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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 3

up vote 11 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
$1=1+0\cdot x+0\cdot x^2+\dots$ –  Julián Aguirre Nov 30 '12 at 17:12
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
@Michael: That problem is an example of why the poster needs to correct the gap in his knowledge, but I certainly read the question as wanting to correct this gap in his knowledge, rather than a request to solve this specific problem, which he already implies he could do if he didn't have this gap in knowledge. And even if the poster was asking to solve that differential equation with the explanation that his own attempt failed because he doesn't know the power series for $x$, I would still think that this answer is the correct one, since it addresses what the questioner needs to know. –  Hurkyl Nov 30 '12 at 18:15

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

$$ \begin{align} -y & =\sum_{n=0}^\infty -a_n x^n\tag{starting with 0} \\[15pt] 2xy' & = 2x\sum_{n=1}^\infty a_n n x^{n-1}\tag{starting with 1} \\ & = \sum_{n=1}^\infty 2a_n n x^n \tag{starting with 1} \\[15pt] y'' & =\sum_{n=2}^\infty a_n n(n-1)x^{n-2}\tag{starting with 2} \\ & = \sum_{n=0}^\infty a_{n+2} (n+2)(n+1) x^n.\tag{starting with 0} \end{align} $$

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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