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History of Mathematics Paul Yiu Department of Mathematics Florida Atlantic University Spring 2014 10A: Newton s binomial series

Issac Newton (1642-1727) obtained the binomial theorem and the area of a segment of the circle by interpolation. 1

Newton to Oldenberg, 1676 At the beginning of my mathematical studies, when I had met with the works of our celebrated Wallis, on considering the series by the intercalation of which he himself exhibits the area of the circle and the hyperbola, the fact that, in the series of curves whose common base or axis is x and the ordinates (1 x 2 ) 0 2, (1 x 2 ) 1 2, (1 x 2 ) 2 2, (1 x 2 ) 3 2, (1 x 2 ) 4 2, (1 x 2 ) 5 2, etc., if the areas of every other of them, namely, x, x 1 3 x3, x 2 3 x3 + 1 5 x5, x 2 3 x3 + 3 5 x5 1 7 x7, etc., could be interpolated, we should have the areas of the intermediate ones, of which the first (1 x 2 ) 1 2 is the circle: 2

Newton s discovery of the area of a segment of a circle in order to interpolate these series I noted that in all of them the first term was x, and that the second terms 0 3 x3, 1 3 x3, 2 3 x3, 3 3 x3, etc., were in arithmetical progression, and hence that the first two terms of the series to be intercalated ought to be x 1 3 (1 2 x3 ), x 1 3 (3 2 x3 ), x 1 3 (5 2 x3 ), etc. To intercalate the rest I began to reflect that the denominators 1, 3, 5, 7, etc. were in arithmetic progression, so that the numerical coefficients of the numerators only were still in need of investigation. But in the alternately given areas these were the figures of powers of the number 11, namely of these, 11 0, 11 1, 11 2, 11 3, 11 4, that is, first 1, then 1,1; thirdly, 1,2,1; fourthly 1,3,3,1; fifthly 1,4,6,4,1,etc. 3

Newton s discovery of the area of a segment of a circle And so I began to inquire how the remaining figures in these series could be derived from the first two given figures, and I found that on putting m for the second figure, the rest would be produced by continual multiplication of the terms of this series, m 0 m 1 m 2 m 3 m 4, etc. 1 2 3 4 5 For example, let m =4, and 4 1 2 (m 1), that is 6 will be the third term, and 6 1 3 (m 2), that is 4 the fourth, and 4 1 4m 3, that is 1 the fifth, and 1 1 5 (m 4), that is 0 the sixth, at which term in this the case the series stops. 4

Newton s discovery of the binomial series Accordingly, I applied this rule for interposing series among series, and since, for the circle, the second terms was 1 3 (1 2 x3 ), I put m = 1 2, and the terms arising were 1 1 2 2 1 2, or 1 1 8, 1 8 2 2 3, or + 1 1 16, 1 16 2 3 4, or 5 128, and so to infinity. Whence I came to understand that the area of the circular segment which I wanted was 1 2 x x3 1 3 8 x5 1 5 16 x7 5 7 128 x9 etc. 9 5

Newton s discovery of the area of a segment of a circle Accordingly, I applied this rule for interposing series among series, and since, for the circle, the second terms was 1 3 (1 2 x3 ), I put m = 1 2, and the terms arising were 1 1 2 2 1 2, or 1 1 8, 1 8 2 2 3, or + 1 1 16, 1 16 2 3 4, or 5 128, and so to infinity. Whence I came to understand that the area of the circular segment which I wanted was 1 2 x x3 1 3 8 x5 1 5 16 x7 5 7 128 x9 etc. 9 1 1 2 x2 dx = x x3 1 3 8 x5 1 5 16 x7 5 7 128 x9 +. 9 6

Newton discovery of the binomial series And by the same reasoning the areas of the remaining curves, which were to be inserted, were likewise obtained: as also the area of the hyperbola and of the other alternate curves in this series (1 + x 2 ) 0 2, (1 + x 2 ) 1 2, (1 + x 2 ) 2 2, (1 + x 2 ) 3 2, etc. And the same theory serves to intercalate other series, and that through intervals of two or more terms when they are absent at the same time. This was my first entry upon these studies, and it had certainly escaped my memory, had I not a few weeks ago cast my eye back on some notes. 7

Newton s discovery of the binomial series But when I had learnt this, I immediately began to consider that the terms that is to say, (1 x 2 ) 0 2, (1 x 2 ) 1 2, (1 x 2 ) 2 2, (1 x 2 ) 3 2, etc., 1, 1 x 2, 1 2x 2 + x 4, 1 3x 2 +3x 4 x 6, etc. could be interpolated in the same way as the areas generated by them: and that nothing else was required for this purpose but to omit the denominators 1,3,5,7. etc., which are in the terms expressing the areas; 8

