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Worked solution

Match a product-series coefficient

The coefficient of x2x^2 in the series expansion of 1ex(1+ax)\dfrac1{e^x(1+ax)} is 1/41/4. Find the constant aa. You may use standard results from MF27.

[2016/CJC/JC2/CA1/3(b)]

CJC · Discussion · Question 5

Answer

a=−1/2a=-1/2.

Expand the two factors

1ex(1+ax)=e−x(1+ax)−1=(1−x+x2/2+O(x3))(1−ax+a2x2+O(x3))\dfrac1{e^x(1+ax)}=e^{-x}(1+ax)^{-1}=(1-x+x^2/2+O(x^3))(1-ax+a^2x^2+O(x^3)). The coefficient of x2x^2 is 1/2+a+a21/2+a+a^2.

How did this part go?

Solve the coefficient condition

a2+a+1/2=1/4a^2+a+1/2=1/4, so a2+a+1/4=(a+1/2)2=0a^2+a+1/4=(a+1/2)^2=0. Thus a=−1/2a=-1/2.

How did this part go?

Common mistakes

Do not omit the product of the two linear terms.

Read the hint