Re: [數值]梯形法的誤差估計推導
※ 引述《drinks9216 (drinks)》之銘言:
: 各位前輩與同好午安 ~
: 小弟前幾天看梯形法的誤差估計推導 , 中間有點卡住
: 想了好幾天都想不出來 , 所以PO上來問大家
: 中間有個式子 ,
: x_{i+1}
: f"(c_i)∫ {(x-x_i)(x-x_i+1)}/2! dx = -h^3/12 (f"(c_i))
: x_i
: 其中 h=(b-a)/n ; a=x_0 < x_1 < .... < x_n-1 < x_n=b
x_{i+1}
f"(c_i)∫ {(x-x_i)(x-x_{i+1})}/2! dx
x_i
f"(c_i) x_i+h f"(c_i) x_i+h 2
=---------∫ (x-x_i)(x-(x_i+h))dx=---------∫ (x-x_i) -h(x-x_i)dx
2! x_i 2! x_i
3 2
f"(c_i) (x-x_i) |x=x_i+h h(x-x_i) |x=x_i+h
=---------[-------- | - ----------| ]
2 3 |x=x_i 2 |x=x_i
3 3
f"(c_i) h h -f"(c_i) 3
=---------[----- - ---------]=----------h
2 3 2 12
: 主要是積分那 不知道怎麼變成 -h^3/12
: 所以PO上來問大家 , 感謝 <(__)>
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※ 編輯: jacky7987 來自: 111.251.146.209 (07/25 16:55)
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