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pretend infinity border is 10 grid and derivative #4 Textbook 2nd ed. : TRUE CALCULUS; without the phony limit concept
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Archimedes Plutonium
May 24, 2013, 12:16:34 AM5/24/13
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Alright, in the next edition I need to use the 10 grid more than the
10^-603 grid so that High School students have an easy time with True
Calculus.
Now I have the four functions:
y = 3 (type of a box function)
y = x (identity function)
y = x^2
y = 1/x (similar to a type of a logarithm function)
their derivates are respectively (y' denotes derivative)
y' = 0
y' = 1
y' = 2x
y' = -x^-2
Now let me focus on the function y= 1/x and its derivative using the
10 grid. The 10 grid is where infinity borderline is 10 and thus the
smallest nonzero number is 1/10 and where all the numbers that exist
for mathematics in the 1st quadrant are these:
0, 1/10, 2/10, 3/10, 4/10, . . on up to 10
Those are 100 individual numbers and since the y-axis has the same
amount, there are 100X100 = 10,000 numbers in all.
Now let me focus in on the first four numbers of this 10 grid and
since I cannot ascii art 100 rows, let me mark the first four
coordinate points:
for x=0, it is undefined for 1/x
for x=.1 then we have (.1, 10)
for x=.2 then we have (.2, 5)
for x=.3 then we have (.3, 3.333..)
for x=.4 then we have (.4, 2.5)
for x=.5 then we have (.5, 2)
Now here is a truncated version to get a sense of the picture but the
student is asked to plot it in pencil on their graph paper to use an
erasure so you can use over again.
. x . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . x . . .
. . . . . .
x
. . . . . .
x
. . . . . x
. . . . . .
. . . . . .
0 .1 .2 .3 .4 .5
Now we can draw a picketfence in the interval of dx for .3, .4, .5.
That interval of dx involves .1 + .1 = .2
That interval of dy involves 3.33.. - 2 = 1.33..
So that dy/dx is 1.33../.2 = 6.65 slope and negative
Whereas the derivative of y = 1/x is -1/x^2 which is
-1/.16 which is - 100/16 = -6.25 slope
Now let me draw the two picketfences
. . . . . .
x
. . . . . .
x
. . . . . x
. . . . . .
. . . . . .
. . . . . .
|\
. . . |. \ . .
| x
. . . |____\
. . . . . .
. . . . . .
Now the interval .3, .4, .5 involves two picketfences
in this picture, but in True Calculus we end up with just one
picketfence that models the derivative and integral. The limit concept
of Old Math is the use of two picketfences as the delta gets smaller
and smaller. However, as the above shows, no matter how small, the two
picketfences do not give the proper correct value. What does give the
proper and correct value is that we use just one picketfence and we
have that empty space in which the angle of the hypotenuse can become
the exact angle for the slope or tangent at the given point.
Now let me do another example in the 10^603 grid instead of the
pretend 10 grid. In this example we can also see how the Weierstrass
function and the sin(1/x) function are differentiable and continuous
everywhere except perhaps 0 since division by zero is undefined.
. x . x . x
. . . . . .
. . . . . .
break in scale
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
x . x . x .
0 a b c d e
Now a is 1*10^-603 and b is 2*10^-603 and c is 3*10^-603 etc etc
and the x's on top are where y = 10^603 and the x's on bottom are
where y= 0.
So the function here is F(x) = 0 for even x and F(x) = 10^603 for odd
x. Now this graph is like the world's steepest mountain range or it is
a sawblade. It has the steepest angle except for 90 degrees itself, so
the angle involved is (90- 1*10^-603). Now in Old Math, this function
would be a step function and discontinuous everywhere and
differentiable nowhere. In Old Math the Weierstrass function is
continuous everywhere but differentiable nowhere. In New Math this
sawblade function is continuous everywhere and differentiable
everywhere. In New Math the function sin(1/x) is discontinuous at 0
but is continuous everywhere else and differentiable everywhere else.
So let us examine the sawblade function just described. If we draw a
straight line connecting the three points (0, 0), (1*10^-603, 10^603),
(2*10^-603, 0) we get our first tooth, a sharp tooth of the sawblade
for it stretches from 0 to the infinity borderline of 10^603. Now Old
Math with their limit concept has no way of handling or dealing with
such a function, not to mention the Weierstrass or sin(1/x) but New
Math handles every function that exists in Nature because we can list
all the grid points and examine any neighborhood of grid points.
Now the picketfence of this function is rather different from most
picketfences because it is all one long slender triangle and not a
rectangle involved. Most derivatives and integrals involve the
rectangle with triangle sitting atop:
/|
||
||
||
Here there is no rectangle portion but just the triangle, and it has
width 1*10^-603 and it has height of 1*10^603 so it has area and it
has a hypotenuse and thus it has a derivative.
Now we can do the same for Weierstrass function and see that it is
indeed continuous and differentiable everywhere and we can do the same
for y= sin(1/x) although discontinuous at 0, it is continuous
everywhere else and differentiable everywhere else.
--
More than 90 percent of AP's posts are missing in the Google
newsgroups author search archive from May 2012 to May 2013. Drexel
University's Math Forum has done a far better job and many of those
missing Google posts can be seen here:
http://mathforum.org/kb/profile.jspa?userID=499986
Archimedes Plutonium
http://www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
10^-603 grid so that High School students have an easy time with True
Calculus.
Now I have the four functions:
y = 3 (type of a box function)
y = x (identity function)
y = x^2
y = 1/x (similar to a type of a logarithm function)
their derivates are respectively (y' denotes derivative)
y' = 0
y' = 1
y' = 2x
y' = -x^-2
Now let me focus on the function y= 1/x and its derivative using the
10 grid. The 10 grid is where infinity borderline is 10 and thus the
smallest nonzero number is 1/10 and where all the numbers that exist
for mathematics in the 1st quadrant are these:
0, 1/10, 2/10, 3/10, 4/10, . . on up to 10
Those are 100 individual numbers and since the y-axis has the same
amount, there are 100X100 = 10,000 numbers in all.
Now let me focus in on the first four numbers of this 10 grid and
since I cannot ascii art 100 rows, let me mark the first four
coordinate points:
for x=0, it is undefined for 1/x
for x=.1 then we have (.1, 10)
for x=.2 then we have (.2, 5)
for x=.3 then we have (.3, 3.333..)
for x=.4 then we have (.4, 2.5)
for x=.5 then we have (.5, 2)
Now here is a truncated version to get a sense of the picture but the
student is asked to plot it in pencil on their graph paper to use an
erasure so you can use over again.
. x . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . x . . .
. . . . . .
x
. . . . . .
x
. . . . . x
. . . . . .
. . . . . .
0 .1 .2 .3 .4 .5
Now we can draw a picketfence in the interval of dx for .3, .4, .5.
That interval of dx involves .1 + .1 = .2
That interval of dy involves 3.33.. - 2 = 1.33..
So that dy/dx is 1.33../.2 = 6.65 slope and negative
Whereas the derivative of y = 1/x is -1/x^2 which is
-1/.16 which is - 100/16 = -6.25 slope
Now let me draw the two picketfences
. . . . . .
x
. . . . . .
x
. . . . . x
. . . . . .
. . . . . .
. . . . . .
|\
. . . |. \ . .
| x
. . . |____\
. . . . . .
. . . . . .
