Prove the approximate formula nth root of (a^n+x) ≈ a+

Prove the approximate formula $\sqrt[n]{a^{n}+x} \approx a+\dfrac{x}{na^{n-1}}$, where $a>0$ and $|x|$ is small compared with $a^{n}$.

Step 1: Set up the linearisation

Regard the left side as a function of the small quantity $x$: $f(x) = \bigl(a^{n}+x\bigr)^{1/n}, \qquad f(x) \approx f(0)+f'(0)\,x$.

Step 2: The value at $x = 0$

\[ f(0) = \bigl(a^{n}\bigr)^{1/n} = a \qquad (a>0). \]

Step 3: The derivative at $x = 0$

\[ f'(x) = \frac1n\bigl(a^{n}+x\bigr)^{\frac1n-1} = \frac1n\bigl(a^{n}+x\bigr)^{\frac{1-n}{n}}, \]

\[ f'(0) = \frac1n\bigl(a^{n}\bigr)^{\frac{1-n}{n}} = \frac1n\,a^{\,n\cdot\frac{1-n}{n}} = \frac1n\,a^{1-n} = \frac{1}{na^{\,n-1}}. \]

Step 4: A second proof, by factoring out $a$

Pulling $a^{n}$ out of the root reduces the statement to the binomial rule of Problem 256: \[\sqrt[n]{a^{n}+x} = \sqrt[n]{a^{n}\left(1+\frac{x}{a^{n}}\right)} = a\left(1+\frac{x}{a^{n}}\right)^{1/n}.\]

With $\varepsilon = \dfrac{x}{a^{n}}$ small and $\bigl(1+\varepsilon\bigr)^{1/n}\approx1+\dfrac{\varepsilon}{n}$: $\sqrt[n]{a^{n}+x} \approx a\left(1+\frac{x}{na^{n}}\right) = a+\frac{x}{na^{\,n-1}}$.

This form also shows precisely what "small" must mean: it is $x/a^{n}$, not $x/a$, that has to be negligible.

Step 5: The error term

The next binomial term is $\dfrac{\frac1n\bigl(\frac1n-1\bigr)}{2}\varepsilon^{2}$, so

\[ \text{error} \approx a\cdot\frac{1-n}{2n^{2}}\cdot\frac{x^{2}}{a^{2n}} = \frac{\bigl(1-n\bigr)x^{2}}{2n^{2}a^{\,2n-1}}, \]

negative for every $n>1$: the estimate overshoots, as it must for a concave root function.

Step 6: Verification

\[n = 2,\ a = 10,\ x = 1:\quad \sqrt{101} \approx 10+\frac{1}{20} = 10.05 \text{ vs } 10.0498756 \;\;\checkmark\]

\[n = 3,\ a = 2,\ x = 1:\quad \sqrt[3]{9} \approx 2+\frac{1}{12} = 2.08\overline{3} \text{ vs } 2.0800837 \;\;\checkmark\]

\[\text{Error formula, } n=3:\ \frac{(1-3)\cdot1}{2\cdot9\cdot2^{5}} = -\frac{2}{576} = -0.00347, \text{ actual } 2.080084-2.083333 = -0.00325 \;\;\checkmark\]