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Do we have any idea of how Newton proved the generalised binomial theorem?

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  • $\begingroup$ Related Math SE post: How did Sir Isaac Newton develop and formulate the famous binomial theorem?. $\endgroup$ – user1709 Jul 16 '16 at 18:30
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    $\begingroup$ I think it was Niels Abel who was the first to prove it. $\endgroup$ – Omar Nagib Sep 7 '16 at 21:29
  • $\begingroup$ There's actually nothing to prove in the binomial theorem (I take it we're talking about the cases when the index is not a positive integer, so that we have an infinite series) other than that the series developed is well-defined. Newton did not prove this, but used a combination of physical insight and blind faith to work out when the series makes sense. In general, apart from issues of convergence, the binomial theorem is actually a definition -- namely an extension of the case when the index is a positive integer. As you may know, the latter case can be proved by induction. $\endgroup$ – Allawonder May 4 at 10:40
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According to William Dunham in his Journey Through Genius

[Newton's Binomial Theorem is] not a "theorem" in the sense of Euclid or Archimedes in that Newton did not furnish a complete proof. Yet his insight and intuition served him well enough to devise a germane formula... Further, his inherent belief in the persistence of patterns suggested to him that the formula that correctly generated coefficients for binomial powers like $(a + b)^2$ or $(a + b)^3$ should work just as well for powers like $(a + b)^{\frac12}$ or $(a + b)^{-3}$.

Although Newton did not prove this result, we do have some idea of how he arrived at it. Newton's explanation is presented in a pair of letters sent to Leibniz dated 1676 and delivered via Henry Oldenburg of the Royal Society. In it, he explains how he arrived at his result by a somewhat indirect route. He was investigating the areas under the curves $y = (1-x^2)^{\frac{n}{2}}$ from $0$ to $x$ for $n = 0, 1, 2, 3,\dots$. These are easy to calculate for even $n$. By observing patterns and interpolating, Newton was able to guess the answer for odd values of $n$. Then he realised that he could get the same answers by expressing $(1-x^2)^{\frac{n}{2}}$ as an infinite series.

Newton wrote : $$(P + PQ)^{\frac{m}{n}} = P^{\frac{m}{n}} + \frac{m}{n}AQ + \frac{m-n}{2n}BQ + \frac{m-2n}{3n}CQ + \dots$$ where $P+PQ$ is the binomial to be considered; where $\frac{m}{n}$ is the power to which we shall raise the binomial "whether that power is integral or (so to speak) fractional, whether positive or negative"; and where $A,B,C, \dots$ represent the immediate preceding term in the expansion. (See TL;DR)

This may look unfamiliar when compared to the modern form. When you factor out the $P^{\frac{m}{n}}$ term from the right hand side and cancel you arrive at the more familiar form :

$$(1+Q)^{\frac{m}{n}} = 1 + \frac{m}{n}Q + \frac{(m/n)(m/n-1)}{2}Q^2 + \frac{(m/n)(m/n-1)(m/n-2)}{2\times3}Q^3 + \frac{(m/n)(m/n-1)(m/n-2)(m/n-3)}{2\times3\times4}Q^4 + \dots$$


TL;DR

Thus, in Newton's notation, $$A = P^{\frac{m}{n}}$$ $$B = \frac{m}{n}AQ = \frac{m}{n}P^{\frac{m}{n}}Q$$ $$C = \frac{(m-n)}{2n} BQ = {\frac{(m-n)m}{(2n)n}}P^{\frac{m}{n}}Q^2 = {\frac{(m/n)(m/n-1)}{2}}P^{\frac{m}{n}}Q^2$$ etc.

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