If x is so small that x^3 and higher powers o | vedclass

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(A) Using the binomial expansion $(1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2$ for small $x$:$(1+x)^{3/2} \approx 1 + \frac{3}{2}x + \frac{\frac{3}{2

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$-\frac{3}{8}x^2$

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(A) Using the binomial expansion $(1+x)^n \approx 1 + nx + \frac{n(n-1)}{2}x^2$ for small $x$:$(1+x)^{3/2} \approx 1 + \frac{3}{2}x + \frac{\frac{3}{2} \cdot \frac{1}{2}}{2}x^2 = 1 + \frac{3}{2}x + \frac{3}{8}x^2$$(1 + \frac{1}{2}x)^3 \approx 1 + 3(\frac{1}{2}x) + \frac{3 \cdot 2}{2}(\frac{1}{2}x)^2 = 1 + \frac{3}{2}x + \frac{3}{4}x^2$$(1-x)^{-1/2} \approx 1 + (-\frac{1}{2})(-x) = 1 + \frac{1}{2}x$Substituting these into the expression:$\frac{(1 + \frac{3}{2}x + \frac{3}{8}x^2) - (1 + \frac{3}{2}x + \frac{3}{4}x^2)}{1} \approx (\frac{3}{8} - \frac{6}{8})x^2 = -\frac{3}{8}x^2$Since we neglect $x^3$ and higher powers,the denominator $(1-x)^{1/2}$ effectively acts as $1$ in the product with the $x^2$ term.

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

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