Very important, help me plz, i'm in trouble

[takayuki]

Hi guyz,,,

I had a math exam and my mark was 15/20
The two questions that the teacher marked them for me as wrong are still not getting in my mind that I’m wrong, I’d like to post the 2 questions with my answers and plz tell me whether it’s right or wrong….

Question2: (b)
Show that any fixed point of the one-point iteration function
Xn+1= 4 / (Xn-5)=g(Xn); Xn is not equal to 5 ; n=0,1,2,……
Will be a root of the equation X^2 – 5X – 4=0

My Solution was:
I solved the quadratic equation and I got the roots, X=5.702 X=-0.702
Then I took X0=1 as a starting point to iterate the Xn+1 to show that it has a root of the quadratic equation’s roots

X1=-1.000 X2=-0.667 X3=-0.706
X4=-0.701 X5=-0.702 X6=-0.702=X7=……..

In this way I showed that this one-point iteration function has a root of the quadratic equation’s roots

The instructor gave me 0 out of 3 on this question !!!!

Question2: ( c )
Show that the iteration function in part (b) will exhibit oscillatory convergence if the starting value X0 (X0 is not equal to 5) satisfies the inequality X0^2 – 10X0 +21 > 0

My solution:
First I equalized the inequality to zero and I found the zeros which were equal to
3 and 7
Then I found that X0 should not belong to the closed period [3,7] in order to satisfy the inequality

I differentiated g(Xn)
I got g’(Xn)= -4 / (Xn-5)2; and as I chose X0=8 which doesn’t belong to [3,7]
And I substituted in g’(Xn) to show that it exhibits oscillatory

I got g’(8) = -0.444 ; This means that -1<g’(X0)<0
Which shows and proves that it converges oscillatory

I got 1 out of 2 on this question!!!

Plz can you tell me what is my mistake and it doesn’t make sense to me,,,, plz help me in this trouble
My regards

[HallsofIvy]

Show that any fixed point of the one-point iteration function
Xn+1= 4 / (Xn-5)=g(Xn); Xn is not equal to 5 ; n=0,1,2,……
Will be a root of the equation X^2 – 5X – 4=0

My Solution was:
I solved the quadratic equation and I got the roots, X=5.702 X=-0.702
Then I took X0=1 as a starting point to iterate the Xn+1 to show that it has a root of the quadratic equation’s roots

X1=-1.000 X2=-0.667 X3=-0.706
X4=-0.701 X5=-0.702 X6=-0.702=X7=……..

In this way I showed that this one-point iteration function has a root of the quadratic equation’s roots
No, you haven't. What you've shown is that the numbers appear, to three decimal places, to be approaching one of the roots. You haven't shown that they really do. Strictly speaking, even if you had shown that the limit actually was one of the roots, that would not show that it was a fixed point or that it was the only fixed point.
Unfortunately, you also have not shown that you have any idea what a "fixed point" is. If X is a fixed point, Then each iteration must give X again: g(X)= X or 4/(X-5)= X. Multiply that last equation out and see what you get.


Question2: ( c )
Show that the iteration function in part (b) will exhibit oscillatory convergence if the starting value X0 (X0 is not equal to 5) satisfies the inequality X0^2 – 10X0 +21 > 0

My solution:
First I equalized the inequality to zero and I found the zeros which were equal to
3 and 7
Then I found that X0 should not belong to the closed period [3,7] in order to satisfy the inequality

I differentiated g(Xn)
I got g’(Xn)= -4 / (Xn-5)2; and as I chose X0=8 which doesn’t belong to [3,7]
And I substituted in g’(Xn) to show that it exhibits oscillatory

I got g’(8) = -0.444 ; This means that -1<g’(X0)<0
Which shows and proves that it converges oscillatory

I got 1 out of 2 on this question!!!
Your teacher was generous. By taking x0= 8, you showed that that particular number gives oscillatory behaviour. You did NOT show that that was true for any other x0< 3 or x0> 7.