Reason for multiplication of function with step size (and subsequent addition) in Euler method

Question

What is the reason behind the multiplication of the function's derivative with the step size (and the subsequent addition) in the numerical Euler method? $$ y_{n+1} = y_n + hf(t_n, y_n) $$ I can't figure out why exactly this would work for generating a new value. How can scaling the output of the function $f$ and subsequently adding it to the prior value $y_n$ approximate a new value? Is there some formal or graphical explanation of this? The formula seems somewhat counterintuitive.

Answer 1

I would say Euler's method is actually the most intuitive of the numerical methods!

We have a function $f$ that tells us the derivative of a solution to the ODE passing through our current point $(t_n, y_n)$. What is the derivative? In a small enough interval around $y_n$, it's approximately the "rise over run", so if $h = t_{n+1} - t_n$ is 'small' then we can use this approximation:

$$ \frac{y(t_{n+1}) - y_n}{t_{n+1} - t_n} \simeq f(t_n, y_n)$$

$$ \implies y(t_{n+1}) \simeq y_n + h \cdot f(t_n, y_n)$$

Answer 2

$f(t_n,y_n)$ is the (approximation of) the derivative of $y$ at $t_n$. $h$ is the time step, so the change in $y$ over the time step is the length of the time step times the slope of the tangent at that point. For example, if $f(t,y)=3y$ and we start from $t=1,y=1$ we have $f(1,1)=3$. If our timestep is $0.1$ we have $h=0.1, y(1.1)\approx 1+0.1 \cdot 3 = 1.3$ The true solution is $y=e^{3t}-e^3+1$ and the true $y(1.1)=1.286$ to three places