 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2019/1/solution

The successive derivatives are

f^{(n)} (x) = (- 6)^{n} e^{- 6 x}

, so

f^{(n)} (- 4) = (- 6)^{n} e^{24}

. Substitution in the Taylor series formula about

a = - 4

gives

e^{- 6 x} = e^{24} n = 0 \sum \infty \frac{( - 6 ) ^{n} ( x + 4 ) ^{n}}{n !} = e^{24} [1 - 6 (x + 4) + 18 (x + 4)^{2} - 36 (x + 4)^{3} + \dots] .

The ratio test shows convergence for every finite real or complex

x

: the absolute ratio of successive terms is

6∣ x + 4∣/ (n + 1) \to 0

. This is also the usual exponential function series applied to

e^{24} e^{- 6 (x + 4)}

, proving that the series equals the stated function everywhere.

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