Another math problem! (Not HW)

[Overburn]

My math is pretty limited, so I'm just kinda thinking of loopholes for this more than figuring it out, but could he meant the net distance is 1? As in If you go under the Y axis, it's negative. I vaguely remember one my math teachers making us calculate total distances and net distances at some point. I don't know if that'd make an answer possible or whatever.

[Eliseos]

This.

All he did in class was draw an x,y plane, put a point at 0,0 and another at 1,1, and said to make a path that connects the two, is continuous, and has length of 1.

This is just kinda a bonus question, if anyone gets it right they don't have to do the next lab and get an auto 20/20 on it <.<

Careful not to turn into this guy Dead Fly + Dare - 'A' Grade = Angry Student - cbs13.com

[Kaylia]

My math is pretty limited, so I'm just kinda thinking of loopholes for this more than figuring it out, but could he meant the net distance is 1? As in If you go under the Y axis, it's negative. I vaguely remember one my math teachers making us calculate total distances and net distances at some point. I don't know if that'd make an answer possible or whatever.

Nah, no matter how far he goes in the negative, he will always have to come back to 0 to eventually reach 1. The net sum for x and y will be 1 in both case.


You can demonstrate this using triangle inequality
d(x, z) ≤ d(x, y) + d(y, z)
distance between 2 points is always smaller or equals to the distance it takes if you stop by a 3rd point (which is what you're doing if you go in the negative). In this case, the straigth line is sqrt(2), so no matter where you go before reaching (1,1), the net distance will always be bigger than sqrt(2)

[joft]

sounds weird, I can't help but think it's going to be some stupid trick like using the 1-norm instead of the 2-norm

the only actual interesting possibility i can imagine would be something like
Cantor function - Wikipedia, the free encyclopedia

except that thing has arc length 2

yeah any path between (0,0) and (1,1) is, by definition, at least as long as the shortest path-- the geodesic with length sqrt2

[Kaylia]

sounds weird, I can't help but think it's going to be some stupid trick like using the 1-norm instead of the 2-norm

the only actual interesting possibility i can imagine would be something like
Cantor function - Wikipedia, the free encyclopedia

except that thing has arc length 2

yeah any path between (0,0) and (1,1) is, by definition, at least as long as the shortest path-- the geodesic with length sqrt2

Function like this scare the fuck out of me. I love it. I wish I would have went for maths instead of physics at time.

Weird function are bounc by the same metric. You will find many that have arc length equals to infinity, or somewhere inbetween sqrt(2) and infinity, but can you really find a continuous one under this value.

[Mojo]

You and two friends arrive at a hotel that charges $30 for the night. You split it evenly down the middle. Later, the clerk realizes he overcharged you $5. He figures he should just refund you $1 each to make it more even, thus you only paid $9 each and he keeps $2. So now your friends have paid a total of $27(3x9=27), and the clerk has $2. Where did the extra dollar go?

[Kaylia]

What kind of shitty hotel charge you 25$ for the night, no wonder the clerk stole your 2$.

[Korietsu]

You and two friends arrive at a hotel that charges $30 for the night. You split it evenly down the middle. Later, the clerk realizes he overcharged you $5. He figures he should just refund you $1 each to make it more even, thus you only paid $9 each and he keeps $2. So now your friends have paid a total of $27(3x9=27), and the clerk has $2. Where did the extra dollar go?

The 33 cents a person that was not split right.

25/3 = 8.33 ea. + 1 = 9.33 ea. x3 = 28 +2 = 30.

[Kaylia]

You gave away 27$ (30-3), the clerk has 2$, the hotel 25$.


Isn't it easier to word it like this? Trying to add up money to 30$ isn't very intuitive here, since it's just a fake number in the problem. If we want to know how the 30$ was split:
25$ hotel + 2$ clerk + 3$ in your pocket = 30$

[Trajan]

sounds weird, I can't help but think it's going to be some stupid trick like using the 1-norm instead of the 2-norm

the only actual interesting possibility i can imagine would be something like
Cantor function - Wikipedia, the free encyclopedia

except that thing has arc length 2

yeah any path between (0,0) and (1,1) is, by definition, at least as long as the shortest path-- the geodesic with length sqrt2

If you cannot change from Euclidean to Non-Euclidean geometry for this problem, which I think someone has stated, this seems to be the only option. I was thinking something almost like a ladder, something tiled horizontally only. I think that can still be continuous. Only question is how to calculate its length and see if it equals 1. I would def like to see the solution to this problem.

Edit: Now that I am thinking about this, why wouldn't it work? I.e. put a tile from (0,0) to (.1,0), the length is .1. Then put another tile from (.1,.1) to (.2,.1), the length is again .1. Repeat that 10 times to get a ladder, that should be continuous, that is length 1.

