 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-1/9/c/solution

e^{- (x - a)^{2}} = e^{- a^{2}} e^{2 a x} e^{- x^{2}}

. Multiplying the two exponential series gives

e^{- (x - a)^{2}} = e^{- a^{2}} [1 + 2 a x + (2 a^{2} - 1) x^{2} + (\frac{4}{3} a^{3} - 2 a) x^{3} + O (x^{4})] .

For

a \neq = 0, \pm 1/ 2

, its first three nonzero terms are

e^{- a^{2}} [1 + 2 a x + (2 a^{2} - 1) x^{2}] .

a = 0

, the first three nonzero terms are

1 - x^{2} + x^{4} /2

. If

a = \pm 1/ 2

, they are

e^{- 1/2} (1 + 2 a x - \frac{4 a}{3} x^{3}) .

The latter expression retains the cubic term because the quadratic coefficient is zero.

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