Theoretical Procedure for Power Series Equation:

Question

If I have the following equation: \begin{equation}2c_0(x-1)+\sum_{k=2}^\infty[(c_{k-2}+2c_{k-1})(x-1)^k]+\sum_{k=0}^\infty[(c_{k+2}(k+2)(k+1)+kc_k+(k+1)c_{k+1}+c_k)(x-1)^k]=0 \end{equation} I was wondering if it would reasonable that I can do the following: \begin{align}c_0&=0\\c_{k-2}+2c_{k+2}&=0\\ c_{k+2}(k+2)(k+1)+kc_k+(k+1)c_{k+1}+c_k&=0 \end{align} Or do I need to combine the two summations first?

Answer 1

From the series we set $k=1$ and $k=0$

$$\sum_{k=0}^\infty[(c_{k+2}(k+2)(k+1)+kc_k+(k+1)c_{k+1}+c_k)(x-1)^k] $$ For $k=1$ we have the coefficient for $x-1$ $$(x-1)(6c_3+2c_1+2c_2)$$ You have to add that to the term $2c_0(x-1)$. So that: $$(x-1)(6c_3+2c_1+2c_2+2c_0)=0$$ $$ (6c_3+2c_1+2c_2+2c_0)=0$$ $$\implies 3c_3+c_1+c_2+c_0=0$$

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$$\sum_{k=0}^\infty[(c_{k+2}(k+2)(k+1)+kc_k+(k+1)c_{k+1}+c_k)(x-1)^k] $$ For $k=0$ $$2c_2+c_1+c_0=0$$

To summarize you have two equalities for the coefficients. You can easily deduce that: $$ 3c_3+c_1+c_2+c_0=0$$ $$ 3c_3-2c_2+c_2=0$$ $$c_2=3c_3$$ These are your equalities: $$ \begin{cases} c_2=3c_3 \\ 2c_2+c_1+c_0=0 \end{cases} $$ And the series: $$\sum_{k=2}^\infty[(c_{k-2}+2c_{k-1}+c_{k+2}(k+2)(k+1)+(k+1)c_{k+1}+c_k(k+1))(x-1)^k]=0 $$