How many of you self-study mathematics from textbooks - Page 2

A Beautiful Mind · #16 $Old$ October 9th, 2009, 11:13 AM

Oh, I misread something in that post. The factors of 15 was about his thread, not my own. But still...not that much difference.

Plus or minus: $\frac{1, 2, 3, 4, 6, 12}{1,2,4}$

$1,2,3,4,6,12,\frac{1}{2}, \frac{1}{4}, \frac{3}{2}, \frac{3}{4}$ ...plus or minus. I think I got them all.

CaptainBlack · #17 $Old$ October 9th, 2009, 11:18 AM

Quote:

Originally Posted by A Beautiful Mind $View Post$

CaptainBlack, can you show an example problem using that technique? One more worked out than the couple of sentences in the article on wordpress? Some of that stuff (I'd say most) becomes a lot more clearer once I see a fully-worked out solution.

There is literaly no more to it than a few lines:

The Cauchy bounds for the roots of:

$P(x)=a_nx^n+a_{n-1}x^{n-1}+ ... + a_0$

are given by:

$\frac{|a_0|}{|a_0|+\max_{i=1,..,n}|a_i|}\le |x| \le \frac{|a_n|+\max_{i=0,..,n-1}|a_i|}{|a_n|}$

Now consider the polynomial I posted earlier:

$P(x)=x^6+12.4x^3+2x^2+17.23x+23$

Then as $\max_{i=0,..,n-1}|a_i|=23$ and $\max_{i=1,..,n}|a_i|=17.23$

we have any root $x$ of $P$ satisfies:

$\frac{23}{23+17.23}\le |x| \le \frac{1+23}{1}$

or:

$0.5717.. \le |x| \le 24$

(and this is true for all the roots not just the real ones)

(and you should not be so dismissive of those few words in the Blog article, that article has been accepted for publication in the Mathematical Gazette as a comment on the note referenced from MG - now MG is not a prestigious journal but nor is it the Daily Mail)

(I must say it is nice to be able to just cut LaTeX from the WordPress article and paste it here without changes, it almost makes the effort of getting LaTeX to work in WordPress)

CB

A Beautiful Mind · #18 $Old$ October 9th, 2009, 11:30 AM

Quote:

Originally Posted by CaptainBlack $View Post$

There is literaly no more to it than a few lines:

The Cauchy bounds for the roots of:

$P(x)=a_nx^n+a_{n-1}x^{n-1}+ ... + a_0$

are given by:

$\frac{|a_0|}{|a_0|+\max_{i=1,..,n}|a_i|}\le |x| \le \frac{|a_n|+\max_{i=0,..,n-1}|a_i|}{|a_n|}$

Now consider the polynomial I posted earlier:

$P(x)=x^6+12.4x^3+2x^2+17.23x+23$

Then as $\max_{i=0,..,n-1}|a_i|=23$ and $\max_{i=1,..,n}|a_i|=17.23$

we have any root $x$ of $P$ satisfies:

$\frac{23}{23+17.23}\le |x| \le \frac{1+23}{1}$

or:

$0.5717.. \le |x| \le 24$

(and this is true for all the roots not just the real ones)

(and you should not be so dismissive of those few words in the Blog article, that article has been accepted for publication in the Mathematical Gazette as a comment on the note referenced from MG - now MG is not a prestigious journal but nor is it the Daily Mail)

(I must say it is nice to be able to just cut LaTeX from the WordPress article and paste it here without changes, it almost makes the effort of getting LaTeX to work in WordPress)

CB

Er...I was saying I just wanted a better understanding of what was going on through a more worked out example, not like I was attacking the way it was written.

CaptainBlack · #19 $Old$ October 9th, 2009, 11:38 AM

Quote:

Originally Posted by A Beautiful Mind $View Post$

Er...I was saying I just wanted a better understanding of what was going on through a more worked out example, not like I was attacking the way it was written.

It is a theorem, work through the proof/s (Those posts are there because it is quite difficult to find a proof on-line).

