# Math Help - Iterative schemes of the form x=g(x)

1. ## Iterative schemes of the form x=g(x)

I've done all bar four questions for an assignment, which I'm stuck on; if anyone could help me or point me in the right direction then I'll be in your debt.

Question 23

The equation of ((e^x) - (x^5)) = 0.395 has a root around to 1.3. Rearrange this equation into the iterative schemes of the form of x = g(x). Find one arrangement that is convergent to the root close to 1.3 and one arrangement that does not converge to this region. Calculate the root close to 1.3 to 3 decimal places and for the second arrangement to complete at least six iterations.

2. Originally Posted by Trapper Dave
I've done all bar four questions for an assignment, which I'm stuck on; if anyone could help me or point me in the right direction then I'll be in your debt.

Question 23

The equation of ((e^x) - (x^5)) = 0.395 has a root around to 1.3. Rearrange this equation into the iterative schemes of the form of x = g(x). Find one arrangement that is convergent to the root close to 1.3 and one arrangement that does not converge to this region. Calculate the root close to 1.3 to 3 decimal places and for the second arrangement to complete at least six iterations.
One rearrangement is:

$x=(e^x-0.395)^{1/5}$

This has $g'(x)<1$ near 1.3 so this should converge there.

Another rearrangement is:

$x=\ln(0.395+x^5)$

This has $g'(x)>1$ near 1.3 so this should diverge there.

The first code block below shows the result of itterating the first of the
above formulae and we see that the itteation converges to ~1.255

Code:
n    x_n     (exp(x)-0.395)^(1/5)
===================================
0  2          1.475522452
1  1.475522452 1.318074455
2  1.318074455 1.272861115
3  1.272861115 1.260020489
4  1.260020489 1.256384493
5  1.256384493 1.255355754
6  1.255355754 1.255064759
7  1.255064759 1.254982451
8  1.254982451 1.254959171
9  1.254959171 1.254952586
10 1.254952586 1.254950724
11 1.254950724 1.254950197
12 1.254950197 1.254950048
13 1.254950048 1.254950006
14 1.254950006 1.254949994
15 1.254949994 1.254949991
16 1.254949991 1.25494999
The following code block shows some itterations of the second formulae

Code:
n       x_n     ln(0.395+x^5)
===========================
0  1.3         1.412919252
1  1.412919252 1.796085993
2  1.796085993 2.948962056
3  2.948962056 5.409035881
4  5.409035881 8.440439638
5  8.440439638 10.66518121
RonL

3. I'm probably being dumb as per, but where did you get the value of 2 from to use as the value of x in the convergent?

4. Originally Posted by Trapper Dave
I'm probably being dumb as per, but where did you get the value of 2 from to use as the value of x in the convergent?
Just picked at random while experimenting, probably should have used 1.3.

RonL