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Agincourt

Prove me wrong please

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Agincourt    122
This post is a little off topic of game math, but it''s been plagueing me for awhile now. And if my math proffesor spoke any english I''d ask him. Here it is - if 4 + 4 = 8; divide each 4 in half like this - (2 + 2) + (2 + 2) = 8; and repeat - 1+1+1+1+1+1+1+1=8; and repeat some more until you reach 1/infinite. so the problem looks like 1/inf + 1/inf + ... inifinite time = 8; so, (and this is where i''m probably wrong) can''t you equate this to - (1/inf)^inf = 8 ?? and 1/inf = 0, right? so if you multiply 0 by itself infinite times, can it not equal any number? obviously not, or I''d just right 0^inf for all my homework problems :-) so which assumption am I wrong on? I''m a little unclear on all the nuances of the infinite concept. Use the WriteCoolGame() function Works every time

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Dredge-Master    175
no, 1/inf IS NOT = to 0

1/inf APPROACHES 0. ie; it is so bloody close but it isn''t quite there yet.


a rule:
Any finite number divided by infinity is a number infinitesimally larger than, but never equal to, zero.


Oh yeah, when dealing with infinite numbers, use a pen and paper and remember that you have limits in it such as x -> infinity. Luckily I don''t have to deal with them or imaginary numbers anymore. Atleast I hope not. Unfortunately I still have to do statistics.



Beer - the love catalyst
good ol'' homepage

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medovids    122
You can't repeat the step so much that it reaches 1/inf that is impossible it will never be 1/inf. By the way 1/inf does equall zero. If you have 0.0000000 and the zeros go on forever where is the space for that last "1"?

Edited by - medovids on January 10, 2002 12:21:55 AM

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Graylien    160
As Dredge-Master points out, the limit of 1/n as n approaches positive infinity is zero. Not everyone has a background in Calculus, but this result itself is not very hard to derive.

The original question, however, is malformed. When you have the equation 4 + 4 = 8, the transformation (2 + 2) + (2 + 2) = 8 is equivalent to 2 * (4 / 2) + 2 * (4 / 2) = 8. Taking this one step further, we get (2/2)*4 + (2/2)*4 = 4 + 4 = 8.

What exactly are you doing when you "divide each 4 in half"? Not much, I''m afraid. In essence, you''re replacing one expression (which evaluates to 4) with another expression (which also evaluates to 4). This reasoning alone should be sufficient to convince you that there is nothing wrong.

You needn''t resort to a (Calculus-based) proof of convergence.

medovids: The value (1 / infinity), taken just like that, is undefined. I''m afraid the problem is more difficult than you think.

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LilBudyWizer    491
Infinity is not a number. As an example the interval (0,1) has an infinite number of numbers as does the interval (0,2), but (0,2) has more numbers than (0,1), i.e. (0,2) is a larger interval than (0,1). Infinity is not equal to infinity. As another example the limit as n goes to infinity of 2^n and 3^n are both infinity but they approach infinity at differant rates, i.e. as n increases by one one of them doubles and the other triples.

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Brother Bob    10344
The actual fault in the text by Agincourt has already been mentioned. He/she assumes 1/inf = zero, when it''s infact undefined. It aproaches zero.

Anyways, here''s a general description of the problem, and a "proof" that the result is always 8.

First the basic case. Rewrite it as a sum.
  
1
2
---
\ 4
4 + 4 = | ----- = 2 * 4 = 8
/ 0
--- 2
n=1


Then make a general case, where you divide the 4 by two, and sum it two times as many times.

  
k
2
---
\ 4 k 4 1 4
| ----- = 2 * ----- = 2 * ----- = 2 * 4 = 8
/ k-1 k-1 0
--- 2 2 2
n=1


As you can see, the result is always 2*4, and is independent of k (which is the number of times you divide the original 4).

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Tjoppen    122
Hey...CanLt ‡ be negative??? in that case, u r 0wnzed...

