Airy's equation bounces back

[logistic_guy]

I got

$$\begin{array}{lll} \begin{bmatrix} b_{2n-1}\\b_{2n} \\b_{2n+1} \\ \end{bmatrix}& =\underbrace{\left(\prod_{k=1}^{n-1}\begin{bmatrix}0&0&1\\2k&1&0\\0&2k+1&1\end{bmatrix}\right)}_{k=n-1\text{ most left}} \begin{bmatrix} b_{1}\\b_0 \\b_0+b_1 \\ \end{bmatrix} \end{array}$$

Thanks professor Stefan for passing by. My idea is to write the sum $\displaystyle y(x) = \sum_{n=0}^{\infty}a_n(x - 1)^n$ in terms of the arbitrary constants $\displaystyle a_0$ and $\displaystyle a_1$. And I was able to do that with the help of the recursion formula. What about you? What is your idea of all of this?

My final touches are:

$\displaystyle y(x) = a_0\left( 1 + \frac{1}{2}(x - 1)^2 + \frac{1}{6}(x - 1)^3 - \frac{1}{24}(x - 1)^4 + \frac{1}{60}(x - 1)^5 + \frac{1}{144}(x - 1)^6 + \cdots\right) + a_1\left((x - 1) - \frac{1}{6}(x - 1)^3 + \frac{1}{12}(x - 1)^4 + \frac{1}{120}(x - 1)^5 - \frac{1}{120}(x - 1)^6 + \cdots\right)$

 

[fresh_42]

Thanks professor Stefan for passing by. My idea is to write the sum $\displaystyle y(x) = \sum_{n=0}^{\infty}a_n(x - 1)^n$ in terms of the arbitrary constants $\displaystyle a_0$ and $\displaystyle a_1$. And I was able to do that with the help of the recursion formula. What about you? What is your idea of all of this?

My final touches are:

$\displaystyle y(x) = a_0\left( 1 + \frac{1}{2}(x - 1)^2 + \frac{1}{6}(x - 1)^3 - \frac{1}{24}(x - 1)^4 + \frac{1}{60}(x - 1)^5 + \frac{1}{144}(x - 1)^6 + \cdots\right) + a_1\left((x - 1) - \frac{1}{6}(x - 1)^3 + \frac{1}{12}(x - 1)^4 + \frac{1}{120}(x - 1)^5 - \frac{1}{120}(x - 1)^6 + \cdots\right)$

My formula delivers all coefficients $a_k$ via matrix multiplication

$$\begin{bmatrix} (2k-1)!\,a_{2k-1}\\(2k)!\,a_{2k} \\(2k+1)!\,a_{2k+1} \\ \end{bmatrix}= \begin{bmatrix} 0& 0& 1\\2k-2 &1 &0 \\0 &2k-1 &1 \\ \end{bmatrix} \begin{bmatrix} 0& 0& 1\\2k-4 &1 &0 \\0 &2k-3 &1 \\ \end{bmatrix}\cdots \begin{bmatrix} 0&0&1\\4&1&0\\0 &5&1 \end{bmatrix} \begin{bmatrix} 0&0 &1 \\ 2& 1& 0\\ 0& 3&1 \\ \end{bmatrix} \begin{bmatrix} a_1\\a_0 \\a_0+a_1 \\ \end{bmatrix}$$
The problem is, that it gets inconvenient pretty fast (see my post #20 where I wrote the matrix to determine $a_9,a_{10},a_{11}$). But you can theoretically write a program to calculate any coefficient depending only on the two initial values $a_0,a_1.$ My method is in post #18.

My solutions for $a_2, \ldots,a_8$ are:

$$\begin{array}{lll} a_2&=\dfrac{a_0}{2!}\\[18pt] a_3&=\dfrac{a_0+a_1}{3!}\\[18pt] a_4&=\dfrac{a_1+a_2}{12}=\dfrac{a_1}{12}+\dfrac{a_0}{24}=\dfrac{a_0+2a_1}{4!}\\[18pt] a_5&=\dfrac{a_2+a_3}{20}=\dfrac{a_0}{40}+\dfrac{a_0}{120}+\dfrac{a_1}{120}=\dfrac{4a_0+a_1}{5!}\\[18pt] a_6&=\dfrac{a_3+a_4}{30}=\dfrac{a_0+a_1}{180}+\dfrac{a_0+2a_1}{720}=\dfrac{5a_0+6a_1}{6!}\\[18pt] a_7&=\dfrac{a_4+a_5}{42}=\dfrac{a_0+2a_1}{24\cdot 42}+\dfrac{4a_0+a_1}{120\cdot 42}=\dfrac{9a_0+11a_1}{7!}\\[18pt] a_8&=\dfrac{a_5+a_6}{56}=\dfrac{24a_0+6a_1+5a_0+6a_1}{40320}=\dfrac{29a_0+12a_1}{8!} \end{array}$$

 

[logistic_guy]

The matrix formula allows you to calculate all the $a_n$ but you should use a computer program. Already small values become inconvenient:
$$\begin{pmatrix}b_9\\b_{10}\\b_{11}\end{pmatrix}=\begin{pmatrix}9!\,a_9\\10!\,a_{10}\\11!\,a_{11}\end{pmatrix}= \begin{pmatrix}24&15&29\\82&83&18\\42&186&119\end{pmatrix} \begin{pmatrix}a_1\\a_0\\a_0+a_1\end{pmatrix}$$

Yeah, I agree that it is a fancy way to find the coefficients by Matrix, but using the recursion formula directly is straightforward.

 

[fresh_42]

Yeah, I agree that it is a fancy way to find the coefficients by Matrix, but using the recursion formula directly is straightforward.

The recursion formula is

$$a_{n+2}=\dfrac{a_{n-1}+a_n}{(n+1)(n+2)}=\dfrac{n!}{(n+2)!}(a_{n-1}+a_n)\, , \,n\geq 1$$
which cannot be resolved easily and does not show how they depend on $a_0,a_1.$ The matrix formula gives a calculation device from $a_0,a_1$ up to any index you want. And it avoids adding nasty quotients which we would have otherwise.