Binomial Series — Question 4

Question 4 – Solution

Step 1: Expand in an intermediate variable.

$$(1+u)^{3/2}=1+\frac32u+\frac38u^2-\frac1{16}u^3+\frac3{128}u^4+\cdots.$$

Step 2: Substitute $u=-2x$.

$$\begin{align*} (1-2x)^{3/2}&=1+\frac32(-2x)+\frac38(4x^2)-\frac1{16}(-8x^3)+\frac3{128}(16x^4)+\cdots\\&=\boxed{1-3x+\frac32x^2+\frac12x^3+\frac38x^4+\cdots}. \end{align*}$$ Each degree-$n$ coefficient is multiplied by $(-2)^n$.

Step 3: Determine convergence.

The condition $|u|<1$ gives $|x|<1/2$, so $R=1/2$. For exponent $3/2$, coefficient magnitudes are order $n^{-5/2}$, so both endpoint series converge absolutely.

Conclusion.

The interval is $\boxed{[-1/2,1/2]}$.