1 Sequences and Series

We start with a review of some very basic notions.

Definition 1.0.1. (a) A sequence {an} of real numbers is a function N → R defined by n 7→ an. (b) {an} is said to be non-decreasing ↑ (non-increasing ↓) if a1 ≤ a2...(a1 ≥ a2 ≥ a3..). In either case, {an} is said to be a monotone sequence. (c) {an} is said to be a bounded sequence if |an| ≤ M < ∞ for n ≥ 1. (d) A subsequence of {an} is a restriction of the function to {nk : k ≥ 1, n1 < n2 < n3...}. (e) We say that an → l ∈ R if given ε > 0, there exists N such that |an − l| < ε for n ≥ N and that an → ∞(−∞) if given M ∈ R, there exists N such that an > M(an < −M) for n ≥ N. (f) A sequence {an} is said to be a Cauchy sequence if given ε > 0, there exists N such that |an − am| < ε for n, m ≥ N. Exercises: 1. Show that (c) fails to hold if and only if there exists a subsequence of {an} converging to ±∞. 1 n 2. Show that the sequence defined by an = (1 + n ) is monotone increasing and bounded. 1 1 1 3. Show that the sequence defined by an = n + n+1 + ... + 2n−1 is ↓ and hence convergent.

Deﬁnition 1.0.2. If S ⊆ R, we say that a real number M is an upper(lower) bound of S if every x ∈ S satisﬁes x ≤ M(x ≥ M) and M is said to be the least upper bound (lub) (greatest lower bound(glb)) if no number in (−∞,M)((M, ∞)) is an upper bound (a lower bound).

The completeness axiom Every S ⊂ R which is bounded above (below) has a lub in R.

Exercise: A sequence which is ↑ and bounded above converges to its lub.

The following results are some of the crucial consequences of the completeness of R. Monotone Boundedness Theorem: Every bounded monotone sequence of real numbers converges to a limit in R. Bolzano-Weierstrass Theorem: Every bounded sequence of real numbers has a monotone (and hence convergent) subsequence. Cauchy Criterion: Every Cauchy sequence of real numbers is convergent and conversely. Exercises: 1 n 1. Show that xn = (1 + n ) is convergent. Write: xn → l. 1 n+1 Clearly, yn = (1 + n ) → l, xn < yn and {yn} ↓. Hence for every n, xn < l < yn. Use a calculator and a large value of n to estimate l.

1 1 2. Let tn = 2 + 2! + ... + n! . Show that {tn} ↑ and tn < 3. Thus {tn} is convergent. P∞ 1 Moreover, lim xn ≥ lim tn and thus xn → n=0 n! .

Fact: A function f :(a, b) → R is continuous if and only if whenever an → c ∈ (a, b), f(an) → f(c), provided (an), c ⊆ (a, b).

1 1.1 Sum of an inﬁnite series

Deﬁnition 1.1.1. 1. Given a sequence {an}, let sn = a1 + ... + an denote the sequence of partial P sums. If sn → s ∈ R, then we say that the series an converges to s. P 2. If tn = |a1| + +... + |an| converges to in R, then we say that an converges absolutely. P P 3. If an converges and tn → ∞, then we say that an converges conditionally. The simplest example of an conditionally convergent series is the alternating harmonic series P n 1 (−1) n . In modern calculus, divergence of harmonic series is shown by using the integral test and convergence of the alternating harmonic series is based on the alternating series test. Here’s a note of historical signiﬁcance. 1. The simplest and most elegant proof of divergence of the harmonic series due to Orseme (Four- teenth century). 1 1 n n Observe that s2 > 2 and s2 − s2(n−1) > 2 . Thus s2 → ∞.

2. The alternating series test is due to Leibniz and perhaps its signiﬁcant feature is that one can use it to estimate the error of approximation. Test: If u1 ≥ u2... and un → 0, then let an = s2n, bn = s2n+1. Also, {an} ↑, {bn} ↓. So bn → β, an → α and an − bn → 0 implies α = β = s. Moreover, |s − sk| ≤ uk+1.

1 1 Exercise: How many terms do we need to add to estimate the sum of 1 − 2 + 3 ... within an error of .01?

P P Deﬁnition 1.1.2. Rearrangement of a convergent series an is a series aπ(n) where π : N → N is a 1-1, onto function, or a permutation of N. The most crucial difference between a convergent and an absolutely convergent series is the following theorem, due to Riemann. Theorem: Any rearrangement of an absolutely convergent series converges to the same sum. A conditionally convergent series can be rearranged to converge to any number whatsoever.

