I was reading an applied mathematics e-Book today and I read the Disclaimer section which I had to take a screen shot of. The picture is attached to this post. What do you think of the last two items at the bottom of the page?
Attachments:
Posted on 2007-01-10 13:33:09 by XCHG
I think it says that you'd better be capable of learning math or your butt is going to be sore :lol:
Posted on 2007-01-10 15:44:37 by SpooK
Heh, interesting disclaimer if a bit unprofessional - I guess this is a free eBook? Might be interesting for others, post URL if you can?
Posted on 2007-01-10 16:49:16 by f0dder
Yeah it's a little unprofessional but the exact opposite of the content of the book. So far, it is one of the best mathematics books I have ever read, specially the Calculus chapter. And yes the book is for free and you can download it from here. It is 9.685 megabytes and has 2321 pages.
Posted on 2007-01-10 21:38:10 by XCHG
Interesting, I think I've seen that one before, but I can't find it on my harddrive - weird. Thanks for sharing! :)
Posted on 2007-01-11 03:48:28 by f0dder
The guy asks for constructive criticism.

Tell him you want the best for him. It's good he has a sense of humor but...there is nothing wrong with getting some counseling.
:-)
Posted on 2007-01-11 09:03:31 by skywalker
I think it's great that someone would not only learn and share mathmatics, but also spend time developing themselves @ the comedy club. The old saying of not looking a gift horse in the mouth really applies here. To me it's worth the bandwidth to download, and if it's worth the paper, I'll print it out. If not, there's always that key that says "del."  ;)
Posted on 2007-01-11 09:26:26 by Jeronimo0d0a
Mmmm...fiber *drools like Homer Simpson*

Thanks for sharing :-)
Posted on 2007-01-11 09:53:47 by d00zer
To the Ineffable All,

    Interesting "book".  I perused and analyzed the material and concluded that it covered a great deal of material.  Having said that, I also observed that it is a survey of some selected branches of mathematics, and not an in depth coverage.  If it was, it would have to be many time larger.  For instance, in the calculus section, no mention is made of the important L'Hospital's theorem.  In one section, the author needs an epiphany with respect to the square root of -1.  On page 181 he says, "We cannot solve x^2 = -1 because SQRT(-1) is undefined".  Well, the square root of -1 is a definite unique number.  In polar form, it is 1/_90 degrees.  Also he says that i is the representation of the square root of -1.  That is wrong, because  i (or j) is a mathematical operator, not a constant.  So, a few lapses and mistakes do not make it an unsatisfactory book.  There is a great deal of material to study and most of it is very interesting.  Just be cognizant of what it is.  Ratch
Posted on 2007-01-11 17:17:12 by Ratch

On page 181 he says, "We cannot solve x^2 = -1 because SQRT(-1) is undefined".  Well, the square root of -1 is a definite unique number.  In polar form, it is 1/_90 degrees.  Also he says that i is the representation of the square root of -1.  That is wrong, because i (or j) is a mathematical operator, not a constant.

You are very imprecise here. First you say "the square root of -1 is a definite unique number". In what field/ring/algebra? In the field of complex numbers there are two different numbers that give -1 when squared, in the quaternion algebra there are many more.
As for the claim that i is an operator, not a constant - well, we can make every number correspond to some operator, like multiplication by it, however I'm more used to the convention that i is an element of complex plane.
Posted on 2007-01-25 05:59:52 by Tomasz Grysztar
Tomasz Grysztar,

You are very imprecise here. ...


    Not really, I gave its principal value.  There are actually an infinite number of values, not just two values,  whose square is -1, like 1/_90, 1/_270, 1/_450 ... etc.  Similarly, when you ask for the Arcsin of .5, you expect to receive 30 degrees, not 150, 390, 510 ... etc.  Get the idea?  Anyway, the point I wanted to make is that sqrt(-1) is certainly defined.  Many claim it is not.

As for the claim that i is an operator, not a constant - well, we can make every number correspond to some operator, like multiplication by it


    Perhaps, but why define an infinite number of operators?  The only reason "i" works as a constant is because it has conformal similiarity.  Defining "i" as a 90 rotational operator gives a better conception of what is going on.

...I'm more used to the convention that i is an element of complex plane.


    That is one commonly used method for graphing a number with an orthogonal component, but it is not part of its definition.  Ratch

   
Posted on 2007-01-25 10:25:08 by Ratch

Not really, I gave its principal value.  There are actually an infinite number of values, not just two values,  whose square is -1, like 1/_90, 1/_270, 1/_450 ... etc.  Similarly, when you ask for the Arcsin of .5, you expect to receive 30 degrees, not 150, 390, 510 ... etc.  Get the idea?  Anyway, the point I wanted to make is that sqrt(-1) is certainly defined.  Many claim it is not.

I have to repeat: in which field/algebra is it defined?
There are algebras with exactly one square root of -1, the fields of real/rational numbers have none. What you're speaking about seems to me to be a more an "engineering" knowledge than mathematical one.
And in the field of complex numbers, if you're speaking about it, there are exactly two different numbers that become -1 when squared (moreover, for any natual number n there are always exactly n different complex numbers that solve the equation z^n=a). On the space of polar coordinates you need to define equivalence classes to get the complex numbers space.

Perhaps, but why define an infinite number of operators?

You might want to make a space/algebra out of them. Quite frequently used concept in the abstract algebra and algebraic topology.


...I'm more used to the convention that i is an element of complex plane.

That is one commonly used method for graphing a number with an orthogonal component, but it is not part of its definition.  Ratch

The field of complex numbers was defined over a set RxR (with R being the real number field) by the most of the sources of my education. It hasn't much to do with graphing.
Posted on 2007-01-26 13:12:15 by Tomasz Grysztar
Tomasz Grysztar,

I have to repeat: in which field/algebra is it defined?


    I don't know what you are asking.  Why don't you list some of the selections.

There are algebras with exactly one square root of -1


    And what is that "algebra" called?

the fields of real/rational numbers have none.


    I would not call real/rational numbers a "field".  I would call it a set.  Yep, duplex (complex) numbers cannot be expressed with a single simplex number.

What you're speaking about seems to me to be a more an "engineering" knowledge than mathematical one.


    My training has been in engineering, but that does not change the mathematics.

And in the field of complex numbers, if you're speaking about it, there are exactly two different numbers that become -1 when squared (moreover, for any natual number n there are always exactly n different complex numbers that solve the equation z^n=a).


    All right, I will agree with the above statement.  I was thinking of only the first square root of -1.

On the space of polar coordinates you need to define equivalence classes to get the complex numbers space.


    What does that mean?

You might want to make a space/algebra out of them. Quite frequently used concept in the abstract algebra and algebraic topology.


    Fine and good.  But my engineering outlook asks what practical use is it?

The field of complex numbers was defined over a set RxR (with R being the real number field) by the most of the sources of my education. It hasn't much to do with graphing.


    "RxR"?  Whats that?  Ratch






Posted on 2007-01-26 18:51:33 by Ratch
I think the argument here is pointless as it ends up being a fight between a mathematican and an engineer.

I would like to request both parties to stop replying to this thread.
Posted on 2007-01-26 23:11:04 by roticv