$A'(x) = 2xA(x)$. Find a closed formula for $A(x)$.

Question

Let $A(x) = \sum_{n\geq0} a_nx^n$ that satisfies $A'(x) = 2xA(x)$. Find a closed formula for $A(x)$.

My Attempt:

$A'(x)$ = $\sum_{n=0}^{\infty} a_nnx^{n-1}$ $\Rightarrow$ $\sum_{n\geq0} a_nnx^{n-1}$ $=\sum_{n\geq0} a_n2x^{n+1}$

Now I thought about doing this:

$\Rightarrow$ $\sum_{n\geq1} a_{n+1}(n+1)x^{n}$ + $a_0$ $=\sum_{n\geq0} a_n2x^{n+1}$

But this does not look right. Can I increase n by $1$ and manipulate the summation. How should I approach this?

Answer 1

Hint If you're looking for a closed form, you can just solve the (separable) equation $A'(x) = 2 x A(x)$ directly: Separating gives $$\frac{dA}{A} = 2 x \,dx ,$$ and both integrations are straightforward.

To carry out the computation in series form instead, we after reindexing the equation becomes $$\sum_{n = 0}^\infty (n + 1) a_{n + 1} x^n = \sum_{n = 1}^\infty a_{n - 1} x^n.$$ Comparing like coefficients (after reindexing once more) gives the relations $$a_1 = 0, \qquad a_n = \frac{2 a_{n - 2}}{n}, \quad n \geq 2 .$$ Induction immediately gives that

• $a_n = 0$ for all odd $n$, and
• $a_{2m} = \frac{a_0}{m!}$ for all even $n$, so $$A(x) \sim a_0 \sum_{m = 0}^\infty \frac{x^{2 m}}{m!} .$$ Can you recognize the series as the Taylor series of a familiar function?

Additional hint If not, rewrite the series in terms of $u := x^2$.

Answer 2

I agree that you can solve the separable equation. However, if you don't want to, you can also take the following roundabout way.

$A'(x) = \sum_{n\geq 0} a_nnx^{n-1} = \sum_{n\geq 1} a_nnx^{n-1} = \sum_{n\geq 0} a_{n+1}(n+1)x^{n}$.

We also know that $A'(x) = 2xA(x) = \sum_{n\geq 0} 2a_nx^{n+1}$, so $\sum_{n\geq 0} a_{n+1}(n+1)x^{n} = \sum_{n\geq 0} 2a_nx^{n+1}$. Looking at the first few terms, we see that $a_1+2a_2x+3a_3x^2+\cdots = 2a_0x+2a_1x^2+2a_2x^3+\cdots$. Clearly, $a_1 = 0$. In fact for any $a_{2n+1}$, we see that because $a_{2n+1}(2n+1)x^{2n} = 2a_{2n-1}x^{2n}$, we inductively know that $a_{2n+1} = 0$. Hence, $a_1, a_3, a_5, \dots = 0$.

We similarly look at $a_0, a_2, a_4, a_6, \dots$, and we find that $a_{2n}=\frac{a_0}{n!}$.

Thus, $A(x) = \sum_{n\geq 0} \frac{a_0}{n!}x^{2n}= a_0\sum_{n\geq 0} \frac{x^{2n}}{n!}$ for any $a_0\in\mathbb{R}$.

Answer 3

By the chain rule, $A(x)=ke^{x^2}$ is a solution for any constant $k$.

So, substituting $x^2$ for $x$ in the well-known power series formula for $e^x$, one gets $$ke^{x^2}=\sum_{n=0}^{\infty} \frac{k}{n!} x^{2n}.$$

So, $a_{2n+1}=0$ and $a_{2n}=\frac{k}{n!}$.

To solve directly in the form of a power series, let $A(x)=a_0+a_1x+a_2x^2+...$.

Then, $A'(x)=a_1+2a_2x+3a_3x^2+...$ and $2xA(x)=2a_0x+2a_1x^2+2a_2x^3+...$.

So, $a_1=0$ and for all $n \ge 0$, $(n+2)a_{n+2}=2a_n$. This means that $a_{n+2}=\frac{2}{n+2}a_n$. By induction, it follows that $a_n=0$ for all odd $n$. Also, for $n=2m$, $a_{2m+2}=\frac{2}{2m+2}a_{2m}=\frac{1}{m+1}a_{2m}$. So, by induction, it follows that $a_{2m}=\frac{1}{m!}a_0$ for all $m \ge 0$. The constant term $a_0$, in fact, is just the coefficient $k$ in $A(x)=ke^{x^2}$.