Linear approximation of $\sqrt[7]{e}$ 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?
 A: 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 $0.00002$.
A: Welcome.
Aside from Taylor expansion, another thing you can do (which is quite special to $e$) is use the infinite simple continued fraction expansion for $e^{1/n}$. I want to share this because I find it remarkable, even if it's a little off-track: $$e^{1/n}=1+\cfrac{1}{n-1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{3n-1+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{5n-1+\cdots}}}}}}}$$For integers $n\ge2$.
Let $n=7$ and take the first few terms: $$e^{1/7}\approx1+\frac{1}{7-1+\frac{1}{1+\frac{1}{1}}}=\frac{15}{13}$$
Which has error less than $3\cdot10^{-4}$. Taking the partial fraction up to $3n-1$, we have: $$e^{1/7}\approx1+\frac{1}{7-1+\frac{1}{1+\frac{1}{1+\frac{1}{20}}}}=\frac{308}{267}$$Which has error less than $7\cdot10^{-6}$.
A: $[k,k]$ Padé approximants $P_k$ of $e^{\frac{1}{n}}$ are not bad
$$P_1=\frac{2 n+1}{2 n-1}\qquad P_2=\frac{12 n^2+6 n+1}{12 n^2-6 n+1} \qquad P_3=\frac{120 n^3+60 n^2+12 n+1}{120 n^3-60 n^2+12 n-1}$$ and so on are more then decent. For $n=7$ they generate
$$\left(
\begin{array}{ccc}
k & P_k & \log_{10} \left(\left| P_k-e^{\frac{1}{7}}\right| \right)\\
 1 & \frac{15}{13} & -3.55 \\
 2 & \frac{631}{547} & -7.02 \\
 3 & \frac{44185}{38303} & -10.86 \\
 4 & \frac{4330761}{3754241} & -14.95 \\
 5 & \frac{545720071}{473072669} & -19.24 \\
 6 & \frac{84045221695}{72856945267} & -23.68 \\
\end{array}
\right)$$
