I need to find a linear approximation $\sqrt[7]{e}$. I know that for $x_0 = a + \Delta x$
$$
f(x_0) \approx f(a)+f'(a)\cdot \Delta x
$$
Thus my inital function is $f(x) = \sqrt[7]{x}$ and $e = 1 + (e – 1)$, so
$$
f(e) \approx \sqrt[7]{1} + \frac{1}{7}1^{\frac{-6}{7}}\cdot(e – 1) = \frac{6}{7} + \frac{e}{7} \approx 1.2455.
$$
However, I can't say that I'm entirely pleased by the result. Even though one could argue that $f(x) = \sqrt[7]{x}$ grows in such slow manner that approximating $f(e)$ by $f(1)$ is good enough for linear approximation, I'm not convinced. So I started to wonder: does there exist a function $g(x): f(g(x)) = \sqrt[7]{g(x)}$ and $f(g(c)) = \sqrt[7]{g(c)}= \sqrt[7]{e}$ that could provide a more precise approximation?
Best Answer
You are missing the fact that for the exponential you have a good approximation by definition. You have $$ \sqrt[7]{e}=e^{1/7}. $$ Expanding the exponential around $0$ you have $$\tag1 \sqrt[7]{e}=e^{1/7}\simeq1+\frac17=1.1\overline6 $$ which is fairly close to the actual value $1.1535\ldots$ (consistent with the fact that the error in $(1)$ is of order $\frac1{7^2}\simeq 0.02$).
If you want to improve your approximation you can take more terms in the series. The quadratic approximation is $$\tag2 \sqrt[7]{e}=e^{1/7}\simeq1+\frac17+\frac12\,\frac1{7^2}=1.15306\ldots $$ and now you have three good decimals. Things then improve fast: the cubic approximation is $$\tag3 \sqrt[7]{e}=e^{1/7}\simeq1+\frac17+\frac12\,\frac1{7^2}+\frac16\,\frac1{7^3}=1.153547\ldots $$ where the error is less than $0.00002$.