Hence, if we divide a positive number by a negative, we get a negative in return...If you add upp all the values, you get -8...

So the answer is that the sum = +-8

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grhodes_at_work    1385
Normally I disapprove of homework/school related posts that are not related to game development.

This post is acceptable for two reasons. First, Agincourt showed that he had thought about the problem himself and was not merely looking for an answer. Second, it provoked some interesting theoretical discussion. I believe the responses to this post could help Agincourt build his problem solving skills and I do approve of that sort of discussion here.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

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BS-er    181
LilBudyWizer said it right. The = operation isn''t even appropriate in such a case. Neither is dividing a number by a non-number. You might as well talk about 1/banana.

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Scarab0    122
quote:
Original post by BS-er
LilBudyWizer said it right. The = operation isn''t even appropriate in such a case. Neither is dividing a number by a non-number. You might as well talk about 1/banana.


Yes but everyone knows that bananas aren''t numbers (I hope ), but not everyone knows that infinity isn''t.

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Dredge-Master    175
quote:
Original post by Tjoppen
Hey...CanLt ‡ be negative??? in that case, u r 0wnzed...

Hence, if we divide a positive number by a negative, we get a negative in return...If you add upp all the values, you get -8...

So the answer is that the sum = +-8


if this "Lt ‡ " is meant to be infinity then no it isn't negative. It is a LARGE number, it gets bigger. To get negative infinity you put a minus sign infront of it like this

-infinity.



Beer - the love catalyst
good ol' homepage

Edited by - Dredge-Master on January 13, 2002 6:28:14 PM

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Timkin    864
quote:
Original post by BS-er
[...] Neither is dividing a number by a non-number. You might as well talk about 1/banana.



Actually, you can divide 1 by a banana, so long as you also define the result.

Timkin

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Beer Hunter    712
quote:
Original post by LilBudyWizer
Infinity is not a number.


Exactly. All sorts of gibberish occur when you assume that infinity is a number.



Let x = 1 + 2 + 4 + 8 + 16 + 32 + ...

2x = 2 + 4 + 8 + 16 + 32 + ...
x - 1 = 2 + 4 + 8 + 16 + 32 + ...
2x = x - 1
x = -1
-1 = 1 + 2 + 4 + 8 + 16 + 32 + ...


x = 1 - 1 + 1 - 1 + 1 - 1 + ...
x = (1 - 1) + (1 - 1) + (1 - 1) + ...
x = 0 + 0 + 0 + ...
x = 0


x = 1 - 1 + 1 - 1 + 1 - 1 + ...
x = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ...
x = 1 + 0 + 0 + 0 + ...
x = 1


x = 1 - 1 + 1 - 1 + 1 - 1 + ...
1 - x = 1 - (1 - 1 + 1 - 1 + 1 - 1 + ...)
1 - x = 1 - 1 + 1 - 1 + 1 - 1 + ...
1 - x = x
x = 0.5

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sjelkjd    171
quote:
Original post by Agincourt
This post is a little off topic of game math, but it''s been plagueing me for awhile now. And if my math proffesor spoke any english I''d ask him. Here it is - if 4 + 4 = 8; divide each 4 in half like this - (2 + 2) + (2 + 2) = 8; and repeat - 1+1+1+1+1+1+1+1=8; and repeat some more until you reach 1/infinite. so the problem looks like 1/inf + 1/inf + ... inifinite time = 8; so, (and this is where i''m probably wrong) can''t you equate this to - (1/inf)^inf = 8 ?? and 1/inf = 0, right? so if you multiply 0 by itself infinite times, can it not equal any number? obviously not, or I''d just right 0^inf for all my homework problems :-) so which assumption am I wrong on? I''m a little unclear on all the nuances of the infinite concept.

Use the WriteCoolGame() function
Works every time


Actually all you guys are wrong:
1/inf+1/inf+...+1/inf != (1/inf)^inf
1/inf+1/inf+...+1/inf == (1/inf)*inf

since we can''t use infinity, we can replace it with x and we get

limit x-> inf 1/x*x
which is equal to 1(by L''Hopital''s).