Now the interval .3, .4, .5 involves two picketfences
in this picture, but in True Calculus we end up with just one
picketfence that models the derivative and integral. The limit concept
of Old Math is the use of two picketfences as the delta gets smaller
and smaller. However, as the above shows, no matter how small, the two
picketfences do not give the proper correct value. What does give the
proper and correct value is that we use just one picketfence and we
have that empty space in which the angle of the hypotenuse can become
the exact angle for the slope or tangent at the given point.
Now let me do another example in the 10^603 grid instead of the
pretend 10 grid. In this example we can also see how the Weierstrass
function and the sin(1/x) function are differentiable and continuous
everywhere except perhaps 0 since division by zero is undefined.
. x . x . x
. . . . . .
. . . . . .
break in scale
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
. . . . . .
x . x . x .
0 a b c d e
Now a is 1*10^-603 and b is 2*10^-603 and c is 3*10^-603 etc etc
and the x's on top are where y = 10^603 and the x's on bottom are
where y= 0.
So the function here is F(x) = 0 for even x and F(x) = 10^603 for odd
x. Now this graph is like the world's steepest mountain range or it is
a sawblade. It has the steepest angle except for 90 degrees itself, so
the angle involved is (90- 1*10^-603). Now in Old Math, this function
would be a step function and discontinuous everywhere and
differentiable nowhere. In Old Math the Weierstrass function is
continuous everywhere but differentiable nowhere. In New Math this
sawblade function is continuous everywhere and differentiable
everywhere. In New Math the function sin(1/x) is discontinuous at 0
but is continuous everywhere else and differentiable everywhere else.
So let us examine the sawblade function just described. If we draw a
straight line connecting the three points (0, 0), (1*10^-603, 10^603),
(2*10^-603, 0) we get our first tooth, a sharp tooth of the sawblade
for it stretches from 0 to the infinity borderline of 10^603. Now Old
Math with their limit concept has no way of handling or dealing with
such a function, not to mention the Weierstrass or sin(1/x) but New
Math handles every function that exists in Nature because we can list
all the grid points and examine any neighborhood of grid points.
Now the picketfence of this function is rather different from most
picketfences because it is all one long slender triangle and not a
rectangle involved. Most derivatives and integrals involve the
rectangle with triangle sitting atop:
/|
||
||
||
Here there is no rectangle portion but just the triangle, and it has
width 1*10^-603 and it has height of 1*10^603 so it has area and it
has a hypotenuse and thus it has a derivative.
Now we can do the same for Weierstrass function and see that it is
indeed continuous and differentiable everywhere and we can do the same
for y= sin(1/x) although discontinuous at 0, it is continuous
everywhere else and differentiable everywhere else.
--
More than 90 percent of AP's posts are missing in the Google
newsgroups author search archive from May 2012 to May 2013. Drexel
University's Math Forum has done a far better job and many of those
missing Google posts can be seen here:
http://mathforum.org/kb/profile.jspa?userID=499986
Archimedes Plutonium
http://www.iw.net/~a_plutonium
whole entire Universe is just one big atom
where dots of the electron-dot-cloud are galaxies
bacle...@gmail.com
May 24, 2013, 1:11:38 AM5/24/13
to
Why this is marked as abuse? It has been marked as abuse.
Report not abuse
On Friday, May 24, 2013 12:16:34 AM UTC-4, Archimedes Plutonium wrote:
> Alright, in the next edition I need to use the 10 grid more than the
>
> 10^-603 grid so that High School students have an easy time with True
>
> Calculus.
Asshole: how about a $100 bet that less than 3 people will buy your toilet paper textbook.
> Alright, in the next edition I need to use the 10 grid more than the
>
> 10^-603 grid so that High School students have an easy time with True
>
> Calculus.
bacle...@gmail.com
May 24, 2013, 1:13:54 AM5/24/13
to
Why this is marked as abuse? It has been marked as abuse.
Report not abuse
On Friday, May 24, 2013 12:16:34 AM UTC-4, Archimedes Plutonium wrote:
> Alright, in the next edition I need to use the 10 grid more than the
>
> 10^-603 grid so that High Ami - Trumpet Lights (official video HD)
>
MediaProMusic·501 videos
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Published on Sep 12, 2012
Ami - Trumpet Lights (official video HD)
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Carlos Llopis 1 month ago
Romancele sunt cele mai bune din lume :P Sunt din Spania si gagica mea e bucuresteanca si e perfecta :))
Iubesc Romania :))
Reply · 88
lejenamure 4 weeks ago
Im from Poland, living in Deutchland married in UK and working in Sweden, so tell me about racism. She's cute and song is awesome. greets
Reply · 30
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Zdravko Zafirov 11 hours ago
Awesome music, definitely inspire me to dance; greetings from Bulgaria :)
Reply ·
Olaru Bogdan 3 days ago
uneori da si uneori nu dar acum cel putin azi m-am topit :D
Reply · in reply to Carlos Llopis (Show the comment)
ramaz727272 3 days ago
Y Like
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ramaz727272 3 days ago
super
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CaTa Lin 4 days ago
Nice :x
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nevodari monica 4 days ago
1% muzica si 99% porno
Reply ·
Alexandru Messi 5 days ago
Multe hituri scoase de ziua mea :D
Reply ·
Erik Horváth 5 days ago
nem birnak velee..:D
Reply ·
Andrada Debby 6 days ago
Y Like
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soo you are living in Deutchland and working in Sweden?
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MediaProMusic·501 videos
Subscribe
328,192 6,945,227
31,569 791
Like About Share Add to
Published on Sep 12, 2012
Ami - Trumpet Lights (official video HD)
Be our friend: http://www.facebook.com/MediaProMusic...
con...@mediapromusic.ro
Find us on the web:
http://www.mediapromusic.ro
http://www.facebook.com/MediaProMusic...
http://www.youtube.com/MediaProMusic
booking: Marius China 0723315180 email:off...@sprintmedia.ro
(C) & (P) MediaPro Music Entertainment 2012
All rights reserved. Unauthorized reproduction is a violation of applicable laws
Show more
Top Comments
Carlos Llopis 1 month ago
Romancele sunt cele mai bune din lume :P Sunt din Spania si gagica mea e bucuresteanca si e perfecta :))
Iubesc Romania :))
Reply · 88
lejenamure 4 weeks ago
Im from Poland, living in Deutchland married in UK and working in Sweden, so tell me about racism. She's cute and song is awesome. greets
Reply · 30
All Comments (4,729)
click to leave a comment
Zdravko Zafirov 11 hours ago
Awesome music, definitely inspire me to dance; greetings from Bulgaria :)
Reply ·
Olaru Bogdan 3 days ago
uneori da si uneori nu dar acum cel putin azi m-am topit :D
Reply · in reply to Carlos Llopis (Show the comment)
ramaz727272 3 days ago
Y Like
Reply ·
ramaz727272 3 days ago
super
Reply ·
CaTa Lin 4 days ago
Nice :x
Reply · in playlist Favorite Tania Song
nevodari monica 4 days ago
1% muzica si 99% porno
Reply ·
Alexandru Messi 5 days ago
Multe hituri scoase de ziua mea :D
Reply ·
Erik Horváth 5 days ago
nem birnak velee..:D
Reply ·
Andrada Debby 6 days ago
Y Like
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Anna Maggie 1 week ago
soo you are living in Deutchland and working in Sweden?
