Numerical Methods: Newton-Raphson, Euler, and Runge-Kutta

Numerical methods are algorithms for approximate answers. Newton-Raphson is used to find a root of an equation such as $f(x)=0$, while Euler and Runge-Kutta are used to approximate solutions of differential equations.

If you only need the quick distinction, it is this: Newton-Raphson updates a guess for $x$; Euler and Runge-Kutta step a solution forward in time. Whether they work well depends on conditions such as a sensible starting guess, a usable derivative, or a step size $h$ that is small enough for the problem.

What each numerical method is for

Newton-Raphson: find a root

If you want a value of $x$ such that $f(x)=0$, Newton-Raphson updates a guess by following the tangent line:

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$

The intuition is simple: if the graph is smooth near the root, the tangent line is a local linear model, and its intercept can be a better guess than the current point.

This tends to work well when $f$ is differentiable, $f'(x_n) \ne 0$, and the initial guess is already close to a simple root. If those conditions fail, the method can stall, jump away from the root, or diverge.

For example, with $f(x)=x^2-2$ and $x_0=1.5$,

$$x_1 = 1.5 - \frac{1.5^2 - 2}{2(1.5)} = 1.4167$$

and one more step gives about $1.4142$, which is already close to $\sqrt{2}$.

Euler method: one slope, one step

For an initial-value problem

$$y' = f(t,y), \qquad y(t_0)=y_0,$$

Euler's method uses the current slope to step forward:

$$y_{n+1} = y_n + h f(t_n, y_n)$$

It is the simplest approximation: move ahead by step size $h$ using the slope you know right now. That makes Euler easy to learn and implement, but its error can grow quickly if $h$ is too large or the solution changes rapidly.

Runge-Kutta method: several slope checks in one step

Runge-Kutta methods improve on Euler by sampling slope information more than once inside the same step. In introductory courses, "Runge-Kutta" often means the classical fourth-order method RK4:

$$k_1 = f(t_n, y_n), \qquad k_2 = f\left(t_n + \frac{h}{2}, y_n + \frac{h}{2}k_1\right),$$ $$k_3 = f\left(t_n + \frac{h}{2}, y_n + \frac{h}{2}k_2\right), \qquad k_4 = f(t_n + h, y_n + hk_3)$$ $$y_{n+1} = y_n + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4)$$

RK4 takes a weighted average of several slope estimates, so it usually tracks the curve much better than Euler for the same step size.

Worked example: Euler vs. Runge-Kutta on the same ODE

Take

$$y' = y, \qquad y(0)=1$$

and use one step of size $h=0.1$ to estimate $y(0.1)$.

Euler step

At $t=0$, the current value is $y_0=1$, so the slope is

$$f(0,1)=1$$

Euler gives

$$y_1 = 1 + 0.1(1) = 1.1$$

RK4 step

Now use the same problem with RK4:

$$k_1 = 1$$ $$k_2 = 1 + \frac{0.1}{2}(1) = 1.05$$ $$k_3 = 1 + \frac{0.1}{2}(1.05) = 1.0525$$ $$k_4 = 1 + 0.1(1.0525) = 1.10525$$

So

$$y_1 = 1 + \frac{0.1}{6}(1 + 2(1.05) + 2(1.0525) + 1.10525)$$ $$y_1 \approx 1.105170833$$

For this equation, the exact value is $e^{0.1} \approx 1.105170918$, so the RK4 step is much closer than the Euler step.

That is the main lesson. Euler uses the slope only at the left endpoint. RK4 samples how the slope changes during the step, so it usually gives a better local picture.

When to use Newton-Raphson, Euler, or Runge-Kutta

Use Newton-Raphson when the job is to solve a nonlinear equation and you can compute or approximate the derivative. Use Euler when you want the basic idea of stepping through an ODE or need a quick baseline.

Use Runge-Kutta, especially RK4, when you want a practical accuracy upgrade without changing the problem setup. If the ODE is stiff, though, neither Euler nor classical RK4 is always a good choice; the method has to match the equation.

Common mistakes in numerical methods

Mixing up the problem types

Newton-Raphson is for roots of equations. Euler and Runge-Kutta are for differential equations. If you choose the wrong method family, the setup is wrong before you even calculate.

Assuming the method will always converge

Newton-Raphson can fail if the starting guess is poor or if $f'(x)$ is very small near the iterate. Euler and RK methods can behave badly if the step size is too large for the problem.

Treating the step size as a minor detail

For ODE methods, the step size $h$ is part of the method, not an afterthought. A smaller $h$ often improves accuracy, but it also increases cost, and for some difficult problems you may need methods designed for stiffness rather than just a smaller step.

Forgetting that the answer is approximate

A numerical output with many digits is not automatically more trustworthy. The useful question is whether the approximation is stable, converging, and accurate enough for the purpose.

Where numerical methods are used

Numerical methods show up whenever the model is clear but an exact symbolic answer is inconvenient or unavailable. That includes physics, engineering, optimization, finance, and scientific computing.

The common pattern is practical rather than theoretical: you want an answer that is accurate enough for the decision you need to make. That is why checking convergence, step size effects, or sensitivity to the starting guess matters as much as writing down the formula.

Try a similar problem

Try the same ODE example with $h=0.05$ instead of $0.1$ and compare the Euler answer with the RK4 answer again. Then try Newton-Raphson on $f(x)=x^2-3$ starting from $x_0=2$ and see how fast the iterates move toward $\sqrt{3}$.