10.1 – Trapezregel, Simpsonregel, 3/8-Regel

 

Leiten Sie die Trapezregel, die Keplersche Fassregel (Simpsonregel) und die 3/8-Regel her, indem Sie das entsprechende Interpolationspolynom integrieren.

Lösung

Trapezregel:

\int_a^b {f\left( x \right)dx} \approx \frac{1}{2}\left( {f\left( a \right)+f\left( b \right)} \right)\left( {b-a} \right)

trapezregel

p\left( x \right) = {c_0}+{c_1}x

p\left( a \right) = f\left( a \right)

p\left( b \right) = f\left( b \right)

\int_a^b {p\left( x \right)dx} = \int_a^b {\left( {{c_0}+{c_1}x} \right)dx} = \left[ {{c_0}x+\frac{1}{2}{c_1}{x^2}} \right]_a^b

= {c_0}\left( {b-a} \right)+f\frac{1}{2}{c_1}\left( {{b^2}-{a^2}} \right)

= \frac{1}{2}\left( {b-a} \right)\left( {2 \cdot {c_0}+{c_1}\left( {b+a} \right)} \right)

= \frac{1}{2}\left( {b-a} \right)\left( {\underbrace {{c_0}+{c_1}a}_{ = p\left( a \right)}+\underbrace {{c_0}+{c_1}b}_{ = p\left( b \right)}} \right)

= \frac{1}{2}\left( {b-a} \right)\left( {p\left( a \right)+p\left( b \right)} \right)

= \frac{1}{2}\left( {b-a} \right)\left( {f\left( a \right)+f\left( b \right)} \right)

Simpsonregel:

\int_a^b {f\left( x \right)dx} \approx \frac{1}{6}\left( {b-a} \right)\left( {f\left( a \right)+f\left( a \right)+4f\left( {\frac{{a+b}}{2}} \right)+f\left( b \right)} \right)

Wir haben einen quadratischen Interpolanten mit Stützstellen a,\:\:\frac{{a+b}}{2},\:\:b, das heißt es ist

p\left( x \right) = {c_0}+{c_1}x+{c_2}{x^2}

Dabei müssen {c_i} so gewählt werden, dass

p\left( a \right) = f\left( a \right)

p\left( b \right) = f\left( b \right)

p\left( {\frac{{a+b}}{2}} \right) = f\left( {\frac{{a+b}}{2}} \right)

Einsetzen:

\int_a^b {p\left( x \right)dx} = \int_a^b {{c_0}+{c_1}x+{c_2}{x^2}dx}

= \left[ {{c_0}x+\frac{1}{2}{c_1}{x^2}+\frac{1}{3}{c_2}{x^3}} \right]_a^b

= {c_0}\left( {b-a} \right)+\frac{1}{2}{c_1}\left( {{b^2}-{a^2}} \right)+\frac{1}{3}{c_2}\left( {{b^3}-{b^3}} \right)

= \frac{1}{6}\left( {b-a} \right)\left( {6{c_o}+3{c_1}\left( {b+a} \right)+2{c_2}\left( {{a^2}+ab+{b^2}} \right)} \right)

= \frac{1}{6}\left( {b-a} \right)\left( {6{c_0}+3{c_1}b+3{c_1}a+2{c_2}{a^2}+2{c_2}ab+2{c_2}{b^2}} \right)

= \frac{1}{6}\left( {b-a} \right)\left( {\left( {{c_0}+{c_1}a+{c_2}{a^2}} \right)+\left( {{c_0}+{c_1}b+{c_2}{b^2}} \right)+4{c_0}+2{c_1}a+2{c_1}b+{c_2}{b^2}+{c_2}{a^2}+2{c_2}ab} \right)

= \frac{1}{6}\left( {b-a} \right)\left( {p\left( a \right)+p\left( b \right)+4\underbrace {\left( {{c_0}+\frac{1}{2}\left( {a+b} \right){c_1}+\frac{1}{4}{{\left( {a+b} \right)}^2}{c_2}} \right)}_{ = p\left( {\frac{{a+b}}{2}} \right)}} \right)

= \frac{1}{6}\left( {b-a} \right)\left( {f\left( a \right)+4f\left( {\frac{{a+b}}{2}} \right)+f\left( b \right)} \right)

Die 3/8-Regel ergibt sich analog.

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