 One of the key discoveries in the invention of calculus was the generalized binomial theorem. Isaac Newton claimed to have discovered it around 1664, but didn't describe the binomial theorem itself until the June 13, 1676 letter to Henry Oldenburg, secretary of the Royal Society. And in a letter of October 24, 1676, Newton explained how he discovered it. Newton considered the areas under a sequence of curves. 1 minus x squared to the 0, 1 minus x squared to the 1, 1 minus x squared to the 2nd, and so on, then began looking for patterns to find 1 minus x squared to power 1 half. So in modern terms we might note that the area under the curve 1 minus x squared to the 0 is 1 minus x squared to the 1 is 1 minus x squared squared, and so on. But now let's look carefully at our results. So the first thing we notice is the first term in every expansion is x. And so this suggests the first term in the expansion of 1 minus x squared to power 1 half should also be x. Next, notice the denominators and signs of all of the terms in all of our series form a predictable pattern. So we can focus on the unsigned numerators. So let's consider our numerators of the x-cube terms of 1 minus x squared to power n. If n is equal to 0, there's no x-cube term, so our numerator of the coefficient is 0. If n equals 1, our numerator is 1. If n equals 2, our numerator is 2. If n equals 3, 4, or 5, our numerators are 3, 4, or 5. And it's pretty easy to see that there's this nice pattern emerging. The numerators are n, the exponent. And this suggests that for 1 minus x squared to power 1 half, the x-cube term will be negative 1 half over 3 x cubed, negative 1 6 x cubed. Now let's look at the unsigned numerators of the x to the 5 terms of our expansions. So if our exponent is 0 or 1, the x to the 5th term has coefficient 0. If n equals 2, the coefficient has numerator 1. And for 3, 4, and 5, the coefficients are 3, 6, and 10. Now remember Newton is learning mathematics at a time where these number properties are part of everybody's education. So Newton actually recognizes this series. These numerators are the triangular numbers, which are generated by the formula k times k plus 1 over 2. Now we have to do a slight reparameterization here, since n equals 2 corresponds to the triangular number k equal to 1, we can let k equal n minus 1 and generate our numerators by the formula n minus 1 times n over 2. And so if n equals 1 half, our numerator will be, and so the x to the 5th term, notice they all have denominator 5 and an x to the 5th, and the numerator is the triangular number, so the x term will be minus 1 8th over 5 x to the 5th, or minus 1 40th x to the 5th. Consider the unsigned numerators of the x to the 7th terms of 1 minus x squared to the n. These are. Now you might have recognized the previous sequence as the set of triangular numbers. This one's a little bit more obscure, and Newton actually knew these as what are known as the pyramidal numbers, which could be generated by the formula k times k plus 1 times k plus 2 over 1 times 2 times 3. And again, since n equals 3 corresponds to the k equals 1 pyramidal number, then k is equal to n minus 2, and we can generate our numerators by the formula n minus 2 times n minus 1 times n over 1 times 2 times 3. And so if n equals 1 half, our numerator should be, and we see that our x to the 7th terms are negative and denominator 7 with numerator 116. At this point, we look for a pattern within the pattern. So remember, our first numerator was just n. So n, well, that's equal to n over 1. Our second numerator came from the triangular numbers n minus 1 times n over 2. Well, that's just n over 1 times n minus 1 over 2. Our third numerator came from the pyramidal numbers n minus 2 n minus 1 times n over 1 times 2 times 3. But we can rewrite that as, which is a really remarkable pattern. Remember, these give us the numerators of our coefficients. The denominators and the signs are going to be based on the term. So what's a mathematician who doesn't make conjectures? Based on this, Newton claimed that the successive numerators could be found in the partial products of, and it was only later that Newton realized he could interpolate the terms of the series itself. To verify this work, Newton considered the expansion of 1 minus x squared to power 1 half. So the kth term in the expansion will have the form ak1 to power 1 half minus k times minus x squared to power k, where ak is the product of the first k factors of this thing, starting with k equal to 0. Now, for n equals 1 half and k equals 0, we define our first coefficient to be a0 equal to 1. And so our first term is 1, the coefficient, 1 to power 1 half minus 0 times minus x squared to power 0 or 1. The next coefficient will be the first factor, 1 half over 1 or 1 half. And so our next term will be, so for the next term, the next coefficient will be the product of the first two factors here. Again, n equals 1 half, so that's 1 half over 1 times 1 half minus 1 over 2, which is, and so our next term will be minus 1 eighth times minus x squared squared or minus 1 eighth x to the fourth. The third coefficient will be the product of the first three factors. And so our next term will be, the next coefficient will be the product of the first four factors, which gives us, and so on. And Newton verified that this was actually the case by considering what happens when we square the series. And if we do that, we find we do get 1 minus x squared with the remaining terms vanishing by continuation of the series to infinity. In other words, while if we do this expansion, we do get additional terms. If we include the next term of our series, we'll eliminate some of these extra terms. Now remember, Newton communicated these results by letter in 1676, and so the first actual appearance of the binomial theorem was in the treatise on algebra 1685 by John Wallace, where Wallace includes Newton's result.