Why is $ A_1 x + ... + A_n x^n $ a solution of $ \sum_0^{n} (-1)^n \frac{x^n}{n!} \frac{d^n y}{d x^n} = 0 $?

Question

I was playing(/fiddling) around with some maths and I saw this pattern( where $ A_n $ is a constant.):

$ A_1 x $ is a soultion of: $$ \frac{y}{x} - \frac{dy}{dx} = 0 $$

$ A_1 x + A_2 x^2 $ is a solution of: $$ \frac{y}{x} - \frac{dy}{dx} + \frac{x}{2!} \frac{d^2y}{dx^2} =0 $$

$ A_1 x + A_2 x^2 + A_3 x^3 $ is a solution of: $$ \frac{y}{x} - \frac{dy}{dx} + \frac{x}{2!} \frac{d^2y}{dx^2} - \frac{x^2}{3!} \frac{d^3y}{dx^3} =0 $$

It continues so on. Can someone prove the solution of $ A_1 x + A_2 x^2 + A_3 x^3 + ... + A_n x^n $ is:

$$ \sum_{k=0}^{n} (-1)^k \frac{x^{k-1}}{k!} \frac{d^k y}{d x^k} = 0 $$

Answer 1

Use induction. Suppose the property holds for

$$ A_1 x + A_2 x^2 + A_3 x^3 + ... + A_n x^n $$

Prove that it works for $ n+1$ terms by writing the expression, use the fact that $\dfrac{\mathrm d^{n+1}y}{\mathrm dx^{n+1}}f = \dfrac{\mathrm d^{n}y}{\mathrm dx^{n}}\left(\dfrac{\mathrm dy}{\mathrm dx}f\right)$ and then show that your new equation is equivalent to one with $n$ terms.

Answer 2

To briefly outline one way of solving this (by induction), assume that $y = x^i$ are solutions for $i=1,...,n$ (where $n$ is the order of the particular ODE), and then note that in the $(n+1)$th case the above solutions are already covered as the $y^{(n+1)}$ terms disappear, and $y=x^{n+1}$ is a solution as after substituting this in, you're left with what ends up being the binomial expansion of $(1-1)^{n+1}$.

You could also try using the sub $z = \log(x)$, but this gets a lot nastier.

Answer 3

Let $z=\frac yx$. Then we have

$$\dfrac{d^nz}{dx^n}=\dfrac{d^n(yx^{-1})}{dx^n}=\sum_{k=0}^n\binom nk\dfrac{d^ky}{dx^k}\dfrac{d^{n-k}(x^{-1})}{dx^{n-k}}=$$ $$\sum_{k=0}^n\binom nk\dfrac{d^ky}{dx^k}(-1)^{n-k}(n-k)!x^{-1-n+k}=$$ $$\sum_{k=0}^n(-1)^{n-k}\dfrac{n!}{k!}x^{-1-n+k}\dfrac{d^ky}{dx^k}$$

So we have

$$(\pm1)n!x^{n+1}\dfrac{d^nz}{dx^n}=\sum_{k=0}^n(-1)^k\dfrac{x^k}{k!}\dfrac{d^ky}{dx^k}$$

Setting this to $0$, since $x=0$ won't work, we have $\dfrac{d^nz}{dx^n}=0$. Integrate $n$ times, then $y=zx$.

Answer 4

First, note that $$\displaystyle \sum_{k=0}^n \frac{(-x)^k}{k!} \dfrac{d^k y}{dx^k}=0$$ is a linear ODE, i.e. if $y_1,y_2$ are solutions then so is any linear combination of them. (This is the same ODE as above except for an overall factor of $x$.) Hence it suffices to show that $x^m$ is a solution for any positiveinteger $m\leq n$. But

\begin{align} \sum_{k=0}^n \frac{(-x)^k}{k!}\dfrac{d^k}{dx^k}(x^m) &=\sum_{k=0}^n \frac{(-x)^k}{k!}\cdot\dfrac{m!}{(m-k)!}x^{m-k}\\ &=\left[\sum_{k=0}^n (-1)^k\binom mk\right]x^m\\ \end{align} (The first equality is valid for $m\geq 1$.) So all we need to show is that this sum vanishes. But $\binom{m}{k}=0$ if $k\geq m$, and since $m\leq n$ by assumption the sum is $\displaystyle \sum_{k=0}^m (-1)^k\binom mk=(1+(-1))^k=0$ by the binomial theorem. Thus we conclude that $\{x^m\}_{1\leq m \leq n}$ is indeed a set of solutions, and by linearity of the ODE it follows that so is $\displaystyle \sum_{k=1}^n A_k x^k$ for arbitrary $\{A_k\}$.

Answer 5

For any polynomial $P(x)$ of degree at most $n$, we have $\frac{d^k}{dx^k}P(x) = 0$ for any $k > n$. As a result, for any constant $y$, we have

$$\sum_{k=0}^{n} \frac{(y-x)^k}{k!} \frac{d^k}{d x^k}P(x) = \sum_{k=0}^{\infty} \frac{(y-x)^k}{k!} \frac{d^k}{d x^k}P(x) = P(y)$$

The last equality is true because the expression in the middle is nothing but the Taylor expansion of $P(y)$ with respect to the point $x$.

Substitute $y = 0$, we find any polynomial $P(x)$ with $P(0) = 0$ satisfy the ODE

$$\sum_{k=0}^n \frac{(-x)^k}{k!} \frac{d^k}{dx^k} P(x) = P(0) = 0$$