VincentTam

由於 (1 + 1/n)ⁿ 是二項式的 n 次方，要了解它的特性（如剛才展示的上下限），需要用上二項式定理。上面不等式最具技巧的一步在於藍色分母從 n! 變成 2ⁿ⁻¹。 小問題：為何上標有 "-1"？ 重點在於 n! = 1 ⋅ 2 ⋅ 3 ⋯ (n - 1) n 中，當 n! 變成 (n + 1)! 時，額外乘上的數字 n + 1 是隨 n 而增長，但 2ⁿ 變成 2ⁿ⁺¹ 時，額外乘上的數字永遠是 2，所以當 n「足夠大」（n > 3）時，n! > 2ⁿ。 取倒數後不等式反轉，於是出現 1/n! < 1/2ⁿ，然後就可以運用等比數列和公式，為 (1 + 1/n)ⁿ 給出上限。

上面實驗發現 n 越大，(1 + 1/n)ⁿ 也越大。即若 m < n，(1 + 1/m)ᵐ < (1 + 1/n)ⁿ。這種數列我們叫__嚴格遞增數列__。（「嚴格遞增」比「單調遞增」「嚴格」，只接受「>」，不接受「=」。） 練習：試證 (1 + 1/n)ⁿ < (1 + 1/(n + 1))ⁿ⁺¹ 提示：

1. 如上面般展開 (1 + 1/n)ⁿ，分離最簡單兩項。
2. 其餘 n - 1 項中，調整分子分母配對，先把 1 / k! 放在左邊。

VincentTam

# 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 <florent.chauveau@gmail.com>
# 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

VincentTam

Interpretation of exponentation and permutation

Question

I have $3^3$, can I call it a permutation/permutation with repetition?

As written in here:

VincentTam

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

VincentTam

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)

VincentTam

# 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

VincentTam

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
 

VincentTam

My answer to a deleted question Interpretation of exponentation and permutation on math.SE

WayneXMayersX asked this question.

Question

I have $3^3$, can I call it a permutation/permutation with repetition?

VincentTam

# 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

VincentTam

# 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);