Roman C. answered • 04/04/16

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Masters of Education Graduate with Mathematics Expertise

Let y = x - 1.

Then 3x/(2 - x) = (3y + 3)/(1 - y) and the latter is to be written as a power series centered at 0.

(3y + 3)/(1 - y) = -3 + 6/(1 - y)

Use the geometric series to get

-3 + (1 + y + y

^{2}+ y^{3}+ ...) = -2 + y + y^{2}+ y^{3}+ ... = -2 + ∑_{n=1,...,∞}y^{n}So the original expression's power series centered at 1 is

3x/(2 - x) = -2 + ∑

_{n=1,...,∞}(x - 1)^{n}Douglas N.

In my comment above,

3x/(2-x) = -2 + ∑

_{n>=1}6(x - 1)^{n}should be

3x/(2-x) = 3 + ∑

_{n>=1}6(x - 1)^{n}
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04/04/16

Roman C.

tutor

Thanks for the correction. I rushed because there was only a short time window for me at that moment.

You are right that we should report a convergence interval.

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04/06/16

Douglas N.

^{2}+ y^{3 }+ ... ), it equals 6(1 + y + y2 + y3 + ... ).^{ }_{n>=1}6(x - 1)^{n}_{n>=0}6(x - 1)^{n}04/04/16