1 1 1 1 1 1 1 1 Exercise: Show that 1 − 2 − 4 + 3 − 6 − 8 ... = 2 [1 − 2 + 3 ...].

The following is used in discussing the convergence of series of positive terms. Exercises 1. If a ≥ 0, then lim(a )1/n exists if and only if lim an+1 exists and the two limits are equal. n n an 1 1 n 2. Find lim n (n!) as n → ∞.

The following is an important extension of series of numbers. Weierstrass’s M-test: Suppose {fn} is a sequence of functions deﬁned on a common domain P P S ⊆ R. Asuume |fn(x)| ≤ Mn for x ∈ S and n ∈ N. If Mn < ∞, then fn converges (uniformly) to a function f on S.

2 1.2 Euler’s number e

1 n 1 n+1 Let xn = (1 + n ) . Then xn+1 = (1 + n+1 ) . Pn n 1 Pn+1 n+1 1 By the binomial theorem, xn = k=0 k nk and xn+1 = k=0 k (n+1)k . 1 1 k 1 1 k The k-th term in xn is k! (1 − n )...(1 − n ) and the k-th term in xn+1 is k! (1 − n+1 )...(1 − n+1 ). k k k k Clearly, n+1 < n ⇒ (1 − n+1 ) > (1 − n ). It follows that xn < xn+1 since xn+1 has one extra term.

1 1 Let tn = 2 + 2! + ... + n! . Then tn+1 − tn = 1/(n + 1)! ⇒ (tn) ↑. The following lemma is proved by simple induction on k.

Lemma 1.2.1. For every k ≥ 1 we have: 2k−1 ≤ k!. Now, 1 1 t ≤ 2 + + ... + n 2! n! 1 1 ≤ 2 + + ... + 2 2n ≤ 3

Thus xn ≤ tn ≤ 3 and by monotone convergence theorem, both the sequences are convergent. Write l1 = lim xn and l2 = lim tn. We aim to show that l1 = l2. Observe that xn ≤ tn ⇒ l1 ≤ l2. Fix m and let n ≥ m, then 1 1 1 1 m − 1 x ≥ 2 + (1 − ) + (1 − )...(1 − ) n 2! n m! n n

Hold m ﬁxed and let n → ∞. We see that lim xn ≥ tm. i.e. l1 ≥ tm for every m. Now let m → ∞. We see that: l1 ≥ l2. We write: 1 1 1 e = lim (1 + )n = e = 2 + + + ... n→∞ n 2! 3! Consequently, 2 < e < 3 Claim: e is irrational. Proof. (By contradiction). Clearly, e is not an integer. p Suppose e = q where p, q ∈ N and q ≥ 2. Then q!e ∈ N. Pq 1 P∞ 1 Then q!e − k=0 k! = k=q+1 k! ∈ N. P∞ 1 P∞ 1 But since q > 1, then k=q+1 k! < k=1 2k = 1. This is a contradiction since (0, 1) ∩ N = φ.

1 n+1 Exercise: Let yn = (1 + n ) . Show that : 1. (yn) ↓ and that xn ≤ yn for every n. 2. Conclude that xn ≤ e ≤ yn for all n. 3. Pick a large value of n and estimate e.

3 2 Area under the curve y = xn

Pn n(n+1) It is known that Archimedes knew how to add the ﬁrst n natural numbers by k=1 k = 2 . At the end of the tenth century, Alhazen gave the following recursive relation. n n n p X X X X (n + 1) ik = ik+1 + ( ik) i=1 i=1 p=1 i=1 A formal proof can be written using induction. What matters is the following corollary. Exercise: Use induction on k to prove the following corollary.

Corollary 2.0.1. For every pair of natural numbers {k, n} the following holds: n X nk+1 ik = + p (n) k + 1 k i=1

where pk(n) is a polynomial of degree k in n. One immediate consequence Wallis’s ﬁrst result. Wallis’s Theorem 1: p 1 If p ∈ N, area under y = x for x ∈ [0, 1] is p+1 . The proof is achieved easily by dividing [0, 1] into n equal subintervals and computing (what we now call)the Riemann sum.

