Explain the error term in Euler method Task: I had to find out some estimates for M and L to make sure the proportional accucrazy is not above $10^{-4}$ in the Euler method with the problem below.
I am trying to understand the page 672 on this book here. The book provides the formula 
$$\left|y(x_{n})-y_{n}\right|\leq \left(\frac{M}{2L}\right)\left(e^{L(x_{n}-x_{0})}-1\right)h$$
about the error where you can see the M and L (which looks like some use of Lambert function is needed or rough estimate for the upper bound, look at the $L$ term). There is also an example where it finds some upper bounds and claims some rough estimate. More precisely, I am trying to apply the method of deducing the error term on pages 673-674 for the problem 2 on page 676.
M
I cannot yet understand why the second derivative is used as an estimate for the $M$. On page 673, it just claims that assume that $|y''(x)|\leq M$ but cannot find any premise for it, the 2 in the above. This point about M on pages 673-674 is something black magic to me. Please, explain.
L
The L is apparently just length of the interval.
 A: 
Definitions
L
$L$ is not the length of the interval, but a Lipschitz constant: some
  number such that  $$|f(x,y_1) - f(x,y_2)| \le L|y_1 - y_2|$$ 
  satisfied for all $y_1,y_2,x$ (in some interval of interest).
M
Similarly, $M$ is any number such that the assumption $|y''(x)| \le M$
  is satisfied. The error analysis is also explained in this article
  here (reference taken from the Wikipedia article on the Euler
  method).

Example
Suppose the equation $y' = -\frac12 y$ with initial condition $y(0) = 1$ with the Euler method on the interval $x \in [0,1]$. We actually know the exact solution in this case, $y(x) = \exp(\frac12x)$.
Finding L
We note that $f(x,y) = \frac12 y$ in the example, so the equation
$$|f(x,y_1) - f(x,y_2)| \le L|y_1 - y_2|$$ 
becomes 
$$|\frac12 y_1 - \frac12 y_2| \le L |y_1 - y_2|.$$
This equation is satisfied for $L = \frac12$, so this is the value of $L$ we will take (we can also have taken a bigger value for $L$, like $L=1$, in which case we'll get a result which is also valid but not as sharp).
Finding M
We need it to satisfy $|y''(x)| \le M$. Since we know the exact solution, $y(x) = \exp(\frac12x)$, we know that $y''(x) = \frac14\exp(\frac12x)$ and so $|y''(x)| \le \frac14$ on the interval $x \in [0,1]$. However, for most equations we do not know the exact solution, so let us pretend we do not. Then what we can do is the following: differentiate the differential equation $y'(x) = \frac12 y$ to get $y''(x) = \frac12 y'$, and then substitute the differential equation in it to get 
$$y''(x) = \frac12 \cdot \frac12 y = \frac14y.$$ 
We do know that the numerical solution given by the Euler method is a decreasing sequence for this example, so $y\le1$ and thus $|y''(x)| \le \frac14$ (as we found before).
Conclusion
So, we can take $L = \frac12$ and $M = \frac14$ in the error bound.