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LilBudyWizer    491
Well, when it comes to homework as long as the result is real you could just say it is between positive and negative infinity. Hum, there''s a thought, an imaginary number tending towards an imaginary infinity.

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Dredge-Master    175
quote:
Original post by sjelkjd
Actually all you guys are wrong:
1/inf+1/inf+...+1/inf != (1/inf)^inf
1/inf+1/inf+...+1/inf == (1/inf)*inf

since we can''t use infinity, we can replace it with x and we get

limit x-> inf 1/x*x
which is equal to 1(by L''Hopital''s).




Again, not quite


your limit is x->c, where c is infinity

if it was 1/x*x, this is 1/x/(1/x).
This means that f(x)=g(x)=1/x so it makes it easy
f''(x) = g''(x)=-1/x^2

Now as far as you have gotten with L''Hopital''s is that
f(x)/g(x) = f''(x)/g''(x).
This is true as both ratios are 1.


This is the part of L''Hopital''s rule which you forgot
f(c)=g(c)=0

It only works if the above is true.
(also, as a side note g''(x) may never be equal to 0)

Now this is where it fails in this case

f(c)=f(inf) == 1/inf

and as we know, 1/inf is not equal to 0, so you can not use L''Hopital''s Rule.



Beer - the love catalyst
good ol'' homepage

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Guest Anonymous Poster   
Guest Anonymous Poster
Well I ve read a lot of things here but none really answers to this poor Agincourt. Brother bob explains that he describes a serie constantly equal to 8, for any n. Which is quite true !ROFLMAO

Lim 8*(x/x) = 8 ! Nice
x->+inf

Lim 8*x = inf
Lim x = inf
x->inf

cant lead to
Lim 8*x/x = inf/inf. Basically because this notation is illegal. Beware the exact conditions of the theorems ! As it means nothing to 1 or 8 if you want it will just make a bigger incorrect statement.

Infinity is not a unique concept in maths. Kantor describes different infinities. For instance what LilBudyWizer says is totally false. The weird thing about |R is that any non singleton interval of |R contains as many ( an infinite number of ) elements as |R itself. Dont mislead distances and cardinal ! Nothing in common.

For instance Card(A=[0,1])=Card(B=[0,2]) is really easy to prove because f: x -> 2*x is a bijection between A and B. Each x of A is associated to a unique y of B => same cardinal or "number of elements" if you prefer. Likewise [-1,1] can be mapped to |R with g: x -> x/((1-x)*(1+x))

Different infinities : Card(|N) = Card(|Q) < Card(|R) < Card(|C)


Now 1/inf = 0 ?

One could say it''s obvious because 1/x -> 0 when x -> inf

The problem is that 1/0 is undefined. Is it +inf or -inf ? Dont forget that 1/-inf=0 too. A number must have only one inverse ! And +inf and -inf are quite distant arent they ?

Thus really no ? No in the field of reals. But yes in the field of extended reals, |R*, in the projective closure of |R. ?)
BTW projective / 3D visuals, cant you see the link ?

-inf=inf, both infinites are merged.

x+inf=inf+x=inf if x!=inf
x*inf=inf*x=inf if x!=0
x/inf=0 if x!=inf
x/0=inf if x!=0

Thus 1/inf=0 and 1/0=inf. All right.

But as you can see even here inf/inf and 0*inf are still illegal notations.

Beware that |R* does not have exactly the same properties as |R !
Too theoric ? You should know that the FPU of your Pentium or PPC can use |R or |R* ! (As far as I remember.)

I wont go further just let U think that maths are only bothered by their internal efficiency and total coherence. The fact that they succeed to model reality is a bit of mistery cause they dont always match our common sense.

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sjelkjd    171
quote:
Original post by Dredge-Master
[quote]Original post by sjelkjd
Actually all you guys are wrong:
1/inf+1/inf+...+1/inf != (1/inf)^inf
1/inf+1/inf+...+1/inf == (1/inf)*inf

since we can''t use infinity, we can replace it with x and we get

limit x-> inf 1/x*x
which is equal to 1(by L''Hopital''s).




