\partial x}(x,r)+q_2(x){\partial y\over\partial r}(x,r)=p'_0(r)x^r+p_0(r) x^r \ln x.
\end{eqnarray*}

By changing the order of differentiation in the first two terms on the left we can rewrite this as

\begin{eqnarray*}
x^2q_0(x){\partial^3 y\over \partial x^2\partial r}(x,r) +xq_1(x){\partial^2 y\over \partial x\partial r}(x,r)+q_2(x){\partial y\over \partial r}(x,r)=p'_0(r)x^r+p_0(r) x^r \ln x,
\end{eqnarray*}

or

\begin{eqnarray*}
x^2q_0(x){\partial^2\over \partial x^2} \left({\partial y\over\partial r}(x,r)\right) +xq_1(x){\partial\over\partial r}\left({\partial y\over\partial x}(x,r)\right) +q_2(x){\partial y\over\partial r}(x,r)= p'_0(r)x^r+p_0(r) x^r \ln x,
\end{eqnarray*}

which is equivalent to \eqref{eq:3.6.6}.

## Theorem $$\PageIndex{2}$$

Let $$L$$ be as in Theorem $$(3.6.1)$$ and suppose the indicial equation $$p_0(r)=0$$ has a repeated real root $$r_1.$$ Then

\begin{eqnarray*}
y_1(x)=y(x,r_1)=x^{r_1}\sum_{n=0}^\infty a_n(r_1)x^n
\end{eqnarray*}

and

\label{eq:3.6.8}
y_2(x)={\partial y\over\partial r}(x,r_1)=y_1(x)\ln x+x^{r_1}\sum_{n=1}^\infty a_n'(r_1)x^n

form a fundamental set of solutions of $$Ly=0.$$

Proof

Since $$r_1$$ is a repeated root of $$p_0(r)=0$$, the indicial polynomial can be factored as

\begin{eqnarray*}
p_0(r)=\alpha_0(r-r_1)^2,
\end{eqnarray*}

so

\begin{eqnarray*}
p_0(n+r_1)=\alpha_0n^2,
\end{eqnarray*}

which is nonzero if $$n>0$$. Therefore the assumptions of Theorem $$(3.6.1)$$ hold with $$r=r_1$$, and \eqref{eq:3.6.4} implies that $$Ly_1=p_0(r_1)x^{r_1}=0$$. Since

\begin{eqnarray*}
p_0'(r)=2\alpha(r-r_1)
\end{eqnarray*}

it follows that $$p_0'(r_1)=0$$, so \eqref{eq:3.6.6} implies that

\begin{eqnarray*}
Ly_2=p_0'(r_1)x^{r_1}+x^{r_1}p_0(r_1)\ln x=0.
\end{eqnarray*}

This proves that $$y_1$$ and $$y_2$$ are both solutions of $$Ly=0$$. We leave the proof that $$\{y_1,y_2\}$$ is a fundamental set as an exercise (Exercise $$(3.6E.53)$$).

## Example $$\PageIndex{1}$$

Find a fundamental set of solutions of

\label{eq:3.6.9}
x^2(1-2x+x^2)y''-x(3+x)y'+(4+x)y=0.

Compute just the terms involving $$x^{n+r_1}$$, where $$0\le n\le4$$ and $$r_1$$ is the oot of the indicial equation.

For the given equation, the polynomials defined in Theorem $$(3.6.1)$$ are

\begin{eqnarray*}
\begin{array}{lllll}
p_0(r)&=&r(r-1)-3r+4&=&(r-2)^2,\\
p_1(r)&=&-2r(r-1)-r+1&=&-(r-1)(2r+1),\\
p_2(r)&=&r(r-1).
\end{array}
\end{eqnarray*}

Since $$r_1=2$$ is a repeated root of the indicial polynomial $$p_0$$, Theorem $$(3.6.2)$$ implies that

\label{eq:3.6.10}

form a fundamental set of Frobenius solutions of \eqref{eq:3.6.9}. To find the coefficients in these series, we use the recurrence formulas from Theorem $$(3.6.1)$$:

\label{eq:3.6.11}
\begin{array}{ccl}
a_0(r)&=&1,\\
a_1(r)&=&-\displaystyle{p_1(r)\over p_0(r+1)} =-\displaystyle{(r-1)(2r+1)\over(r-1)^2} =\displaystyle{2r+1\over r-1},\\
a_n(r)&=&-\displaystyle{p_1(n+r-1)a_{n-1}(r)+p_2(n+r-2)a_{n-2}(r)\over p_0(n+r)}\\
&=&\displaystyle{(n+r-2)\left[(2n+2r-1)a_{n-1}(r) -(n+r-3)a_{n-2}(r)\right]\over(n+r-2)^2}\\
&=&\displaystyle{{(2n+2r-1)\over(n+r-2)}a_{n-1}(r)- {(n+r-3)\over(n+r-2)}a_{n-2}(r)},n\ge2.
\end{array}

Differentiating yields

\label{eq:3.6.12}
\begin{array}{ccl}
a'_1(r)&=&-\displaystyle{3\over (r-1)^2},\\
a'_n(r)&=&\displaystyle{{2n+2r-1\over n+r-2}a'_{n-1}(r)-{n+r-3\over n+r-2}a'_{n-2}(r)}\\
\end{array}

Setting $$r=2$$ in \eqref{eq:3.6.11} and \eqref{eq:3.6.12} yields

\label{eq:3.6.13}
\begin{array}{ccl}
a_0(2)&=&1,\\
a_1(2)&=&5,\\
\end{array}

and

\label{eq:3.6.14}
\begin{array}{ccl}
a_1'(2)&=&-3,\\
a'_n(2)&=&\displaystyle{{2n+3\over n}a'_{n-1}(2)-{n-1\over n}a'_{n-2}(2) -{3\over n^2}a_{n-1}(2)-{1\over n^2}a_{n-2}(2)},\quad n\ge2.
\end{array}

Computing recursively with \eqref{eq:3.6.13} and \eqref{eq:3.6.14} yields

\begin{eqnarray*}
a_0(2)=1,\,a_1(2)=5,\,a_2(2)=17,\,a_3(2)={143\over3},\,a_4(2)={355\over3},
\end{eqnarray*}

and

\begin{eqnarray*}
a_1'(2)=-3,\,a_2'(2)=-{29\over2},\,a_3'(2)=-{859\over18}, \,a_4'(2)=-{4693\over36}.
\end{eqnarray*}

Substituting these coefficients into \eqref{eq:3.6.10} yields

\begin{eqnarray*}
y_1=x^2\left(1+5x+17x^2+{143\over3}x^3 +{355\over3}x^4+\cdots\right)
\end{eqnarray*}

and

\begin{eqnarray*}
y_2=y_1 \ln x -x^3\left(3+{29\over2}x+{859\over18}x^2+{4693\over36}x^3 +\cdots\right).
\end{eqnarray*}

Since the recurrence formula \eqref{eq:3.6.11} involves three terms, it's not possible to obtain a simple explicit formula for the coefficients in the Frobenius solutions of \eqref{eq:3.6.9}. However, as we saw in the preceding sections, the recurrence formula for $$\{a_n(r)\}$$ involves only two terms if either $$\alpha_1=\beta_1=\gamma_1=0$$ or $$\alpha_2=\beta_2=\gamma_2=0$$ in \eqref{eq:3.6.1}. In this case, it's often possible to find explicit formulas for the coefficients. The next two examples illustrate this.

## Example $$\PageIndex{2}$$

Find a fundamental set of Frobenius solutions of

\label{eq:3.6.15}
2x^2(2+x)y''+5x^2y'+(1+x)y=0.

Give explicit formulas for the coefficients in the solutions.

For the given equation, the polynomials defined in Theorem $$(3.6.1)$$ are

\begin{eqnarray*}
\begin{array}{ccccc}
p_0(r)&=&4r(r-1)+1&=&(2r-1)^2,\\
p_1(r)&=&2r(r-1)+5r+1&=&(r+1)(2r+1),\\
p_2(r)&=&0.
\end{array}
\end{eqnarray*}

Since $$r_1=1/2$$ is a repeated zero of the indicial polynomial $$p_0$$, Theorem $$(3.6.2)$$ implies that