[Sylvrdragon]

Solved:

Spoiler: show

http://i31.photobucket.com/albums/c384/Sylvr/Duh.jpg

[Kaylia]

^nothing you can't fix with portal

If you cannot change from Euclidean to Non-Euclidean geometry for this problem, which I think someone has stated, this seems to be the only option. I was thinking something almost like a ladder, something tiled horizontally only. I think that can still be continuous. Only question is how to calculate its length and see if it equals 1. I would def like to see the solution to this problem.

Edit: Now that I am thinking about this, why wouldn't it work? I.e. put a tile from (0,0) to (.1,0), the length is .1. Then put another tile from (.1,.1) to (.2,.1), the length is again .1. Repeat that 10 times to get a ladder, that should be continuous, that is length 1.

Ladder has a length of two because you go up, than go right (1+1). Any ladder function will have a length of 2 here.


About your edit, doing this 10 time would only get you halfway

[Sylvrdragon]

This problem isn't solvable by 'inside-the-box' math. The prof said "Path" not "Line" right? He can't have meant line, cause it couldn't be both 'lenth=1' and 'continuous'. At first, I was thinking of a Circle with Diameter of 1, but laying that inside the square wouldn't touch the corners.

How about a Square with length=1? As an enclosed geometric shape, it is by definition 'continuous', and if you follow it's edges, it makes a 'path' from 0,0 to 1,1. It may be a technicality, but it does have a "Length" of 1.

[Trajan]

This problem isn't solvable by 'inside-the-box' math. The prof said "Path" not "Line" right? He can't have meant line, cause it couldn't be both 'lenth=1' and 'continuous'. At first, I was thinking of a Circle with Diameter of 1, but laying that inside the square wouldn't touch the corners.

How about a Square with length=1? As an enclosed geometric shape, it is by definition 'continuous', and if you follow it's edges, it makes a 'path' from 0,0 to 1,1. It may be a technicality, but it does have a "Length" of 1.

If you did that the length would be 1 ( (0,0)->(1,0) ) + 1 ( (1,0)->(1,1) ) = 2

[Korietsu]

If you cannot change from Euclidean to Non-Euclidean geometry for this problem, which I think someone has stated, this seems to be the only option. I was thinking something almost like a ladder, something tiled horizontally only. I think that can still be continuous. Only question is how to calculate its length and see if it equals 1. I would def like to see the solution to this problem.

Edit: Now that I am thinking about this, why wouldn't it work? I.e. put a tile from (0,0) to (.1,0), the length is .1. Then put another tile from (.1,.1) to (.2,.1), the length is again .1. Repeat that 10 times to get a ladder, that should be continuous, that is length 1.

Actually, this gave me a great idea.

delta(t-1), or an individual impulse delta at x = 1, with magnitude 1.

[Kaylia]

This problem isn't solvable by 'inside-the-box' math. The prof said "Path" not "Line" right? He can't have meant line, cause it couldn't be both 'lenth=1' and 'continuous'. At first, I was thinking of a Circle with Diameter of 1, but laying that inside the square wouldn't touch the corners.

How about a Square with length=1? As an enclosed geometric shape, it is by definition 'continuous', and if you follow it's edges, it makes a 'path' from 0,0 to 1,1. It may be a technicality, but it does have a "Length" of 1.

But the line -is- the shortest path in eucledian space, with normal metric. Sticking a square with an half length of 1 over your first space isn't mathematics at all, because you're using an object from a different space without establishing a relation between both.

In the end, what you said is the equivalent of saying "a length of sqrt(2) is equal to 1 in a different space where S1 -> sqrt(2)* S2

[Sylvrdragon]

If you did that the length would be 1 ( (0,0)->(1,0) ) + 1 ( (1,0)->(1,1) ) = 2

That's the length of the path, not the length of the square. Though I just re-read the OP and, if it was quoted correctly, then that solution would indeed not work. I don't know though... if it's just a bonus question, then I wouldn't put it past him to put the actual "Problem" in the wording rather than the math.

If you can't come up with anything else, try the square thing. Or print out the Portal pic and hope he's a fan. Maybe you can amuse him enough to give you credit!

[Kaylia]

Actually, this gave me a great idea.

delta(t-1), or an individual impulse delta at x = 1, with magnitude 1.

If it's continuous between 0 and 1, your length is going to be 2.

[Trajan]

^nothing you can't fix with portal


Ladder has a length of two because you go up, than go right (1+1). Any ladder function will have a length of 2 here.


About your edit, doing this 10 time would only get you halfway

I am only talking about tiling horizontally, not vertically. It should look (something) like this: (each ___ has a length of .1, ten tiles in total)

Edit: my make shift picture didn't come out well. But yes if you only tile horizontally (and not vertically) you will get a continuous ( but not differentiable) path of length one.

[Korietsu]

Wouldn't the individual delta impuse not exist anywhere else on the graph elsewhere? Since the impulse extends to infinity, but only has magnitude of 1, at least, from a signals/systems point of view.