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This may not be a very good question, but I'm totally stumped.

I need to know the power series representation of $x$, or if there even is one.

I'll show you why:

I am trying to solve $y''+2xy'-y=x$ using power series.

I know that $y''=\displaystyle\sum\limits_{n=2}^\infty a_n n(n-1)x^{n-2}$, $y'=\displaystyle\sum\limits_{n=1}^\infty a_n n x^{n-1}$, and $y=\displaystyle\sum\limits_{n=0}^\infty a_n x^n$, but I don't know what the power series representation of $x$ is.

I can solve the homogeneous equation no problem by setting $y''+2xy'-y=0$, but I do not feel this is correct.

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3 Answers

up vote 9 down vote accepted

$$ x=0+x+0\cdot x^2+0\cdot x^3+\dots =\sum_{n=0}^\infty a_nx^n $$ where $$ a_1=1,\quad a_n=0\text{ if }n\ne1. $$

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You seem to be construing the subject line literally rather than reading the body of the question. –  Michael Hardy Nov 30 '12 at 16:56
    
In other words, I knew it the whole time.. What about the power series representation of a constant? Say 1? –  ground.clouds1 Nov 30 '12 at 17:09
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$1=1+0\cdot x+0\cdot x^2+\dots$ –  Julián Aguirre Nov 30 '12 at 17:12
    
So you just leave it as 1? There is no power series representation? Sorry, its been so long since i've done power series –  ground.clouds1 Nov 30 '12 at 17:14
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The power series representation is just $1=1$. The coefficients are all equal to $0$ except for he first one. You can see this using the formula for the coefficients in terms of the derivatives. More generally, any polynomial ($1$ and $x$ are polynomials) has a power series expansion which is identical to the polynomial, and has only a finite number of coefficients different from $0$. –  Julián Aguirre Nov 30 '12 at 17:19
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Note that $y''$ and $y$ both have a degree zero term in their series expansions, but that $2xy'$ does not. If you put your various power series together, then, you'll have $$\begin{align}y''+2xy'-y &= 2a_2-a_0+\sum_{n=3}^\infty a_nn(n-1)x^{n-2}+\sum_{n=1}^\infty 2a_nnx^n-\sum_{n=1}^\infty a_nx^n\\ &= 2a_2-a_0+\sum_{n=1}^\infty a_{n+2}(n+2)(n+1)x^n+\sum_{n=1}^\infty a_n(2n-1)x^n\\ &= 2a_2-a_0+\sum_{n=1}^\infty\bigl[a_{n+2}(n+2)(n+1)+a_n(2n-1)\bigr]x^n\\ &=\sum_{n=0}^\infty b_nx^n,\end{align}$$ where $\sum b_nx^n$ is the power series expansion of $x$. But this means that $b_n=0$ for $n\neq 1$ and $b_1=1$. Thus, what you need is: $$\begin{cases}0=2a_2-a_0 & \\1=6a_3+a_1 & \\0=a_{n+2}(n+2)(n+1)+a_n(2n-1) & \text{for all }n\geq 2\end{cases}$$

Hopefully that's enough to get you moving in the right direction for a general solution. You may even be able to get a closed (non-recursive) form for your $a_n$ in terms of $a_0$ (for even $n$) or $a_1$ (for odd $n$).

Alternately, you could use your general solution to the corresponding homogeneous equation, and use this (or some other) method to find a particular solution to $y''+2xy'-y=x$, then add the two to get your general nonhomogeneous solution.

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I'm having a hard time following what you mean when you say "But this means $b_n=0$ for $n\neq 1$ and $b_1=1..." –  ground.clouds1 Nov 30 '12 at 17:34
    
Recall that two power series are identical if and only if their coefficients all match. In particular, since we chose $$\sum_{n=0}^\infty b_n x^n=x,$$ then we have $b_n=0$ for all $n\neq 0$ (since $x$ has no terms of any power other than $1$) and $b_1=1$. Does that make sense? –  Cameron Buie Nov 30 '12 at 17:38
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$$ \begin{align} -y & =\sum_{n=0}^\infty -a_n x^n\tag{starting with 0} \\[15pt] 2xy' & = 2x\sum_{n=1}^\infty a_n n x^{n-1}\tag{starting with 1} \\ & = \sum_{n=1}^\infty 2a_n n x^n \tag{starting with 1} \\[15pt] y'' & =\sum_{n=2}^\infty a_n n(n-1)x^{n-2}\tag{starting with 2} \\ & = \sum_{n=0}^\infty a_{n+2} (n+2)(n+1) x^n.\tag{starting with 0} \end{align} $$

For $n=0$, the sum of the coefficients of the $x^n$ terms is $-a_0+a_2\cdot2\cdot1$ on the left side, and $0$ on the right side.

For $n=1$, the sum of the coefficients of the $x^n$ terms is $-a_1+2a_1\cdot1 + a_3\cdot3\cdot2$ on the left side and $1$ on the right side.

For $n\ge 2$, the sum of the coefficients of the $x^n$ terms is $-a_n + 2a_n\cdot n +a_{n+2} (n+2)(n+1)$ on the left side and $0$ on the right side.

Thus one must solve the recursion $(n+2)(n+1)a_{n+2} +(n-2)a_n=0$ with the initial conditions implied by the above.

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You lost a $2$, there (in the middle sum). –  Cameron Buie Nov 30 '12 at 18:02
    
Fixed. Thanks. ${{{{}}}}$ –  Michael Hardy Nov 30 '12 at 18:14
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