2-group: Difference between revisions

From formulasearchengine
Jump to navigation Jump to search
en>Zundark
fix a link
 
en>Yobot
m WP:CHECKWIKI error #38 + general fixes using AWB (8888)
Line 1: Line 1:
Mexico... The spot to celebration. It&quot;s the best location to relax on the beach, barter with locals, drink tequila, and get a tan. For the most element, Mexico is considered a pretty protected location to go to if you will be in the security of an all inclusive resort. This was the attitude of Pamela Woods, just three months ago, on her trip to Mexico which was drastically various than she expected.<br><br>I dont want to be costing the resort all kinds of funds and give them a poor reputation, so I will maintain the name of the resort confidential. But what I will inform you is the story of Pamela Woods.<br><br>Pam is a household girl, as she is married to a good doctor, and has two wonderful daughters. Her life went in a spin 3 months ago, for the duration of her pay a visit to to an all inclusive resort in mexico. 1 of her daughers, and her husband, drove down from canada, to meet the other daughter and herself at the resort. All was nicely when the household was with each other again, but the second evening of mexico wasn&quot;t so spectacular.<br><br>Pamela Woods, mother of two, was beaten and raped on the shores of the resort, by a nearby mexican about 7:30pm. Navigating To [http://www.slideshare.net/pleasurepriciple02 understandable] possibly provides suggestions you might give to your cousin. This stylish [https://vimeo.com/channels/cancunescorts Cancun Escorts on Vimeo] wiki has a myriad of dynamite aids for the inner workings of it. She left her sandals at the beach that afternoon, and remembered to go and get them right after dinner. When she did not come back to the room, the children and husband started to panic.<br><br>The threesome set out in search for her, and located her lying on the beach, bloody, bruised, and naked. This was a devistating trip for the Woods household.<br><br>This story is not supposed to be a scary or disgusting, it is intended to inform you vacationers of the threats that are there, even in a 5 star resort. I&quot;m not saying "Stay away from mexico", since it can be a fun time just take into consideration that the worst CAN occur, and it did to Pam Woods. We discovered [http://massageplanet.net/forum/showthread.php?90389-Pleasure-Principle-Spa-Cancun-Escorts cancun escort agencies] by browsing Google Books. Travel with each other, and by no means go out alone..<br><br>If you have any sort of inquiries pertaining to where and the best ways to make use of [http://www.blogigo.com/woebegonegather23 health insurance cheap], you can call us at the web page.
{{Multiple issues
|orphan = February 2009
|unreferenced = February 2009
}}
In [[calculus]],
the '''racetrack principle''' describes the movement and growth of two functions in terms of their [[derivative]]s.
 
This principle is derived from the fact that if a horse named Frank Fleetfeet always runs faster than a horse named Greg Gooseleg, then if Frank and Greg start a race from the same place and the same time, then Frank will win.  More briefly, the horse that starts fast and stays fast wins.
 
In symbols:
:if <math>f'(x)>g'(x)</math> for all <math>x>0</math>, and if <math>f(0)=g(0)</math>, then <math>f(x)>g(x)</math> for all <math>x>0</math>.  
or, substituting ≥ for > produces the theorem
:if <math>f'(x) \ge g'(x)</math> for all <math>x>0</math>, and if <math>f(0)=g(0)</math>, then <math>f(x) \ge g(x)</math> for all <math>x>0</math>.  
which can be proved in a similar way
 
 
==Proof==
This principle can be proven by considering the function h(x) = f(x) - g(x). If we were to take the derivative we would notice that for x>0
 
:<math> h'= f'-g'>0.</math>
 
Also notice that h(0) = 0. Combining these observations, we can use the [[mean value theorem]] on the interval [0, x] and get
 
:<math> h'(x_0)= \frac{h(x)-h(0)}{x-0}= \frac{f(x)-g(x)}{x}>0.</math>
 
Since x > 0 for the mean value theorem to work then we may conclude that f(x) - g(x) > 0. This implies f(x) > g(x).
 
==Generalizations==
 
The statement of the racetrack principle can slightly generalized as follows;
:if <math>f'(x)>g'(x)</math> for all <math>x>a</math><!--where a≥0 **don't need this restriction**-->, and if <math>f(a)=g(a)</math>, then <math>f(x)>g(x)</math> for all <math>x>a</math>.
 
as above, substituting ≥ for > produces the theorem
:if <math>f'(x) \ge g'(x)</math> for all <math>x>a</math><!--where a≥0 **don't need this restriction**-->, and if <math>f(a)=g(a)</math>, then <math>f(x) \ge g(x)</math> for all <math>x>a</math>.
 
