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Update 2024-03-01-数学学习.md
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nooblong committed Jul 15, 2024
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Expand Up @@ -203,9 +203,13 @@ ln 答案 = $ln(\lim_{n \to \infty} (1 + \frac{1}{n})^n)$ = 1

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$f(x) \approx f(x) + f(x)' f(x - x0)$
$f(x) \approx f(x_0) + f'(x_0)*(x - x_0)$

$\frac{\Delta f} {\Delta x} \approx f'(x_0)$ ($\Delta x = x - x_0$)
$\frac{\Delta f} {\Delta x} \approx f'(x_0)$ ($\Delta x = x - x_0$) 当x=x0时,和上面公式一样

ln 1.1 = ?

x0 在 0 时,ln(1+x) ~= f(0) + f'(0)f(x-x0) ~= ln(1) + 1/1+0 * (x - 0) ~= x

x0 在 0 时,(1+x)^r ~= 1 + r(1+0)^r-1 * (x - 0) ~= 1 + rx

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