Iterative Methods: cobweb and staircase diagrams

[Monkeyseat]

Hi,

I've made a copy of the question because I thought typing the subscript would make things confusing.

Question

Working (there is meant to be some subscript here, I think you can figure it out though e.g. with xn+1 the n+1 should be subscript)

a)

x1 = 2.5
x2 = 2.45
x3 = 2.4005
x4 = 2.3525

b)

I didn't know what to write here - I've kind of shown this in part (a) haven't I? If xn < 3 then xn+1 will be even smaller e.g. if x1 = 2.9, x2 = 2.882. By squaring xn, xn+1 will continue to get smaller, therefore, xn+1 will always be less than 3 if xn < 3.

c)

i)
L = (1/5)(L^2 + 6)
5L = L^2 + 6
L^2 - 5L + 6 = 0
ii)
(L-2)(L-3) = 0

Reject L = 3, since xn+1 < 3. Therefore L = 2 (and you can see this from part (a)).

d)

This is the part that's just confusing.

Here's my graph:

Here's the book's graph:

By the look of it, the book has drawn it converging to 3, rather than 2. Also, their values of x2, x3 etc. are bigger than x1 which I think is wrong (see part a). It has to converge to 2 though doesn't it (because x1 is 2.5, so xn+1 < 3)?

I tried to varify my answer using this this tool, but I don't know whether it's right and it's not very clear:

I just wanted to get a second opinion on part (d). Have I done it wrong - is the limit 3? I'm just not sure. The reason I ask is that I can't believe that this would be a mistake, it seems a bit odd, although I know the book has been wrong a few times. This book can be very frustrating.

Thanks.

By the way, sorry if this is the wrong forum, I didn't know where to put this thread.

 

[Deleted member 4993]

Hi,

I've made a copy of the question because I thought typing the subscript would make things confusing.

Question

Working (there is meant to be some subscript here, I think you can figure it out though e.g. with xn+1 the n+1 should be subscript)

a)

x1 = 2.5
x2 = 2.45
x3 = 2.4005
x4 = 2.3525

b)

I didn't know what to write here - I've kind of shown this in part (a) haven't I? If xn < 3 then xn+1 will be even smaller e.g. if x1 = 2.9, x2 = 2.882. By squaring xn, xn+1 will continue to get smaller, therefore, xn+1 will always be less than 3 if xn < 3.

No - you have only shown special case. You need to show this in general.

Find the expression for y = (x[sub:1790inu1]n+1[/sub:1790inu1] - x[sub:1790inu1]n[/sub:1790inu1]) - and you'll see that for x[sub:1790inu1]n[/sub:1790inu1] < 3 (and >2), y is negative.

c)

i)
L = (1/5)(L^2 + 6)
5L = L^2 + 6
L^2 - 5L + 6 = 0
ii)
(L-2)(L-3) = 0

Reject L = 3, since xn+1 < 3. Therefore L = 2 (and you can see this from part (a)).

d)

This is the part that's just confusing.

Here's my graph:

Here's the book's graph:

By the look of it, the book has drawn it converging to 3, rather than 2. Also, their values of x2, x3 etc. are bigger than x1 which I think is wrong (see part a). It has to converge to 2 though doesn't it (because x1 is 2.5, so xn+1 < 3)?

I tried to varify my answer using this this tool, but I don't know whether it's right and it's not very clear:

I just wanted to get a second opinion on part (d). Have I done it wrong - is the limit 3? I'm just not sure. The reason I ask is that I can't believe that this would be a mistake, it seems a bit odd, although I know the book has been wrong a few times. This book can be very frustrating.

Thanks.

By the way, sorry if this is the wrong forum, I didn't know where to put this thread.

As far as I can see your work is correct (except as noted above).

 

[Monkeyseat]

Hi, thanks for checking and confirming that the diagram is correct. Sorry for my very late reply.

My book does not cover what you pointed out for part (b) so I'm not entirely sure how I would go about doing what you suggested. The book seems to explain things in words. For a similar question (not this one) where it asked "Show that all values of xn are positive", it just said "Since x1 is positive and x2 is the positive square root of a positive number, x2 is also positive. SImilarly, if xn is positive then xn+1 is positive. Therefore, all values for xn are positive." That's why I did it the way I did.

Trying your method:

xn+1 = (1/5)(xn^2 + 6)
5(xn+1) = xn^2 + 6
0 = (xn^2) - 5(xn+1) + 6

I don't think that is correct? Please could you elaborate? I do not know how to get it in the form you stated.

Regarding what you said "for xn < 3 (and >2), y is negative", when x1 = 2.5, x2 = positive - I am just confused where "y" came from and what it would mean if negative.

Sorry, what you are saying is new to me!

Thanks.