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Old 04-09-2005, 04:36 PM   #1 (permalink)
Hain
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Rapidly finding terms of Taylor polynomial

w = INTEGRAL e<sup>sin(t)</sup> dt
w has no indefinite form. But a Taylor polynomial can be generated for
w(x) = INTEGRAL<sub>0</sub><sup>x</sup> e<sup>sin(t)</sup> dt
Where it is centered around zero. However, to do this rapidly there needs to be a quick way to determine the nth derivative of w(x).
I have noted that the nth derivative always has the term e<sup>sin(t)</sup>, so I have started with letting the nth derivative be equal to:
w<sup>(n)</sup>(x) = f<sub>n</sub>(x)*e<sup>sin(t)</sup>
Where:
f<sub>1</sub>(x) = 1 ;
f<sub>2</sub>(x) = cos(x) ;
f<sub>3</sub>(x) = cos<sup>2</sup>(x) - sin(x) ; so on and so forth if you wish to find each derivative.
If the derivative of w(x) continues then the pattern that f<sub>n</sub>(x) follows is:
f<sub>n</sub>(x) = f<sub>n-1</sub>(x)*cos(x) + <sup>d</sup>/<sub>dx</sub> f<sub>n-1</sub>(x)
Because the Taylor function we desire is centered at 0 let:
g(n) = f<sub>n</sub>(0)
The Taylor polynomial then becomes
SUM<sub>n=1</sub><sup>T</sup> g(n)*x<sup>n</sup>/n!
My question is: what is g(n) so that a single summation formula can be made to describe INTEGRAL<sub>0</sub><sup>x</sup> e<sup>sin(t)</sup> dt?
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Last edited by Hain; 04-11-2005 at 07:55 AM.. Reason: Thank you CSFlim
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