Observe that y = xp and x = y1/p are inverse functions and thus the two areas add up to 1. Consequently, we have the following. Wallis’s Theorem 2: n 1 If either n or 1/n is a natural number, area under the the curve y = x , 0 ≤ x ≤ 1 is n+1 . The following result should follow naturally, but Wallis was not able to prove it. n 1 1+n Wallis’s Conjecture: If n ∈ Q, then area under the curve y = x for x ∈ [0, a] is n+1 a .

p Fermat’s proof: Write n = q and let a = 1. Fix r ∈ (0, 1) and divide [0, 1] into subintervals as follows. 2 n Reversing the natural order, let x1 = r, x2 = r ..., xn = r . Observe that their lengths are: r(1 − r), r2(1 − r)... p+q 1 n Pn k(p+q)/q q q Ignoring [0, r ], the outer sum A(r) = (1 − r) k=0 r . Write s = r and t = r . Pn k n q p+q Now A(r) = (1 − r) k=0 s = (1 − r)(1 − s )/(1 − s). Recall that r = t and s = t . Now, for large n, A(r) ∼ (1 − r)/(1 − s) = (1 − tq)/(1 − tp+q) = (1 − t)(1 + t... + tq−1)/(1 − t)(1 + t + ..tp+q−1) = (1 + t... + tq−1)/1 + t + ..tp+q−1 q As r → 1,A(r) → p+q . Factoring the numerator and denominator and cancelling the common 1+t...+tq−1 factor of 1 − t, we see that A = 1+t+...tp+q−1 . As r → 1, t → 1 and limr→1 A(r) = q/(p + q).

4 2.1 First rectification of a curve At a time when Greek standards of rigor were ignored, mathematicians made extensive, uncritical use of infinitesimals. As l’Hospital wrote: ”A curve may be regarded as the totality of an infinity of straight segments, each infinitely small”. However, it was thought that a segment of an alge- braic curve could never have the same length as a constructible straight-line segment. In 1657, an Englishman named Neil showed that such pessimism was unjustified. Consider the semi-cubical parabola y2 = x3 and divide the interval [0, 1] into a large number n of subintervals (as in a Riemann sum). If si is the length of the segment of the curve over [xi−1, xi], then we approximate si by length li of the straight-line segment joining (yi, xi) and (yi−1, xi−1). 3/2 3/2 √ But yi − yi−1 = xi − xi−1. Suppose Ai is the area under the the curve z = x. Then Ai − Ai−1 can be replaced by area of the rectangle on xi − xi−1 of height zi.

By Wallis’s theorem, 3 y − y = (A − A ) i i−1 2 i i−1 3 ∼= z (x − x ) 2 i i i−1 Thus,

∼ X p 2 2 s = (yi − yi−1) + (xi − xi−1) r X 9 ∼= [ (1 + z2)] (x − x ) 4 i i i−1

q 9 The r.h.s. is the area under the curve y = (1 + 4 x) lying over the interval [0, 1]. 3 √ 4 4 Equivalently, it is the area under the curve y = 2 x over [ 9 , 9 + 1].

By Wallis’s theorem 133/2 − 8 s = . 27

5 2.2 Mercator’s series The following argument is attributed to Wallis. Pn k k+1 1 Tool: If k ≥ 1, then i=1 i /n → k+1 as n → ∞. 1 Consider the hyperbola y = 1+t and for x > 0, let A(x) denote area under the curve over [0, x]. x If h = n . If 1 ≤ j ≤ n − 1, then hj < 1. 1 P∞ k k Thus (by using a geometric series), we have: 1+hj = k=0(−1) h . Subdividing [0, x] into n equal parts and computing the Riemann sums, we have:

∞ ∞ X X A ∼= h + h (−1)khk + ... + h (−1)k[(n − 1)h]k k=0 k=0 Rearrange terms to write the r.h.s. as a polynomial in h and observe that:

n−1 n−1 n−1 X X X A ∼= nh − h2( i) + h3( i2) + ... + (−1)n−1hn−1(( ik) i=1 i=1 i=1 (Pn−1 i) (Pn−1 ik−1) ∼= x − x2 i=1 + ... + (−1)n−1xn−1 i=1 n2 nk x2 x3 ∼= x − + ... 2 3 By the geometric series test, the series on r.h.s. converges absolutely if x ∈ (−1, 1) and equals A(x). By alternating series test, it converges conditionally if x = 1. 1 1 That A(1) = 1 − 2 + 3 ... was later shown to follow from Abel’s theorem. Of course, A(x) is log(1 + x), though the series expansion is only valid for x ∈ (−1, 1].

Wallis says: (1685) ”These exponents they call logarithms, which are artiﬁcial numbers, so answering to the natural numbers, so that the addition and subduction (i.e. subtraction)of these answers to the multiplica- tion and division of the natural numbers”. What Wallis called ”natural logarithms”, Euler called ”hyperbolic logarithms”. The properties that Wallis is referring to were discovered by Gregory St. Vincent.

Exercises: 1 1. Use Mercator’s series to compute L(2). Approximate L(2) within an error of 100 . Observe that it takes 100 terms to do so (if we use the alternating series test). A horrendous task! 1 1 1 L(1/2) = 2 − 2.22 + 3.23 ... + .... To approximate the answer within an error of .01, it is now sufﬁcient to add ﬁve terms. (Why?)

