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Yet more errata... March 23, 2008 05:49PM |
Registered: 13 years ago Posts: 39 Rating: 0 |

Ok, spent the whole day trying to work through Scaled Partial Pivoting, but struggling to get a hold on it. Along the way, I did pick up these errors.

On page 102 (7th international ed.), second matrix on the page, row 1, column 2. The value in the book is 0.01. I'm saying that it should be 0.02

Second paragraph after that, they introduce R = [ 0.03,__-0.03__, 0.5]. I reckon it should be R = [ 0.03, __-0.01__, 0.5].

After that, I don't get anywhere near to the two matrices on the bottom of page 102. If I work through my stuff, I do get the correct results, namely x= [1,1,1], but why does it differ so much from what's in the book. It will most likely be either that I'm doing it incorrectly, or there are printed errors in the book. I'm going with the latter.

I've sent all of this off in an email to the lecturers, and will update once I get a response. In the meantime, if you're bored or if you're also struggling with this, I've attached__my version of the example__ in the textbook. If you see something glaringly wrong, please let me know.

On page 102 (7th international ed.), second matrix on the page, row 1, column 2. The value in the book is 0.01. I'm saying that it should be 0.02

Second paragraph after that, they introduce R = [ 0.03,

After that, I don't get anywhere near to the two matrices on the bottom of page 102. If I work through my stuff, I do get the correct results, namely x= [1,1,1], but why does it differ so much from what's in the book. It will most likely be either that I'm doing it incorrectly, or there are printed errors in the book. I'm going with the latter.

I've sent all of this off in an email to the lecturers, and will update once I get a response. In the meantime, if you're bored or if you're also struggling with this, I've attached

Re: Yet more errata... June 17, 2008 11:37AM |
Registered: 12 years ago Posts: 74 Rating: 0 |

Sorry for the long posting but please see the helpful/useful e-mail correspondence that I had with one of our lecturers. Plz read from bottom up for errata.

Regards,

Richard

=============================================

This message (and attachments) is subject to restrictions and a disclaimer. Please refer to http://www.unisa.ac.za/disclaimer for full details.

Dear student

You are right. The formula in the textbook is not correct.

on page 264, h^(n+1) must be included in the expression of the error and

there is no obvious objection on the suggested solution for your

question no.3

We wish you good luke for the preparation of your assignments and exams.

Regards

Cos233-8 Team

-----Original Message-----

From: Leeman, Richard (Selestia) [mailto:Richard.Leeman@skandia.co.uk]

Sent: 13 June 2008 11:33 AM

To: C Dongma

Cc: Viljoen, Elizabeth

Subject: RE: Query for APM213X - 34567771

Dear lecturer,

Thank you for your reply.

Regarding my question 2) I am referring to the formula which is as

follows:

Error = [Product(s - j) from j = 0 to n] * (f^(n+1)(Greek Xi)/(h^(n-1) *

(n+1)!), Greek Xi in [x, x1, ..., xn]

on page 264 for Wheatley (seventh edition, ISBN 0-321-13304-8)

From 5.6 we have:

Error = (x-x0)(x-x1)...(x-xn)(f^(n+1)(Greek Xi)/(n+1)!)

Taking s = (x - x0)/h

i.e. x - x0 = s.h

(sh)(h(s-1))(h(s-2))...(h(s-n))(Greek Xi)/(n+1)!)

= h^(n+1)[Product(s - j) from j = 0 to n] * (f^(n+1)(Greek Xi)/((n+1)!)

= [Product(s - j) from j = 0 to n] * (h^(n+1) * f^(n+1)(Greek

Xi)/((n+1)!)

Which is not the same as the formula given 3/4 of the way down the page

on page 264 just above the paragraph above (5.8). Mine has h^(n+1) in

the numerator, the text has h^(n-1) in the denominator. I therefore

believe that the formula was meant to have h^(n+1) in the numerator -

i.e. the text is wrong.

Note that I took s = (x - x0)/h. If I had used s = (x - xi)/h then I get

a different product term if i <> 0. However I suspect that the product

term given above is the worst case scenario (where i = 0) which seems

fair enough.

When I look at 5.9 on page 265 where the they have s = 0: the term (x -

xi) in 5.6 drops out leaving h^n in the numerator in 5.9 not h^(n+1)

which is correct. Also this leaves only n (x - xj) terms and hence the

(-1)^n.

