Zeros Localisation Thm. Could someone check

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October 24th 2008, 05:23 AM
John_maths

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Zeros 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.
October 24th 2008, 05:42 AM
Opalg

Quote:

Originally Posted by John_maths

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.

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 $M = \max\{|a_{n-1}|,\ldots,|a_1|,|a_0|\}$ . Then a couple of lines later he defines $M = 1+ \max\{|a_{n-1}|,\ldots,|a_1|,|a_0|\}$ .

If you take the first of those definitions for M, and then ask that $|x|>M+1$ , the proof makes sense and becomes correct.

[I'll have to tease David about that next time I see him. (Giggle) ]
October 24th 2008, 07:40 AM
John_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.
October 24th 2008, 10:17 AM
Opalg

Quote:

Originally Posted by John_maths

I see M defined only one way as in the boxed theorem 4. as M=1+max(|asub(n-1)|..|asub0|)

I hadn't noticed that M is defined in the statement of the theorem. In that case, the way to correct the proof is to replace M by M-1 at the point where it first appears in the proof. That will lead to the inequality $|r(x)| < \frac{M-1}{|x|-1}$ , which will give $|r(x)|<1$ whenever $|x|\geqslant M$ .
October 24th 2008, 02:24 PM
John_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.