\label{eq:3.6.16}
y_1=x^{1/2}\sum_{n=0}^\infty a_n(1/2)x^n

and

\label{eq:3.6.17}
y_2=y_1\ln x+x^{1/2}\sum_{n=1}^\infty a_n'(1/2)x^n

form a fundamental set of Frobenius solutions of \eqref{eq:3.6.15}. Since $$p_2\equiv0$$, the recurrence formulas in Theorem $$(3.6.1)$$ reduce to

\begin{eqnarray*}
\begin{array}{ccl}
a_0(r)&=&1,\\
a_n(r)&=&-\displaystyle{p_1(n+r-1)\over p_0(n+r)}a_{n-1}(r),\\
&=&-\displaystyle{(n+r)(2n+2r-1)\over(2n+2r-1)^2}a_{n-1}(r),\\
\end{array}
\end{eqnarray*}

We leave it to you to show that

\label{eq:3.6.18}

Setting $$r=1/2$$ yields

\label{eq:3.6.19}
\begin{array}{ccl}
a_n(1/2)&=&(-1)^n\displaystyle\prod_{j=1}^n\displaystyle{j+1/2\over2j}= (-1)^n\prod_{j=1}^n\displaystyle{2j+1\over4j},\\
\end{array}

Substituting this into \eqref{eq:3.6.16} yields

\begin{eqnarray*}
y_1=x^{1/2}\sum_{n=0}^\infty{(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!}x^n.
\end{eqnarray*}

To obtain $$y_2$$ in \eqref{eq:3.6.17}, we must compute $$a_n'(1/2)$$ for $$n=1$$, $$2$$, $$\dots$$. We'll do this by logarithmic differentiation. From \eqref{eq:3.6.18},

\begin{eqnarray*}
\end{eqnarray*}

Therefore

\begin{eqnarray*}
\ln |a_n(r)|=\sum^n_{j=1} \left(\ln |j+r|-\ln|2j+2r-1|\right).
\end{eqnarray*}

Differentiating with respect to $$r$$ yields

\begin{eqnarray*}
{a'_n(r)\over a_n(r)}=\sum^n_{j=1} \left({1\over j+r}-{2\over2j+2r-1}\right).
\end{eqnarray*}

Therefore

\begin{eqnarray*}
a'_n(r)=a_n(r) \sum^n_{j=1} \left({1\over j+r}-{2\over2j+2r-1}\right).
\end{eqnarray*}

Setting $$r=1/2$$ here and recalling \eqref{eq:3.6.19} yields

\label{eq:3.6.20}
a'_n(1/2)={(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!}\left(\sum_{j=1}^n{1\over j+1/2}-\sum_{j=1}^n{1\over j}\right).

Since

\begin{eqnarray*}
{1\over j+1/2}-{1\over j}={j-j-1/2\over j(j+1/2)}=-{1\over j(2j+1)},
\end{eqnarray*}

\eqref{eq:3.6.20} can be rewritten as

\begin{eqnarray*}
a'_n(1/2)=-{(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!} \sum_{j=1}^n{1\over j(2j+1)}.
\end{eqnarray*}

Therefore, from \eqref{eq:3.6.17},

\begin{eqnarray*}
y_2=y_1\ln x-x^{1/2}\sum_{n=1}^\infty{(-1)^n\prod_{j=1}^n(2j+1)\over4^nn!} \left(\sum_{j=1}^n{1\over j(2j+1)}\right)x^n.
\end{eqnarray*}

## Example $$\PageIndex{3}$$

Find a fundamental set of Frobenius solutions of

\label{eq:3.6.21}
x^2(2-x^2)y''-2x(1+2x^2)y'+(2-2x^2)y=0.

Give explicit formulas for the coefficients in the solutions.

