Solve the iterative problem below

[chwala]

Homework Statement

see attached

Relevant Equations

numerical analysis

I need insight on $Q.2 (ii)$ part only,

find mark scheme here;

How do we determine suitable equation, $x=x...$?

ok, just looking at the solution, we shall have;
$2x^3+50 = x^3+100$
$x^3=50$
→$x$=$\sqrt[3]50$≡$3.68403$ which is the value of $α$ ( convergent value as indicated in previous step).

I think it was an error to indicate $3\sqrt 50$ ...am assuming it was a typo error which is a bit misleading...been cracking my head trying to figure out on this.

 

[Mark44]

Homework Statement:: see attached
Relevant Equations:: numerical analysis

I need insight on $Q.2 (ii)$ part only,

View attachment 291842

find mark scheme here;
View attachment 291843

How do we determine suitable equation, $x=x...$?

ok, just looking at the solution, we shall have;
$2x^3+50 = x^3+100$
$x^3=50$
→$x$=$\sqrt[3]50$≡$3.68403$ which is the value of $α$ ( convergent value as indicated in previous step).

From the recurrence relation, you can write this:
$$x = \frac{x^4 + 100x}{2x^3 + 50} \Rightarrow 2x^4 + 50 x = x^4 + 100x$$
IOW, all I did was replace $x_n$ and $x_{n + 1}$ by x.
The latter equation simplifies to $x(x^3 - 50) = 0$, so x = 0 or $x = \sqrt[3]{50}$

I think it was an error to indicate $3\sqrt 50$ ...am assuming it was a typo error which is a bit misleading...been cracking my head trying to figure out on this.

Yes, it was a typo in the problem statement. They wrote $3\sqrt{50}$ when they should have written $\sqrt[3]{50}$; i.e., the cube root of 50, not $3 * \sqrt{50}$.

 

[chwala]

Nice Mark, but you also realize that we could divide both sides of the equation by $x$... to realize the required solution...
$x=0$ in this case would not apply.

 

[Mark44]

Nice Mark, but you also realize that we could divide both sides of the equation by $x$... to realize the required solution...
$x=0$ in this case would not apply.

It's almost never a good idea to divide both sides of an equation by the variable. x = 0 is a solution of the equation $2x^4 + 50x = x^4 + 100x$, and is in fact a solution of the iterative formula $x_{n+1} = \frac{x_n}2 \cdot \frac{x_n^3 + 100}{x_n^3 + 25}$

 

[chwala]

It's almost never a good idea to divide both sides of an equation by the variable. x = 0 is a solution of the equation $2x^4 + 50x = x^4 + 100x$, and is in fact a solution of the iterative formula $x_{n+1} = \frac{x_n}2 \cdot \frac{x_n^3 + 100}{x_n^3 + 25}$

Ok cheers Mark...agreed in that case, we may say that the mark scheme was not conclusive on the possibilities of $x$. They ought to have given provision for two solutions, then indicate that solution $x=0$ is "unsuitable".