# Convergence of the secant method.¶

In this section of Lecture 24, we'll see the convergence rate of the secant method for finding the root of a scalar nonlinear function $f$.

Let $f : \mathbb{R} \to \mathbb{R}$, we want to find $\alpha$ such that $f(\alpha) = 0$.

Recall that the secant method begins with two iterates: $x_0, x_1$ and proceeds by finding the interpolanting line and moving to the root of that line. The interpolanting line in Newton form is $p(x) = f(x_0) + \frac{f(x_k) - f(x_{k-1})}{x_{k} - x_{k-1}} (x - x_k)$. Solving for the update yields

$$x_{k+1} = x_{k} - \frac{x_{k} - x_{k-1}}{f(x_k) - f(x_{k-1})} f(x_k)$$

We can visualize this method.

In [9]:
using Plots
plot();

In [10]:
Fun = x -> (x-5.).*(x-2.)
xprev = 2.5
x = 3.
xx = linspace(0,6,500)
nsteps = 5
xhist = zeros(nsteps+2)
xhist[1] = xprev
xhist[2] = x
anim = @animate for i=3:nsteps+2
xnext = x - ((x-xprev)/(Fun(x)-Fun(xprev)))*Fun(x)
Line = z-> Fun(x) + ((Fun(x) - Fun(xprev))./(x-xprev)).*(z-x)
plot(xx,Fun(xx),leg=false)
hline!([0.],linewidth=3.,linecolor=colorant"black")
plot!(xx,Line(xx))

# Arg, Julia anonymous functions don't capture the current values.
xprev,x = x,xnext
xhist[i] = x

scatter!(xhist[1:i],Fun(xhist[1:i]))
ylims!(-10,10)

end
gif(anim,fps=1)

INFO: Saved animation to /Users/dgleich/Dropbox/courses/cs514-2016/web/input/julia/tmp.gif

Out[10]:

To study the convergence, we make use of a few ideas that use divided differences. Let $\alpha$ be the limit point of the sequence $x_k$. Then note that

$$x_{k+1} - \alpha = (x_{k} - \alpha)(x_{k-1} - \alpha) \frac{f[x_{k-1},x_k,\alpha]}{f[x_{k-1},x_k]}$$

(This is not easy to work out, but the book works through it.

Then as $x_k \to \alpha$, note that $\frac{f[x_{k-1},x_k,\alpha]}{f[x_{k-1},x_k]} \to 1/2 f''(\alpha) / f'(\alpha)$. So what happens is that

$$\lim_{k \to \infty} \frac{x_{k+1} - \alpha}{x_k - \alpha} = (x_{k-1} - \alpha) \frac{f[x_{k-1},x_k,\alpha]}{f[x_{k-1},x_k]} \to 0$$

this means that the method converges superlinearly. Let's see a plot of it's error.

In [11]:
Fun = x -> (x-5.).*(x-2.)
xprev = 2.5
x = 3.
Fun(5.)
xx = linspace(0,6,500)
nsteps = 15
xhist = zeros(nsteps+2)
xhist[1] = xprev
xhist[2] = x
anim = @animate for i=3:nsteps+2
xnext = x - ((x-xprev)/(Fun(x)-Fun(xprev)))*Fun(x)
# Arg, Julia anonymous functions don't capture the current values.
xprev,x = x,xnext
xhist[i] = x
end
plot(abs(xhist-2.),marker=:circle)
yaxis!(:log10)

Out[11]:

Our theory predicts trouble with roots whose derivative is zero. We see this too.

In [12]:
Fun = x -> (x-5.).*(x-2.).*(x-2.)
xprev = 2.5
x = 3.
Fun(5.)
xx = linspace(0,6,500)
nsteps = 5
xhist = zeros(nsteps+2)
xhist[1] = xprev
xhist[2] = x
anim = @animate for i=3:nsteps+2
xnext = x - ((x-xprev)/(Fun(x)-Fun(xprev)))*Fun(x)
Line = z-> Fun(x) + ((Fun(x) - Fun(xprev))./(x-xprev)).*(z-x)
plot(xx,Fun(xx),leg=false)
hline!([0.],linewidth=3.,linecolor=colorant"black")
plot!(xx,Line(xx))

# Arg, Julia anonymous functions don't capture the current values.
xprev,x = x,xnext
xhist[i] = x

scatter!(xhist[1:i],Fun(xhist[1:i]))
ylims!(-10,10)

end
gif(anim,fps=1)

INFO: Saved animation to /Users/dgleich/Dropbox/courses/cs514-2016/web/input/julia/tmp.gif

Out[12]:
In [17]:
Fun = x -> (x-5.).*(x-2.).*(x-2.)
xprev = 2.5
x = 3.
Fun(5.)
xx = linspace(0,6,500)
nsteps = 100
xhist = zeros(nsteps+2)
xhist[1] = xprev
xhist[2] = x
anim = @animate for i=3:nsteps+2
xnext = x - ((x-xprev)/(Fun(x)-Fun(xprev)))*Fun(x)
# Arg, Julia anonymous functions don't capture the current values.
xprev,x = x,xnext
xhist[i] = x
end

In this case that the derivative is not zero, the actual rate of convergence is based on the Golden Ratio. There is a neat proof of this that proceeds by setting: $$e_k = x_k -\alpha$$ and then noting that, in the limit, $$e_{k+1} = e_{k} e_{k-1} C$$ for some constant $C$ that is given by the ratio of determinants. To see why the Golden Ratio arises, note that $$C e_{k+1} = C e_k C e_{k-1}$$ and so if $E_k = C e_k$, then $$E_{k+1} = E_k E_{k-1}.$$
We are almost there, the final step is to take logs, in which case $$\log E_{k+1} = \log E_{k} + \log E_{k-1}$$ and we see the Fibonacci-like series emerge.