Given that x' = tx, x(0) = 1, find x(1), and use Euler’s method with n steps to find an approximation to x(1). (You will end up with a product of n terms whose limit (as n → ∞) is $sqrt(e)$)
I calculated x(1) using the standard way of solving the ODE by integration and managed to find $sqrt(e)$ but I'm having trouble finding this using Euler. What I did is:
Xn+1 = Xn + X'h where h=$\frac{1}{N}$
substituting the X' by tx I have
Xn+1 = Xn + tx($\frac{1}{N}$) = Xn(1+$\frac{t}{N}$)
So Xn = $(1+\frac{t}{N})^N$ and simplifying this I have Xn = $((1+\frac{1}{N})^N)^t$ and taking the limit as n → ∞ I only get $e^t$. Where did I make a mistake? I don't know where does the square root comes from.
Your approach is more or less correct, but there are some indexing mistakes which may be leading to confusion. Consider for instance that $N$ is a fixed step size, whereas the update rule applies for any iteration of the algorithm. So the rule is more properly written as $$X_{i+1} = X_i + \frac{t}{N}X_i.$$ Now, what is $t$ at iteration $i$? $\frac{i}{N}$, of course. So $$X_{i+1} = \left(1 + \frac{i}{N^2}\right)X_i$$ and $$X_N = \prod_{i=0}^{N-1} \left(1+\frac{i}{N^2}\right).$$