===Proof===
This generalization can be proved from the racetrack principle as follows:
 
Given <math>f'(x)>g'(x)</math> for all <math>x>a</math> where a≥0, and <math>f(a)=g(a)</math>,  
 
Consider functions <math>f_2(x)=f(x-a)</math> and <math>g_2(x)=g(x-a)</math>
 
<math>f_2'(x)>g_2'(x)</math> for all <math>x>0</math>, and <math>f_2(0)=g_2(0)</math>, which by the proof of the racetrack principle above means <math>f_2(x)>g_2(x)</math> for all <math>x>0</math> so <math>f(x)>g(x)</math> for all <math>x>a</math>.
 
==Application==
The racetrack principle can be used to prove a [[lemma (mathematics)|lemma]] necessary to show that the [[exponential function]] grows faster than any power function. The lemma required is that
:<math> e^{x}>x </math>
for all real x. This is obvious for x<0 but the racetrack principle is required for x>0. To see how it is used we consider the functions
:<math> f(x)=e^{x}</math>
and
:<math> g(x)=x+1.</math>
Notice that f(0) = g(0) and that
:<math> e^{x}>1</math>
because the exponential function is always increasing ([[monotonic]]) so <math>f'(x)>g'(x)</math>. Thus by the racetrack principle f(x)>g(x). Thus,
:<math> e^{x}>x+1>x</math>
for all x>0.
 
==External links==
*[http://www.matheverywhere.com/mei/candm/race.php3 Usage of Racetrack Principle] (Math Everywhere)
 
[[Category:Calculus]]
[[Category:Pedagogy]]
[[Category:Mathematical principles]]

Revision as of 09:16, 8 February 2013

Template:Multiple issues In calculus, the racetrack principle describes the movement and growth of two functions in terms of their derivatives.

This principle is derived from the fact that if a horse named Frank Fleetfeet always runs faster than a horse named Greg Gooseleg, then if Frank and Greg start a race from the same place and the same time, then Frank will win. More briefly, the horse that starts fast and stays fast wins.

In symbols:

if f(x)>g(x) for all x>0, and if f(0)=g(0), then f(x)>g(x) for all x>0.

or, substituting ≥ for > produces the theorem

if f(x)g(x) for all x>0, and if f(0)=g(0), then f(x)g(x) for all x>0.

which can be proved in a similar way


Proof

This principle can be proven by considering the function h(x) = f(x) - g(x). If we were to take the derivative we would notice that for x>0

h=fg>0.

Also notice that h(0) = 0. Combining these observations, we can use the mean value theorem on the interval [0, x] and get

h(x0)=h(x)h(0)x0=f(x)g(x)x>0.

Since x > 0 for the mean value theorem to work then we may conclude that f(x) - g(x) > 0. This implies f(x) > g(x).

Generalizations

The statement of the racetrack principle can slightly generalized as follows;

if f(x)>g(x) for all x>a, and if f(a)=g(a), then f(x)>g(x) for all x>a.

as above, substituting ≥ for > produces the theorem

if f(x)g(x) for all x>a, and if f(a)=g(a), then f(x)g(x) for all x>a.

Proof

This generalization can be proved from the racetrack principle as follows:

Given f(x)>g(x) for all x>a where a≥0, and f(a)=g(a),

Consider functions f2(x)=f(xa) and g2(x)=g(xa)

f2(x)>g2(x) for all x>0, and f2(0)=g2(0), which by the proof of the racetrack principle above means f2(x)>g2(x) for all x>0 so f(x)>g(x) for all x>a.

Application

The racetrack principle can be used to prove a lemma necessary to show that the exponential function grows faster than any power function. The lemma required is that

ex>x

for all real x. This is obvious for x<0 but the racetrack principle is required for x>0. To see how it is used we consider the functions

f(x)=ex

and

g(x)=x+1.

Notice that f(0) = g(0) and that

ex>1

because the exponential function is always increasing (monotonic) so f(x)>g(x). Thus by the racetrack principle f(x)>g(x). Thus,

ex>x+1>x

for all x>0.

External links