 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2/10/b/ii/solution

Subtract the first four terms from the given full p-series sums. Write

S_{2} = \frac{π ^{2}}{6} - (1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16}) \approx 0.2213229557,

and

S_{4} = \frac{π ^{4}}{90} - (1 + \frac{1}{16} + \frac{1}{81} + \frac{1}{256}) \approx 0.0035713047.

Solving

I_{p} (k_{p}) = S_{p}

yields

k_{p} = 5 - [(p - 1) S_{p}]^{- 1/ (p - 1)}

. Therefore

k_{2} = 5 - S_{2}^{- 1} \approx 0.482, k_{4} = 5 - (3 S_{4})^{- 1/3} \approx 0.464.

The unrounded shifts are approximately

0.4817158633

and

0.4638861447

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