Comparing (5.9) with the formula on page 264 (3/4 of the way down the

page just above the paragraph above (5.8)) they are inconsistent,

because whereas the former has the h^ factor in the numerator the latter

has it in the denominator.

Hopefully I have been able to put my workings across - please let me

know if you would like me to call if that would be better. Otherwise I

would be grateful if you could let me know if I am right or if not where

I have gone wrong.

Regarding my question 3):

I found

http://www.physics.arizona.edu/~restrepo/475A/Notes/sourcea/node41.html

which appears to answer my question. Could you confirm please that I am

on the right track? Our own text Wheatley seems to gloss over the

complexity of integrating the error terms!

Thank you for your help. I hope that you have a good day.

Kind regards,

Richard

PS I am enjoying this course but errata in the prescribed text book have

caused me some frustration!

-----Original Message-----

From: C Dongma [mailtoongmc@unisa.ac.za]

Sent: 13 June 2008 09:49

To: Leeman, Richard (Selestia)

Subject: RE: Query for APM213X - 34567771

This message (and attachments) is subject to restrictions and a

disclaimer. Please refer to http://www.unisa.ac.za/disclaimer for full

details.

Dear student,

you are right, on page 261, in the expression of the formula (5.7) 1

must be replaced by i.

For your question no.2, on the spicified page 264, we cannot find the

expression "h^(n-1)" you are referring to. Can you then check once more

or provide more informations about the error so that we can complete

this question and move to the last one? for instance, if you are trying

to develop the formula, let us know the process you are following so

that we can be able to help you accordingly.

Regards

Cos233-8 Team

-----Original Message-----

From: Leeman, Richard (Selestia) [mailto:Richard.Leeman@skandia.co.uk]

Sent: 11 June 2008 12:50 PM

To: Viljoen, Elizabeth

Cc: C Dongma; Pretorius, Laurette; richard@leeman.org

Subject: Query for APM213X - 34567771

Dear Mrs Viljoen,

Thank you for your help previously.

Please could you help me with the following queries on chapter 5 of

Wheatley? Note that I use <> for not equal to, ^ for to the power of,

and Greek Xi to represent the Greek letter Xi in the formulae in

Wheatley.

1) Page 261, formula 5.7. Surely this should be j <> i, not j <> 1?

Possibly simple typo.

2) Page 264, about three-quarters of the way down the page, the Error is

given with h^(n-1) in the denominator. Surely this should be h^(n+1) in

the *numerator*? Otherwise it appears to be inconsistent with 5.6 and

also with 5.9.

3) Page 274. I cannot get to the Error expression roughly two-thirds

down the page i.e. -(1/12)(h^3)f"(Greek Xi). I have tried integration by

parts but the difficulty is dealing with the f"(Greek Xi) term which I

understand is dependent on s. Please could you provide me with or give

me a reference to a proof for this? I would really like to understand

the error term in greater depth to build my understanding of this.

Please us richard@leeman.org for correspondence.

With best wishes from London,

Richard Leeman

Regards,

Richard

=============================================

This message (and attachments) is subject to restrictions and a disclaimer. Please refer to http://www.unisa.ac.za/disclaimer for full details.

Dear student

You are right. The formula in the textbook is not correct.

on page 264, h^(n+1) must be included in the expression of the error and

there is no obvious objection on the suggested solution for your

question no.3

We wish you good luke for the preparation of your assignments and exams.

Regards

Cos233-8 Team

-----Original Message-----

From: Leeman, Richard (Selestia) [mailto:Richard.Leeman@skandia.co.uk]

Sent: 13 June 2008 11:33 AM

To: C Dongma

Cc: Viljoen, Elizabeth

Subject: RE: Query for APM213X - 34567771

Dear lecturer,

Thank you for your reply.

Regarding my question 2) I am referring to the formula which is as

follows:

Error = [Product(s - j) from j = 0 to n] * (f^(n+1)(Greek Xi)/(h^(n-1) *

(n+1)!), Greek Xi in [x, x1, ..., xn]

on page 264 for Wheatley (seventh edition, ISBN 0-321-13304-8)

From 5.6 we have:

Error = (x-x0)(x-x1)...(x-xn)(f^(n+1)(Greek Xi)/(n+1)!)

