Functions 6 PDF

## Chapter 6: Series and Convergence ### 6.1 Infinite Sums We consider the infinite sum S = Σ_n=0^∞ a_n What does this mean? The approach we take is to consider a **truncation** to: S_N = Σ_n=0^N a_n and to exam whether lim...

## Chapter 6: Series and Convergence ### 6.1 Infinite Sums We consider the infinite sum S = Σ_n=0^∞ a_n What does this mean? The approach we take is to consider a **truncation** to: S_N = Σ_n=0^N a_n and to exam whether lim_N->∞ S_N exists, that is to say, has a finite value. If so, we say the series S **converges**. If not, then the series **diverges**, and the infinite sum then has no meaning. We note that divergence can involve |S_N| increasing without bound as N->∞ or that S_N just does not approach a finite limit eg. it could just oscillate. eg. (-1)ⁿ oscillates So = 1 , S₁ = 0, S₂ = 1, etc. #### Examples **(i)** The geometric series: S_N = 1 + x + x² + ... + x^N Evidently: xS_N = x + x² + ... + x^N + x^N+1 ⇒ S_N(1-x) = 1- x^N+1 and S_N = 1 - x^N+1 / 1 - x So S = lim_N->∞ S_N exists **ONLY** for |x| < 1. For |x| ≥ 1 the series **diverges**. Hence our Maclaurin series expansion of f(x) = 1 + x + x² + ... is **ONLY** valid for |x| < 1. **(ii)** Consider Σ_n=1^∞ 1/n (The Harmonic Series) This series **diverges**. **Proof** 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ... ⇒ Σ_n=1^∞ 1/2ⁿ > 1 + 1/2 + 1/2 + 1/2 + ... So Σ_n=1^∞ 1/2ⁿ ->∞ as N->∞ ### 6.2 More Definitions and Theorems An infinite series S = Σ_n=0^∞ a_n is said to be **absolutely convergent** if Σ_n=0^∞ |a_n| is **convergent**. This is useful because of the theorem: **Absolute Convergence -> Convergence**, and can be applied when some a_n may be negative or complex #### Examples **(i)** S = 1/2 - 1/4 + 1/8 - 1/16 + ... is absolutely convergent because S = 1/2 + 1/4 + 1/8 + 1/16 + ... (geometric progression) = 1/2 / 1- 1/2 = 1. Here the convergence of S -> Convergence of S. [In fact, it is easy to see S = 1/2 / ( 1 - (1/2)² = 2/3]. If Σ_n=0^∞ |a_n| converges and Σ_n=0^∞ a_n does not, then we say Σ_n=0^∞ a_n is **conditionally convergent**. **(ii)** S= 1 -1/2 + 1/3 - 1/4 ... is **conditionally convergent** since it converges (to ln 2), but of course 1 + 1/2 + 1/3 ... diverges. ### 6.3 Convergence Tests Four simple ones are (i) Σ_n=0^∞ a_n **(a)** ** Necessary** but not **sufficient** for convergence is that lim_n->∞ a_n = 0. If this limit is NOT zero, then the series **diverges**. If it's zero, then the series may still **diverge**, e.g. a_n = 1/n. **(ii)** Σ_n=1^∞ cos (1/n) is **divergent** since cos(1/n) ->1 as n->∞ **(b) Comparison Test ** If a_n is given and we can find a **converging series** Σ_n=0^∞ b_n with b_n non-negative such that |a_n| ≤ b_n for all n, then Σ_n=0^∞ a_n is **absolutely convergent** - and so is convergent. Similarly, if we can find a **diverging series** Σ_n=0^∞ b_n with b_n non-negative such that a_n ≥ b_n for all n, then Σ_n=0^∞ a_n is **divergent** **(iii)** Σ_n=1^∞ 1/n² is divergent because n ≤ n² for all n and we know Σ_n=1^∞ 1/n is divergent. **(c) Alternating Series - Leibniz Test** If we have Σ_n=0^∞ (-1)ⁿ a_n with: (i) positive a_n (ii) a_n decreasing a_n+1 ≤ a_n (iii) a_n -> 0 as n ->∞, then our series **converges**. **(iv)** 1 - 1/3 + 1/5 - 1/7 + ... is **convergent**. It's sum is actually tan^-1(1) = π_/4. [It should be noted that this convergence is **VERY SLOW** - there are much better ways of approximating π ...]