(a) Use differentiation to find power series representation for f(x)=1/(1+x)^2
What is the radius of convergence?
(b) Use part (a) to find a power series for f(x)=1/(1+x)^3
(c) Use part (b) to find a power series for f(x)=x^2/(1+x)^3
I found part (a) which was Σ n=0 to ∞ (1)^n(n+1)(x)^n, but I'm not sure how to use this for part (b). I thought something along the lines of:
1/(1+x)^3 = 1/(1+x)^2 * 1/(1+x)
and replace "1/(1+x)^2" with "Σ n=0 to ∞ (1)^n(n+1)(x)^n" Is this how I'm suppose to solve the problem?
Yes, your approach for part (b) is correct. To find the power series for f(x) = 1/(1+x)^3 using the power series representation of f(x) = 1/(1+x)^2 from part (a), you can multiply the two power series together.
So, we have:
1/(1+x)^3 = (1/(1+x)^2) * (1/(1+x))
Replacing the first factor with its power series representation gives:
1/(1+x)^3 = (Σ n=0 to ∞ (1)^n(n+1)(x)^n) * (1/(1+x))
To multiply the power series, you can use the multiplication rules for power series. Multiply each term of the first series by the entire second series and then simplify:
(Σ n=0 to ∞ (1)^n(n+1)(x)^n) * (1/(1+x))
= Σ n=0 to ∞ (1)^n(n+1)(x)^n * 1/(1+x)
Now, simplify the expression (1/(1+x)) term by term using long division or partial fraction decomposition. The result will be a new power series.
Finally, you will have the power series representation for f(x) = 1/(1+x)^3. The radius of convergence for the power series is the same as the radius of convergence for the original function f(x).