Reply · in reply to lejenamure (Show the comment)
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Britain's Got Taln lume :P Sunt din Spania si gagica mea e bucuresteanca si e perfecta :))
Iubesc Romania :))
Reply · 88
lejenamure 4 weeks ago
Im from Poland, living in Deutchland married in UK and working in Sweden, so tell me about racism. She's cute and song is awesome. greets
Reply · 30
All Comments (4,729)
click to leave a comment
Zdravko Zafirov 11 hours ago
Awesome music, definitely inspire me to dance; greetings from Bulgaria :)
Reply ·
Olaru Bogdan 3 days ago
uneori da si uneori nu dar acum cel putin azi m-am topit :D
Reply · in reply to Carlos Llopis (Show the comment)
ramaz727272 3 days ago
Y Like
Reply ·
ramaz727272 3 days ago
super
Reply ·
CaTa Lin 4 days ago
Nice :x
Reply · in playlist Favorite Tania Song
nevodari monica 4 days ago
1% muzica si 99% porno
Reply ·
Alexandru Messi 5 days ago
Multe hituri scoase de ziua mea :D
Reply ·
Erik Horváth 5 days ago
nem birnak velee..:D
Reply ·
Andrada Debby 6 days ago
Y Like
Reply ·
Anna Maggie 1 week ago
soo you are living in Deutchland and working in Sweden?
Reply · in reply to lejenamure (Show the comment)
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>
>
>
> y = 3 (type of a box ent Season 2 Funny Auditions Part 2
by Kallum Smart
637,268 views
14:59
Romanian Dance time with True
>
> Calculus.
>
>
>
> Now I have Ami - Trumpet Lights (official video HD)
MediaProMusic·501 videos
Subscribe
328,192 6,945,227
31,569 791
Like About Share Add to
Published on Sep 12, 2012
Ami - Trumpet Lights (official video HD)
Be our friend: http://www.facebook.com/MediaProMusic...
con...@mediapromusic.ro
Find us on the web:
http://www.mediapromusic.ro
http://www.facebook.com/MediaProMusic...
http://www.youtube.com/MediaProMusic
booking: Marius China 0723315180 email:off...@sprintmedia.ro
(C) & (P) MediaPro Music Entertainment 2012
All rights reserved. Unauthorized reproduction is a violation of applicable laws
Show more
Top Comments
Carlos Llopis 1 month ago
Romancele sunt cele mai bune di)
>
> y = x (identity function)
>
> y = x^2
>
> y = 1/x (similar to a type of a ent Season 2 Funny Auditions Part 2
> y = x (identity function)
>
> y = x^2
>
by Kallum Smart
637,268 views
14:59
Romanian Dance time with True
>
> Calculus.
>
>
>
> Now I have Ami - Trumpet Lights (official video HD)
MediaProMusic·501 videos
Subscribe
328,192 6,945,227
31,569 791
Like About Share Add to
Published on Sep 12, 2012
Ami - Trumpet Lights (official video HD)
Be our friend: http://www.facebook.com/MediaProMusic...
con...@mediapromusic.ro
Find us on the web:
http://www.mediapromusic.ro
http://www.facebook.com/MediaProMusic...
http://www.youtube.com/MediaProMusic
booking: Marius China 0723315180 email:off...@sprintmedia.ro
(C) & (P) MediaPro Music Entertainment 2012
All rights reserved. Unauthorized reproduction is a violation of applicable laws
Show more
Top Comments
Carlos Llopis 1 month ago
Romancele sunt cele mai bune di function)
bacle...@gmail.com
May 24, 2013, 1:15:48 AM5/24/13
to
Why this is marked as abuse? It has been marked as abuse.
Report not abuse
On Friday, May 24, 2013 1:11:38 AM UTC-4, bacle...@gmail.com wrote:
> On Friday, May 24, 2013 12:16:34 AM UTC-4, Archimedes Plutonium wrote:
>
> > Alright, in the next edition I need to use ent Season 2 Funny Auditions Part 2
> On Friday, May 24, 2013 12:16:34 AM UTC-4, Archimedes Plutonium wrote:
>
by Kallum Smart
637,268 views
14:59
Romanian Dance time with True
637,268 views
14:59
>
> Calculus.
>
>
>
> Now I have Ami - Trumpet Lights (official video HD)
MediaProMusic·501 videos
Subscribe
328,192 6,945,227
31,569 791
Like About Share Add to
Published on Sep 12, 2012
Ami - Trumpet Lights (official video HD)
Be our friend: http://www.facebook.com/MediaProMusic...
con...@mediapromusic.ro
Find us on the web:
http://www.mediapromusic.ro
http://www.facebook.com/MediaProMusic...
http://www.youtube.com/MediaProMusic
booking: Marius China 0723315180 email:off...@sprintmedia.ro
(C) & (P) MediaPro Music Entertainment 2012
All rights reserved. Unauthorized reproduction is a violation of applicable laws
Show more
Top Comments
Carlos Llopis 1 month ago
Romancele sunt cele mai bune di 10 grid more than the
MediaProMusic·501 videos
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Published on Sep 12, 2012
Ami - Trumpet Lights (official video HD)
Be our friend: http://www.facebook.com/MediaProMusic...
con...@mediapromusic.ro
Find us on the web:
http://www.mediapromusic.ro
http://www.facebook.com/MediaProMusic...
http://www.youtube.com/MediaProMusic
booking: Marius China 0723315180 email:off...@sprintmedia.ro
(C) & (P) MediaPro Music Entertainment 2012
All rights reserved. Unauthorized reproduction is a violation of applicable laws
Show more
Top Comments
Carlos Llopis 1 month ago
>
>
>
>
>
>
>
>
>
> >
>
> > 10^-603 grid so that High School students have an easy time with True
>
> >
>
> > Calculus.
>
>
>
> Asshole: how about a $100 bet that less than 3 people will buy your toilet paper textbook.
>
> >
>
> >
>
> >
>
> > Now I have the four functions:
>
> >
>
> >
>
> >
>
> > y = 3 (type of a box function)
>
> >
>
> > y = x (identity function)
>
> >
>
> > y = x^2
>
> >
>
> > y = 1/x (similar to a type of a logarithm function)
>
> >
>
> >
>
> >
>
> > their derivates are ent Season 2 Funny Auditions Part 2
>
>
>
>
>
>
>
>
> >
>
> > 10^-603 grid so that High School students have an easy time with True
>
> >
>
> > Calculus.
>
>
>
> Asshole: how about a $100 bet that less than 3 people will buy your toilet paper textbook.
>
> >
>
> >
>
> >
>
> > Now I have the four functions:
>
> >
>
> >
>
> >
>
> > y = 3 (type of a box function)
>
> >
>
> > y = x (identity function)
>
> >
>
> > y = x^2
>
> >
>
> > y = 1/x (similar to a type of a logarithm function)
>
> >
>
> >
>
> >
>
by Kallum Smart
637,268 views
14:59
Romanian Dance time with True
637,268 views
14:59
>
> Calculus.
