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I'm trying to understand the development of the calculus. Does this sound right as one of the stages?

Newton knows the binomial theorem, which gives

$$(x+y)^n={n\choose0}x^ny^0+...\;\;\;\;\;\;\;\;\;\;(1)$$

Letting $y=δ_x$ (I realise this is Leizbniz's notation but I find it easier to follow than Newton's $o$), we get

$$(x+δ_x)^n={n\choose0}x^nδ_x^0+...\;\;\;\;\;\;\;\;\;\;(2)$$

Considering $δ_x$ as the base of a differential triangle under a curve, the vertical of the triangle is given by $(x+δ_x)^n-x^n$, which gives us

$$(x+δ_x)^n-x^n={n\choose0}x^nδ_x^0+...-x^n\;\;\;\;\;\;\;\;\;\;(3)$$

But ${n\choose0}x^nδ_x^0=x^n$, so the first part of the expansion disappears and everything else moves up one place to the left and we get

$$(x+δ_x)^n-x^n={n\choose1}x^{n-1}δ_x^1+...\;\;\;\;\;\;\;\;\;\;(4)$$

Now, the vertical $(x+δ_x)^n-x^n$ can be called $δ_y$, so now we can write

$$\frac{δ_y}{δ_x}=\frac{(x+δ_x)^n-x^n}{δ_x}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\\ = \frac{1}{δ_x}{n\choose1}x^{n-1}δ_x^1+...\;\;\;\;\;\;\;\;\;\;(5)$$

And now we can use the biniomial theorem to make a differential triangle under any polynomial; and as $δ_x$ gets small, the gradient approaches instantaneity.

However, that would also make $\frac{1}{δ_x}$ get large, so I guess I've made a mistake somewhere.

Edit: mrtaurho points out that the $δ_x^1$ term cancels the $\frac{1}{δ_x}$.

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  • $\begingroup$ Doesn't the $\delta_x^1$ term cancels with the $\frac1{\delta_x}$ leaving only a constant factor of $1$? However, I don't see why this is posted on History of Science and Mathematics and not on Mathematics. $\endgroup$ – mrtaurho Aug 21 at 18:59
  • $\begingroup$ @mrtaurho Yes, that seems like it could work. I posted it in HSM because it's a question about the history of the calculus as developed by Newton. $\endgroup$ – mjc Aug 21 at 19:01
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    $\begingroup$ Still this looks more like a specific issue of mathematical manipulation of equations which doesn't seem to fit within the scope of this site. See here. $\endgroup$ – mrtaurho Aug 21 at 19:04
  • $\begingroup$ I didn't spot any reference, positive or negative, to equation manipulation on the page you linked. If you think that's frowned upon here, you may be right (I'm fairly new here); my feeling is that the historical component is strong enough to make it out of place in general mathematics. $\endgroup$ – mjc Aug 21 at 19:10
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    $\begingroup$ If you would have asked: how did Newton resolved this issue (which is not really an issue, anyway) I would say it is on-topic here. Like it is stated right now I would redirect you to the particular site of the stackexchange network designed for such questions: Mahtematics. IMO it is better suited there than here. But that is only my personal opinion. $\endgroup$ – mrtaurho Aug 21 at 19:16
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What you stated is not Newton's binomial theorem. Binomial theorem for integer exponent was known long before Newton. Newton discovered the binomial theorem for non-integer exponent (an infinite series which is called the binomial series nowadays).

If you wish to understand what is the relation to Calculus, I advise reading Newton's Mathematical papers, or at least his two letters to Leibniz where he described the essence of his discovery.

The main contribution of Newton to Calculus is his realization of importance of power series, and systematic use of them for evaluation of integrals and solving differential equations, of which the binomial series is an example.

Reading Newton is difficult, but there are good modetn expositions of the topic, for example, Arnold's little but very informative book "Huygens and Barrow, Newton and Hooke".

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