Again, not quite


your limit is x->c, where c is infinity

if it was 1/x*x, this is 1/x/(1/x).
This means that f(x)=g(x)=1/x so it makes it easy
f''(x) = g''(x)=-1/x^2

Now as far as you have gotten with L''Hopital''s is that
f(x)/g(x) = f''(x)/g''(x).
This is true as both ratios are 1.


This is the part of L''Hopital''s rule which you forgot
f(c)=g(c)=0

It only works if the above is true.
(also, as a side note g''(x) may never be equal to 0)

Now this is where it fails in this case

f(c)=f(inf) == 1/inf

and as we know, 1/inf is not equal to 0, so you can not use L''Hopital''s Rule.



Beer - the love catalyst
good ol'' homepage

What fucked up school taught you calculus?
1/x*x = x/x
Since this is a form of infinity/infinity, you can use L''Hopital''s and say:
lim f(x)/g(x) = lim f''(x)/g''(x)
or
lim x/x = lim 1/1 = 1

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jermz    122
Dredge-Master was trying to point out that there are some conditions upon using L^Hopital''s Rule, one of them being that:

f(x) -> 0
g(x) -> 0
as x -> c (c in R*)

but he forgot that L^Hopital''s is also applicable if:

g(x) -> inf
as x -> c

This is the form you were thinking of sjelkjd. However you made a mistake initially in assuming that you can just "replace" the 1/inf * inf with 1/x * x. As people have stated, infinity is NOT a number and we don''t know the rates at which it was reached. We should say 1/inf * inf = 1/g(x) * f(x) where f,g -> inf as x -> c. For Agincourts problem f() = 8 * g().

Incidentally 1/x * x is also (1/x) / (1/x) as well as x/x.

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Dredge-Master    175
quote:

g(x) -> inf
as x -> c



Only problem is in this case it is still ignored as
g(x)->1/inf, not the other was as the function is 1/x. so it gets ignored anyway.


Thanks for backing me up though.

Complex/imaginary numbers and infinity may be fun, but it sure as hell is still confusing after four years at uni.




Beer - the love catalyst
good ol'' homepage

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sjelkjd    171
quote:
Original post by jermz

This is the form you were thinking of sjelkjd. However you made a mistake initially in assuming that you can just "replace" the 1/inf * inf with 1/x * x.




Hmm...let''s look at my original post:
quote:

since we can''t use infinity, we can replace it with x and we get

limit x-> inf 1/x*x



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Guest Anonymous Poster   
Guest Anonymous Poster
Take for example the equation (2x+4)/(3x+1) as x->infinity. If you plug in inf. then (2(inf)+4)/(3(inf)+1) well, 2*inf and 3*inf = inf and inf+4 = inf and so on. So inf/inf, and so = 1

But there''s lots of paradoxes associated with infinity. So now, let''s do it another way:

Divide both sides by x = (2 + (4/x))/(3 + (1/x)) as x->inf, then 4/x, 1/x -> 0 so (2+0)/(3+0) = 2/3, which is not equal to 1.

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Dobbs    164
quote:
Take for example the equation (2x+4)/(3x+1) as x->infinity. If you plug in inf. then (2(inf)+4)/(3(inf)+1) well, 2*inf and 3*inf = inf and inf+4 = inf and so on. So inf/inf, and so = 1


You''ve got to be kidding me. For startes that''s not even an equation. And you can''t just "plug in infinity," and 2*inf doesn''t equal 3*inf nor does it equal inf+4 because those values are undefined. Infinity is not a number. Your limit is wrong.

quote:
Divide both sides by x = (2 + (4/x))/(3 + (1/x)) as x->inf, then 4/x, 1/x -> 0 so (2+0)/(3+0) = 2/3, which is not equal to 1.


Both sides? Both sides of what?

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