>
>
>
> Now I have Ami - Trumpet Lights (official video HD)
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Carlos Llopis 1 month ago
Romancele sunt cele mai bune di (y' denotes derivative)
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Carlos Llopis 1 month ago
>
> >
>
> >
>
> >
>
> > y' = 0
>
> >
>
> > y' = 1
>
> >
>
> > y' = 2x
>
> >
>
> > y' = -x^-2
>
> >
>
> >
>
> >
>
> > Now let me focus on the function y= 1/x and its derivative using the
>
> >
>
> > 10 grid. The 10 grid is where infinity borderline is 10 and thus the
>
> >
>
> > smallest nonzero number is 1/10 and where all the numbers that exist
>
> >
>
> > for mathematics in the 1st quadrant are these:
>
> >
>
> >
>
> >
>
> > 0, 1/10, 2/10, 3/10, 4/10, . . on up to 10
>
> >
>
> >
>
> >
>
> > Those are 100 individual numbers and sinceent Season 2 Funny Auditions Part 2
> >
>
> >
>
> >
>
> > y' = 0
>
> >
>
> > y' = 1
>
> >
>
> > y' = 2x
>
> >
>
> > y' = -x^-2
>
> >
>
> >
>
> >
>
> > Now let me focus on the function y= 1/x and its derivative using the
>
> >
>
> > 10 grid. The 10 grid is where infinity borderline is 10 and thus the
>
> >
>
> > smallest nonzero number is 1/10 and where all the numbers that exist
>
> >
>
> > for mathematics in the 1st quadrant are these:
>
> >
>
> >
>
> >
>
> > 0, 1/10, 2/10, 3/10, 4/10, . . on up to 10
>
> >
>
> >
>
> >
>
by Kallum Smart
637,268 views
14:59
Romanian Dance time with True
637,268 views
14:59
>
> Calculus.
>
>
>
> Now I have Ami - Trumpet Lights (official video HD)
MediaProMusic·501 videos
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Published on Sep 12, 2012
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Carlos Llopis 1 month ago
Romancele sunt cele mai bune dithe y-axis has the same
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Show more
Top Comments
Carlos Llopis 1 month ago
>
> >
>
> > amount, there are 100X100 = 10,000ent Season 2 Funny Auditions Part 2
by Kallum Smart
637,268 views
14:59
Romanian Dance time with True
637,268 views
14:59
>
> Calculus.
>
>
>
> Now I have Ami - Trumpet Lights (official video HD)
MediaProMusic·501 videos
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Published on Sep 12, 2012
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All rights reserved. Unauthorized reproduction is
>
> >
>
Subscribe
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All rights reserved. Unauthorized reproduction is
>
> >
>
> > for x=.4 then we have (.4, 2.5)
>
> >
>
> > for x=.5 then we have (.5, 2)
>
> >
>
> >
>
> >
>
> > Now here is a truncated version to get a sense of the picture but the
>
> >
>
> > student is asked to plot it in pencil on their graph paper to use an
>
> >
>
> > erasure so you can use over again.
>
> >
>
> >
>
> >
>
> > . x . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . x . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> > x
>
> >
>
> > . . . . . .
>
> >
>
> > x
>
> >
>
> > . . . . . x
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> > 0 .1 .2 .3 .4 .5
>
> >
>
> >
>
> >
>
> >
>
> >
>
> >
>
> >
>
> > Now we can draw a picketfence in the interval oficial video HD)
>
> >
>
> > for x=.5 then we have (.5, 2)
>
> >
>
> >
>
> >
>
> > Now here is a truncated version to get a sense of the picture but the
>
> >
>
> > student is asked to plot it in pencil on their graph paper to use an
>
> >
>
> > erasure so you can use over again.
>
> >
>
> >
>
> >
>
> > . x . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . x . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> > x
>
> >
>
> > . . . . . .
>
> >
>
> > x
>
> >
>
> > . . . . . x
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> > 0 .1 .2 .3 .4 .5
>
> >
>
> >
>
> >
>
> >
>
> >
>
> >
>
> >
>
MediaProMusic·501 videos
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Published on Sep 12, 2012
Ami - Trumpet Lights (official video HD)dx for .3, .4, .5.
Subscribe
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Published on Sep 12, 2012
>
> >
>
> >
>
> >
>
> > That interval of dx involves .1 + .1 = .2
>
> >
>
> >
>
> >
>
> > That interval of dy involves 3.33.. - 2 = 1.33..
>
> >
>
> >
>
> >
>
> > So that dy/dx is 1.33../.2 = 6.65 slope and negative
>
> >
>
> >
>
> >
>
> > Whereas the derivative of y = 1/x is -1/x^2 which is
>
> >
>
> > -1/.16 which is - 100/16 = -6.25 slope
>
> >
>
> >
>
> >
>
> > Now let me draw the two picketfences
>
> >
>
> > . . . .a violation of applicable laws
> >
>
> >
>
> >
>
> > That interval of dx involves .1 + .1 = .2
>
> >
>
> >
>
> >
>
> > That interval of dy involves 3.33.. - 2 = 1.33..
>
> >
>
> >
>
> >
>
> > So that dy/dx is 1.33../.2 = 6.65 slope and negative
>
> >
>
> >
>
> >
>
> > Whereas the derivative of y = 1/x is -1/x^2 which is
>
> >
>
> > -1/.16 which is - 100/16 = -6.25 slope
>
> >
>
> >
>
> >
>
> > Now let me draw the two picketfences
>
> >
>
Show more
Top Comments
Carlos Llopis 1 month ago
Romancele sunt cele mai bune dinumbers in all.
Top Comments
Carlos Llopis 1 month ago
>
> >
>
> >
>
> >
>
> >
>
> >
>
> > Now let me focus in on the first four numbers of this 10 grid and
>
> >
>
> > since I cannot ascii art 100 rows, let me mark the first four
>
> >
>
> > coordinate points:
>
> >
>
> >
>
> >
>
> > for x=0, it is undefined for 1/x
>
> >
>
> > for x=.1 then we have (.1, 10)
>
> >
>
> > for x=.2 then we have (.2, 5)
>
> >
>
> > for x=.3 then we have (.3, 3.333..). .
> >
>
> >
>
> >
>
> >
>
> >
>
> > Now let me focus in on the first four numbers of this 10 grid and
>
> >
>
> > since I cannot ascii art 100 rows, let me mark the first four
>
> >
>
> > coordinate points:
>
> >
>
> >
>
> >
>
> > for x=0, it is undefined for 1/x
>
> >
>
> > for x=.1 then we have (.1, 10)
>
> >
>
> > for x=.2 then we have (.2, 5)
>
> >
>
>
> >
>
> > x
>
> >
>
> > . . . . . .
>
> >
>
> > x
>
> >
>
> > . . . . . x
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> > |\
>
> >
>
> > . . . |. \ . .
>
> >
>
> > | x
>
> >
>
> > . . . |____\
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > Now the interval .3, .4, .5 involves two icial video HD)
> >
>
> > x
>
> >
>
> > . . . . . .
>
> >
>
> > x
>
> >
>
> > . . . . . x
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> > |\
>
> >
>
> > . . . |. \ . .
>
> >
>
> > | x
>
> >
>
> > . . . |____\
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
> > . . . . . .
