A first course in Mathematical Analysis D. Brannan.

if |r(x)|< M/(|x|-1). Does it follow that,if |x|>=M, then |r(x)|<1 ?

I say no! Please see attachment in pdf and post your comment. I think the book might be in error.

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- Oct 24th 2008, 06:23 AMJohn_mathsZeros Localisation Thm. Could someone check
A first course in Mathematical Analysis D. Brannan.

if |r(x)|< M/(|x|-1). Does it follow that,if |x|>=M, then |r(x)|<1 ?

I say no! Please see attachment in pdf and post your comment. I think the book might be in error. - Oct 24th 2008, 06:42 AMOpalg
Haha, yes, it looks as though you have caught David Brannan out there. (Evilgrin)

He seems to have defined M in two contradictory ways. First, he defines . Then a couple of lines later he defines .

If you take the first of those definitions for M, and then ask that , the proof makes sense and becomes correct.

[I'll have to tease David about that next time I see him. (Giggle) ] - Oct 24th 2008, 08:40 AMJohn_maths
Thx for the speedy response. Yet I see M defined only one way as in the boxed theorem 4. as M=1+max(|asub(n-1)|..|asub0|) and this is consistent with the previous page where the zeros Localisation Theorem is introduced. Indeed David used M with this formula in an example. I don't see it defined anywhere without the 1.

P.S. if you see David maybe you could ask if there's an errata for the book (I have spotted a few more unconfirmed errors including the analysis of the blancmange function which is used on the front cover of the book) I have searched the publishers web site with no joy. - Oct 24th 2008, 11:17 AMOpalg
- Oct 24th 2008, 03:24 PMJohn_maths
Thx very much for all of your help. I see it clearly now that you have explained it. I hope you wont mind if I get stuck again, that I call on your assistance. I'm not a student but just someone that is trying to better their understanding of mathematics. I'm reading the book doing the problems in it alone.