Addressing Errors in Apostol's Calculus & Linear Algebra: Volume 1, 2nd Edition

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This is page #293 of the 1-st volume from 2-nd edition of Apostol's "Calculus & Linear Algebra".

Well, the question goes like this:

Can you find a flagrant error in this page...?

Daniel.
 
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This is page #371 from the V-th edition of Arfken & Weber's "Mathematical Methods for Physicists".

The question is the same, but this time the error is a bit harder to spot.

Daniel.
 
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the only "error" I could see in apostol was the failure to check in the first example that it had the form 0/0 before applying l'hopital. but he did say it after the fact, namely in example 2, that it "also" had this form, implying it for example 1.
 
dextercioby said:
This is page #371 from the V-th edition of Arfken & Weber's "Mathematical Methods for Physicists".

The question is the same, but this time the error is a bit harder to spot.

Daniel.

There is at least one error.

In exercise 5.9.10 part (b): the integrand function is negative in ]0,1[ thus the risult can't be positive. The correct expression is

[tex]\lim_{a \rightarrow 1} \int_0^a \frac{\ln(1-x)}{x} dx = - \zeta(2)[/tex] .
 
WigneRacah is absolutely right for the second one, the one from Arfken & Weber. Indeed, there's a minus where he said it was.

As for apostol's book, there's something more about circular logics than anything else.

Daniel.
 
I understand now. While I don't own a copy of Apostol, I can guess that the problem is in example 1.

The typical proof that the derivative of sin(x) is cos(x), that is usually presented in introductory calculus textbooks (if at all), requires as one step to evaluate the limit: [tex]\lim_{ x\to 0 } \frac{ \sin x } { x }[/tex], which is of course 1.

However, if you wish to prove this limit, you cannot use l'Hospital's rule, because such a rule would require the derivative of sin(x) to be known already (which is what you're trying to prove) so it's a circular argument.

This assumes, though, that Apostol does not present an alternative proof that the derivative of sin(x) is cos(x) which does not require application of l'Hospital.
 
this is not the way he does it but you could define e^z by a powers eries, then let cos and sin be the real and imaginary parts. their derivatives follow immediately.

or one could define them as independent solutions of asecond order ode... but as i recall apostol does it by a more unique method.
 
Apostol defines them geometrically and does prove sin'(x)=cos(x) using the limit of sin(x)/x. He did prove this limit in chapter 3 from the inequality cos(x)<sin(x)/x<1/cos(x) for 0<x<pi/2, which followed from his geometirc construction.

So nothing circular. At worst unnecessary since he had already evaluated the limit, but not a bad plan to have a back up way of 'deriving' it if part of your memory fails you.
 
I've always understood "circular logic" to imply there is some sort of error in your argument and it was important that you were trying to assume an unproven result to prove this same unproven result. I'd be willing to accept that's not the normal usage if you want to apply the term here.

That's just semantics though, there is nothing I'd call an error at all here as he had proven the derivative of sin was cos earlier. He wasn't getting a new result, but that doesn't make it wrong. He could have cut out l'hopital and the derivative middle men and said "use example 4, section 3.4 to prove lim sin(x)/x=1" and I still wouldn't call it an error. Silly, yes, but not an error.
 
dextercioby said:
It is circular, as you can't prove A is right by using it as being right already.

Think about it.

Daniel.

You can prove A is right if you've already proven it's right. Just because Apostol is pro doesn't mean you need to be jealous :-p