Newton s discovery of the binomial series this means that the coefficients of the terms of the quantity to be intercalated (1 x 2 ) 1 2, (1 x 2 ) 3 2, or in general, (1 x 2 ) m, arise by the continued multiplication of the terms of this series so that (for example) m m 1 2 m 2 3 m 3, etc., 4 (1 x 2 ) m = ( ) m ( x 2 ) k. k (1 x 2 ) 1 2 was the value of 1 1 2 x2 1 8 x4 1 16 x6, etc., (1 x 2 ) 3 2 of 1 3 2 x2 + 3 8 x4 + 1 16 x6, etc., and (1 x 2 ) 1 3 was the value of 1 1 3 x2 1 9 x4 5 81 x6, etc. 10

Newton s discovery of the binomial series So then the general reduction of radicals into infinite series by that rule, which I laid down at the beginning of my earlier letter, became known to me, and that before I was acquainted with the extraction of roots. But once this was known, that other could not long remain hidden from me. For in order to test these processes, I multiplied 1 1 2 x2 1 8 x4 1 16 x6, etc. into itself; and it became 1 x 2, the remaining terms vanishing by the continuation of the series to infinity. And even so 1 1 3 x2 1 9 x4 5 81 x6, etc. multiplied twice into itself also produced 1 x 2. 11

Newton s discovery of the binomial series And as this was not only sure proof of these conclusions so too it guided me to try whether, conversely, these series, which it thus affirmed to be roots of the quantity 1 x 2, might not be extracted out of it in an arithmetical manner. And the matter turned out well. 1 1 D.J.Struik, A Source Book of Mathematics, Princeton, 1967. 12

Newton s derivation of the exponential series ( ) Start with 1 1+t =1 t + t2 t 3 + +( 1) n t n +. (2) Integration yields x 1 log(1 + x) = 0 1+t dt = x x2 2 + x3 3 x4 4 + + ( 1)n x n+1 n +1 (3) Put y =log(1+x) so that x = e y 1, and try to solve for x in terms of y from the infinite series y = x x2 2 + x3 3 + + ( 1)n 1 x n n +. +. ( ) 13

Newton s derivation of the exponential series (3) Put y =log(1+x) so that x = e y 1, and try to solve for x in terms of y from the infinite series y = x x2 2 + x3 3 + + ( 1)n 1 x n +. ( ) n Put x = a 1 y + a 2 y 2 + a 3 y 3 + + a n y n +. Note that in ( ) only the term x on the right hand side contains y. From this, a 1 =1. Again, in ( ), only the first two terms on the right hand side contain y 2. By comparing coefficients of y 2,wehave 0=a 2 1 2 a2 1 = a 2 1 2 = a 2 = 1 2. Once more, in ( ), only the first three terms on the right hand side contain y 3. 0=a 3 1 2 2a 1a 2 + 1 3 a3 1 = a 3 1 2 + 1 3 = a 3 1 6 = a 3 = 1 6. Exercise. Proceed one step further to show that a 4 = 1 24. 15

Newton s derivation of the exponential series (4) Newton found the first few terms x = y + 1 2 y2 + 1 6 y3 + 1 24 y4 + 1 120 y5 + and confidently concluded that a n = 1 n! after the roots have been extracted to a suitable period, they may sometimes be extended at pleasure by observing the analogy of the series. Exercise. Write down the series for 1 1 t 2 =(1 t 2 ) 1 2 using the binomial theorem. Integrate to obtain the series for arcsin x: arcsin x = x + 1 2 x3 3 + 1 2 3 4 x5 5 + 1 2 3 4 5 6 x7 7 + Then invert to form the series sin y = y 1 3! y3 + 1 5! y5 1 7! y7 + + ( 1)n (2n +1)! y2n+1 + 16

History of Mathematics Paul Yiu Department of Mathematics Florida Atlantic University Spring 2014 10B: Jacob Bernoulli s summation of the powers of natural numbers

Jakob Bernoulli Ars Conjectandi, 1713 Jakob Bernoulli arranged the binomial coefficients in the table below and made use of them to sum the powers of natural numbers. 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 2 1 0 0 0 0 1 3 3 1 0 0 0 1 4 6 4 1 0 0 1 5 10 10 5 1 0 1 6 15 20 15 6 1 1 7 21 35 35 21 7 Let the series of natural numbers 1, 2, 3, 4, 5, etc. up to n be given, and let it be required to find their sum, the sum of the squares, cubes, etc. Bernoulli then gave a simple derivation of the formula n = 1 2 n2 + n, for the sum of the first n natural numbers. He then continued. 1

