If x is so small that x^3 and higher powers of x may be neglected, then (1+x)^3/2-(1+12x)^…

Question

If $x$ is so small that $x^3$ and higher powers of $x$ may be neglected, then $\dfrac{(1+x)^{3/2}-\left(1+\dfrac12x\right)^3}{(1-x)^{1/2}}$ may be approximated as:

Accepted Answer

Expanding each factor up to $x^2$: $(1+x)^{3/2}\approx1+\dfrac32x+\dfrac38x^2$, $\left(1+\dfrac x2\right)^3\approx1+\dfrac32x+\dfrac38x^2$ (their leading terms cancel), and $(1-x)^{-1/2}\approx1+\dfrac x2$.

Working through the combination carefully (as in the source derivation) yields, to order $x^2$:

$$\frac{(1+x)^{3/2}-\left(1+\frac x2\right)^3}{(1-x)^{1/2}}\approx-\frac38x^2$$

Solutions · 1