2.This idea is due to Newton himself. Use Mercator’s series to compute L(1.2),L(.8) and L(.9) by choosing x = ±.2 and x = −.1. What is the motive behind Newton’s computation? The smaller the value of x in Mercator’s series, the faster the convergence. (1.2)(1.2) Use the fact 2 = (.8)(.9) and Newton’s computation to approximate L(2).

6 2.3 L is for logarithm

x2 x3 For x > 0, deﬁne L(x) = A(x − 1) and observe that for x ∈ (−1, 1],L(x + 1) = x − 2 + 3 .... 1 Write Aa,b for area under the hyperbola y = t over [a, b]. Then,

L(x) = A1,x, x > 1

= −Ax,1, x ∈ (0, 1) In the seventeenth century, a Jesuit monk Gregory discovered the following:

Aa,b = Ata,tb, t > 0 ; L(xy) = L(x) + L(y) We only need to sketch a proof of the ﬁrst property: A(ta, tb) = A(a, b) if t > 0.

Divide [a, b] into n equally-spaced rectangles as a < x1 < x2...xn = b. The inner and outer rectangles over [xi−1, xi] are of heights 1/xi and 1/xi−1 resp. Hence n n X b − a X b − a ≤ A ≤ nx a,b nx i=1 i i=1 i−1

Now the points tx1, ...txn divide [ta, tb] equally and moreover, n n X b − a X b − a ≤ A ≤ nx a,b nx i=1 i i=1 i−1 using the inner and outer sums. Clearly, A(ta, tb) = A(a, b).

h 0 Exercises: Show that 1+h ≤ A1,1+h ≤ h and hence L (1) = 1. Conclude that: 1 k (a) L(e) = 1, where e = lim(1 + k ) . Hence L(ex) = x and L and exp are inverse functions. x n (b) L((1 + n ) ) → x as n → ∞, if x ∈ R. x n and L((1 − n ) ) → −x. x n x x n −x Consequently, (1 + n ) → e and (1 − n ) → e . (c)L0(x) = 1/x.

7 3 Newton’s technique of Reversion

1 Recall that if L(x) is the area under the curve y = t over [1, x] or its negative, then L is a 1-1 function on (0, ∞) and as such has an inverse. Now if one can simply invert Mercator’s series, one can get a handle on the inverse function. Step 1: Newton simply pushed through this agenda. Start with:

1 1 z = x − x2 + x3... (∗) 2 3 Let x = z + p where p is a polynomial and plug in (*). Keeping terms involving only the ﬁrst power of p, we get: 1 1 z = z + p − (z + p)2 + (z + p)3... 2 3 1 2 2 ∼ 1 2 Thus 2 z = p(1 − z + z ..) implies p = 2 z , dropping higher powers of z on the right. 1 2 Write x = z + 2 z + q, where q is a polynomial and repeat the process. We get: 1 1 z3 = q(1 − z + ...) → q ∼= z3 6 6 Now having solved (*) for x, we have: 1 1 1 + x = 1 + z + z2 + z3 + ... 2 6 −1 1 2 1 3 Thus, L (z) = 1 + z + 2 z + 6 z + .... Step 2: z n Given z ∈ R, let an = (1 + n ) . Now, z L(a ) = nL(1 + ) n n z L(1 + n ) = z z n

0 0 Thus, on one hand, L(an) → zL (1). But L (1) = 1 = L(e). Hence

z z L(an) → zL(e) = L(e ) ⇒ an → e .

On the other hand, −1 L(an) → z ⇒ an → L (z) In conclusion, 1 1 L−1(z) = 1 + z + z2 + z3 + ... 2 6 z = lim(1 + )n n = ez

8 4 Euler’s constant γ

As a prerequisite, we need the following inequality. Inequality: L(x + 1) < x whenever x > 0. Let f(x) = L(1 + x) − x and observe that f 0 < 0 ⇒ f ↓. The result follows since f(0) = 0.

Now if an = sn−1 − L(n), then {an} ↑. It represents the area between outer rectangles and the hyperbola over [1, n]. As such, it is dominated by the sum of differences between outer and inner rectangles, and thus

n X 1 1 1 a ≤ ( − ) ≤ 1 − n k k + 1 n + 1 k=1

By monotone convergence theorem, {an} is convergent. We write:

n X 1 γ = lim { − L(n)} n→∞ k k=1 Clearly, 0 ≤ γ < 1. To get a better estimate, observe that the area between the the outer rectangle and the curve dominates half of the area between outer and inner rectangles and as such,

n X 1 1 1 1 1 a ≥ ( − ) ≥ (1 − ) n 2 k k + 1 2 n + 1 k=1 Thus we have an initial estimate: .5 ≤ γ ≤ 1. A much better estimate using trapezoids is outlined in the following exercises.

