Neural ODE forward Euler

Background

A Neural ODE treats a network's depth as continuous: instead of stacking discrete layers, it defines the hidden state's rate of change $\frac{dh}{dt}=f(h,t)$ and integrates it with an ODE solver. The simplest solver is forward Euler, which takes fixed steps $h_{n+1}=h_n + \Delta t\,f(h_n, t_n)$. This connects deep residual networks (one Euler step ≈ one residual block) to the rich theory of differential equations.

Problem statement

Implement neural_ode_euler(f, h0, t0, t1, steps) that integrates $\frac{dh}{dt}=f(h,t)$ from t0 to t1 with steps equal forward-Euler steps:

$$\Delta t = \frac{t_1 - t_0}{\text{steps}}, \qquad h_{n+1} = h_n + \Delta t\, f(h_n, t_n), \qquad t_{n+1} = t_n + \Delta t$$

Return the final state $h$ at $t_1$.

Input

• f — callable f(h, t) returning $dh/dt$ (same shape as h).
• h0 — initial state, np.ndarray.
• t0, t1 — float start/end times.
• steps — int, number of Euler steps.

Output

An np.ndarray (same shape as h0): the integrated state at t1.

Examples

Example 1

Input:  f(h, t) = h, h0 = [1.0], t0 = 0, t1 = 1, steps = 1000
Output: ~ [2.7196]   (approaches e ≈ 2.71828)

Explanation: $dh/dt = h$ has exact solution $h(t)=h_0 e^{t}$, so $h(1)=e$. Forward Euler with 1000 steps approximates it closely; more steps reduce the discretization error.

Constraints

• Use a fixed step size $\Delta t = (t_1 - t_0)/\text{steps}$.
• Update both the state and the time each step.
• Evaluate f at the current state and time (explicit Euler).

Notes

• One forward-Euler step h + dt*f(h) is exactly the form of a residual block h + g(h) — Neural ODEs make that analogy precise.
• Euler's error per step is $O(\Delta t^2)$; higher-order solvers (RK4, adaptive) reach the same accuracy with far fewer steps.