Question

(a) Prove that the rate of convergence of Secant Method is 1.618.​

Answer 1

I have a function
()=−(2−)3
f
(
x
)
=
e
x
−
(
2
−
x
)
3

And I am using the Secant method to find a root between 0
0
and 5
5
.

I know that the value of this root is =0.7261444…
t
=
0.7261444
…

Here's my output from matlab for 14 iterations where I show


k
,
x
k
,|−|
|
t
−
x
k
|
,|−|1.68
|
t
−
x
k
|
1.68
,|−|2
|
t
−
x
k
|
2

k | x(k) | |t-x(k)| | |t-x(k)|^1.68 | |t-x(k)|^2
1 | 0.00000000e+000 | 7.261e-001 | |
2 | 5.00000000e+000 | 4.274e+000 | |
3 | 1.91872122e-001 | 5.343e-001 | 3.489e-001 | 2.854e-001
4 | 3.17334769e-001 | 4.088e-001 | 2.225e-001 | 1.671e-001
5 | 6.42309464e-001 | 8.384e-002 | 1.554e-002 | 7.028e-003
6 | 7.12431582e-001 | 1.371e-002 | 7.420e-004 | 1.880e-004
7 | 7.25679662e-001 | 4.648e-004 | 2.518e-006 | 2.160e-007
8 | 7.26141901e-001 | 2.565e-006 | 4.049e-010 | 6.580e-012
9 | 7.26144465e-001 | 4.793e-010 | 2.205e-016 | 2.297e-019
10 | 7.26144466e-001 | 5.551e-016 | 2.347e-026 | 3.081e-031
11 | 7.26144466e-001 | 0.000e+000 | 0.000e+000 | 0.000e+000
12 | 7.26144466e-001 | 1.110e-016 | 1.571e-027 | 1.233e-032
13 | 7.26144466e-001 | 1.110e-016 | 1.571e-027 | 1.233e-032
14 | NaN | NaN | NaN | NaN
Now I am told in a question that the 'rate of convergence of a sequence {0,1,2,…}
{
ϵ
0
,
ϵ
1
,
ϵ
2
,
…
}
is
q
if

lim→∞+1=
lim
k
→
∞
ϵ
k
+
1
ϵ
k
q
=
u

for some constant
u
.

And I am told to make the program output |−|
|
t
−
x
k
|
,|−|1.68
|
t
−
x
k
|
1.68
,|−|2
|
t
−
x
k
|
2
and hence deduce that the rate of converge is greater than 1
1
, less than 2
2
and roughly (1+5‾√)/2=1.618
(
1
+
5
)
/
2
=
1.618
, the golden mean.

I can't see how I supposed to deduce this...I tried getting a few values for
|−+1||−|1.68
|
t
−
x
k
+
1
|
|
t
−
x
k
|
1.68
... and this seems to go to 0
0
and then the precision becomes too small for the computer to handle. I don't really know what I am doing here trying to deduce that the convergence is 1.6.18. Can anyone explain how I am supposed to be making this deduction?

EDIT: If I check =(1+5‾√)/2≈1.618
ϕ
=
(
1
+
5
)
/
2
≈
1.618
, |−+1|/|−|
|
t
−
x
k
+
1
|
/
|
t
−
x
k
|
ϕ
up to iteration 9 I get

3 | 5.094e-002
4 | 1.127e+000
5 | 3.564e-001
6 | 7.569e-001
7 | 4.802e-001
8 | 6.331e-001
9 | 5.330e-001