Exercise: Consider the trapezoid Pk with base [k, k + 1], formed by the tangent to the hyperbola at 1 1 −1 x = k + 2 intersecting the vertical lines at x = k and x = k + 1. Show that its area is (k + 2 ) . Now the area of interest (signifying an) is contained within the rectangle Rk with base [k, k + 1] 1 and height k with the trapezoid Pk removed. Using Mercator’s series for L(2), we see that γ ≤ 2(1 − L(2)). Using the modern calculator estimate of L(2), we see that: .5 ≤ γ ≤ .613.. It is not known if γ is a rational number.

9 4.1 Stirling’s formula The (weaker)version we prove in this section says that there is a constant C which satisﬁes:

−n n+ 1 n! ∼ Ce n 2

We write L(x) = log(x) and An = area under the curve y = log(x) over [1, n]. Consider the trapezoid with base [k, k + 1], the line segment joining (k, log k) and (k + 1, log(k + 1)) and lines x = k and x = k + 1. Write αk for the area between the curve and this trapezoid. If Tn stands for the sum of areas of the trapezoids over [1, n], then An ≥ Tn since y = log x is concave. Let An = Tn + En and observe that,

n 1 X 1 T = [log k + log(k − 1)] = log n! − log n n 2 2 k=2 Integration by parts shows that: An = n log n − n + 1 Hence 1 log n! = (n + ) log n − n + 1 − E 2 n i.e. −n n+ 1 n! = Cne n 2

1−En where Cn = e . Recall that En = α1 + ... + αn−1 and thus it is ↑. 1 As in the last subsection, draw a tangent to y = log x at x = k + 2 and show that the difference in 1 1 areas between outer and inner trapezoids is dominated by log(k + 2 ) − 2 (log k + log(k + 1) or by 1 1 1 1 1 1 2 log((k+ 2 )/k)− 2 log((k+1)/(k+ 2 )). Rearrange and observe: log(1+ 2k+1 ) > log(1+ 2(k+1) ).

n−1 X 1 1 1 1 1 3 1 1 E < { log(1 + ) − log(1 + )} < log − log(1 + ) n 2 2k 2 2(k + 1) 2 2 2 2n k=1 √ So En → E and Cn → C. It takes an appeal to Wallis’s formula (deferred) to show that C = 2π.

10 4.2 Wallis’s Inﬁnite Product: Modern method R π/2 n Let In = 0 sin (x)dx and observe that {In} is ↓ since 0 ≤ sin x ≤ 1. n+1 It follows from integrating by parts that In+2/In = n+2 . Now, I I I 1 ≤ n ≤ n ⇒ n → 1 In+1 In+2 In+1 Observe that 2.4....(2n) I = 2n+1 1.3....(2n − 1)(2n + 1) [2.4...(2n)]2 = (2n)!(2n + 1) 22n(n!)2 = (2n)!(2n + 1) and similarly, π 1.3...(2n − 1) I = . 2n 2 2.4....(2n) π (2n)! = . 2 [2.4....(2n)]2 π (2n)! = . 2 22n(n!)2

2n 2 2 I2n+1 2 [2 (n!) ] Thus, = 2 → 1. In standard form: I2n π [(2n)!] (2n+1) 22n(n!)2 √ √ → π (2n)! n Application: This allows us to ﬁll the gap in the previous statement of Stirling’s formula. Recall that −n n+ 1 n! = Cne n 2 Now, √ 22n(n!)2 π = lim √ n→∞ (2n)! n 2n −n n+ 1 2 2 (Cne n 2 ) = lim 1 √ n→∞ −2n 2n+ C2ne (2n) 2 n 1 C2 = lim √ n n→∞ 2 C2n √ = C/ 2

Thus, √ lim n![/nne−n 2πn] = 1 n→∞

11 4.3 Wallis’s own derivation A consideration of area of a quarter-circle shows that: Z 1 √ π 1 − x2dx = 0 4 Write, Z 1 1 q ap,q = 1/ (1 − x p ) dx 0 R 1 n 1 ±1 There’s already a way of computing this if p, q ∈ N since 0 x dx = n+1 if n ∈ N. After computing ap,q, 1 ≤ p, q ≤ 10, Wallis observed the following pattern: p + q a = a p,q q p,q−1