Sums of squares of integers 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 2 1 0 0 0 0 1 3 3 1 0 0 0 1 4 6 4 1 0 0 1 5 10 10 5 1 0 1 6 15 20 15 6 1 1 7 21 35 35 21 7 A term in the third column is generally taken to be (n 1)(n 2) 1 2 = n2 3n +2, 2 and the sum of all terms (that is, of all n2 3n+2 2 )is n(n 1)(n 2) 1 2 3 = n3 3n 2 +2n, 6 and 1 2 n2 = n3 3n 2 +2n 3 + 6 2 n 1; 2

Sums of squares of integers 1 but 3 2 n = 3 2 3 2 n 2 n2 = n3 3n 2 +2n + 1; 6 n = 3 4 n2 + 3 4 n and 1=n. Substituting, we have 1 2 n2 = n3 3n 2 +2n 6 = 3n2 +3n 4 n = 1 6 n3 + 1 4 n2 + 1 12 n, of which the double n 2 (the sum of the squares of all n) 2 = 1 3 n3 + 1 2 n2 + 1 6 n. 2 More generally, ( ) ( k k + k+1 ) ( k + + n ( k) = n+1 k+1). This identity can be established by considering the number of (k +1) element subsets of {1, 2, 3,...,n+1}, noting that the greatest element m of each subset must be one of k +1, k +2,...,n +1, and that there are exactly ( ) m 1 k subsets with greatest element m. 3

Sum of cubes of integers 1 0 0 0 0 0 0 1 1 0 0 0 0 0 1 2 1 0 0 0 0 1 3 3 1 0 0 0 1 4 6 4 1 0 0 1 5 10 10 5 1 0 1 6 15 20 15 6 1 1 7 21 35 35 21 7 A term of the fourth column is generally (n 1)(n 2)(n 3) 1 2 3 and the sum of all terms is n(n 1)(n 2)(n 3) 1 2 3 4 it must certainly be that 1 11 6 n3 n 2 + 6 n = n3 6n 2 +11n 6, 6 = n4 6n 3 +11n 2 6n. 24 1= n4 6n 2 +11n 2 6n. 24 Hence, 1 6 n3 = n4 6n 3 +11n 2 6n + 24 n 2 11 6 n + 1. 4

Sums of cubes of integers 1 6 n3 = n4 6n 3 +11n 2 6n 24 + n 2... When all substitutions are made, the following results: n 3 = 1 4 n4 + 1 2 n3 + 1 24 n2. 11 6 n + 1. 5

Thus, we can step by step reach higher and higher powers and with slight effort form the following table: n 1 = 1 2 n2 + 1 2 n, n 2 = 1 3 n3 + 1 2 n2 + 1 6 n, n 3 = 1 4 n4 + 1 2 n3 + 1 4 n2 n 4 = 1 5 n5 + 1 2 n4 + 1 3 n3 1 30 n n 5 = 1 6 n6 + 1 2 n5 + 5 12 n4 1 12 n2 n 6 = 1 7 n7 + 1 2 n6 + 1 2 n5 1 6 n3 + 1 42 n n 7 = 1 8 n8 + 1 2 n7 + 7 12 n6 7 24 n4 + 1 12 n2 n 8 = 1 9 n9 + 1 2 n8 + 2 3 n7 7 15 n5 + 2 9 n3 1 30 n n 9 = 1 10 n10 + 1 2 n9 + 3 4 n8 7 10 n6 + 1 2 n4 3 20 n2 n 10 = 1 11 n11 + 1 2 n10 + 5 6 n9 n 7 +n 5 1 2 n3 + 5 66 n. Whoever will examine the series as to their regularity may be able to continue the table. 6

Sums of powers of integers Taking c to be the power of any exponent, the sum of all n c,or n c 1 = c +1 nc+1 + 1 2 nc + c c(c 1)(c 2) 2 Anc 1 + Bn c 3 2 3 4 c(c 1)(c 2)(c 3)(c 4) + Cn c 5 2 3 4 5 6 c(c 1)(c 2)(c 3)(c 4)(c 5)(c 6) + Dn c 7 +, 2 3 4 5 6 7 8 the exponents of n continually decreasing by 2 until n or n 2 is reached. The capital letters A, B, C, D denote in order the coefficients of the last terms in the expressions for n 2, n 4, n 6, n 8 etc., namely, A = 1 6, B = 1 30, C = 1 42, D = 1 30. The coefficients are such that each one completes the others in the same expression to unity. Thus, D must have the value 1 30, because 1 9 + 1 2 + 2 3 7 15 + 2 9 1 30 =1. 7

