Newton-Raphson methodEdexcel A-Level Maths: Revision notes

The Newton-Raphson method improves an estimate $x_n$ of a root of $f(x)=0$: $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}.$ Example: $f(x)=x^3-2x-5$, $f'(x)=3x^2-2$, and $x_0=2$. Then $f(2)=-1$, $f'(2)=10$, so $x_1=2+0.1=2.1$. Next $f(2.1)=0.061$ and $f'(2.1)=11.23$, so $x_2=2.0946$. You must be able to find $f'$, substitute carefully and keep full calculator accuracy until the end. The formula is in the formulae booklet.

For $f(x)=\mathrm{e}^x-3x$, $f'(x)=\mathrm{e}^x-3$, and $x_0=0.5$ gives $x_1=0.6101$ and $x_2=0.6190$. For $f(x)=x^2-7$, $f'(x)=2x$, so $x_{n+1}=x_n-\frac{x_n^2-7}{2x_n}=\frac12\left(x_n+\frac{7}{x_n}\right)$. From $x_0=3$: $x_1=\frac83$. Newton-Raphson is a special recurrence relation $x_{n+1}=g(x_n)$ with $g(x)=x-\frac{f(x)}{f'(x)}$, and it usually converges faster than a simple rearrangement.

The tangent to $y=f(x)$ at $x_n$ has equation $y-f(x_n)=f'(x_n)(x-x_n)$. It meets the $x$-axis where $x=x_n-\frac{f(x_n)}{f'(x_n)}=x_{n+1}$. So $x_{n+1}$ is the $x$-intercept of the tangent at $x_n$. Example: for $y=x^2-7$ at $x=3$, the tangent is $y=6x-16$, which meets the axis at $x=\frac83=x_1$. If the curve stays close to its tangent between $x_n$ and the root, $x_{n+1}$ is much closer to the root.

The method fails or is unreliable when $f'(x_n)$ is zero or small.

• $f'(x_n)=0$: the tangent is horizontal and never meets the axis. For $f(x)=\mathrm{e}^x-3x$, $f'(\ln3)=0$, so $x_0=\ln3$ gives no $x_1$.
• $f'(x_n)$ small: the tangent meets the axis far away. For $f(x)=x^3-2x+2$, $f'(0.8)=-0.08$ gives $x_1=12.2$.
• Cycling: for the same $f$, $x_0=0$ gives $x_1=1$, $x_2=0$, and so on, never converging.
• Converging to a different root, if the start is nearer another root. Before using the method, locate the root by a sign change and choose $x_0$ close to it, away from stationary points.