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Topic: new tutor here
Replies: 151   Last Post: Jul 17, 2013 4:09 AM

 Messages: [ Previous | Next ]
 Joe Niederberger Posts: 4,657 Registered: 10/12/08
Re: new tutor here
Posted: Jun 19, 2013 1:44 PM

R Hansen (quoted at length):
"What bugs me is that you are saying that something exists but you can't explain it. That you "visually thought" through a problem, but not in any order and that, according to your last post, you weren't even thinking about the original problem, but solved it nonetheless. I guess that means that someone doesn't even have to tell you the problem ahead of time. You solve the problem first and then I reveal to you what the problem was. Ok, I am thinking of a new problem. Go.

This is exactly what I meant when I said that people get overly enthralled with the elegance of visual proofs and they forget that what goes into them is the EXACT same thing that goes into non visual proofs or any proof. Because they didn't use symbols or words it must not be reasoning. It is the same reasoning. You used visuals instead of symbols but it still has to be justified with reasoning. Besides, didn't you just ask the thread to figure out how you could use this picture to solve the original problem? Why would you ask that? I mean, that sounded a lot like you asked us to THINK. "

- -----

Ahem, take it easy.

[1] A lot of things you think I said sound absolutely crazy after you're done with them.

[2] If you think my approach to solving that problem *has* to be exactly the same as any other way, up to some sort of "mental isomorphism" (sorry, I can't make any other sense of your saying "it's the same reasoning" - to me its different), well we'll just disagree.

Here's the solution chosen as "best" at Yahoo answers:

Consider the function f(x) = x / ln(x). Its derivative is
f ' (x) = 1*(1/ln(x)) + x[ -(ln(x))^-2 ]*(1/x)
f ' (x) = (1/ln(x)) - 1/(ln(x))^2
f ' (x) = (ln(x) - 1) / [ln(x)]^2
Setting this equal to 0 shows that it reaches a local minimum when x=e. So since e < pi, that must mean f(e) < f(pi). This means:

e / ln(e) < pi / ln(pi)
e < pi/ln(pi)
e ln(pi) < pi
ln(pi^e) < pi
ln(pi^e) < pi * 1
ln(pi^e) < pi * ln(e)
ln(pi^e) < ln(e^pi)
pi^e < e^pi

So e^pi is greater.
- -----------

Whew! - I'll stick with my nice simple pic.