For \eqref{eq:3.6.21}, the polynomials defined in Theorem $$(3.6.1)$$ are

\begin{eqnarray*}
\begin{array}{ccccc}
p_0(r)&=&2r(r-1)-2r+2&=&2(r-1)^2,\\
p_1(r)&=&0,\\
p_2(r)&=&-r(r-1)-4r-2&=&-(r+1)(r+2).
\end{array}
\end{eqnarray*}

As in Section $$3.5$$, since $$p_1\equiv0$$, the recurrence formulas of Theorem $$(3.6.1)$$ imply that $$a_n(r)=0$$ if $$n$$ is odd, and

\begin{eqnarray*}
\begin{array}{ccl}
a_0(r)&=&1,\\
a_{2m}(r)&=&-\displaystyle{p_2(2m+r-2)\over p_0(2m+r)}a_{2m-2}(r)\\
&=&\displaystyle{(2m+r-1)(2m+r)\over2(2m+r-1)^2}a_{2m-2}(r)\\
\end{array}
\end{eqnarray*}

Since $$r_1=1$$ is a repeated root of the indicial polynomial $$p_0$$, Theorem $$(3.6.2)$$ implies that

\label{eq:3.6.22}
y_1=x\sum_{m=0}^\infty a_{2m}(1)x^{2m}

and

\label{eq:3.6.23}
y_2=y_1\ln x+x\sum_{m=1}^\infty a'_{2m}(1)x^{2m}

form a fundamental set of Frobenius solutions of \eqref{eq:3.6.21}. We leave it to you to show that

\label{eq:3.6.24}
a_{2m}(r)={1\over2^m}\prod_{j=1}^m{2j+r\over2j+r-1}.

Setting $$r=1$$ yields

\label{eq:3.6.25}
a_{2m}(1)={1\over2^m}\prod_{j=1}^m{2j+1\over2j} ={\prod_{j=1}^m(2j+1)\over4^mm!},

and substituting this into \eqref{eq:3.6.22} yields

\begin{eqnarray*}
y_1=x\sum_{m=0}^\infty{\prod_{j=1}^m(2j+1)\over4^mm!}x^{2m}.
\end{eqnarray*}

To obtain $$y_2$$ in \eqref{eq:3.6.23}, we must compute $$a_{2m}'(1)$$ for $$m=1$$, $$2$$, $$\dots$$. Again we use logarithmic differentiation. From \eqref{eq:3.6.24},

\begin{eqnarray*}
|a_{2m}(r)|={1\over2^m}\prod_{j=1}^m{|2j+r|\over|2j+r-1|}.
\end{eqnarray*}

Taking logarithms yields

\begin{eqnarray*}
\ln |a_{2m}(r)|=-m\ln2+ \sum^m_{j=1} \left(\ln |2j+r|-\ln|2j+r-1|\right).
\end{eqnarray*}

Differentiating with respect to $$r$$ yields

\begin{eqnarray*}
{a'_{2m}(r)\over a_{2m}(r)}=\sum^m_{j=1} \left({1\over 2j+r}-{1\over2j+r-1}\right).
\end{eqnarray*}

Therefore

\begin{eqnarray*}
a'_{2m}(r)=a_{2m}(r) \sum^m_{j=1} \left({1\over 2j+r}-{1\over2j+r-1}\right).
\end{eqnarray*}

Setting $$r=1$$ and recalling \eqref{eq:3.6.25} yields

\label{eq:3.6.26}
a'_{2m}(1)=\displaystyle{{\prod_{j=1}^m(2j+1)\over4^mm!} \sum_{j=1}^m\left({1\over2j+1}-{1\over2j}\right)}.

Since

\begin{eqnarray*}
{1\over2j+1}-{1\over2j}=-{1\over2j(2j+1)},
\end{eqnarray*}

\eqref{eq:3.6.26} can be rewritten as

\begin{eqnarray*}
a_{2m}'(1) = -\displaystyle{{\prod_{j=1}^m(2j+1)\over2 \cdot 4^mm!} \sum_{j=1}^m \left({1 \over (2j+1)}\right)}.
\end{eqnarray*}

Substituting this into \eqref{eq:3.6.23} yields

\begin{eqnarray*}
y_2=y_1\ln x-{x\over2}\sum_{m=1}^\infty{\prod_{j=1}^m(2j+1)\over4^mm!} \left(\sum_{j=1}^m{1\over j(2j+1)}\right)x^{2m}.
\end{eqnarray*}

If the solution $$y_1=y(x,r_1)$$ of $$Ly=0$$ reduces to a finite sum, then there's a difficulty in using logarithmic differentiation to obtain the coefficients $$\{a_n'(r_1)\}$$ in the second solution. The next example illustrates this difficulty and shows how to overcome it.

