MTH1030: Assignment 3, 2016 Ramanujan Due in hardcopy form with your support class leader on Wednesday, 12. October 2016, 5 p.m. The Rules of the Game ... are the same as those for the first two assignments. This assignment is worth a total of 100 marks. 1 Convergence at the speed of light [50 Marks, each subquestion is worth 10 marks] The Indian mathematician Srinivasa Ramanujan1 discovered the formula 1 π = 1 X n =0 p8(4n)!(1103 + 26390n) 9801(n!)43964n : Note that the series starts with n = 0. This means that in the following it makes sense to talk about the 0th term and the 0th partial sum of this series. This series converges at an amazing speed and was used in 1985 to compute the first 17,526,100 digits of π, which was the world record at the time. To prove this identity is quite tricky, so let's do some things with it that are within our reach. a) Use the 0th term of Ramanujan's series to approximate π. How many correct digits do you get (counting the 3 at the start as the first digit)? Do the same using the 0th and the 1st term of the series.2 1This is mathematician that the recent movie \The man who knew infinity" is all about. 2To coax Mathematica into displaying the first 40 digits of some number after the decimal point use the command N[number, 40]. For example, to display 40 digits of π use N[Pi, 40]. 1b) Using the ratio test, verify that the series is convergent. c) Let an denote the nth term of this series, starting with 0th term a0. It turns out that an+1 < Lan for n ≥ 1, where L is the limit that you calculated under b). Show that this implies that an < Ln−1a1 for n ≥ 2. d) Let Sn be the nth partial sum of our series starting with the 0th partial sum S0 = a0. Show that (for n ≥ 1) 0 < 1 π − Sn < a1Ln 1 − L : e) Use this estimate to find as small an n as possible such that the first 17,526,100 digits of Sn coincide with those of 1 π : Divide 17,526,100 by your number n to figure out how many additional correct digits of 1=π you get on average by adding one more term of the series. (Note that the reciprocal of an approximation of 1=π that shares 17,526,100 leading digits with 1=π will not necessarily have 17,526,100 leading digits in common with π. A little bit more work is needed to be able to guarantee that.) 2 Ramanujan's impossible sum [50 marks] Watch the Mathologer video Ramanujan: Making sense of 1+2+3+... = -1/12 and Co. https://youtu.be/jcKRGpMiVTw. a) In part 1 of the video it is shown how two infinite series can be multiplied. This way of multiplying is called the Cauchy product. It turns out that the Cauchy product of two absolutely convergent series is convergent and its sum is the product of the sums of the two series that get multiplied. Use this information to answer the following question. What infinite series do you arrive at when you square the geometric series 1 + 1 2 + 1 4 + 1 8 + · · ·? 2What is the sum of this new series? To double-check your result, use Mathematica to calculate the first 40 digits after the decimal point of the sum of the first 100 terms of the new series.3 [20 marks] b) In part 3 of the video the Cesaro sum of a series is introduced. Calculate the Cesaro sums of the series 1 − 1 + 0 + 1 − 1 + 0 + 1 − 1 + 0 + 1 − 1 + 0 + · · · 1 − 1 + 0 + 0 + 1 − 1 + 0 + 0 + 1 − 1 + 0 + 0 + · · · Both series are generated from the series 1 −1+1−1+· · · by inserting 0s into this sum. As you will see, by doing so the Cesaro sum changes. On the other hand, given a convergent (in the standard way) series, will it's sum change if you insert 0s into it? Explain. [20 marks] c) Guess the answer to the first of the hard homework assignments in part 4 of the video, that is, sum the series 1 − 22 + 32 − 42 + 52 − · · · based on the sequence of averages of averages of averages of the partial sums of this series. Base your guess on the first 5000 elements of this sequence calculated using Mathematica. Too devilish? Okay, here is some Mathematica code (just copy and paste into a Mathematica notebook) that you can modify and extend to calculate those 5000 elements (you only have to record the last few of those 5000 elements in your write-up). It calculates and prints the first 5000 elements of the sequence of averages of the partial sums of the series 1−1+1−1+· · · : [10 marks] ClearAll PartialSum[0] = 0; For[i = 0, i < 5000, i++, PartialSum[i + 1] = PartialSum[i] + (i + 1)^0 (-1)^i ] SumOfPartialSum[0] = 0; For[i = 0, i < 5000, i++, SumOfPartialSum[i + 1] = SumOfPartialSum[i] + PartialSum[i + 1]; Average[i + 1] = SumOfPartialSum[i + 1]/(i + 1); Print[N[Average[i + 1]]]] 3Hint: Use the command SUM (google the syntax) and display the first 40 digits using N[your sum, 40]. 3