If $x=e^{\frac{t}{1+x}}$ and $t>0$, find $x(t)$.
My solution: Solving it numerically (Newton-Raphson), I found that it is almost linear for the given range of $t$. See the plot below:
Based on the figure, we can have two conjectures: 1) It is almost linear for all $t$; 2) The linear term of the Taylor expansion of $m(t)$ is dominant.
So, I calculated the Taylor series of $x(t)$ using Maple:
$$1+\frac{t}{2}-\frac{t^3}{96}+\frac{t^4}{192}-\frac{13 t^5}{7680}+O(t^6)$$
Then, I plotted the numerical solution (red curve) and the linear term of Taylor expansion (blue curve), see the curve below. They are different. Why? Given these curves, what can we say about the function $m(t)$? How (analytically) we can derive the linear approximation that is consistent with the numerical result?
Concerning your plots : the problem is that the expansions you made $$x=\sum_{i=0}^n a_i t^i$$ would be valid for small values ot $t$ (which is not the case). Increasing $n$ (I did it) would lead to much more monstrous results. $$x=1+\frac{t}{2}-\frac{t^3}{96}+\frac{t^4}{192}-\frac{13 t^5}{7680}+\frac{13 t^6}{46080}+\frac{19 t^7}{215040}-\frac{179 t^8}{1720320}+\frac{19487 t^9}{371589120}-\frac{28997 t^{10}}{1857945600}+\frac{28493 t^{11}}{65399685120}+\frac{78877 t^{12}}{28028436480}+\cdots$$ would lead to $x\approx 2.74476\times 10^6$ for $t=10$ and to $x\approx 2.81422\times 10^{42}$ for $t=10000$ !!! In fact, even for very small values of $t$, this expansion does not approximate the solution.
If fact, more or less (as shown below), $x$ is almost linear with $\frac t {\log(t)}$. So, if you want to perform series expansions, they would be looking like $$x=\sum_{i=0}^n a_i \left(\frac t {\log(t)} \right)^i$$ but I am afraid that tha calculation of the coefficients would be rather difficult.
Basically, you are looking for the zero of function $$f(x)=(1+x)\log(x)-t$$ It is easy to show that
$$ x \log(x) < (x+1)\log(x) < (x+1) \log(x+1)$$ which makes the solution to be
$$y=\frac{t}{W(t)}-1 < x < \frac{t}{W(t)}=z=y+1$$ where $W(t)$ is Lambert function and, in practice, the solution is close to the left bound.
Using, for large values of $t$, the expansion $$W(t)=L_1-L_2+\frac{L_2}{L_1}+\frac{L_2(L_2-2)}{2L_1^2}+\frac{L_2(6-9L_2+2L_2^2)}{6L_1^3}+\cdots$$ where $L_1=\log(t)$ and $L_2=\log(L_1)$ will give you an extremely good approximation for Newton method as shown below $$\left( \begin{array}{ccc} t & y & x \\ 10 & 4.72893 & 5.12180 \\ 20 & 8.07028 & 8.39834 \\ 30 & 11.0519 & 11.3498 \\ 40 & 13.8323 & 14.1115 \\ 50 & 16.4771 & 16.7431 \\ 60 & 19.0214 & 19.2775 \\ 70 & 21.4868 & 21.7352 \\ 80 & 23.8881 & 24.1299 \\ 90 & 26.2354 & 26.4718 \\ 100 & 28.5366 & 28.7683 \\ 200 & 49.8939 & 50.0987 \\ 300 & 69.4961 & 69.6876 \\ 400 & 88.0932 & 88.2763 \\ 500 & 106.001 & 106.178 \\ 600 & 123.393 & 123.565 \\ 700 & 140.374 & 140.543 \\ 800 & 157.018 & 157.184 \\ 900 & 173.377 & 173.540 \\ 1000 & 189.491 & 189.651 \\ 2000 & 341.658 & 341.804 \\ 3000 & 484.096 & 484.235 \\ 4000 & 620.827 & 620.961 \\ 5000 & 753.584 & 753.715 \\ 6000 & 883.326 & 883.455 \\ 7000 & 1010.66 & 1010.78 \\ 8000 & 1135.99 & 1136.11 \\ 9000 & 1259.62 & 1259.74 \\ 10000 & 1381.77 & 1381.89 \end{array} \right)$$
In practice, starting from the right bound for esthetical reasons (and from the fact that $f(z)f''(z)>0$ will avoid an overshoot of the solution - by Darboux theorem), the first iterate of Newton method would be $$x=\frac {t \left(\frac{t}{W(t)}-W(t)+t+1\right) } {(t+1) W(t)+t }$$ which, for $t=10,100,1000,10000$ would lead to $x=5.13116, 28.7705, 189.651, 1381.89$ which are almost the solutions (compare to the values of $x$ given in the table above).