## Example $$\PageIndex{4}$$

Find a fundamental set of Frobenius solutions of

\label{eq:3.6.27}
x^2y''-x(5-x)y'+(9-4x)y=0.

Give explicit formulas for the coefficients in the solutions.

For \eqref{eq:3.6.27} the polynomials defined in Theorem $$(3.6.1)$$ are

\begin{eqnarray*}
\begin{array}{ccccc}
p_0(r)&=&r(r-1)-5r+9&=&(r-3)^2,\\
p_1(r)&=&r-4,\\
p_2(r)&=&0.
\end{array}
\end{eqnarray*}

Since $$r_1=3$$ is a repeated zero of the indicial polynomial $$p_0$$, Theorem $$(3.6.2)$$ implies that

\label{eq:3.6.28}
y_1=x^3\sum_{n=0}^\infty a_n(3)x^n

and

\label{eq:3.6.29}
y_2=y_1\ln x+x^3\sum_{n=1}^\infty a_n'(3)x^n

are linearly independent Frobenius solutions of \eqref{eq:3.6.27}. To find the coefficients in \eqref{eq:3.6.28} we use the recurrence formulas

\begin{eqnarray*}
\begin{array}{ccl}
a_0(r)&=&1,\\
a_n(r)&=&-\displaystyle{p_1(n+r-1)\over p_0(n+r)}a_{n-1}(r)\\
\end{array}
\end{eqnarray*}

We leave it to you to show that

\label{eq:3.6.30}
a_n(r)=(-1)^n\prod_{j=1}^n{j+r-5\over(j+r-3)^2}.

Setting $$r=3$$ here yields

\begin{eqnarray*}
a_n(3)=(-1)^n\prod_{j=1}^n{j-2\over j^2},
\end{eqnarray*}

so $$a_1(3)=1$$ and $$a_n(3)=0$$ if $$n\ge2$$. Substituting these coefficients into \eqref{eq:3.6.28} yields

\begin{eqnarray*}
y_1=x^3(1+x).
\end{eqnarray*}

To obtain $$y_2$$ in \eqref{eq:3.6.29} we must compute $$a_n'(3)$$ for $$n=1$$, $$2$$, $$\dots$$. Let's first try logarithmic differentiation. From \eqref{eq:3.6.30},

\begin{eqnarray*}
\end{eqnarray*}

so

\begin{eqnarray*}
\ln |a_n(r)|=\sum^n_{j=1} \left(\ln |j+r-5|-2\ln|j+r-3|\right).
\end{eqnarray*}

Differentiating with respect to $$r$$ yields

\begin{eqnarray*}
{a'_n(r)\over a_n(r)}=\sum^n_{j=1} \left({1\over j+r-5}-{2\over j+r-3}\right).
\end{eqnarray*}

Therefore

\label{eq:3.6.31}
a'_n(r)=a_n(r) \sum^n_{j=1} \left({1\over j+r-5}-{2\over j+r-3}\right).

However, we can't simply set $$r=3$$ here if $$n\ge2$$, since the bracketed expression in the sum corresponding to $$j=2$$ contains the term $$1/(r-3)$$. In fact, since $$a_n(3)=0$$ for $$n\ge2$$, the formula \eqref{eq:3.6.31} for $$a_n'(r)$$ is actually an indeterminate form at $$r=3$$.

We overcome this difficulty as follows. From \eqref{eq:3.6.30} with $$n=1$$,

\begin{eqnarray*}
a_1(r)=-{r-4\over (r-2)^2}.
\end{eqnarray*}

Therefore

\begin{eqnarray*}
a_1'(r)={r-6\over(r-2)^3},
\end{eqnarray*}

so

\label{eq:3.6.32}
a_1'(3)=-3.

From \eqref{eq:3.6.30} with $$n\ge2$$,

\begin{eqnarray*}
a_n(r)=(-1)^n (r-4)(r-3)\,{\prod_{j=3}^n(j+r-5)\over\prod_{j=1}^n(j+r-3)^2} =(r-3)c_n(r),
\end{eqnarray*}

where

\begin{eqnarray*}
\end{eqnarray*}

Therefore

\begin{eqnarray*}
which implies that $$a_n'(3)=c_n(3)$$ if $$n\ge3$$. We leave it to you to verify that