>
> >
>
> >
>
> >
>
MediaProMusic·501 videos
Subscribe
328,192 6,945,227
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Published on Sep 12, 2012
Ami - Trumpet Lights (official video HD)
>
> >
>
Subscribe
328,192 6,945,227
31,569 791
Like About Share Add to
Published on Sep 12, 2012
Ami - Trumpet Lights (official video HD)
>
> >
>
> > in this picture, but in True Calculus we e
>
>
> > picketfences do not give the proper correct value. What does give the
>
> >
>
> > proper and correct value is that we nd up with just one
>
> >
>
>
> >
>
> > picketfence that models the derivative and integral. The limit concept
>
> >
>
> > of Old Math is the use of two picketfences as the delta gets smaller
>
> >
>
> > and smaller. However, as the above shows, no matter how small, the two
>
> >
>
> > picketfence that models the derivative and integral. The limit concept
>
> >
>
> > of Old Math is the use of two picketfences as the delta gets smaller
>
> >
>
> > and smaller. However, as the above shows, no matter how small, the two
>
bacle...@gmail.com
May 24, 2013, 1:16:08 AM5/24/13
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bacle...@gmail.com
May 24, 2013, 1:16:17 AM5/24/13
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On Friday, May 24, 2013 1:13:54 AM UTC-4, bacle...@gmail.com wrote:
bacle...@gmail.com
May 24, 2013, 1:16:23 AM5/24/13
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On Friday, May 24, 2013 1:13:54 AM UTC-4, bacle...@gmail.com wrote:
Archimedes Plutonium
May 24, 2013, 1:52:05 AM5/24/13
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Alright for integration, I am going to pare down to these three
functions and not use the y= 1/x.
y = 3
y = x
y = x^2
there derivates are respectively
Int = 3x
Int = 1/2x^2
Int = 1/3x^3
Antiderivative Technique
(1) for the derivative of a function x^n the derivative is
n(x)^(n-1).
(2) for the integral the antiderivative works backward. So for x^n,
the antiderivative is (1/(n+1)(x^(n+1))
Now we know the derivative was the slope or tangent or rate of change,
and we know the integral is the area under the curve with respect to
the x-axis.
Now we have the antiderivative technique that gives us the derivative
and integral for a class of functions of form y = x^n.
Now we have the technique and we know the integral is the area under
the graph of the function and we want to see how that technique gives
us that area. That technique was known by Newton and Leibniz circa
about 1675. And both Newton and Leibniz probably understood the
antiderivative by examining two of our three functions, the identity
function along with what I call a box function y=3. I do not know if
Newton or Leibniz had in their writings on how they came upon the
Antiderivative Technique
but I would hazard to guess that they discovered it from a box like
function of y=3 and the identity function.
If you look at y=3 its intervals for integration are squares or
rectangles and the triangle top of the picketfence has no triangle
for the derivative is 0. And the area of a rectangle is length by
width. So the area under the graph of the function y=3 for interval 0
to 2 would be 2x3 or area 6 just as the antiderivative as integral
gives us Int = 3x and that is also 6. When x is 3 we have a square
box and thus the area is 3x3 =9. And then Newton and Leibniz probably
noticed that the identity function, y= x is a equilateral triangle
itself with the dy and dx and the area of an equilateral triangle is
1/2x^2 or 1/2 of a square box. So that the entire identity function
is the magnified tiny triangle atop the picketfence for the function
y=x.
So I reckon that both Newton and Leibniz analyzed and saw this box
function and identity function and then discovered the Antiderivative
Technique. I do not know for sure how Newton and Leibniz discovered
the Antiderivative technique but the above is a good guess.
Now let me use the Pretend Infinity borderline 10-grid where the
smallest nonzero number is 1/10 for the Identity function y= x.
. . . x . .
quadrant. It is a fine function to illustrate the Picketfence model.
So for the interval .4 to .5 for x then
y is .4 and .5. And the picketfence looks like this:
/|
||
||
||
||
where its rectangle width is 1/10 and its height is 4/10 and the
triangle atop has width 1/10 and height 1/10. So an area is found for
this picketfence.
Now here is problem for Old Math, in that the area is found for
rectangles alone and the limit is used as the tool to make the
rectangle width as small as possible to gain the most accurate area.
That is the problem, the only width to gain a precise accurate area is
for the limit to go to zero and thus the width of the rectangle is
also zero. So in Old Math, they wanted you to believe, not that the
Moon is made of cheese, but something almost as bad, that rectangles
of 0 width can have area.
In New Math, we escape that contradiction of integration of rectangles
with 0 width, because in New Math, we have holes and gaps of 1/10 in
the 10 grid and of 1*10^-603 in the 10^603 grid. These empty spaces
between successive numbers insures that the integral of myriad
picketfences each have an internal nonzero width and thus a true
internal area.
So let me summarize why Old Math fails to have a Calculus with a limit
concept and why New Math succeeds in having a Calculus. Because
successive numbers have a hole or gap or empty space between them,
they allow enough freedom to form any angle between 0 and 90 degrees
for the hypotenuse of the triangle atop the picketfence, and because
the empty space has a width that gives the picketfence internal area
allows the formation of the integral.
In Old Math, the limit concept was just a yakkity yak time filling
exercise, for in Old Math, they needed something to talk about in
deriving the derivative and integral and the limit became just a
mouthpiece. Sort of like in medicine when we do not know what cancer
truly is, yet we have to talk about something to fill up the time, so
we spend all the lecture time on talking of how various drugs have
effective use. In New Math, we see that it is the empty space between
two successive numbers that allows a derivative and integral to form
and that they are related so that if you change one, you change the
other.
functions and not use the y= 1/x.
y = 3
y = x
y = x^2
there derivates are respectively
y' = 0
y' = 1
y' = 2x
there integrals (abbreviated Int) are respectively
y' = 1
y' = 2x
Int = 3x
Int = 1/2x^2
Int = 1/3x^3
Antiderivative Technique
(1) for the derivative of a function x^n the derivative is
n(x)^(n-1).
(2) for the integral the antiderivative works backward. So for x^n,
the antiderivative is (1/(n+1)(x^(n+1))
Now we know the derivative was the slope or tangent or rate of change,
and we know the integral is the area under the curve with respect to
the x-axis.
Now we have the antiderivative technique that gives us the derivative
and integral for a class of functions of form y = x^n.
Now we have the technique and we know the integral is the area under
the graph of the function and we want to see how that technique gives
us that area. That technique was known by Newton and Leibniz circa
about 1675. And both Newton and Leibniz probably understood the
antiderivative by examining two of our three functions, the identity
function along with what I call a box function y=3. I do not know if
Newton or Leibniz had in their writings on how they came upon the
Antiderivative Technique
but I would hazard to guess that they discovered it from a box like
function of y=3 and the identity function.
If you look at y=3 its intervals for integration are squares or
rectangles and the triangle top of the picketfence has no triangle
for the derivative is 0. And the area of a rectangle is length by
width. So the area under the graph of the function y=3 for interval 0
to 2 would be 2x3 or area 6 just as the antiderivative as integral
gives us Int = 3x and that is also 6. When x is 3 we have a square
box and thus the area is 3x3 =9. And then Newton and Leibniz probably
noticed that the identity function, y= x is a equilateral triangle
itself with the dy and dx and the area of an equilateral triangle is
1/2x^2 or 1/2 of a square box. So that the entire identity function
is the magnified tiny triangle atop the picketfence for the function
y=x.
So I reckon that both Newton and Leibniz analyzed and saw this box
function and identity function and then discovered the Antiderivative
Technique. I do not know for sure how Newton and Leibniz discovered
the Antiderivative technique but the above is a good guess.
Now let me use the Pretend Infinity borderline 10-grid where the
smallest nonzero number is 1/10 for the Identity function y= x.
