Given a postive real number n, we are asked to evaluate the following integral:
where: the symbol "" is the floor function,
and "" is the round (to the nearest integer) function.
We may rewrite the above function in Matlab as:
>> S = @(n) round(integral(@(x) sqrt(floor(x))-floor(sqrt(x)),0,n));
Therefore, for , we have:
>> s = S(10*pi)
s =
12
Be careful though, in using the Matlab integral function, as it is only an approximation. For example if :
>> s = S(100000)
Warning: Reached the limit on the maximum number of intervals in use. Approximate bound on error is
8.0e+01. The integral may not exist, or it may be difficult to approximate numerically to the
requested accuracy.
> In integralCalc/iterateScalarValued (line 372)
In integralCalc/vadapt (line 132)
In integralCalc (line 75)
In integral (line 87)
In @(n)round(integral(@(x)sqrt(floor(x))-floor(sqrt(x)),0,n))
s =
49841
The correct answer is . The integral function is off by only 2 units here, but the discrepancy could be even greater for other values of n. integral function can also be quite slow. The challenge is to find an efficient and more accurate algorithm to evaluate the integral.
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Last Solution submitted on May 18, 2023

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David Hill

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