How to generalize a sequence of terms/patterns in this case

Question

$y_{0}=0$

$y_{1}=\frac{-1}{2}x^2$

$y_{2}=\frac{-1}{2}x^2 - \frac{1}{10}x^5$

$y_{3}=\frac{-1}{2}x^2 - \frac{1}{10}x^5- \frac{1}{80}x^8$

$y_{4}=\frac{-1}{2}x^2 - \frac{1}{10}x^5- \frac{1}{80}x^8- \frac{1}{880}x^{11}$

Now I am trying to figure out a general pattern. I can see that the powers follow an arithmetic sequence 3 is added each time. The denominator is the current power multiplied by its previous power. so for example $10 = 5 \times 2$ and $80 = 8 \times 5 \times 2$.

My problem is that I am not sure how to generalise this.

I am thinking $(\frac{1}{\prod^{n}_{k = 0} 3n-1}\times x^{3n-1}$). But not sure if this is correct? How can I prove by induction in this case? I have not done anything with the $\prod$ symbol before.

EDIT:

Thank you @mathlove for the detailed and thoughtful answer.

For the proving part, I forgot to write up the whole problem. But here it is: This problem about applying the methods of successive approximations ${φ0(x)=0, φ1(x), φ2(x), . . . , φn(x)}$ to the initial value problem.

$$\frac{dy}{dx} = -x + x^2 y, \space y(0) = 0$$

where $φ_{n+1}(x) = \int _{0} ^{x} f(s, φ_{n+1}(s)) ds$

Right now, I am having issues with verifying final form with induction.

Answer 1

I am thinking $(\frac{1}{\prod^{n}_{k = 0} 3n-1}\times x^{3n-1}$).

You are almost there.

We can guess that for $n\ge 1$, $$y_n=-\sum_{\color{red}k=1}^{n}x^{3\color{red}k-1}\prod_{m=1}^{\color{red}k}\frac{1}{3m-1}\tag1$$

For $n=5$, for example, $(1)$ gives $$\begin{align}y_5&=-\sum_{k=1}^{5}x^{3k-1}\prod_{m=1}^{k}\frac{1}{3m-1} \\\\&=-\bigg(x^{2}\prod_{m=1}^{1}\frac{1}{3m-1}+x^{5}\prod_{m=1}^{2}\frac{1}{3m-1}+x^{8}\prod_{m=1}^{3}\frac{1}{3m-1}+x^{11}\prod_{m=1}^{4}\frac{1}{3m-1}+x^{14}\prod_{m=1}^{5}\frac{1}{3m-1}\bigg) \\\\&=-\frac{x^2}{2}-\frac{x^5}{2\times 5}-\frac{x^8}{2\times 5\times 8}-\frac{x^{11}}{2\times 5\times 8\times 11}-\frac{x^{14}}{2\times 5\times 8\times 11\times 14} \\\\&=-\frac{x^2}{2}-\frac{x^5}{10}-\frac{x^8}{80}-\frac{x^{11}}{880}-\frac{x^{14}}{12320}\end{align}$$

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However, note that we cannot prove $(1)$ since from the given conditions, $y_n\ (n\ge 5)$ are not determined.

There are infinitely many examples which satisfy the given conditions.

For example, for $n\ge 1$, $$y_n=-\sum_{k=1}^{n}\frac{x^{3k-1}}{f(k)}$$ satisfies the given conditions for any $a$ where $$f(k)=ak^4+\bigg(\frac{334}{3}-10a\bigg)k^3+(35a-637)k^2+\bigg(\frac{3419}{3}-50a\bigg)k+24a-612$$

Note that for any $a$, we have $$f(1)=2,f(2)=10,f(3)=80,f(4)=880,f(5)=24a+3078$$

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Added :

Let us prove by induction that for $n\ge 1$, $$φ_n(x)=-\sum_{k=1}^{n}x^{3k-1}\prod_{m=1}^{k}\frac{1}{3m-1}\tag2$$

Proof :

For $n=1$, $(2)$ holds since $$φ_1(x)=\int_0^x(-s)ds=-\frac 12x^2$$

Suppose that $(2)$ holds for $n$.

