人工智能教程 - 數學基礎課程1.1 - 數學分析(一)8-9 導數的應用,線性近似(逼近),二階近似,凹凸性

導數的應用 Applications of diff

線性近似(逼近)

Linear Approximations

f(x)f(x0)+f(x0).(xx0){\color{Red} f(x) \approx f(x_0)+f'(x_0).(x-x_0)}

General:

ln(1+x)xln(1+x) \approx x

(1+x)r1+rx(1+x)^r \approx 1+rx

f’ f(0) f’(0)
1/(1+x) 0 1
r(1+x)^(r-1) 1 r
EX2:

ln1.1110ln1.1 \approx \frac{1}{10}

ln(1+x)xln(1+x) \approx x

EX3:Find Linear
approx near x=0 of

e3x1+x=e3x.(1+x)12\frac{e^{-3x}}{\sqrt{1+x}} = e^{-3x}.(1+x)^{\frac{1}{2}}

(13x).(112x)\approx (1-3x).(1-\frac{1}{2}x)

13x12x+32x2\approx 1-3x-\frac{1}{2}x+\frac{3}{2}x^2

172x\approx1-\frac{7}{2}x

drops  x2 term(s) +x3 and higherdrops \ \ x^2 \ term(s) \ +x^3 \ and \ higher

EX4:

satellite
時間膨脹(timedilation)

T=T1v2t2T' = \frac{T}{\sqrt{1-\frac{v^2}{t^2}}}

(1u)121+12u\because (1-u)^{-\frac{1}{2}} \approx 1+\frac{1}{2}u

TT.(1+12. v2c2)\therefore T' \approx T.(1+\frac{1}{2}. \ \frac{v^2}{c^2})

二階近似

(Quadratic approximation)

f(x)f(x0)+f(x0).(xx0)+f(x0)2(xx0)2{\color{Red} f(x) \approx f(x_0) +f'(x_0).(x-x_0)+\frac{f''(x_0)}{2}(x-x_0)^2}

x0x \approx 0,則:

sinxxsinx \approx x

cosx112x2cosx \approx 1-\frac{1}{2}x^2

ex1+x+12x2e^x \approx 1+x+\frac{1}{2}x^2

f’’ f’’(0)
-sinx 0
-cosx -1
exe^x 1

Quadratic approx

(use these when linear is not enough)

important rule!!!

f(x)f(0)+f(0)x+f(0)xx2{\color{Red} f(x) \approx f(0)+f'(0)x+\frac{f''(0)}{x}x^2 }

逼近 rate of convergence

(lnak)10(ln{a_k})-1\rightarrow 0

曲線構圖 Curve sketching
Goal: draw graph of f using f’,f’’,positive/negative

f>0f is increasingf'>0\Rightarrow f \ is \ increasing
f<0f is decreasingf'<0\Rightarrow f \ is \ decreasing
f>0f is increasingf''>0\Rightarrow f' \ is \ increasing

f concave up 凸

f<0f is decreasingf''<0\Rightarrow f' \ is \ decreasing

f concave down 凹
  • 駐點 critical point
  • 駐點值 critical value
  • 極值點 turning points
  • 拐點 inflection point
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