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Mathematica Solutions to the ISSAC '97 Systems
Challenge
Wolfram Research, Inc.
Problem 2
What is the value of to 7 significant digits?
Result
Method 1: Use a built-in function.
The most natural approach gives the correct result but with a warning
message.
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr32.gif]](ISSACChallengegr32.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr33.gif]](ISSACChallengegr33.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr34.gif]](ISSACChallengegr34.gif)
To guarantee the required digits and to avoid the message, we use
arbitrary-precision arithmetic, allow more recursive subdivisions at the
right endpoint, and use a more appropriate integration method.
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr35.gif]](ISSACChallengegr35.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr36.gif]](ISSACChallengegr36.gif)
Method 2: Substitute to get a nicer integrand.
We make the substitution . Using ,
this is equivalent to ![[Graphics:ISSACChallengegr39.gif]](ISSACChallengegr39.gif) .
This makes the integrand much better behaved, getting rid of any
subdivision problems.
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr41.gif]](ISSACChallengegr41.gif)
![[Graphics:ISSACChallengegr7.gif]](ISSACChallengegr7.gif) ![[Graphics:ISSACChallengegr42.gif]](ISSACChallengegr42.gif)
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