Question

Find the power series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). give the interval of convergence for the resulting series

Answer 1

The power series  is

g(x) = -2x - (2x)2/2 - (2x)3/3 -(2x)4/4 -.......-(2x)n/n - .....

To deduce the power series of g(x) from the power series for f(x) and identify its radius of convergence

The power series for f(x) is just the geometric series derived from 1/1-y ,setting y=2x.

Its radius of convergence is 0.5

Let,

f(x)= 1/1-2x = 1+ (2x) + (2x)2 + .........+(2x)n......+....

The power series expansion (geometric series),

valid for I2xI < 1 , IxI < 0.5

so, radius of convergence = 0.5

The power series for g(x) is found by integrating term by term the power series of f(x) (upto a constant). The radius of converngence of g(d) is the same as that of f(x) (from general theory) =0.5

Now, g(x) = ln(1-2x)

= -2 $\int\limits^a_b {(1/1-2x)} \, dx = -2 \int\limits^a_b {f(x)} \, dx$

=-2 $\int\limits^a_b {[1+(2x)+ (2x)2 +........+ (2x)n+.......]} \, dx$

g(x) = -2x - (2x)2/2 - (2x)3/3 -(2x)4/4 -.......-(2x)n/n - .....

is the power series expansion for g(x).

radius of convergence =0.5

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