Math gurus - tell me why this is wrong.

[BrownTown]

Not a right triangle.

 

[MrChad]

http://mathforum.org/library/drmath/view/55173.html

 

[hypn0tik]

Originally posted by: MrChad
http://mathforum.org/library/drmath/view/55173.html

Thank you!

 

[Furyline]

Originally posted by: MrChad
http://mathforum.org/library/drmath/view/55173.html

yay!

I remember being very confused about this several years ago and never getting a satisfactory answer from friends.

 

[BrownTown]

Just google taxicab geometry if you dont believe me.

 

[her209]

Originally posted by: MrChad
http://mathforum.org/library/drmath/view/55173.html

So why does it not work in this case but works when approximating the area under a "curve" as N approaches 8?

 

[sao123]

Originally posted by: hypn0tik

Originally posted by: MrChad
For N = 1:

L = (a/1) + (b/1)

For N = 2:

L = (a/2) + (b/2) + (a/2) + (b/2)
L = 2*(a/2) + 2*(b/2)

For N = 3:

L = (a/3) + (b/3) + (a/3) + (b/3) + (a/3) + (b/3)
L = 3*(a/3) + 3*(b/3)

For N = n
L = n*(a/n) + n*(b/n)

Now, take the limit as n -> 8

You find that the n's cancel out and you are left with a + b.

Now, I know that's not the expected result, but I can't see why.

For N = n
L = n*(a/n) + n*(b/n)

Now, take the limit as n -> 8



Chances are this is expandable as some infinite series which will lead to an known identity equation, then using some algebra it is morphable into sqrt(a^2+b^2).

 

[dethman]

hilarious how half the people in this thread didn't even know what the fvck was going on here and spouted their babble anyway.

 

[her209]

BUMP!

 

[Syringer]

Originally posted by: dethman
hilarious how half the people in this forum didn't even know what the fvck was going on here and spouted their babble anyway.

Fixed.

 

[Goosemaster]

Wait, so the hypothenuse of a triangle now has infinate length?

Why didn't I get the memo?

 

[her209]

Originally posted by: Goosemaster
Wait, so the hypothenuse of a triangle now has infinate length?

Why didn't I get the memo?

No it is divided into N "steps". What is the length as N approaches 8?

 

[imported_Rat]

Originally posted by: MrChad
For N = 1:

L = (a/1) + (b/1)

For N = 2:

L = (a/2) + (b/2) + (a/2) + (b/2)

For N = 3:

L = (a/3) + (b/3) + (a/3) + (b/3) + (a/3) + (b/3)

Extrapolate that out to N = 8

L = (a/8) + (b/8) + (a/8) + (b/8) + (a/8) + (b/8) + (a/8) + (b/8) + ...

The formula breaks down as N approaches infinity, because essentially you start adding up an infinite number of 0-length sides.

 

[Goosemaster]

Originally posted by: her209

Originally posted by: Goosemaster
Wait, so the hypothenuse of a triangle now has infinate length?

Why didn't I get the memo?

No it is divided into N "steps". What is the length as N approaches 8?

Liek I said, I didn't get the memo so I am jsut winging it at the memo as if I did.

 

[Goosemaster]

Originally posted by: her209

Originally posted by: Goosemaster
Wait, so the hypothenuse of a triangle now has infinate length?

Why didn't I get the memo?

No it is divided into N "steps". What is the length as N approaches 8?

length of the total = sqrt(a^2^ + b^2^) / N

 

[imported_Rat]

Originally posted by: her209

Originally posted by: Goosemaster
Wait, so the hypothenuse of a triangle now has infinate length?

Why didn't I get the memo?

No it is divided into N "steps". What is the length as N approaches 8?

There is no special case as it approaches infinity. For N=any integer, you get a+b as expected. At N=infinity, the length of each step is exactly zero, so you don't add them in the conventional way.

 

[SpecialEd]

Originally posted by: her209

Originally posted by: MrChad
http://mathforum.org/library/drmath/view/55173.html

So why does it not work in this case but works when approximating the area under a "curve" as N approaches 8?

When you take the limit of a sequence a functions, the properites of the those functions will not necessarly be preserved in the limit. Your example is one where the limit of the arclength is not the arclength of the limit.

Even area under the curve is not always preserved under the limit. For example, consider the interal of the constant function 1/n on the entire real line. Clearly the area under the curve is infinite for any positive integer. However as n goes to infinity the limiting function is 0 and the integral of 0 is just 0.

 

[Savij]

Originally posted by: hypn0tik

Originally posted by: MrChad
For N = 1:

L = (a/1) + (b/1)

For N = 2:

L = (a/2) + (b/2) + (a/2) + (b/2)
L = 2*(a/2) + 2*(b/2)

For N = 3:

L = (a/3) + (b/3) + (a/3) + (b/3) + (a/3) + (b/3)
L = 3*(a/3) + 3*(b/3)

For N = n
L = n*(a/n) + n*(b/n)

Now, take the limit as n -> 8

You find that the n's cancel out and you are left with a + b.

Now, I know that's not the expected result, but I can't see why.

Right.

You're on the right track. The A+B that you're measuring is different from the square root that you're expecting. One is measuring the distance from the two points and the other is measuring the path that you took the get there. The n -> infinity case is NOT the squareroot A squared plus B squared because it's measuring path, not straight line distance.