. . . . . x
. . . . x .
. . . x . .
. . x . . .
. x . . . .
x . . . . .
. x . . . .
x . . . . .
0 .1 .2 .3 .4 .5
So the function y= x is a straight line diagonal of 45degrees in 1st
quadrant. It is a fine function to illustrate the Picketfence model.
So for the interval .4 to .5 for x then
y is .4 and .5. And the picketfence looks like this:
/|
||
||
||
||
where its rectangle width is 1/10 and its height is 4/10 and the
triangle atop has width 1/10 and height 1/10. So an area is found for
this picketfence.
Now here is problem for Old Math, in that the area is found for
rectangles alone and the limit is used as the tool to make the
rectangle width as small as possible to gain the most accurate area.
That is the problem, the only width to gain a precise accurate area is
for the limit to go to zero and thus the width of the rectangle is
also zero. So in Old Math, they wanted you to believe, not that the
Moon is made of cheese, but something almost as bad, that rectangles
of 0 width can have area.
In New Math, we escape that contradiction of integration of rectangles
with 0 width, because in New Math, we have holes and gaps of 1/10 in
the 10 grid and of 1*10^-603 in the 10^603 grid. These empty spaces
between successive numbers insures that the integral of myriad
picketfences each have an internal nonzero width and thus a true
internal area.
So let me summarize why Old Math fails to have a Calculus with a limit
concept and why New Math succeeds in having a Calculus. Because
successive numbers have a hole or gap or empty space between them,
they allow enough freedom to form any angle between 0 and 90 degrees
for the hypotenuse of the triangle atop the picketfence, and because
the empty space has a width that gives the picketfence internal area
allows the formation of the integral.
In Old Math, the limit concept was just a yakkity yak time filling
exercise, for in Old Math, they needed something to talk about in
deriving the derivative and integral and the limit became just a
mouthpiece. Sort of like in medicine when we do not know what cancer
truly is, yet we have to talk about something to fill up the time, so
we spend all the lecture time on talking of how various drugs have
effective use. In New Math, we see that it is the empty space between
two successive numbers that allows a derivative and integral to form
and that they are related so that if you change one, you change the
other.
Archimedes Plutonium
May 24, 2013, 3:28:37 AM5/24/13
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Alright, we covered derivative and integral. We understand that the
picketfence model is the essence of the derivative and integral, in
that a empty space is required between successive numbers. The metric
of that empty space is the inverse of the borderline of finite with
infinite. That metric is Floor-pi*10^603 with its inverse abbreviated
as 10^-603.
In Old Math, the limit concept destroys the integral because it
demands the summation of rectangles each having 0 width, in other
words, rectangles that are not rectangles but rather line segments.
Now if you believe a line segment or line can have area, well, then
join the millions of other fools who accept Old Math Calculus.
The limit concept destroys the derivative because it allows no room,
no empty space for the dy/dx to form an angle without the
interference of neighboring points of the function. Now when the
function is a straight line and not a curve like that of y = 1/x, then
you do get a true derivative. But when you have a curve function, your
dy/dx is never able to match the true value. When you have a empty
space, a hole or gap that is 10^-603 from neighboring points, that
empty space gives the function the freedom to form the correct angle
between 0 degrees and 90 degrees (excluding 90 itself for then it is
no longer a function) and not be constrained with the limit delta
interval.
So, a empty space between successive numbers is essential and
required for the Calculus to exist. The empty space gives the
integral a nonzero area for its picketfences that are summed over the
graph of the function, and the empty space gives the derivative room
enough that a correct angle is formed between a point and its
neighboring points.
So that in Old Math with their phony limit concept, they coughed up
silly and fictional functions such as the Weierstrass function. They
did get something correct about the Weierstrass function for they
call it a pathological function as seen in the Wikipedia entry of
this function. In New Math with holes of 10^-603 there are no
pathological functions. In New Math, all continuous functions are
differentiable everywhere.
Now even Strang in his textbook CALCULUS, 1991 on page 86 shows us
what can be considered another pathological function when you do not
have a borderline of finite to infinite and instead use a limit
concept. Strang shows us the function y= sin(1/x). Strang thinks it
is discontinous because it fluctuates so wildly when it comes near 0.
It is discontinuous at zero for division by zero is undefined.
However, it is continuous thereafter.
So that if Weierstrass or Strang had done a Cartesian Coordinate
System Grid of the x-axis and y-axis in the First Quadrant of those
dots with 10^-603 holes, that the Weierstrass function and the
function y= sin(1/x) are simply normal functions.
Now here we have an interesting consequence of the fact that the
Calculus requires a empty space between successive numbers. So that,
by logic we require the existence of Calculus, which then implies,
holes or empty spaces must exist. That then implies a borderline
between finite and infinity must exist and we just have to find it. So
in this textbook, I started out with a borderline between finite and
infinity and then built the Calculus. Be we can start out in the
opposite direction of insisting Calculus exists and then end up
looking for a infinity borderline. Now I wonder if there is any
special function in Calculus of its derivative or integral that would
seek out the number Floor-pi*10^603? When I did the textbook
Correcting Math, the tractrix and pseudosphere were such functions.
But there could well be others that are more simple and easy and
intuitive.
Archimedes Plutonium
May 24, 2013, 3:37:23 AM5/24/13
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Looks like I learned a second new thing in this book, other than a
definition of continuous function. Looks like I may have found a new
means of discovery of the infinity number borderline. In my last post
#6, I discussed a possibility of a function that requires the infinity
borderline to be a specific number or else the Calculus does not
exist. I am always on the look-out for ways of arriving at Floor-
pi*10^603 as the infinity borderline.
If we draw a half-circle function in the 1st quadrant we involve
ourselves with pi. So we look for a number to be the borderline and
then take its inverse to be the width of the holes or empty space
between successive numbers in order to have a Calculus at all. So now,
can the infinity borderline be 10^500 say, and which would make the
size of the holes as 10^-500?
Well, if we look at this function of a half circle
y = sqrt(r^2-x^2), centered at origin so we have a quarter circle in
the 1st quadrant. Now we ask the question what should the size of
holes or empty space between successive numbers be in order for the
Calculus to exist at all? It cannot be 10^500 because we need there to
be a place in the digits of pi where it is evenly divisible by 5! =
120 so that the entire quarter circle picketfences of derivatives
(slopes or tangents) become a truncated regular polyhedra. In other
words, a place in the digits of pi so that the derivative of the
circle function is a truncated regular polyhedra that becomes the
circle. This happens to pi at Floor-pi*10^603. So at each successive
number points on the graph the curve of the circle is replaced by a
straight line segment of the truncated regular polyhedra as the circle
itself. That can only happen first when infinity borderline is Floor-
pi*10^603. It could happen further down in pi digits beyond 10^603 but
it happens first in pi at 10^603 and so that is the infinity
borderline.
definition of continuous function. Looks like I may have found a new
means of discovery of the infinity number borderline. In my last post
#6, I discussed a possibility of a function that requires the infinity
borderline to be a specific number or else the Calculus does not
exist. I am always on the look-out for ways of arriving at Floor-
pi*10^603 as the infinity borderline.
If we draw a half-circle function in the 1st quadrant we involve
ourselves with pi. So we look for a number to be the borderline and
then take its inverse to be the width of the holes or empty space
between successive numbers in order to have a Calculus at all. So now,
can the infinity borderline be 10^500 say, and which would make the
size of the holes as 10^-500?
