文書の過去の版を表示しています。
台形公式の誤差
1
区間[a,b]をN分割した標本点をxk=a+hkとおく、ただし、hは分割幅で h=b−aN。ここで、x0=a、xN=bであることに注意。
また、 I0≡∫baf(x)dx=N−1∑k=0∫xk+1xkf(x)dxIn≡∫baf(n)(x)dx=f(n−1)(b)−f(n−1)(a) と定義する。
f(x)をx=xkの周りでテーラー展開したものを[xk,xk+1]で積分し、和をとると、 f(x)=f(xk)+∞∑n=11n!f(n)(xk)(x−xk)n∫xk+1xkf(x)dx=f(xk)(xk−xk+1)+∞∑n=11n!f(n)(xk)∫xk+1xk(x−xk)ndx=hf(xk)+∞∑n=1hn+1(n+1)!f(n)(xk)I0=hN−1∑k=0f(xk)+∞∑n=1hn+1(n+1)!N−1∑k=0f(n)(xk) 同様にf(x)をx=xk+1の周りでテーラー展開したものを[xk,xk+1]で積分し、和をとると、
I0=hN−1∑k=0f(xk+1)−∞∑n=1(−h)n+1(n+1)!N−1∑k=0f(n)(xk+1)
2
和をとる変数をそろえると、
I0=hN−1∑k=0f(xk)+∞∑n=1hn(n+1)!(hN−1∑k=0f(n)(xk))I0=hN∑k=1f(xk)+∞∑n=1hn(n+1)!((−1)nhN∑k=1f(n)(xk))
二式の平均をとると、12() I0=h2(N−1∑k=0f(xk)+N∑k=1f(xk))+∞∑n=1hn(n+1)!h2(N−1∑k=0f(n)(xk)+(−1)nN∑k=1f(n)(xk))
n=2m−1のとき、 N−1∑k=0f(n)(xk)+(−1)nN∑k=1f(n)(xk)=N−1∑k=0f(2m−1)(xk)−N∑k=1f(2m−1)(xk)=f(2m−1)(x0)−f(2m−1)(xN)=−(f(2m−1)(b)−f(2m−1)(a))=−∫baf(2m)(x)dx=−I2m
n=2mのとき、 h2(N−1∑k=0f(n)(xk)+(−1)nN∑k=1f(n)(xk))=h2(N−1∑k=0f(2m)(xk)+N∑k=1f(2m)(xk))=h2(f(2m)(x0)+2N−1∑k=1f(2m)(xk)+f(2m)(xN))=h(f(2m)(a)2+N−1∑k=1f(2m)(xk)+f(2m)(b)2)≡T2m
I0=T0−∞∑m=1h2m−1(2m)!h2I2m+∞∑m=1h2m(2m+1)!T2m
3
I0=T0−∞∑m=1h2m2(2m)!I2m+∞∑m=1h2m(2m+1)!T2mI2ℓ=T2ℓ−∞∑m=1h2m2(2m)!I2(m+ℓ)+∞∑m=1h2m(2m+1)!T2(m+ℓ)T2ℓ=I2ℓ+∞∑m=1h2m2(2m)!I2(m+ℓ)−∞∑m=1h2m(2m+1)!T2(m+ℓ)
I0=T0−h22(2!)I2+h23!T2−h42(4!)I4+h45!T4−h62(6!)I6+h67!T6−⋯T2=I2+h22(2!)I4−h23!T4+h42(4!)I6−h45!T6+⋯T4=I4+h42(2!)I6−h23!T6+⋯
()
I0=T0−h22(2!)I2+h23!(I2+h22(2!)I4−h23!T4+h42(4!)I6−h45!T6+⋯)−h42(4!)I4+h45!T4−h62(6!)I6+h67!T6−⋯=T0−h212I2+h448I4−7h4360T4+h6360I6−6h67!T6−⋯=T0−h22(2!)I2h23!I2+h23!h22(2!)I4−h23!h23!T4+h23!h42(4!)I6−h23!h45!T6−h42(4!)I4+h45!T4−h62(6!)I6+h67!T6−⋯
T2ℓ=I2ℓ+∞∑m=1h2m2(2m)!I2(m+ℓ)−∞∑m=1h2m(2m+1)!T2(m+ℓ)
2
f(x)=f(xk+1)+∞∑n=11n!f(n)(xk+1)(x−xk+1)n
∫xk+1xkf(x)dx=f(xk+1)h−∞∑n=11(n+1)!f(n)(xk+1)(−h)n+1
I(f)=hN−1∑k=0f(xk+1)+∞∑n=1(−h)n(n+1)!hN−1∑k=0f(n)(xk+1)=hN∑k=1f(xk)+∞∑n=1(−h)n(n+1)!hN∑k=1f(n)(xk)=hN−1∑k=0f(xk)−hf(x0)+hf(xN)+∞∑n=1(−h)n(n+1)!{hN−1∑k=0f(n)(xk)−hf(n)(x0)+hf(n)(xN)}
I(f)=Rh(f)+∞∑n=1hn(n+1)!Rh(f(n))I(f)=Rh(f)+h(f(b)−f(a))+∞∑n=1(−h)n(n+1)!{Rh(f(n))+h(f(n)(b)−f(n)(a))}
I(f)=Rh(f)+∞∑m=1h2m−1(2m)!Rh(f(2m−1))+∞∑m=1h2m(2m+1)!Rh(f(2m))I(f)=Rh(f)+h(f(b)−f(a))−∞∑m=1h2m−1(2m)!{Rh(f(2m−1))+h(f(2m−1)(b)−f(2m−1)(a))}+∞∑m=1h2m(2m+1)!{Rh(f(2m))+h(f(2m)(b)−f(2m)(a))}
I(f)=Rh(f)+h2(f(b)−f(a))−∞∑m=1h2m−1(2m)!{h2(f(2m−1)(b)−f(2m−1)(a))}+∞∑m=1h2m(2m+1)!{Rh(f(2m))+h2(f(2m)(b)−f(2m)(a))}
3
Th(f)≡Rh(f)+h2(f(b)−f(a))I(f(n))=f(n−1)(b)−f(n−1)(a)
I(f)=Th(f)−∞∑m=1h2m(2m)!12I(f(2m))+∞∑m=1h2m(2m+1)!Th(f(2m))
I(f)=Th(f)−h22!12I(f(2))+h23!Th(f(2))−h44!12I(f(4))+h45!Th(f(4))−h66!12I(f(6))+h47!Th(f(6))+⋯−h2m(2m)!12I(f(2m))+h2m(2m+1)!Th(f(2m))+⋯0=h23!I(f(2))−h23!Th(f(2))+h23!h22!12I(f(4))−h23!h23!Th(f(4))+h23!h44!12I(f(6))−h23!h45!Th(f(6))+⋯0=h23!I(f(2))−h23!Th(f(2))+h44!I(f(4))−h43!2Th(f(4))+h63!4!2I(f(6))−h63!5!Th(f(6))+⋯ I(f)=Th(f)−h22!16I(f(2))+h44!12I(f(4))+h43!(120−16)Th(f(4))+h63!4!2I(f(6))−h66!12I(f(6))+h47!Th(f(6))−h63!5!Th(f(6))+⋯+⋯−h2m(2m)!12I(f(2m))+h2m(2m+1)!Th(f(2m))+⋯