CB

Sampras · #20 $Old$ October 9th, 2009, 02:27 PM

Quote:

Originally Posted by CaptainBlack $View Post$

Indeed, but then that is for me (I never asked for help form the teaching assistants and/or lecturers either, but I'm also not going to tell anyone else to do that).

If the stuff appears to have gaps that you can't see how to fill while reading then you need pencil and paper to fill the gaps yourself.

CB

Well I guess you did do problems anyway since you practiced tons for the exams? It's just that you didnt do the problems as you went along? Only when preparing for an exam? But you did go through the books with pen and paper right?

And Ed Witten did all the problems in most books that he studied.

Quote:

Originally Posted by Ed Witten $View Post$

When I was young, I was very thorough in doing every exercise in a book.
Regrettably I've lost that patience over the years.
- Edward

Also reading from multiple sources is beneficial also (e.g. going through similar sections of different books with pen and paper).

e^(i*pi) · #21 $Old$ October 9th, 2009, 02:33 PM

I'm OK with books as long as there are worked examples and answers in the back

CaptainBlack · #22 $Old$ October 9th, 2009, 02:37 PM

Quote:

Originally Posted by Sampras $View Post$

Well I guess you did do problems anyway since you practiced tons for the exams? It's just that you didnt do the problems as you went along? Only when preparing for an exam? But you did go through the books with pen and paper right?

I must say that we did not use the problem sets from books (in fact while there were recommended texts we did not use them for anything other than reference), and I don't recall any of the courses having particularly demanding problem sets (other than the Profs that is).

CB

Sampras · #23 $Old$ October 9th, 2009, 03:06 PM

Quote:

Originally Posted by CaptainBlack $View Post$

I must say that we did not use the problem sets from books (in fact while there were recommended texts we did not use them for anything other than reference), and I don't recall any of the courses having particularly demanding problem sets (other than the Profs that is).

CB

functional analysis?

CaptainBlack · #24 $Old$ October 9th, 2009, 11:44 PM

Quote:

Originally Posted by Sampras $View Post$

functional analysis?

That's what the course was called yes (but its content was significantly different from most texts)

CB

Utterconfusion · #25 $Old$ October 10th, 2009, 08:44 AM

Quote:

Originally Posted by Jameson $View Post$

I think almost anyone can learn to self-study. You absolutely must have the right textbook though that presents the material like a teacher would. Some advanced textbooks are so dense with material and have almost no explanations.

Read the chapter content. Go through the example problems, then do all the practice problems. When you get stuck post your question here on MHF along with your work and you'll get the help you need. Repeat process.

Thanks for the scoop

Utterconfusion · #26 $Old$ October 10th, 2009, 09:06 AM

Quote:

Originally Posted by A Beautiful Mind $View Post$

Oh, I misread something in that post. The factors of 15 was about his thread, not my own. But still...not that much difference.

Plus or minus: $\frac{1, 2, 3, 4, 6, 12}{1,2,4}$

$1,2,3,4,6,12,\frac{1}{2}, \frac{1}{4}, \frac{3}{2}, \frac{3}{4}$ ...plus or minus. I think I got them all.

4x^{4}+28x^{3}+48x^{2}-x^{2}-7x-12

This is the function. Thanks for your response but let me quote my previous request "it would have been swell if the guy in your link used the rational root theorem as well to find the upper and lower bounds as a demonstration....... Are you capable of doing it?"

As far as I can understand, the rational root theorem only lists all the roots that COULD BE rational roots concerning a particular function. Thats pretty much all I know about it(still yet to see the significance of this)
So anyway now I had taught myself how to obtain the list of possible rational roots using the RRT but I don't know how to obtain the upper and lower bounds of a f(x) from that, whch is what I was hoping you'd help me with

Many thanks to captain black for his alternative method. I went through it but it looks rather complicating to me(and I don't know how to interpret that symbology)

Utterconfusion · #27 $Old$ October 10th, 2009, 09:18 AM

$4x^4 + 2x^3 + 47x^2 -7x - 12$

(Testing how to write functions as images)

Utterconfusion · #28 $Old$ October 10th, 2009, 09:42 AM

Ok so can someone tell me how to find the upper and lower bounds of this

$4x^{4}+28x^{3}+47x^{2}-7x-12$

Must the function always have to be factorized like this?

