Logarithm : inverse of exponential function

\[F(x) = 2^x \\ F^{-1}(x) = \log_2 x \\ \\ \log_2 1 = 0 \\ \log_2 2 = 1 \\ \log_2 4 = 2 \\ \log_2 32 = 5 \\ \log_2 \frac{1}{32} = -5 \\\]
  • Common Logarithm: $ \log_{10} x = \log x $ , 常用對數可以省略基底10，不寫出。
  • Natual Logarithm: $ \log_e x $

Logarithm Identities - 對數律

  • $ \log_b (1) = 0 $, b > 0 and b ≠ 1
  • $ \log_b (xy) = \log_b (x) + \log_b (y) $
  • $ \log_b (\frac{x}{y}) = \log_b (x) - \log_b (y) $
  • $ \log_b (x^p) = p \log_b (x) $

Change the base

\[\log_b (a) = \frac{\log_c (a)}{\log_c (b)} , c \gt 0, c \ne 1 \\ \log_{9} 27 = \frac{log_3 27}{log_3 9} = \frac{3}{2} = 1.5\]

Common Logarithm - 常用對數

\[\log (4) = \log (2 \times 2) = \log 2 + \log 2 = 0.301 + 0.301 \\ \log (5) = \log (10 / 2) = \log 10 - \log 2 = 1 - 0.301 \\ \\ \log A^r = r \log A \\ 10^{\log B} = B\]

換算 大數 與 科學符號

Example: 3的100次方為幾位數？其第一位數為?

log 2   = 0.301   >>> 10^0.301   ~= 2
log 3   = 0.47712 >>> 10^0.47712 ~= 3
log 5   = 0.6990  >>> 10^0.699   ~= 5
log 5.1 = 0.7075  >>> 10^0.7075  ~= 5.1
log 5.2 = 0.7160  >>> 10^0.716   ~= 5.2
log 6   = 0.7781  >>> 10^0.7781  ~= 6
\[3^{100} = 10^{\log(3^{100})} = 10^{100 \times \log(3)} = 10^{100 \times 0.47712} = 10^{47.712} \\ = 10^{0.712} \times 10^{47} = 5.1xxx \times 10^{47}\]

指數與對數方程式

Example: 求解 x = ? , $ \log_2(x-3) = log_4(x-1) $

真數必須為正數，所以 x - 3 > 0 & x - 1 > 0 , 可知 x > 3

\[\log_2(x-3) = \frac{log_2(x-1)}{log_2(4)} = \frac{log_2(x-1)}{2} \\ \log_2(x-3)^2 = \frac{log_2(x-1)}{log_2(4)} = log_2(x-1) \\ (x-2)(x-5) = 0 \\ x = 5\]

Euler’s number - 複利

100 元 的存款，年利率 12%, 若每 x 月 計算一次利息，複利計算後，明年初有多少錢？

x = 12    >>>   100 x (1 + 0.12 / 1)^1 = 100 x 1.12^1
x =  6    >>>   100 x (1 + 0.12 / 2)^2 = 100 x 1.06^2
x =  3    >>>   100 x (1 + 0.12 / 4)^4 = 100 x 1.03^4

若是 年利率 100%, 每毫秒計算一次利息:

\[\lim_{n \to \infty}(1 + \frac{1}{n})^n \to e = 2.718282... \\ \begin{align} (1 + \frac{1}{n})^n & = \binom{n}{0} (\frac{1}{n})^0 & + \binom{n}{1} (\frac{1}{n})^1 & + \binom{n}{2} (\frac{1}{n})^2 + \dots \\ & = 1 & + 1 & + \frac{n(n-1)}{2!n^2} + \frac{n(n-1)(n-2)}{3!n^3} + \dots \\ \end{align} \\ \lim_{x \to \infty} 1 + 1 + \frac{n(n-1)}{2!n^2} + \frac{n(n-1)(n-2)}{3!n^3} + \dots = \\ \lim_{x \to \infty} 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \dots = e\]

Monotone convergence theorem

若一數列為 單調上升 ，並為 有上界(Upper Bound) 則該數列必為收斂

若一數列為 單調下降 ，並為 有下界(Lower Bound) 則該數列必為收斂

指數函數的微分

\[f(x) = a^x, f'(x) = ? \\ f'(x) = \lim_{\Delta x \to 0} \frac{a^{x + \Delta x} - a^x}{\Delta x} = a^x \lim_{\Delta x \to 0} \frac{a^{\Delta x} - 1}{\Delta x}\]

假設 a^(Δx) - 1 = Δx, 上面的公式極限就會趨近1，

等同是 a = e，結果引出 $ f’(e^x) = e^x $

下面的證明過程令 n = 1 / Δx , so n → ∞

\[a^{\Delta x} - 1 = \Delta x \\ a^{\Delta x} = 1 + \Delta x \\ a = (1 + \Delta x)^{\frac{1}{\Delta x}} = (1 + \frac{1}{n})^n \\ a = \lim_{n \to \infty} (1 + \frac{1}{n})^n = e\]

Natural Logarithm

Definition : \(f(x) = log_e x = \ln x\)

\[\frac{d}{dx} \ln x = \frac{1}{x}\]

Prove:

\(e^{\log_e x} = x\) \(e^{\ln x} = x\) \(\frac{d}{dx} (e^{\ln x}) = \frac{d}{dx} x\) \(e^{\ln x} \frac{d}{dx} (\ln x) = 1\) \(\frac{d}{dx} (\ln x) = \frac{1}{e^{\ln x}} = \frac{1}{x}\)

對數函數的微分

$ f(x) = log_a x = \frac{\ln x}{\ln a} $, f’(x) = $ \frac{1}{\ln a \ \ x} $