Reorganization Let S k (n) :=1 k +2 k + + n k. We make use of the series expansion of e z : e z =1+z + z2 zk + + 2! k! + and analogous series for e 2z, e 3z,...,e nz : e z = 1+z + z2 zk + + 2! k! + e 2z = 1+2z + 22 z 2 + + 2k z k + 2! k!. e (n 1)z = 1+(n 1)z + (n 1)2 z 2 2! Combining these (with 1=1), we have n + S 1 (n 1)z + S 2(n 1)z 2 for the series expansion of 2! + + (n 1)k z k k! + + S k(n 1)z k k! 1+e z + e 2z + + e (n 1)z = enz 1 e z 1. 8 +. +

The Bernoulli numbers and sums of powers of integers The series expansion of the function on the right hand side can be found from e nz 1 e z 1 = enz 1 z z e z 1. (1) The first factor clearly has series expansion e nz 1 z = n + n2 nk+1 z + + 2! (k +1)! zk + (2) Suppose we write z e z 1 = B 0 + B 1 1! z + B 2 2! z2 + + B k k! zk + These coefficients B 0, B 1,...,B k,... are called the Bernoulli numbers. The product of these two series has beginning term n, and subsequently, for each k =1, 2,..., the coefficient of z k equal to B i i! n j+1 (j +1)! = 1 k ( ) k +1 B i n k+1 i (k +1)! i i+j=k Now comparison gives S k (n 1) = 1 k +1 i=0 i=0 k ( ) k +1 B i n k+1 i. i 9

Bernoulli numbers and sum of powers of integers (1) If we agree, in the binomial expansion of (B +1) k+1, to replace every B i by the number B i, then the above formula can be simply written as S k (n 1) = (B + n)k+1 B k+1. ( ) k +1 (2) This symbolic device can further help computing S k (n 1) easily, without actually calculating the Bernoulli numbers. Comparing ( ) with k S k 1 (n 1) = (B + n) k B k, we note that formally S k (n 1) = ks k 1 (n 1)dn + cn where the constant c so chosen that the sum of the coefficients is equal to 0. 11

Sums of powers of integers (3) An analogous expression holds for S k (n): S k (n) = ks k 1 (n)dn + cn where the constant c so chosen that the sum of the coefficients is equal to 1. Examples: Since S 2 (n) = 1 3 n3 + 1 2 n2 + 1 6 n,wehave S 3 (n) = (n 3 + 3 2 n2 + 1 2 n)dn + c 3n (c 3 appropriately chosen) = 1 4 n4 + 1 2 n3 + 1 4 n2 ; (c 3 =0) S 4 (n) = (n 4 +2n 3 + n 2 )dn + c 4 n (c 4 appropriately chosen) = 1 5 n5 + 1 2 n4 + 1 3 n3 1 30 n; S 5 (n) = (n 5 + 52 n4 + 53 n3 16 ) n dn + c 5 n (c 5 appropriately chosen) = 1 6 n6 + 1 2 n5 + 5 12 n4 1 12 n2 ; (c 5 =0). 12

The Bernoulli numbers The Bernoulli numbers B n, n =0, 1,...,are, by definition, the coefficients in the series expansion of z e z 1 : z e z 1 = B 0 + B 1 z + B 2 2! z2 + B 3 3! z3 + + B k k! zk +. Clearly, B 0 =1. It is quite easy to check that B 1 = 1 2 (exercise). One important observation is that all subsequent odd-indexed Bernoulli numbers are zero. In terms of the even-indexed Bernoulli numbers, we have z e z 1 =1 z 2 + B 2k (2k)! z2k. k=1 Here are the beginning even-indexed Bernoulli numbers: B 2 = 1 6, B 4 = 1 30, B 6 = 1 42, B 8 = 1 30, B 10 = 5 66, B 12 = 691 2730. 13

History of Mathematics Paul Yiu Department of Mathematics Florida Atlantic University Spring 2014 10C: Euler s first calculation of 1+ 1 2 2k + 1 3 2k + 1 4 2k + 1 5 2k +

Euler s 1734 on sums of reciprocals of even powers Paper 41: De Summis Serierum Reciprocarum. Euler begins with the series for sine and cosine. y = s s3 3! + s5 5! s7 7! +, x = 1 s2 2! + s4 4! s6 6! +. If, for a fixed y, the roots of the equation 0=1 s y + s3 3!y s5 5!y + are A, B, C, D, E etc., then the infinite polynomial factors into the product of 1 s A, 1 s B, 1 s C, 1 s D,... 1

