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*From*: David Bowman <David_Bowman@georgetowncollege.edu>*Date*: Tue, 5 Oct 2021 14:17:58 +0000

Regarding KC's comment:

Now there is also mention of doing "a series expansion of the

integral at a=0", and the Taylor series is shown up to O(a^5).

This amounts to doing the series expansion of the integrand

e^(-(a t)) about (a t) = 0, then taking the definite integral

of each term, and simplifying slightly. If this is the

"textbook answer" for the numerical question that John refers

to, I only point out that if 0 << a t1, taking only 5 terms in

the series expansion may not get us very close to the correct

answer, and each term may involve significant errors due to

subtracting large quantities: a^n (t2^n - t1^n).

Note, if one wants to do a term-by-term Taylor evaluation of the definite integral one can factor out a common factor of (t2 - t1) out of each term, leaving polynomials in t1 & t2 with all + signs between them. But one is still left with whatever precision loss may stem from the original (t2 - t1) difference, because it is a common factor out front.

For example, recall

t2^5 - t1^5 = (t2 - t1)*(t2^4 + (t2^3)*t1 + (t2*t1)^2 + t2*(t1^3) + t1^4).

A couple of multiplications could be saved here and there by some further nesting and use of the distributive law.

Dave Bowman

**References**:**[Phys-L] numerical methods***From:*John Denker <jsd@av8n.com>

**Re: [Phys-L] numerical methods***From:*Ken Caviness <caviness@southern.edu>

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