這個數列一直上升，其實很合乎直覺，因為複利過程中，將一週期的本利和更新作下一週期的本金，可以賺更多。n 越大，代表更新本利和次數更多，自然賺更多。 思考：這個（本利和）數列若可以上升至「無限大」（想要多大，就有多大。比如想要存款一億元，就真的有個 n 值，使數列中有一項不少於一億），那會發生甚麼事？ 答：||那還工作幹嘛？存錢進銀行，然後設定 n「無限大」，讓自己躺着賺，想有多少錢花，就有多少錢花——但世上的貨幣是有限的，那麼貨幣供應會短缺。|| 為何上述事情不會發生？ 答：||從上面實驗觀測，當 n 「很大」時，該數列「接近」一個約為「2.71828」的數字。|| 上面的解答只為給大家一個直覺，並非正式答案，而是背後正式數學內容的一個引子。
接下來假設大家已熟悉要用上
| # This is a GitLab CI configuration to build the project as a docker image | |
| # The file is generic enough to be dropped in a project containing a working Dockerfile | |
| # Author: Florent CHAUVEAU <[email protected]> | |
| # Mentioned here: https://blog.callr.tech/building-docker-images-with-gitlab-ci-best-practices/ | |
| # do not use "latest" here, if you want this to work in the future | |
| image: docker:18 | |
| stages: | |
| - build |
| using Primes | |
| plist = primes(1,10000) | |
| sumdigits(n, base=10) = sum(digits(n, base)) | |
| for p in plist[1:100] | |
| sdgit = sumdigits(big(p)^2016) | |
| println("$(p), $(log(p)), $(sdgit)") | |
| if sdgit == 2017 | |
| break | |
| end | |
| end |
| figure | |
| title('A connected not path connected set') | |
| hold on | |
| % Plot set A | |
| for i = 1:10 | |
| plot([0,1], [0,1/i], "linewidth", 2) | |
| end | |
| % Plot set B | |
| plot([0.5,1],[0,0],'--', "linewidth",2) |
| # input | |
| a = int(input("Donner la valeur de a: ")) | |
| b = int(input("Donner la valeur de b: ")) | |
| # avant échange | |
| print("a = ",a) | |
| print("b = ",b) | |
| # échange | |
| a, b = b, a |
| Nvec = 10000:10000:1000000; % no of intervals | |
| rrsumvec = zeros(size(Nvec)); | |
| for ind = 1:length(Nvec) | |
| N = Nvec(ind); | |
| dx = 5 / N; | |
| x = (1:N) * dx; | |
| fx = sqrt(25 - x.^2); | |
| rrsum(ind) = sum(fx * dx); | |
| end | |
| # Show dual simplex tableau | |
| # Prereq: matrix A, vectors b,c, basis | |
| printf("Current basis:"); printf(" %2i", basis); disp(""); | |
| B = A(:,basis); cB = c(basis); | |
| Bm1A = B\A; xB = B\b; zrow = cB'*Bm1A-c'; zval = cB'*xB; | |
| if length(xB(xB<0)) >= 1 | |
| ovp = find(xB==min(xB(xB<0)))(1); | |
| r = zrow./Bm1A(ovp,:); # ratio for display | |
| T = [0:size(A)(2) 0; basis' Bm1A xB; 0 zrow zval; 0 r 0] | |
| # compute min ratio |
| # Show simplex tableau | |
| # Prereq: matrix A, vectors b,c, basis | |
| printf("Current basis:"); printf(" %2i", basis); disp(""); | |
| B = A(:,basis); cB = c(basis); | |
| Bm1A = B\A; x_B = B\b; zrow = cB'*Bm1A-c'; zval = cB'*x_B; | |
| nv = find(zrow==min(zrow(zrow<=0)))(1); | |
| r = x_B./Bm1A(:,nv); | |
| T = [0:size(A)(2) 0 0; basis' Bm1A x_B r; 0 zrow zval 0] | |
| if length(r(r>=0)) >= 1 | |
| ovp = find(r==min(r(r>=0)))(1); |