Well, if we look at this function of a half circle
y = sqrt(r^2-x^2), centered at origin so we have a quarter circle in
the 1st quadrant. Now we ask the question what should the size of
holes or empty space between successive numbers be in order for the
Calculus to exist at all? It cannot be 10^500 because we need there to
be a place in the digits of pi where it is evenly divisible by 5! =
120 so that the entire quarter circle picketfences of derivatives
(slopes or tangents) become a truncated regular polyhedra. In other
words, a place in the digits of pi so that the derivative of the
circle function is a truncated regular polyhedra that becomes the
circle. This happens to pi at Floor-pi*10^603. So at each successive
number points on the graph the curve of the circle is replaced by a
straight line segment of the truncated regular polyhedra as the circle
itself. That can only happen first when infinity borderline is Floor-
pi*10^603. It could happen further down in pi digits beyond 10^603 but
it happens first in pi at 10^603 and so that is the infinity
borderline.
hbe...@gmail.com
May 24, 2013, 7:03:07 AM5/24/13
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On Friday, May 24, 2013 3:28:37 AM UTC-4, Archimedes Plutonium wrote:
> Alright, we covered derivative and integral. We understand that the
>
> picketfence model is thin particular in algebraic topology, differential geometry and algebraic geometry, the Chern classes are characteristic classes associated to complex vector bundles.
> Alright, we covered derivative and integral. We understand that the
>
Chern classes were introduced by Shiing-Shen Chern (1946).
Contents
[hide] 1 Geometric approach 1.1 Basic idea and motivation
1.2 Construction of Chern classes
2 The Chern class of line bundles
3 The Chern class in Chern-Weil theory 3.1 The Chern class of a Hermitian vector bundle on a smooth manifold
3.2 Example: the complex tangent bundle of the Riemann sphere
4 Properties of Chern classes 4.1 Classical axiomatic definition
4.2 Alexander Grothendieck axiomatic approach
4.3 The top Chern class
5 Proximate notions 5.1 The Chern character
5.2 Chern numbers
5.3 The Chern class in generalized cohomology theories
5.4 Chern classes of manifolds with structure
5.5 Chern classes on arithmetic schemes and Diophantine equations
6 See also
7 References
8 External links
Geometric approach [edit]
Basic idea and motivation [edit]
Chern classes are characteristic classes. They are topological invariants associated to vector bundles on a smooth manifold. The question of whether two ostensibly different vector bundles are the same can be quite hard to answer. The Chern classes provide a simple test: if the Chern classes of a pair of vector bundles do not agree, then the vector bundles are different. The converse, however, is not true.
In topology, differential geometry, and algebraic geometry, it is often important to count how many linearly independent sections a vector bundle has. The Chern classes offer some information about this through, for instance, the Riemann-Roch theorem and the Atiyah-Singer index theorem.
Chern classes are also feasible to calculate in practice. In differential geometry (and some types of algebraic geometry), the Chern classes can be expressed as polynomials in the coefficients of the curvature form.
Construction of Chern classes [edit]
There are various ways of approaching the subject, each of which focuses on a slightly different flavor of Chern class.
The original approach to Chern classes was via algebraic topology: the Chern classes arise via homotopy theory which provides a mapping associated to V to a classifying space (an infinite Grassmannian in this case). Any vector bundle V over a manifold may be realized as the pullback of a universal bundle over the classifying space, and the Chern classes of V can therefore be defined as the pullback of the Chern classes of the universal bundle; these universal Chern classes in turn can be explicitly written down in terms of Schubert cycles.
Chern's approach used differential geometry, via the curvature approach described predominantly in this article. He showed that the earlier definition was in fact equivalent to his.
There is also an approach of Alexander Grothendieck showing that axiomatically one need only define the line bundle case.
Chern classes arise naturally in algebraic geometry. The generalized Chern classes in algebraic geometry can be defined for vector bundles (or more precisely, locally free sheaves) over any nonsingular variety. Algebro-geometric Chern classes do not require the underlying field to have any special properties. In particular, the vector bundles need not necessarily be complex.
Regardless of the particular paradigm, the intuitive meaning of the Chern class concerns 'required zeroes' of a section of a vector bundle: for example the theorem saying one can't comb a hairy ball flat (hairy ball theorem). Although that is strictly speaking a question about a real vector bundle (the "hairs" on a ball are actually copies of the real line), there are generalizations in which the hairs are complex (see the example of the complex hairy ball theorem below), or for 1-dimensional projective spaces over many other fields.
See Chern-Simons for more discussion.
The Chern class of line bundles [edit]
For a sheaf theoretic description, see Exponential sheaf sequence.
An important special case occurs when V is a line bundle. Then the only nontrivial Chern class is the first Chern class, which is an element of the second cohomology group of X. As it is the top Chern class, it equals the Euler class of the bundle.
The first Chern class turns out to be a complete invariant with which to classify complex line bundles, topologically speaking. That is, there is a bijection between the isomorphism classes of line bundles over X and the elements of H2(X;Z), which associates to a line bundle its first Chern class. Addition in the second dimensional cohomology group coincides with tensor product of complex line bundles.[citation needed]
In algebraic geometry, this classification of (isomorphism classes of) complex line bundles by the first Chern class is a crude approximation to the classification of (isomorphism classes of) holomorphic line bundles by linear equivalence classes of divisors.
For complex vector bundles of dimension greater than one, the Chern classes are not a complete invariant.
The Chern class in Chern-Weil theory [edit]
The Chern class of a Hermitian vector bundle on a smooth manifold [edit]
Given a complex hermitian vector bundle V of complex rank n over a smooth manifold M, a representative of each Chern class (also called a Chern form) ck(V) of V are given as the coefficients of the characteristic polynomial of the curvature form Ω of V.
\det \left(\frac {it\Omega}{2\pi} +I\right) = \sum_k c_k(V) t^k
The determinant is over the ring of n × n matrices whose entries are polynomials in t with coefficients in the commutative algebra of even complex differential forms on M. The curvature form Ω of V is defined as
\Omega=d\omega+\tfrac{1}{2}[\omega,\omega]
with ω the connection form and d the exterior derivative, or via the same expression in which ω is a gauge form for the gauge group of V. The scalar t is used here only as an indeterminate to generate the sum from the determinant, and I denotes the n × n identity matrix.
To say that the expression given is a representative of the Chern class indicates that 'class' here means up to addition of an exact differential form. That is, Chern classes are cohomology classes in the sense of de Rham cohomology. It can be shown that the cohomology class of the Chern forms do not depend on the choice of connection in V.
Example: the complex tangent bundle of the Riemann sphere [edit]
Let CP1 be the Riemann sphere: 1-dimensional complex projective space. Suppose that z is a holomorphic local coordinate for the Riemann sphere. Let V = TCP1 be the bundle of complex tangent vectors having the form a∂/∂z at each point, where a is a complex number. We prove the complex version of the hairy ball theorem: V has no section which is everywhere nonzero.
For this, we need the following fact: the first Chern class of a trivial bundle is zero, i.e.,
c_1({\mathbf C\mathbf P}^1\times {\mathbf C})=0.
This is evinced by the fact that a trivial bundle always admits a flat connection.
So, we shall show that
c_1(V) \not= 0.
Consider the Kähler metric
h = \frac{dzd\bar{z}}{(1+|z|^2)^2}.
