啟蒙數學

數學歸納法（Mathematical Induction）是數學上的一種證明方法, 其被用於證明某個給定公式或 定理 在整個自然數範圍內成立且不須知道推導公式或定理 卻能證明公式或定理是否正確的方法. 換言之,其為證明函數ƒ(x)的公式是否成立的方法  - 先證起始點是正確, 如命起始點n=1時而ƒ(1) 成立, 並設n=k時ƒ (k) 成立且n=k+1時也成立. 所以此函數公式是成立的.此觀念與骨牌遊戲的原理是一樣. 如n=1成立推至n=k=1 n=k+1=2也成立,現在n=k=2成立推至n=k+1=3也成立,於至列推n=k=任何都成立. 以下是用數學歸納法證明ån² = n(n+1)(2n+1)/6 是否正確,與用數學歸納法推導及證明隸美佛定理的例子. Mathematical induction is a method of mathematical proof that a given formula or theory, whether is true for all natural numbers or not with and without knowing how to derive the formula or theory. It is done by proving that the initial point a formula or theory is true, and then proving that if any one point in the formula is true, then the next one is true, too. For example, let n=1 is true then n=k=1 Þ n=k+1=2　is true.  Since n = 2 is true, then n=k+1=2 Þ n=k+1=3 is also true. By doing so, n= k = any number can be proven to be true.  The following are two examples to demonstrate how to use Mathematical induction to prove that ån² = n(n+1)(2n+1)/6 is true, even one does not know how to derive the formula and also to derive and prove De Moivre's Theorm.

例1: 證明ån² = n(n+1)(2n+1)/6是成立的.

1)     當n=1時, ån² = n(n+1)(2n+1)/6 => 1² = 1 = 1(1+1)(2·1+1)/6是成立.

2)  設n=k時, åk² = k(k+1)(2k+1)/6是成立. k(k+1)(2k+1)/6+(k+1)²= [k(k+1)(2k+1)+6(k+1)²]/6 = (k+1) (k+2)(2k+3)/6= (k+1) [(k+1)+1][(2k+1)+1]/6, 證明n=k+1也成立. 所以ån² = n(n+1)(2n+1)/6是成立的.

Q1.   Prove ån² = n(n+1)(2n+1)/6 is true.

1)     Prove that n=1, ån² = n(n+1)(2n+1)/6 => 1² = 1 = 1(1+1)(2·1+1)/6 Therefore when n = 1 that the formula is true – prove the initial point is true.

2) when n=k then åk² = k(k+1)(2k+1)/6 is true, Need to prove that k(k+1)(2k+1)/6+(k+1)²= [k(k+1)(2k+1)+6(k+1)² ]/6 = (k+1) (k+2)(2k+3)/6= (k+1) [(k+1)+1][(2k+1)+1]/6, Therefore, n=k+1 is also true. So ån² = n(n+1)(2n+1)/6 is true.

例2: 推導及證明隸美佛定理 (z=|a|( cos q+isin q) 則zⁿ= |a| ⁿ (cos nq+isin nq ).

1)     當n =2時, z²=|a|²( cos q+isin q)²=|a|² ( cos²q+2cosq isin q+(isin q)²) = |a|² ( cos²q+2cosq isin q - sin²q) = |a|² [( cos²q - sin²q) +2cosq isin q] = |a|²( cos2q+isin2q) - \ cos2q = cos²q - sin²q 和isin2q= i2cosq sin = 2cosq isin是成立的.

2)     設n=k時z^k=|a|^k( cos q+isin q)^k=|a|^k ( cos kq+isin kq)是成立.  |a|^k ( cos kq+isin kq)·|a|( cos q+isin q) = |a|^k+1( cos kq cosq+cosq isin kq+isin q cos kq + isin kq isin q) = |a| ^k+1 (cos kq cosq+icosq sin kq+ isin q cos kq  - sin kq sin q) = |a| ^k+1  [(cos kq cosq - sin kq sin q ) + i(cosq sin kq+ sin q cos kq)] = |a| ^k+1  ( cos(k+1)q+isin(k+1)q ) - \ cos(kq+q) = cos kq cosq - sin kq sin q 和isin(kq+q) = i(cosq sin kq+ sin q cos kq). 證明n=k+1也成立. 所以隸美佛定理成立.

Q2.  Prove De Moivre's Theorm (z=|a|( cos q+isin q) 則zⁿ= |a| ⁿ (cos nq+isin nq ).

1)     Prove that n =2, z²=|a|²( cos q+isin q)²=|a|² ( cos²q+2cosq isin q+(isin q)²) = |a|² ( cos²q+2cosq isin q - sin²q) = |a|² [( cos²q - sin²q) +2cosq isin q] = |a|²( cos2q+isin2q) - \ cos2q = cos²q - sin²q 和isin2q= i2cosq sin = 2cosq isin. Therefore when n = 2 that the formula is true – prove the initial point is true.

2)     when n=k then z^k=|a|^k( cos q+isin q)^k=|a|^k ( cos kq+isin kq) is true.  Need to prove that |a|^k ( cos kq+isin kq)·|a|( cos q+isin q) = |a|^k+1( cos kq cosq+cosq isin kq+isin q cos kq + isin kq isin q) = |a| ^k+1 (cos kq cosq+icosq sin kq+ isin q cos kq  - sin kq sin q) = |a| ^k+1  [(cos kq cosq - sin kq sin q ) + i(cosq sin kq+ sin q cos kq)] = |a| ^k+1  ( cos(k+1)q+isin(k+1)q ) - \ cos(kq+q) = cos kq cosq - sin kq sin q 和isin(kq+q) = i(cosq sin kq+ sin q cos kq). n=k+1 is also true. Therefore, De Moivre's Theorm is true.

例3:  您是否能證明z=|a|( cos q+isin q) 則z^(n/m)= |a| ^(n/m) (cos (n/m)q+isin  (n/m)q )?

Q2.  Could you prove z=|a|( cos q+isin q) 則z^(n/m)= |a| ^(n/m) (cos (n/m)q+isin  (n/m)q )?