Power Series

[80922908]

If f(x) = (1 + ax)(1 + bx)^-1 can be expanded as a power series f(x) = 1 + (Sigma x = 1 to x = infinity)c_nxⁿ, how would you show that c_n = (-b)^n-1(a - b)?

Also, how would you find the radius of convergence?

 

[DrPhil]

If f(x) = (1 + ax)(1 + bx)^-1 can be expanded as a power series f(x) = 1 + (Sigma x = 1 to x = infinity)c_nxⁿ, how would you show that c_n = (-b)^n-1(a - b)?

Also, how would you find the radius of convergence?

We need to see your work - where did you get stuck?

To find c_n, you need to take the nth derivative and evaluate at x=0

\(\displaystyle \displaystyle f(x) = \dfrac{1+ax}{1+bx}, \ \ \ \ \ \ \ \ \ \ \ c_0 = f(0) = 1 \)

\(\displaystyle \displaystyle f'(x) = \dfrac{a}{1+bx} - \dfrac{(1+ax)b}{(1+bx)^2}, \ \ \ \ c_1 = f'(0) = a - b \)

\(\displaystyle f''(x) = \cdot \cdot \cdot \)

Can you go far enough to find the pattern?