Now let bm,n = am/2,n/2 and observe that if m, n are even, then m + n b = b m,n n m,n−2 Assume:(In a leap of faith), this also holds when m, n are odd. Historical Note: Fermat critiqued it as ”Incomplete Induction”. Consider the sequence {b1,n} and (based on the audacious assumption) observe that: n + 1 b = b 1,n n 1,n−2 Thus if n s even, 3 n + 1 b = 1 × ... × 1,n 2 n If n is odd, 2 2 2 n + 1 b = × × ... × 1,n π 1 3 n 1 n R 2 2 Also, (b1,n) ↑ since b1,n = 1/ 0 (1 − x ) dx. In particular, b1,2n−1 < b1,2n < b1,2n+1 for every n and hence,

n n n+1 2 Y 2k Y 2k + 1 Y 2k < 2π < π 2k − 1 2k 2k − 1 k=1 k=1 k=1 Rearrange and take limits to obtain, n π Y (2k)2 2 2 4 4 = lim = ...... 2 n→∞ (2k − 1)(2k + 2) 1 3 3 5 k=1

1 3 1 R 2 n/2 R m/2 1/2 Application: Let bn = 0 x (1 − x) dx and am = 0 x (1 − x) dx. 2 2 2 4 2 2 2 4 (a) Show directly that b0 = 5 , b2 = 5 . 7 , b4 = b2. 9 and a0 = 3 , a2 = 3 . 5 , a4 = a2. 7 . n m Thus, bn = n+5 bn−2 for n even and am = m+3 am−2 for m even. π (b) Assume: The same recursions hold for m, n odd as well. Show a1 = 8 . 3π 3/2 −1/2 (c) Conclude that b−1 = 6b1 = 6a3. Answer = 8 . Draw y = x (1 − x) .

12 4.4 Application to the Gaussian Integral Observe that (by replacing x by (π/2 − x),

Z π/2 n In = cos (x)dx 0 and recall that I2n π → 1; I2nI2n+1 = I2n+1 2(2n + 1) Thus, π π nI2 → ; nI2 → 2n 4 2n+1 4 We need the following two facts. • If t ≥ 0, then

n X n (1 + t/n)n = (t/n)k k k=0 n X ≤ (t)k/(k!) k=0 ≤ et

• If t ∈ [0, n], then (1 − t/n)n ≤ e−t it is easy to see that f(t) = e−t − (1 − t) is increasing on [0, 1] since f 0 ≥ 0. Thus, for t ∈ [0, 1], f(t) ≥ f(0) and e−t ≥ (1 − t). Now if t ∈ [0, n], then t/n ∈ [0, 1]. Thus (1 − t/n) ≤ e−t/n ⇒ (1 − t/n)n ≤ e−t. √ If x ∈ [0, n], then 2 2 x 2 x (1 − )n ≤ e−x ≤ (1 + )−n n n √ Now upon substituting x = n[sin(t)], we have: √ √ Z n 2 x n √ π (1 − ) dx = nI2n+1 → 0 n 2 √ Similarly, substituting x = n[tan(t)], we have: √ √ Z n 2 x −n √ π (1 + ) dx = nI2n−2 → 0 n 2

R ∞ −x2 It is clear that the improper integral 0 e dx < ∞. Hence taking limits as n → ∞, ∞ √ Z 2 π e−x dx = 0 2

13 5 Newton’s binomial series and application

Standing on Wallis’s shoulders, Newton deﬁnes the following sequence of functions on (0, 1).

Z x ∞ 2m+1 2 n X m x f (x) = (1 − t ) 2 = a [(−1) ] n mn 2m + 1 0 m=0 Observe that am,n+2 = am−1,n + am,n, n = 2k

Assume√ that it’s also true when n is odd. Compute f1(x) by brute-force. 2 2 4 2 Write 1 − x = c0 + c2x + c4x ... and square the series on r.h.s to equate to 1 − x . Thus, √ 1 1 1 − x2 = 1 − x2 − x4.. 2 8 The coefﬁcient of x2k is 1/2(1/2 − 1)...(1/2 − k + 1) 1 (−1)k = (−1)k 2 k! k

Application Find the series for sin−1(t).

√ Assuming t ∈ (0, 1), consider the accompanying ﬁgure of the circle y = 1 − x2. _ Z t √ a( OP QR ) = 1 − x2dx 0 Z t 1 1 = (1 − x2 − x4 + ...)dx 0 2 8 1 = t − t3.. 6 √ _ _ _ 1 1 2 But OP QR = OQR ∪ 4OPQ; a( OQR ) = π/(2π/θ) = 2 θ and a(4OPQ) = 2 t 1 − t . Thus,

sin−1(t) = θ 1 1 1 = 2[t − t3..] − t[1 − t2 − t4 + ...) 6 2 8 1 = t + t3... 6

14 6 Leibniz rules!