Cheers,
Joe N

Date Subject Author
6/12/13 Michael Mossey
6/12/13 Wayne Bishop
6/12/13 Michael Mossey
6/13/13 Wayne Bishop
6/13/13 Dave L. Renfro
6/13/13 Joe Niederberger
6/14/13 Robert Hansen
6/14/13 Joe Niederberger
6/14/13 Robert Hansen
6/14/13 Joe Niederberger
6/14/13 Robert Hansen
6/14/13 Joe Niederberger
6/14/13 Robert Hansen
6/15/13 Louis Talman
6/14/13 Michael Mossey
6/15/13 Robert Hansen
6/14/13 GS Chandy
6/16/13 GS Chandy
6/15/13 Louis Talman
6/15/13 Joe Niederberger
6/16/13 Robert Hansen
6/15/13 Pam
6/16/13 Robert Hansen
6/15/13 Pam
6/16/13 Wayne Bishop
6/16/13 Joe Niederberger
6/16/13 Robert Hansen
6/16/13 Joe Niederberger
6/17/13 Robert Hansen
6/16/13 Michael Mossey
6/17/13 GS Chandy
6/17/13 GS Chandy
6/17/13 Pam
6/17/13 Robert Hansen
6/17/13 Pam
6/17/13 Robert Hansen
6/17/13 Pam
6/17/13 Pam
6/17/13 Joe Niederberger
6/17/13 Robert Hansen
6/17/13 Joe Niederberger
6/17/13 Joe Niederberger
6/18/13 Robert Hansen
6/17/13 GS Chandy
6/18/13 GS Chandy
6/18/13 GS Chandy
6/18/13 GS Chandy
6/18/13 Joe Niederberger
6/18/13 Robert Hansen
6/18/13 Joe Niederberger
6/18/13 Joe Niederberger
6/18/13 Robert Hansen
6/18/13 Joe Niederberger
6/18/13 Joe Niederberger
6/18/13 Robert Hansen
6/18/13 Robert Hansen
6/18/13 Joe Niederberger
6/19/13 Robert Hansen
6/19/13 Robert Hansen
6/21/13 Wayne Bishop
6/21/13 Wayne Bishop
6/19/13 Joe Niederberger
6/19/13 Robert Hansen
6/19/13 Joe Niederberger
6/19/13 Robert Hansen
6/19/13 Joe Niederberger
6/19/13 Robert Hansen
6/19/13 Joe Niederberger
6/19/13 Robert Hansen
6/19/13 Robert Hansen
6/19/13 Joe Niederberger
6/19/13 Joe Niederberger
6/19/13 Joe Niederberger
6/19/13 Robert Hansen
6/21/13 Joe Niederberger
6/21/13 Robert Hansen
6/21/13 Joe Niederberger
6/21/13 Robert Hansen
6/21/13 Joe Niederberger
6/21/13 Robert Hansen
6/21/13 Joe Niederberger
6/21/13 Joe Niederberger
6/21/13 Dave L. Renfro
6/22/13 Robert Hansen
6/22/13 Robert Hansen
6/22/13 Joe Niederberger
6/22/13 Robert Hansen
6/22/13 Robert Hansen
6/23/13 Joe Niederberger
6/23/13 Robert Hansen
6/23/13 Robert Hansen
6/23/13 Joe Niederberger
6/23/13 Robert Hansen
6/24/13 Robert Hansen
6/24/13 Joe Niederberger
6/24/13 Joe Niederberger
6/24/13 Joe Niederberger
6/25/13 GS Chandy
7/10/13 Dave L. Renfro
7/10/13 Robert Hansen
7/10/13 Joe Niederberger
7/10/13 Robert Hansen
7/10/13 Robert Hansen
7/10/13 Robert Hansen
7/11/13 Joe Niederberger
7/11/13 Robert Hansen
7/11/13 Joe Niederberger
7/11/13 Robert Hansen
7/11/13 Robert Hansen
7/11/13 GS Chandy
7/11/13 Joe Niederberger
7/11/13 Robert Hansen
7/11/13 Robert Hansen
7/11/13 Louis Talman
7/11/13 Robert Hansen
7/12/13 Wayne Bishop
7/12/13 Louis Talman
7/12/13 Wayne Bishop
7/12/13 Robert Hansen
7/12/13 Gary Tupper
7/12/13 Wayne Bishop
7/12/13 Louis Talman
7/11/13 Joe Niederberger
7/11/13 Robert Hansen
7/11/13 Joe Niederberger
7/11/13 GS Chandy
7/12/13 Wayne Bishop
7/12/13 Joe Niederberger
7/12/13 Wayne Bishop
7/12/13 GS Chandy
7/12/13 Joe Niederberger
7/13/13 Robert Hansen
7/13/13 GS Chandy
7/13/13 Joe Niederberger
7/13/13 Wayne Bishop
7/13/13 Robert Hansen
7/13/13 Robert Hansen
7/13/13 Peter Duveen
7/14/13 Joe Niederberger
7/14/13 Joe Niederberger
7/14/13 Joe Niederberger
7/14/13 Robert Hansen
7/14/13 GS Chandy
7/15/13 Wayne Bishop
7/15/13 Joe Niederberger
7/15/13 Robert Hansen
7/15/13 Peter Duveen
7/15/13 Joe Niederberger
7/16/13 GS Chandy
7/16/13 Wayne Bishop
7/16/13 GS Chandy
7/17/13 GS Chandy