One readily shows that the curvature 2-form is given by
\Omega=\frac{2dz\wedge d\bar{z}}{(1+|z|^2)^2}.
Furthermore, by the definition of the first Chern class
c_1= \left[\frac{i}{2\pi} \mathrm{tr} \ \Omega\right] .
We must show that this cohomology class is non-zero. It suffices to compute its integral over the Riemann sphere:
\int c_1 dz\wedge d\bar{z} =\frac{i}{\pi}\int \frac{dz\wedge d\bar{z}}{(1+|z|^2)^2}=2
after switching to polar coordinates. By Stokes' theorem, an exact form would integrate to 0, so the cohomology class is nonzero.
This proves that TCP1 is not a trivial vector bundle.
Properties of Chern classes [edit]
Given a complex vector bundle V over a topological space X, the Chern classes of V are a sequence of elements of the cohomology of X. The k-th Chern class of V, which is usually denoted ck(V), is an element of
H2k(X;Z),
the cohomology of X with integer coefficients. One can also define the total Chern class
c(V) = c_0(V) + c_1(V) + c_2(V) + \cdots .
Since the values are in integral cohomology groups, rather than cohomology with real coefficients, these Chern classes are slightly more refined than those in the Riemannian example.
Classical axiomatic definition [edit]
The Chern classes satisfy the following four axioms:
Axiom 1. c_0(V) = 1 for all V.
Axiom 2. Functoriality: If f : Y \to X is continuous and f*V is the vector bundle pullback of V, then c_k(f^* V) = f^* c_k(V).
Axiom 3. Whitney sum formula: If W \to X is another complex vector bundle, then the Chern classes of the direct sum V \oplus W are given by
c(V \oplus W) = c(V) \smile c(W);
that is,
c_k(V \oplus W) = \sum_{i = 0}^k c_i(V) \smile c_{k - i}(W).
Axiom 4. Normalization: The total Chern class of the tautological line bundle over CPk is 1−H, where H is Poincaré-dual to the hyperplane \mathbf{CP}^{k - 1} \subseteq \mathbf{CP}^k.
Alexander Grothendieck axiomatic approach [edit]
Alternatively, Alexander Grothendieck (1958) replaced these with a slightly smaller set of axioms:
Functoriality: (Same as above)
Additivity: If \ 0\to E'\to E\to E''\to 0 is an exact sequence of vector bundles, then c(E)=c(E')\smile c(E'').
Normalization: If E is a line bundle, then c(E)=1+e(E_{\mathbf R}) where e(E_{\mathbf R}) is the Euler class of the underlying real vector bundle.
He shows using the Leray-Hirsch theorem that the total Chern class of an arbitrary finite rank complex vector bundle can be defined in terms of the first Chern class of a tautologically-defined line bundle.
Namely, introducing the projectivization P(E) of the rank n complex vector bundle E → B as the fiber bundle on B whose fiber at any point b\in B is the projective space of the fiber Eb. The total space of this bundle P(E) is equipped with its tautological complex line bundle, that we denote τ, and the first Chern class
c_1(\tau)=: -a
restricts on each fiber P(Eb) to minus the (Poincaré-dual) class of the hyperplane, that spans the cohomology of the fiber, in view of the cohomology of complex projective spaces.
The classes
1, a, a^2, \ldots , a^{n-1}\in H^*(\mathbf{P}(E))
therefore form a family of ambient cohomology classes restricting to a basis of the cohomology of the fiber. The Leray-Hirsch theorem then states that any class in H*(P(E)) can be written uniquely as a linear combination of the 1, a, a2, ..., an−1 with classes on the basis as coefficients.
In particular, one may define the Chern classes of E in the sense of Grothendieck, denoted c_1(E), \ldots c_{n}(E) by expanding this way the class -a^n, with the relation:
- a^n = c_1(E). a^{n-1}+ \ldots c_{n-1}(E) .a + c_{n}(E) .
One then may check that this alternative definition coincides with whatever other definition one may favor, or use the previous axiomatic characterization.
The top Chern class [edit]
In fact, these properties uniquely characterize the Chern classes. They imply, among other things:
If n is the complex rank of V, then c_k(V) = 0 for all k > n. Thus the total Chern class terminates.
The top Chern class of V (meaning c_n(V), where n is the rank of V) is always equal to the Euler class of the underlying real vector bundle.
Proximate notions [edit]
The Chern character [edit]
Chern classes can be used to construct a homomorphism of rings from the topological K-theory of a space to (the completion of) its rational cohomology. For a line bundle L, the Chern character ch is defined by
\operatorname{ch}(L) = \exp(c_{1}(L)) := \sum_{m=0}^\infty \frac{c_1(L)^m}{m!}.
More generally, if V = L_1 \oplus ... \oplus L_n is a direct sum of line bundles, with first Chern classes x_i = c_1(L_i), the Chern character is defined additively
\operatorname{ch}(V) = e^{x_1} + \dots + e^{x_n} :=\sum_{m=0}^\infty \frac{1}{m!}(x_1^m + ... + x_n^m).
Observe that when V is a sum of line bundles, the Chern classes of V can be expressed as elementary symmetric polynomials in the x_i, c_i(V) = e_i(x_1,...,x_n). In particular, on the one hand
c(V) := \sum_{i=0}^n c_i(V),e essence of the derivative and integral, in
hbe...@gmail.com
May 24, 2013, 7:05:32 AM5/24/13
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On Friday, May 24, 2013 12:16:34 AM UTC-4, Archimedes Plutonium wrote:
> Alright, in the next edition I need to use the 10 grid more than the
>
> 10^-603 grid so that High School students have an easy time with True
>
> Calculus.
You won't take my bet, Huh. No one will buy your stupid textbook, you worthless
> Alright, in the next edition I need to use the 10 grid more than the
>
> 10^-603 grid so that High School students have an easy time with True
>
> Calculus.
baboon
>
>
>
>
hbe...@gmail.com
May 24, 2013, 7:05:49 AM5/24/13
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On Friday, May 24, 2013 12:16:34 AM UTC-4, Archimedes Plutonium wrote:
> Alright, in the next edition I need to use the 10 grid more than the
>
> 10^-603 grid so that High Sin particular in algebraic topology, differential geometry and algebraic geometry, the Chern classes are characteristic classes associated to complex vector bundles.
>
hbe...@gmail.com
May 24, 2013, 7:06:35 AM5/24/13
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On Friday, May 24, 2013 7:05:32 AM UTC-4, hbe...@gmail.com wrote:
> On Friday, May 24, 2013 12:16:34 AM UTC-4, Archimedes Plutonium wrote:
>
> > Alright, in the next edition I need toin particular in algebraic topology, differential geometry and algebraic geometry, the Chern classes are characteristic classes associated to complex vector bundles.
> On Friday, May 24, 2013 12:16:34 AM UTC-4, Archimedes Plutonium wrote:
>
hbe...@gmail.com
May 24, 2013, 7:06:44 AM5/24/13
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May 24, 2013, 7:07:10 AM5/24/13
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hbe...@gmail.com
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On Friday, May 24, 2013 7:05:32 AM UTC-4, hbe...@gmail.com wrote:
1treePetrifiedForestLane
May 24, 2013, 5:27:31 PM5/24/13
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> > No one will e'er try to ingest your textbook,
without a good French sauce.
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