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Taylorpolynomier Funktion af flere variable

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<strong>Funktion</strong> givet ved di¤erentialligning<br />

I Find det 2. Taylorpolynomium med udviklingspunkt π 2<br />

for løsningen til di¤erentialligningen<br />

<br />

x 0 (t) = sin t + x (t) 2 π<br />

<br />

med x = 0<br />

2<br />

I Vi skal …nde<br />

P 2 (t) = x π <br />

2 + x<br />

0 π<br />

2 t<br />

π<br />

2 +<br />

1<br />

2 x 00 π <br />

2 t<br />

I Ved indsættelse <strong>af</strong> t = π 2<br />

i di¤erentialligningen fås<br />

x 0 π <br />

2 = sin π<br />

2 + x π 2<br />

2<br />

= sin π <br />

2 = 1.<br />

I Ved di¤erentiation <strong>af</strong> di¤erentialligningen fås<br />

x 00 (t) = cos t + x (t) 2 (1 + 2x (t) x 0 (t)).<br />

<br />

π 2<br />

2<br />

I Ved indsættelse <strong>af</strong> t = π 2<br />

heri fås<br />

x 00 π <br />

2 = cos π<br />

2 + x π 2<br />

2<br />

1 + 2x π <br />

2 x<br />

0 π<br />

2 = 0.<br />

I Altså fås<br />

<br />

π<br />

P 2 (t) = 0 + t<br />

2 +<br />

1<br />

2 0 t <br />

π 2 π<br />

2<br />

= t<br />

2<br />

I som jo er det samme som det første Taylorpolynomium<br />

P 1 (t). Se Maple for P 3 (t).<br />

<strong>Taylorpolynomier</strong>.<br />

<strong>Funktion</strong> <strong>af</strong> ‡ere<br />

<strong>variable</strong><br />

Preben Alsholm<br />

<strong>Taylorpolynomier</strong><br />

De…nition <strong>af</strong><br />

Taylorpolynomium<br />

Udledning <strong>af</strong> formlen<br />

for Taylorpolynomiet<br />

Formlen for<br />

Taylorpolynomiet<br />

Eksempel 4.8.2 i<br />

Adams<br />

<strong>Funktion</strong> givet ved<br />

simpel forskrift<br />

<strong>Funktion</strong> givet ved<br />

di¤erentialligning<br />

Taylors formel med<br />

Lagrange’s restled<br />

Vurdering <strong>af</strong> fejlen<br />

ved Taylors formel I<br />

Vurdering <strong>af</strong> fejlen<br />

ved Taylors formel II<br />

Store O-notationen<br />

<strong>Funktion</strong> <strong>af</strong> ‡ere<br />

<strong>variable</strong>

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