Quote:

Originally Posted by galactus $View Post$

You can use the rational root theorem, but you can rewrite like so:

$4x^{4}+28x^{3}+48x^{2}-x^{2}-7x-12$

Group and Factor:

$4x^{2}(x^{2}+7x+12)-(x^{2}+7x+12)$

$(4x^{2}-1)(x^{2}+7x+12)$

Each of these factors:

$4x^{2}-1=(2x)^{2}-(1)^{2}$

It is the difference of two squares:

$(2x-1)(2x+1)$

The quadratic factors as $(x+3)(x+4)$

So putting it together we get:

$(x+3)(x+4)(2x+1)(2x-1)$

Isnt there a shorter method of finding the bounds?
And I'm wondering how to use the RRT to find the bounds as this guy proposes before his working...

A Beautiful Mind · #29 $Old$ October 10th, 2009, 12:45 PM

Quote:

Originally Posted by Utterconfusion $View Post$

4x^{4}+28x^{3}+48x^{2}-x^{2}-7x-12

This is the function. Thanks for your response but let me quote my previous request "it would have been swell if the guy in your link used the rational root theorem as well to find the upper and lower bounds as a demonstration....... Are you capable of doing it?"

As far as I can understand, the rational root theorem only lists all the roots that COULD BE rational roots concerning a particular function. Thats pretty much all I know about it(still yet to see the significance of this)
So anyway now I had taught myself how to obtain the list of possible rational roots using the RRT but I don't know how to obtain the upper and lower bounds of a f(x) from that, whch is what I was hoping you'd help me with

Many thanks to captain black for his alternative method. I went through it but it looks rather complicating to me(and I don't know how to interpret that symbology)

I think my teacher said that if it's a root, it'll be in there or there aren't any zeros or whatever that are rational.

You get all the possible rational roots from the RRT and you do the synthetic division for them. Once you notice a pattern like if they're alternating in symbols, you've reached the lower-bound and if they're all positive or zero you've reached the upper-bound.

The function you gave. I don't have time to do all of the work right now because I got hotwings almost done lol, but try doing synthetic division with 4 and that polynomial. You will get all positive numbers. That means you do not have to go up any higher. So you test the number below. If that's all positive too, you move down another until I guess you've reached one that isn't all positive. The last one that was all positive would be the upper-bound, I'm pretty sure since you don't need to go any higher than that.

I don't know any efficient way of finding bounds. No one answered a faster way to find zeros when I posed that question, haha. Trial-and-error is your friend, I guess.

CaptainBlack · #30 $Old$ October 10th, 2009, 01:58 PM

Quote:

Originally Posted by A Beautiful Mind $View Post$

I think my teacher said that if it's a root, it'll be in there or there aren't any zeros or whatever that are rational.

You get all the possible rational roots from the RRT and you do the synthetic division for them. Once you notice a pattern like if they're alternating in symbols, you've reached the lower-bound and if they're all positive or zero you've reached the upper-bound.

The function you gave. I don't have time to do all of the work right now because I got hotwings almost done lol, but try doing synthetic division with 4 and that polynomial. You will get all positive numbers. That means you do not have to go up any higher. So you test the number below. If that's all positive too, you move down another until I guess you've reached one that isn't all positive. The last one that was all positive would be the upper-bound, I'm pretty sure since you don't need to go any higher than that.

I don't know any efficient way of finding bounds. No one answered a faster way to find zeros when I posed that question, haha. Trial-and-error is your friend, I guess.

Please tell us what you would do if the polynomial were:

4*x^4+28*x^3+48*x^2-x^2-7*x-11

which has no rational roots?

CB