 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/18/a/iii/solution

Put

D_{n} = (- 1)^{n} e^{π} - 1

. The two linear equations from parts (i) and (ii) give

I_{n} = D_{n} + n J_{n}

and

J_{n} = - n I_{n}

, so

(1 + n^{2}) I_{n} = D_{n}

. Therefore

I_{n} = \frac{( - 1 ) ^{n} e ^{π} - 1}{1 + n ^{2}}, J_{n} = \frac{n [ 1 - ( - 1 ) ^{n} e ^{π} ]}{1 + n ^{2}} .

These formulas include

n = 0

, when

I_{0} = e^{π} - 1

and

J_{0} = 0

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