. A natural question to ask is whether condition (ii) is actually needed - as long as particularly (iii) is satisfied. That it is so is given by: S = √2 - 1 - 1/(√2 + 1) + √3 - 1 - 1/(√3 + 1) + ... [ image of graph ] **Conditions** (i) ✓, (ii) x, (iii) ✓ So **FAIL** for the convergence tests. **BUT** of course we have: S_2k = 2/(√2-1) + 2/(√3-1) + 2/(√4-1) + .... (2k terms) = 2 [ 1 + 1/2 + 1/3 + ... +1/k ] -->∞ as k -> ∞ So that S **DIVERGES** **(d) The Ratio Test** For Σ_n=0^∞ a_n we suppose a_n≠ 0 for all n. Define L = lim_n->∞ |a_n+1 / a_n| Then if: |L| < 1 -> **Absolute Convergence -> Convergence** |L| > 1 -> **Divergence** If L = 1 we have **NOT PROVEN** = Don't know! - at least without further work! ### 6.4 Radius of Convergence of Taylor/Maclaurin Series The ratio test allows us to determine the range of convergence of Taylor/Maclaurin series. **(i)** f(x) = e^x = Σ_n=0^∞ xⁿ / n! = Σ_n=0^∞ a_n So |a_n+1 / a_n| = (xⁿ⁺¹ / (n+1)!) / (xⁿ / n!) = |x| / (n +1). And we have L = lim_n->∞ |a_n+1 / a_n| = lim_n->∞ |x| / (n +1) = 0 for all fixed x. Since L < 1, the Maclaurin series for e^x **converges for all x** [Real and complex! note RADIUS OF CONVERGENCE]. **(ii)** f(x) = sin(x) = Σ_n=0^∞ (-1)ⁿ x²ⁿ⁺¹ / (2n+1)! Here | a_n+1 / a_n| = ( |x|²ⁿ⁺³ / (2n + 3)! ) / ( |x|²ⁿ⁺¹ / (2n+1)! ) = x² / (2n+3)(2n+2) Thus, L = lim_n->∞ x² / (2n + 3)(2n+2) = 0 for all fixed x. And again our Maclaurin series for sin(x) **converges for all x** [Real and complex! note RADIUS OF CONVERGENCE]. **(iii)** f(x) = 1/x = Σ_n=0^∞ xⁿ L = lim_n->∞ | xⁿ⁺¹ / xⁿ | = |x| so convergent for |x| <1. **(iv)** f(x) = ln(1 + x) = Σ_n=0^∞ (-1)ⁿ xⁿ⁺¹ / (n+1) L = lim_n->∞ |( (-1)ⁿ⁺¹ xⁿ⁺² / (n+2) ) / ( (-1)ⁿxⁿ⁺¹ / (n+1) ) | = |x|. So convergence if |x| < 1. ( RADIUS OF CONVERGENCE 1) ** WITHIN** the 'RADIUS OF CONVERGENCE' we have CONVERGENCE. **AT** that value, we have NOT PROVEN = DON'T KNOW! **(v)** Σ_n=1^∞ xⁿ / n = x + x² / 2 + x³ / 3 ... Ratio test: L = lim_n->∞ |(xⁿ⁺¹/(n+1)) / ( xⁿ / n) | = |x| So the series **CONVERGES** if |x| < 1, **DIVERGES** if |x| > 1. ( RADIUS OF CONVERGENCE 1) At x = +1, we have 1 + 1/2 + 1/3 ... **DIVERGES**. At x = -1, we have -1 + 1/2 - 1/3 + 1/4 - ... **CONVERGES** = -ln2. [The word RADIUS in our accounts refers to that of a CIRCLE OF CONVERGENCE in the complex plane]. ### 6.5 The Mysterious Zeta Function* The zeta function of RIEMANN(1859) is the function: ζ(s) = Σ_n=1^∞ 1/n^s Which has attained a mysterious, indeed almost mythical status. ζ(1) is of course the harmonic series, which diverges. ζ(s) can be shown to converge: Re s > 1 [NOT by the tests in 6.3 above, which are inconclusive, but by being bounded by an integral. The tail is bounded above by an integral that is convergent. [image of graph]. ie. 1/2^s + 1/3^s + ... < ∫₁^∞ 1/x^s dx. This argument works for s complex too. We can find: ζ(2) = 1 + 1/2² + 1/3² + ... = π² / 6 = 1.6449... ( The BASEL PROBLEM, not easily by elementary means) ζ(4) x⁴ / 90 etc. ζ(3) = 1.2020569... ( APÉRY, IRRATIONAL 1979) THE RIEMANN HYPOTHESIS is that ALL non-trivial zeros of the ζ function have real part 1/2, ie. s = 1/2 + i.t [This is known as the CRITICAL LINE in the complex s plane]. 1/2 ± 14.135 i, 1/2 ± 21.022 i, 1 / 2 ± 25.01i, .... A very large number have been computed [1 TRILLION]! but so far there is NO PROOF... **MYSTERY** Connections to (a) DISTRIBUTION OF PRIMES (b) STATISTICAL MECHANICS (c) QUANTUM CHAOS WHY? HOW?