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I am reading Euler's Introduction to analysis of the infinite and get stuck here. I wish to ask two questions:

1) The part I don't understand is circled in red box. If I understand correctly, he is trying expand the binomial series for binomial, trinomial and quadrinomial and finally polynomial expression. But I don't understand how did he derive the recurrent relationship between the first and the second coefficients, the second and the third, and so forth.

I have also numbered the red box for easy reference.

For example, for red box number 2, I don't understand where does $\dfrac{2m-n}{n}\beta\cdot L$ come from?

2) I also don't understand the yellow part. Is this the consequence of expanding a trinomial into trinomial series?

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  • $\begingroup$ This is just a guess. From red box 1, he conjectured that $N$ (in red box 2) is of the form $\frac{m-n}{n}\alpha M+(\text{something})\times L$. Calculating $(B-\frac{m-2}{2}\alpha A)/1$, $(C-\frac{m-3}{3}\alpha B)/A$, $(D-\frac{m-4}{4}\alpha C)/B$ and so on, he conjectured that the relation in red box 2 holds. This might be how $\frac{2m-n}{n}\beta$ appears. $\endgroup$ – mathlove Dec 8 at 8:09
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For the first red box, they're getting an expression for the $(m+1)$th (or $n$th) coefficient (defined as $N$) in terms of the $m$th coefficient (defined as $M$).

Note the last two terms in the equations preceding red box 1 and note that in this case $m=2$ and $n=m+1=3$:

$$M = \alpha^2\frac{(m-1)(m-2)}{1\cdot 2}; N = \alpha^3\frac{(m-1)(m-2)(m-3)}{1\cdot 2 \cdot 3},$$

so

$$N=\alpha \frac{m-3}{3}M.$$

The other red boxes are similar redefinition of coefficients to get recurrence formulas.

For the yellow part, these terms come from multiplying out $(\alpha z + \beta z z)^2$:

$$(\alpha z + \beta z z)^2 = \alpha^2z^2+2\alpha\beta z^3+\beta^2z^4.$$

(Only the first two terms are written out explicitly. The last term is a higher-order term covered by the "etc.")

To verify these formulas, it really is just a matter of multiplying the things out and doing all of the bookkeeping, and looking at the patterns. If you pick a particular value of $m$, this will determine a particular value of $n$, which is the highest power of $z$ in the overall expansion. Then just call the other coefficients $A, B, C,$ and so forth, and you'll find that the relations in the red boxes hold for the particular values of $m$ and $n$.

You can start simply with $m=2$. Then there's nothing even to multiply out and you can verify things immediately.

Then move to $m=3$. This will square the polynomial on the left sides of the equations. In section 75, $n=4$ because your highest power is $z^4$. In section 76, $n=6$ because your highest power is $z^6$.

Then try $m=4$, and if you're not tired yet, try $m=5$.

Hope this helps.

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  • $\begingroup$ I still don't understand the yellow part. You said multiplying out. Could you do in an example so I can learn? $\endgroup$ – James Warthington Dec 3 at 0:21
  • $\begingroup$ I edited my answer with some more explanation. $\endgroup$ – John Dec 3 at 15:48
  • $\begingroup$ What about the recurrent formula $B=\dfrac{m-2}{2}\alpha\cdot A+\dfrac{2m-2}{2}\beta$. Where does $\dfrac{2m-2}{2}\beta$ come from? I don't understand where does he get $\dfrac{2m-2}{2}$ $\endgroup$ – James Warthington Dec 3 at 16:00
  • $\begingroup$ All of these formulas are found by applying the binomial expansion to the polynomial. This is what gives you the expansion at the start of each section. From there, multiply out the resulting polynomials in the expansion, and combine like powers of $z$ and observe the patterns. $\endgroup$ – John Dec 3 at 23:21
  • $\begingroup$ I know these formulas used the binomial expansion, what I don't understand is $\dfrac{2m-n}{n}$ and $\dfrac{3m-n}{n}$. Where do they come from. To be more useful, can you just write out in more details how the binomial expansion gives rise to these coefficients? $\endgroup$ – James Warthington Dec 4 at 3:40

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