Then, we have $$\begin{align}φ_{n+1}(x)&=\int_0^x\bigg(-s-s^2\sum_{k=1}^{n}s^{3k-1}\prod_{m=1}^{k}\frac{1}{3m-1}\bigg)ds \\\\&=\int_0^x\bigg(-s-\sum_{k=1}^{n}s^{3k+1}\prod_{m=1}^{k}\frac{1}{3m-1}\bigg)ds \\\\&=\bigg[-\frac 12s^2-\sum_{k=1}^{n}\frac{s^{3k+2}}{3k+2}\prod_{m=1}^{k}\frac{1}{3m-1}\bigg]_0^x \\\\&=-\frac 12x^2-\sum_{k=1}^{n}\frac{x^{3k+2}}{3k+2}\prod_{m=1}^{k}\frac{1}{3m-1} \\\\&=-\sum_{k=1}^{n+1}x^{3k-1}\prod_{m=1}^{k}\frac{1}{3m-1}\end{align}$$

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Added 2 :

How did you combine the $\frac{−1}{2}x^2$ into the summation symbol from $n$ to $n+1$ in the second and second to last line of your answer.

In the following, we prove that $$-\frac 12x^2-\sum_{k=1}^{n}\frac{x^{3k+2}}{3k+2}\prod_{m=1}^{k}\frac{1}{3m-1}=-\sum_{k=1}^{n+1}x^{3k-1}\prod_{m=1}^{k}\frac{1}{3m-1}\tag3$$

Proof :

Let $g(k):=\displaystyle\frac{x^{3k+2}}{3k+2}\prod_{m=1}^{k}\frac{1}{3m-1}$ and $h(k):=x^{3k-1}\displaystyle\prod_{m=1}^{k}\frac{1}{3m-1}$.

Then, we have $$h(k+1)=g(k)\tag4$$ since $$\begin{align}h(k+1)&=x^{3(k+1)-1}\prod_{m=1}^{\color{red}{k+1}}\frac{1}{3m-1} \\\\&=x^{3k+2}\times \frac{1}{3(k+1)-1}\times\prod_{m=1}^{\color{red}k}\frac{1}{3m-1} \\\\&=\frac{x^{3k+2}}{3k+2}\prod_{m=1}^{k}\frac{1}{3m-1} \\\\&=g(k)\end{align}$$

Also, we have $$h(1)=\frac{1}{2}x^{2}\tag5$$

So, using $(4)(5)$, we can prove $(3)$ as follows : $$\begin{align}&-\frac 12x^2-\sum_{k=1}^{n}\frac{x^{3k+2}}{3k+2}\prod_{m=1}^{k}\frac{1}{3m-1} \\\\&=-h(1)-\sum_{k=1}^{n}g(k) \\\\&=-h(1)-\sum_{k=1}^{n}h(k+1) \\\\&=-h(1)-\bigg(h(2)+h(3)+\cdots +h(n+1)\bigg) \\\\&=-\bigg(h(1)+h(2)+\cdots +h(n+1)\bigg) \\\\&=-\sum_{k=1}^{n+1}h(k) \\\\&=-\sum_{k=1}^{n+1}x^{3k-1}\prod_{m=1}^{k}\frac{1}{3m-1}\end{align}$$

Answer 2

You’re thinking is correct. All the $\prod$ symbol means is successive multiplication. You’ve used that symbol implicitly without realizing many times before. For example $x^n = \prod_{k = 1}^n x$ or $n! = \prod_{k = 1}^n k$. It is the same as a $\sum$, but for multiplication instead of addition.

But yes, from those patterns it seems to be exactly what you said you start with 0 then subtract $\frac{x^2}{2}$ then every successive term you subtract by adding 3 to the power of $x$ and multiplying the new power of $x$ to the previous denominator and that is your last part of your next iteration.