Exponential system of equations i struggle with stationary points for this problem, minimize:
$\text{f}(\textbf{x})=x_{1}^{2}-x_{1}x_{2}+2x_{2}^{2}-2x_{1}+\mathrm{e}^{(x_{1}+x_{2})}$
I know that gradient is:
$\nabla f (\textbf{x)}=(2x_{1}-x_{2}-2+\mathrm{e}^{(x_{1}+x_{2})}, -x_{1}+4x_{2}+\mathrm{e}^{(x_{1}+x_{2})})^{T}$
And also if I want local extrema I set them equal to zero:
$2x_{1}-x_{2}-2+\mathrm{e}^{(x_{1}+x_{2})}=0$
$-x_{1}+4x_{2}+\mathrm{e}^{(x_{1}+x_{2})}=0$
And suddenly I stopped because I'm not sure how to solve this kind of system.
Thank you for your hint :)
 A: I don't think this system has solutions in terms of elementary functions. Nevertheless we can manipulate it to attain something a little easier to work with and get approximate solutions for.
So here is our system:
$$
\begin{cases}
2x_{1}-x_{2}-2+\mathrm{e}^{(x_{1}+x_{2})}=0 \\
-x_{1}+4x_{2}+\mathrm{e}^{(x_{1}+x_{2})}=0
\end{cases}
$$
I asked my graphing program to plot both relations to get a sense of what our solutions (if any) would look like. Here are the results

So we should look for positive $x$ and negative $y$, both a little less than $1/2$ in absolute value.
Now for some algebra. The exponential term can be simplified a bit if we make the following substitutions:
$$
\left\{
\begin{aligned}
  u &:= x_1 + x_2 \\
  v &:= x_1 - x_2
\end{aligned}
\right\}
\iff
\left\{
\begin{aligned}
x_1 &= \small{\frac{u+v}{2}} \\
x_2 &= \small{\frac{u-v}{2}}
\end{aligned}
\right\}
$$
Then our system becomes
$$
\left\{
\begin{aligned}
(u+v) - \frac{u-v}{2} + e^u &= 2 \\
-\frac{u+v}{2} + 2(u-v) + e^u &= 0
\end{aligned}
\right.
$$
After some more algebra we get
$$
\left\{
\begin{aligned}
  v &= \frac{1}{3} \left(4-u-2e^u \right) \\
  v &= \frac{1}{5} \left(3u+2e^u \right)
\end{aligned}
\right.
$$
Setting them both equal to each other and going through further simplifications we obtain
$$ 7u + 8e^u = 10 $$
This can't be solved in terms of elementary functions, but a numerical approximation yields
$$ u \approx 0.129 $$
With a good approximation for $u$, we can get a good approximation for $v$ by plugging $u$ in for either of the two formulas for $v$ deduced above:
$$ v \approx 0.532 $$
And it's a quick calculation to get $x_1$ and $x_2$ back using the formulas
\begin{align}
x_1 &= \frac{u+v}{2} \\
x_2 &= \frac{u-v}{2}
\end{align}
A: Changing notations $(x_1\to x,x_2\to y)$ you want to solve $$f'_x=e^{x+y}+2 x-y-2=0\tag 1$$ $$f'_y=e^{x+y}-x+4 y=0\tag 2$$ So $$f'_x-f'_y=3 x-5 y-2=0\implies y=\frac{3 x-2}{5} $$ Replace in $(1)$ and obtain $$\frac{7}{5}x+e^{\frac{8 x-2}{5}}-\frac{8}{5}=0\tag 3$$ which could be very easily solved using Newton method starting with $x_0=0$ as shown below (notice that the derivative of this new function is always positive).
$$\left(
\begin{array}{cc}
 n & x_n \\
 0 & 0 \\
 1 & 0.3760062343 \\
 2 & 0.3313943919 \\
 3 & 0.3304499020 \\
 4 & 0.3304494983 
\end{array}
\right)$$
Sooner or later, you will learn that any equation which can write or rewrite $$A+Bx+C\log(D+Ex)=0$$ has real and/or complex solution(s) in terms of Lambert function (which is not an elementary function).
The Wikipedia page shows many examples of the kind of manipulations to be performed. In your case, the solution is given by$$x=\frac{8}{7}-\frac{5}{8} W\left(\frac{8 }{7}e^{10/7}\right)\qquad y=\frac{2}{7}-\frac{3}{8} W\left(\frac{8 }{7}e^{10/7}\right)$$
Using the expansion given in Wikipedia
$$W(x)={L_1}-{L_2}+\frac{{L_2}}{{L_1}}+\frac{({L_2}-2) {L_2}}{2
   {L_1}^2}+\frac{{L_2} \left(2 {L_2}^2-9 {L_2}+6\right)}{6
   {L_1}^3}+\cdots\tag4$$ where $L_1=\log(x)$ and $L_2=\log(L_1)$ you would find that
$$W\left(\frac{8 }{7}e^{10/7}\right)\approx 1.3061$$ (the exact value should be $\approx 1.2999$) So $$x\approx 0.3266\implies y\approx -0.2041$$ while the exact solution would be $$x\approx 0.3304\implies y\approx -0.2017$$ For sure, adding extra terms to $(4)$ would give something closer.