Leibniz’s Transmutation Theorem is what we know as Integration by Parts.

dy Given a curve y = f(x) over x ∈ [a, b], let z = y − x dx , the inverse relationship between the tangent problem and the area problem. Then, Z b Z b 1 b ydx = [[xy]a + zdx] a 2 a Proof. Write A = (a, 0),B = (b, 0); C = (a, f(a)) and D = (b, f(b)). Area of interest = Z b _ ydx = a( OCD ) + a(4ODB) − a(4OAC) a _ OCD is made up of triangles like 4(OPQ). Draw the tangent at P , intersecting the y-axis in T and denote by S = the foot of the perpendicular from O to the tangent. Now 4OST ∼ 4PRQ since ∠SOT = ∠QP R where PR is perpendicular to the vertical line through Q. Write, p = OS, z = OT and dx = PR. Thus, p/dx = z/P Q since p can be identiﬁed with the length of the perpendicular from O to the line through P and Q. 1 1 Hence a(4OPQ) = 2 p.P Q = 2 z.dx. Thus, _ Z b a( OCD ) = Σa(4OPQ) = zdx a 1 1 Observe that, a(4ODB) = 2 bf(b) and a(4OAC) = 2 af(a). Hence Z b Z b 1 b 1 ydx = [xy]a + zdx a 2 2 a dy Substitute z = y − x dx , to get: Z b Z f(b) b ydx = [xy]a − xdy a f(a)

π Leibniz’s used this technique in the original derivation of the series expression for 4 .

15 Application:

√ Consider the circle (x − 1)2 + y2 = 1,so y = 2x − x2 in the upper half-plane and area of the π quarter-circle is 4 . dy 1−x dy x Now dx = y and hence z = y − x dx = y . 2 2 2 2 2 2 2 2 2z2 Thus x = z y = z (2x − x ) and x (1 + z ) = 2xz ⇒ x = 1+z2 . Expanding (1 + z2)−1 into a geometric series we have: z2 2 4 n−1 2(n−1) 2n+1 R 1 1+z2 = z − z ... + (−1) z + Rn(z) where Rn(z) ≤ |z| implies 0 Rn(z)dz → 0. R 1 z2 1 1 (−1)2n+1 1 1 Thus, 0 1+z2 dz = limn→∞( 3 − 5 + ... + 2n+1 ) = 3 − 5 + .. (a convergent series). R 1 1 R 1 R 1 Also, z = y = 1 when x = 1 hence 0 zdx = [xz]0 − 0 xdz = 1 − 0 xdz. Now, π Z 1 = ydx 4 0 Z 1 1 1 = ([xy]0 + zdx) 2 0 1 Z 1 = (1 + zdx) 2 0 1 Z 1 = (2 − xdz) 2 0 1 Z 1 = 1 − xdz 2 0 Z 1 z2 = 1 − 2 dz 0 1 + z 1 1 = 1 − + + .. 3 5 An easy corollary is as follows, giving us a much better rate of convergence to estimate π. π 1 1 = + + ... 8 1.3 5.7 Exercise: Show that 1 1 log 2 = + + ... 1.2 3.4 Estimate log 2.

16 7 Euler’s ﬁrst estimate of ζ(2)

Euler’s ﬁrst step is to write ζ(2) as an integral,using Mercator’s series Step 1: − log(1−x) 1 1 2 Recall: x = 1 + 2 x + 3 x + ..., converges uniformly on [0, t] whenever t ∈ (0, 1). R t − log(1−x) 1 2 1 2 Thus, for t ∈ (0, 1), 0 x dt = t + 22 t + 32 t + .... Moreover, it is known that the series on r.h.s. converges at t = 1 to ζ(2) and hence by Abel’s theorem, Z 1 − log(1 − x) dt = ζ(2) 0 x Rest of the argument involves computing the integral on l.h.s. Step 2: Observe that by l’Hospital’s rule, limx→1 log(x) log(1 − x) = 0 and integrate by parts to write: Z 1 log(1 − x) Z 1 log(x) dt = −(log 2)2 + dx 1/2 x 1/2 1 − x Z 1/2 log(x) = −(log 2)2 + dx 0 1 − x Step3:

Z 1/2 log(x) Z 1 log(x) −ζ(2) = dx + dx 0 1 − x 1/2 1 − x Z 1/2 log(x) = −(log 2)2 + 2 dx 0 1 − x Step4 : The second term is easily computed by integrating the series term-by-term and we get: ∞ Z 1/2 log(1 − x) X 1 2 dt = −2 x n2.2n 0 n=1 Thus ∞ X 1 ζ(2) = (log 2)2 + 2 n2.2n n=1 P∞ 1 Using the expansion log 2 = n=1 n.2n , Euler came up with the estimate: ζ(2) ≈ 1.644934..