1 s (1 y + s3 3!y s5 5!y + = s )( 1 s ) ( 1 s ). A B D By comparing coefficients, 1 y = 1 A + 1 B + 1 C + 1 D + 2

1 s (1 y + s3 3!y s5 5!y + = s )( 1 s ) ( 1 s ). A B D By comparing coefficients, 1 y = 1 A + 1 B + 1 C + 1 D + If A is an acute angle (smallest arc) with sin A = y, then all the angles with sine equal to y are A, ±π A, ±2π + A, ±3π A, ±4π + A,... The sums of these reciprocals is 1 y. Sum of these reciprocals taking two at a time is 0, taking three at a time is 1 3!y. etc. 3

In general, if a + b + c + d + e + f + = α, ab + ac + bc + = β, abc + abd + bcd + = γ, then a 2 + b 2 + c 2 + = α 2 2β, a 3 + b 3 + c 3 + = α 3 3αβ +3γ, and a 4 + b 4 + c 4 + = α 4 4α 2 β +4αγ +2β 2 4δ. 4

Euler denotes by P the sum of the numbers, Q the sum of the squares, R the sum of cubes, S the sum of fourth powers etc., and writes Now he applies these to P = α, Q = Pα 2β, R = Qα Pβ +3γ, S = Rα Qβ + Pγ 4δ,. α = 1 1, β =0, γ = y 3!y, δ =0, ɛ = 1 5!y,... and obtains P = 1 y, Q = 1 y 2, R = Q y 1 2!y, S = R y P 3!y,... 5

The Leibziz series Now Euler takes y =1. For the equation 1=s s3 3! + s5 5! s7 7! +, all the angles with sines equal to 1 are From these, π 2, π 2, 3π 2, 3π 2, 5π 2, 5π 2, 7π 2, 7π 2, 9π 2, 9π 2,... ( 4 1 1 π 3 + 1 5 1 7 + 1 9 1 ) 11 + =1. This gives Leibniz s famous series 1 1 3 + 1 5 1 7 + 1 9 1 11 + = π 4. 6

Noting that P = α =1, β =0, so that Q = P =1, Euler proceeded to obtain 1+ 1 3 + 1 2 5 + 1 π2 + = 2 72 8. From this, he deduced that 1+ 1 2 + 1 2 3 + 1 2 4 + 1 2 5 + 1 2 6 + 1 π2 + = 2 72 6. 7

Continuing with R = 1 2,S= 1 3,T= 5 24,V= 2 15,W= 6 17,X= 720 315,... Euler obtained 1 1 3 + 1 3 5 1 3 7 + 1 π3 = 3 93 32, 1+ 1 3 + 1 4 5 + 1 4 7 + 1 π4 + = 4 94 96, 1+ 1 2 + 1 4 3 + 1 4 4 + 1 π4 + = 4 54 90, 1 1 3 + 1 5 5 1 5 7 + 1 5π5 = 5 95 1536, 1+ 1 3 + 1 6 5 + 1 6 7 + 1 π6 + = 6 96 960, 1+ 1 2 + 1 6 3 + 1 6 4 + 1 π6 + = 6 56 945,. 8

Euler gave an alternative method of determining these sums. This time, he considered y =0and made use of 0=s s3 3! + s5 5! s7 7! +. Removing the obvious factor s, he noted that the roots of are 0=1 s2 3! + s4 5! s6 7! + ±π, ±2π, ±3π,... so that ) 1 (1 s2 3! +s4 5! s6 7! + = )(1 s2 )(1 s2 )(1 s2 s2 π 2 4π 2 9π 2 16π 2 From these, the sums n=1 1 n 2k can be computed successively for k = 1, 2, 3,... 9

Euler then summarized the formulas which made him famous throughout Europe (for the first time): 1+ 1 2 + 1 2 3 + 1 2 4 + 1 π2 + = 2 52 6, 1+ 1 2 + 1 4 3 + 1 4 4 + 1 π4 + = 4 54 90, 1+ 1 2 + 1 6 3 + 1 6 4 + 1 π6 + = 6 56 945, 1+ 1 2 + 1 8 3 + 1 8 4 + 1 π8 + = 8 58 9450, 1+ 1 2 + 1 10 3 + 1 10 4 + 1 π10 + = 10 510 93555, 1+ 1 2 + 1 12 3 + 1 12 4 + 1 12 5 + = 691π 12 12 6825 93555,. 10

On the other hand, nothing is known about n=1 1 n k when the exponent k is odd. In 1978, L. Apery proved that n=1 1 n is an irrational number. 3 11