Taking s = (x - x0)/h

i.e. x - x0 = s.h

(sh)(h(s-1))(h(s-2))...(h(s-n))(Greek Xi)/(n+1)!)

= h^(n+1)[Product(s - j) from j = 0 to n] * (f^(n+1)(Greek Xi)/((n+1)!)

= [Product(s - j) from j = 0 to n] * (h^(n+1) * f^(n+1)(Greek

Xi)/((n+1)!)

Which is not the same as the formula given 3/4 of the way down the page

on page 264 just above the paragraph above (5.8). Mine has h^(n+1) in

the numerator, the text has h^(n-1) in the denominator. I therefore

believe that the formula was meant to have h^(n+1) in the numerator -

i.e. the text is wrong.

Note that I took s = (x - x0)/h. If I had used s = (x - xi)/h then I get

a different product term if i <> 0. However I suspect that the product

term given above is the worst case scenario (where i = 0) which seems

fair enough.

When I look at 5.9 on page 265 where the they have s = 0: the term (x -

xi) in 5.6 drops out leaving h^n in the numerator in 5.9 not h^(n+1)

which is correct. Also this leaves only n (x - xj) terms and hence the

(-1)^n.

Comparing (5.9) with the formula on page 264 (3/4 of the way down the

page just above the paragraph above (5.8)) they are inconsistent,

because whereas the former has the h^ factor in the numerator the latter

has it in the denominator.

Hopefully I have been able to put my workings across - please let me

know if you would like me to call if that would be better. Otherwise I

would be grateful if you could let me know if I am right or if not where

I have gone wrong.

Regarding my question 3):

I found

http://www.physics.arizona.edu/~restrepo/475A/Notes/sourcea/node41.html

which appears to answer my question. Could you confirm please that I am

on the right track? Our own text Wheatley seems to gloss over the

complexity of integrating the error terms!

Thank you for your help. I hope that you have a good day.

Kind regards,

Richard

PS I am enjoying this course but errata in the prescribed text book have

caused me some frustration!

-----Original Message-----

From: C Dongma [mailtoongmc@unisa.ac.za]

Sent: 13 June 2008 09:49

To: Leeman, Richard (Selestia)

Subject: RE: Query for APM213X - 34567771

This message (and attachments) is subject to restrictions and a

disclaimer. Please refer to http://www.unisa.ac.za/disclaimer for full

details.

Dear student,

you are right, on page 261, in the expression of the formula (5.7) 1

must be replaced by i.

For your question no.2, on the spicified page 264, we cannot find the

expression "h^(n-1)" you are referring to. Can you then check once more

or provide more informations about the error so that we can complete

this question and move to the last one? for instance, if you are trying

to develop the formula, let us know the process you are following so

that we can be able to help you accordingly.

Regards

Cos233-8 Team

-----Original Message-----

From: Leeman, Richard (Selestia) [mailto:Richard.Leeman@skandia.co.uk]

Sent: 11 June 2008 12:50 PM

To: Viljoen, Elizabeth

Cc: C Dongma; Pretorius, Laurette; richard@leeman.org

Subject: Query for APM213X - 34567771

Dear Mrs Viljoen,

Thank you for your help previously.

Please could you help me with the following queries on chapter 5 of

Wheatley? Note that I use <> for not equal to, ^ for to the power of,

and Greek Xi to represent the Greek letter Xi in the formulae in

Wheatley.

1) Page 261, formula 5.7. Surely this should be j <> i, not j <> 1?

Possibly simple typo.

2) Page 264, about three-quarters of the way down the page, the Error is

given with h^(n-1) in the denominator. Surely this should be h^(n+1) in

the *numerator*? Otherwise it appears to be inconsistent with 5.6 and

also with 5.9.

3) Page 274. I cannot get to the Error expression roughly two-thirds

down the page i.e. -(1/12)(h^3)f"(Greek Xi). I have tried integration by

parts but the difficulty is dealing with the f"(Greek Xi) term which I

understand is dependent on s. Please could you provide me with or give

me a reference to a proof for this? I would really like to understand

the error term in greater depth to build my understanding of this.

Please us richard@leeman.org for correspondence.

With best wishes from London,

Richard Leeman

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