Contents

• 1  Secant method
• 2  Exercise
  • 2.1  Approximate, using the secant method $r=\sqrt{3}.$ Use as initial guesses $x_{0}=1$ and $x_{1}=2$. Perform 3 iterations
  • 2.2  Compute the residual
  • 2.3  Use a calculator to estimate the absolute error of the approximation

Secant method

The secant method can be considered a variant of Newton's method in which the derivative is replaced by an approximation.

We can define the derivative of a function $f$ at a point $a$ as

$$f'(a)=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$$

If $x=a+h$ then $h=x-a$

$$f'(a)=\lim_{x\rightarrow a}\frac{f(x)-f(a)}{x-a}$$

also

$$f'(a)=\lim_{x\rightarrow a}\frac{f(a)-f(x)}{a-x}$$

If $a = x_{k-1}$ and $x = x_{k-2}$

$$f'(x_{k-1})=\lim_{x_{k-2}\rightarrow x_{k-1}}\frac{f(x_{k-1})-f(x_{k-2})}{x_{k-1}-x_{k-2}}$$

Removing the limit we obtain

$$f'(x_{k-1})\approx \frac{f(x_{k-1})-f(x_{k-2})}{x_{k-1}-x_{k-2}}$$

If the iteration formula for Newton's method is

$$x_k=x_{k-1}-\dfrac{f(x_{k-1})}{f'(x_{k-1})}$$

and we substitute the derivative by its approximation

$$x_k=x_{k-1}-f(x_{k-1})\dfrac{x_{k-1}-x_{k-2}}{f(x_{k-1})-f(x_{k-2})}$$

Graphically, the secant method can be explained as an iterative method where

• The curve is replaced by a straight line that passes through two of its points (secant line)
• The zero of the curve is approximated with the zero of the straight line.
• The oldest point is discarded (important!), and a new secant line is drawn with the point that remains and the last calculated point.
• We repeat until a stop condition is met.

That is, we start with

• $x_0$ and $x_1$ and we obtain $x_2.$

Next

• $x_1$ and $x_2$ and we obtain $x_3.$

And so on.

Algorithm

• Given two initial guesses $x_0$ and $x_1$
• For $k=2,3\ldots,\mathrm{MaxNumIter}$:

  • Compute $x_k=x_{k-1}-f(x_{k-1})\dfrac{x_{k-1}-x_{k-2}}{f(x_{k-1})-f(x_{k-2})}$
  • If $x_k$ meets the stopping condition, stop.
  • Else, perform another iteration.

Once Loop Reflect

Drawbacks

• We need two initial guesses.
• In general, it converges more slowly than Newton (but faster than bisection).
• If we do not select the initial points well, it may not converge.

Advantages

• We don't need to know the exact derivative, just the value of the function.
• The speed of convergence is relatively fast. The order of convergence is $p\approx1.6$

Exercise

Approximate, using the secant method $r=\sqrt{3}.$ Use as initial guesses $x_{0}=1$ and $x_{1}=2$. Perform 3 iterations

Our equation is

$$x=\sqrt{3} \quad \Rightarrow \quad x^2=3 \quad \Rightarrow \quad x^2-3=0$$

Therefore $f(x)=x^2-3$

The iteration formula is

$$x_k=x_{k-1}-f(x_{k-1})\dfrac{x_{k-1}-x_{k-2}}{f(x_{k-1})-f(x_{k-2})}$$

In this case

$$x_k=x_{k-1}-(x_{k-1}^2-3)\dfrac{x_{k-1}-x_{k-2}}{(x_{k-1}^2-3)-(x_{k-2}^2-3)}$$

Iterations

• $x_0=1$, $x_1=2$ $$x_2=x_{1}-(x_{1}^2-3)\dfrac{x_{1}-x_{0}}{(x_{1}^2-3)-(x_{0}^2-3)}=2-(2^2-3)\dfrac{2-1}{(2^2-3)-(1^2-3)}$$

$$x_2=2-(1)\dfrac{1}{(1)-(-2)}=2-\dfrac{1}{3}=1.666667$$

• $x_1=2$, $x_2=1.666667$, $$x_3=x_{2}-(x_{2}^2-3)\dfrac{x_{2}-x_{1}}{(x_{2}^2-3)-(x_{1}^2-3)}$$

$$x_3=1.666667-(1.666667^2-3)\dfrac{1.666667-2}{(1.666667^2-3)-(2^2-3)}=1.727273$$

• $x_2=1.666667$, $x_3=1.727273$, $$x_4=x_{3}-(x_{3}^2-3)\dfrac{x_{3}-x_{2}}{(x_{3}^2-3)-(x_{2}^2-3)}$$

$$x_4=1.727273-(1.727273^2-3)\dfrac{1.727273-1.666667}{(1.727273^2-3)-(1.666667^2-3)}=1.732143$$

\begin{array}{|cc|} \hline k & x_k \\ \hline 0 & 1.000000\\ 1 & 2.000000\\ 2 & 1.666667\\ 3 & 1.727273\\ 4 & 1.732143\\ \hline \end{array}

Once Loop Reflect

Compute the residual

The value of the function at the root is zero. The value of the function in the approximation is called residual and, if it is a good approximation, it is a small

$$r =|f(x_4)|=|1.732143^2-3|=0.0003$$

Also, the residual in step $k$ is known, while the error is unknown (we need the exact solution, which is what we are looking for).

Use a calculator to estimate the absolute error of the approximation

As $\alpha= 1.732051$ the absolute error is

$$e_a = |x_4-\alpha|=|1.732143-1.732051|=0.000092\approx 0.0001$$