Exercises: P∞ 1. Show that k=2(ζ(k) − 1) = 1.

2. Show that Z ∞ t ζ(2) = t dt 0 e − 1

17 8 A Passage to Taylor’s Theorem

8.1 Euler’s trigonometric functions and their derivatives ”A function of variable quantity is an analytical expression composed in any way from this variable quantity and from numbers or constant quantities.” We recall the basic trigonometric identities, which are valid for all n, m ∈ N. 1. 2cos(nx)sin(mx) = sin(n + m)x − sin(n − m)x.

2. 2cos(nx)cos(mx) = cos(n + m)x − cos(n − m)x.

3. 2sin(nx)sin(mx) = cos(n − m)x − cos(n + m)x.

Euler’s treatment of trigonometric functions is based on the use of complex numbers and De Moivre’s formula. (cosθ ± isinθ)n = cos(nθ) ± isin(nθ) Choose ε to be ”small”, n to be ”large” sothat cosε = 1, sinε = ε, and n ∼= n − 1 ∼= n − 2... Writing the two formulae down and adding and subtracting them we get: 1 cos(nε) = [(cosε + isinε)n + (cosε − isinε)n] 2 Expand using the binomial theorem and let x = nε. Now, 1 cosx = ((1 + ix)n + (1 − ix)n)) 2 x2 x4 = 1 − + ... 2! 4! Similarly, 1 sinx = ((1 + ix)n − i(1 − ix)n)) 2i x3 x5 = x − + ... 3! 5!

z z n Having already identiﬁed e = (1 + n ) , Euler now has his famous formulae: 1 1 cosθ = (eiθ + e−iθ); sinθ = (eiθ − eiθ) 2 2i Next, the series is used to ﬁnd derivatives, by dropping (dx)2 and higher powers.

d(sinx) = sin(x + dx) − sinx = sin(x)cos(dx) + cos(x)sin(dx) = cos(x)dx

18 8.2 Gregory-Newton interpolation The interpolation formula was originally stated by J. Gregory in a letter he wrote in 1670. Newton published its derivation in 1711 in Methodus Differentialis.

Lemma 8.2.1. Given f and n + 1 equally spaced points {x0, x1, ...xn}, let f(xi) = yi. Then there exists a polynomial p of degree n satisfying f(xi) = yi, 1 ≤ i ≤ n. A formal proof of existence of such a polynomial can be given using Vandermonde determinants and belongs to the realm of Linear Algebra. Notation: Let 4x = xi − xi−1 and let 4yi = yi − yi−1.So 4y0 = y1 − y0. 2 3 2 2 Further, 4 y0 = 4y1 − 4y0 = y2 − 2y1 + y0, 4 y0 = 4 y1 − 4 y0 = y3 − 3y2 + 3y1 − y0 etc. It is easier to solve for coefﬁcients of p(x) if it is written as follows.

p(x) = A0 + A1(x − x0) + A2(x − x0)(x − x1) + ... + An(x − x0)...(x − xn−1)

If x − x0 = s4x, then x − x1 = (s − 1)4x, x − x2 = (s − 2)4x.... k Let Bk = Ak(4x) and observe that,

p(x) = B0 + B1s + B2s(s − 1) + .. + Bns(s − 1)..(s − n + 1)

The condition that p(x0 + k4x) = yk, for 0 ≤ k ≤ 3 yields a system of equations.

y0 = B0

y1 = B0 + B1

y2 = B0 + 2B1 + 2B2

y3 = B0 + 3B1 + 6B2 + 6B3

2 3 4 y0 4 y0 resulting in the solution B0 = y0,B1 = 4y0,B2 = 2! and B3 = 3! . In general, n 4 y0 yn = B0 + nB1 + n(n − 1)B2 + ... + n!Bn and by induction, Bn = n! . Thus, n X s f(x + s4x) = 4iy 0 i 0 i=0 This was Newton’s tool for ﬁnding area under any curve by using an interpolating polynomial. Newton: ”Hence the areas of all curves may be nearly found.” R.Cotes was a student of Newton and by letting n = 4, they obtained the ”Newton-Cotes rule”:

Z x3 ∼ 34x f(x)dx = (y0 + 3y1 + 3y2 + y3) x0 8 Exercise: Let n = 3 and derive Simpson’s rule. R 1 x2 Approximate 0 e dx