[Guest]

I thought of something interesting.

Suppose I am at point (0,0) in an (x,y) plane.

I want to travel to the point (1,1).

Suppose I pick a big number N, and decide to move a distance 1/N in the +x direction, then 1/N in the +y direction.

Suppose I continue to alternately move by 1/N in the +x and then +y directions until I reach (1,1).

Let's also say that instead of N being just a big number, let's look at the described situation in the limit as N approaches infinity. Clearly as N gets bigger, this path more closely resembles a direct line between (0,0) and (1,1).

When I reach (1,1), how far have I traveled? (Total distance of path, not displacement between end-points).

Edited October 18, 2010 by mmiguel1

[Guest]

if N = infinity.... distance travelled = sqrt(2)

else distance travelled = 2

[pdqkemp]

if N = infinity.... distance travelled = sqrt(2)

else distance travelled = 2

To be clear, compared to you all I have the mathematical capabilities of a squirrel. With that in mine, here is my question regarding Abhijith's post:

It can't just go from being 2 to being the sqrt (2) when it reaches infinity, can it?

Shouldn't it start approaching sqrt (2) when it starts approaching infinity?

Hmm....then again, at what point in time could it "start" approaching infinity?

The more I think about it, the more I think I'd have to agree with Abhijith's post as is. As long as infinity is never actually reached, which of course it can't be, then it's 2. If you assume infinity, then it's the sqrt (2) like he said.

OK...I'll just look forward to reading the other answers like I usually do.

Good question, mmiguel1!

[Guest]

[

You have to look at the physical shape and size of the vehicle on which you are travelling. It is not possible for an ant, a man, a car, a bus or for that matter, any physical object to take a sharp step shaped path with so many corners when the size of each step (1/N) small. There is always some rounding at the corners. Thus, physically the path becomes shorter as N increases and becomes sqrt(2) when 1/N becomes comparable to the size of the travelling object

[Guest]

for large value of N it is 2 and if N approaches infinity its value is sqrt 2

[Guest]

When N>0 is very large number, distance traveled = 2 units.

When N reaches infinity,

1/infinity = infinitesimally small == 0. so even though it will again travel 2 units to reach (1,1), it will NEVER reach there (distance in x plane = 0; distance in y plane = 0).

So when N is infinity, no distance traveled.

[bonanova]

The path length is 2.

Consider the case in one dimension.

We take a step of length 1/N precisely N times.

The distance traveled is S_N = Sum _i=1^N [1/N] = 1/N + 1/N + ... + 1/N (N times) = N/N = 1.

Letting N increase without bound does not change the distance traveled:

That is, Lim_N->inf S_N = Lim _N->inf N/N = Lim_N->inf 1 = 1.

Now suppose we take two steps of length 1/N precisely N times.

The distance traveled then is S_N = Sum_i=1^N[1/N + 1/N] = 1/N + 1/N + ... + 1/N (2N times) = 2N/N = 2.

Again, Lim_N->inf S_N = Lim _N->inf 2N/N = Lim_N->inf 2 = 2.

Finally, let's suppose we take odd numbered steps in the x direction and even numbered steps in the y direction.

Distance [path length] does not depend on the direction of the path; it remains 2.

[Guest]

Lot's of great thought here to all of you.

When I wrote the question, the answer I had in mind was (and is)

2

The distance traveled does not depend on N.

Although the feature size would be too small to see, there is a greater "density of points" on the zig-zag line we are talking (points per unit displacement) than there are on a perfectly straight path.

In general, if you restrict motions to perpendicular directions, the density of points on your path will be sin(theta) + cos(theta) where 0 <= theta <= pi/4 and theta represents the angle between the direction you are traveling and the nearest axis of motion (or axes if theta = pi/4).

Just using this idea, you can mathematically lengthen your path up by a factor between 1 and sqrt(2) depending on how you choose your axes of restricted motion relative to the direction you wish to travel.

Hard to believe?

It reminds me a bit of fractals:

The Koch Snowflake is a fractal with a finite area, but infinite perimeter.

http://ecademy.agnesscott.edu/~lriddle/ifs/ksnow/ksnow.htm

More interesting in my opinion is the Mandelbrot set:

Nice job everyone, especially bonanova

Edited October 18, 2010 by mmiguel1

[Guest]

seems like square root of 2

[Guest]

even though bonnanova's explanation seems logical why is it that N * sqrt(2/N^2) not the right answer?

[Guest]

Distance along x-axis is number of steps times size of steps = n*1/n; cancels out to 1. No matter the size of n, distance along each axis is 1.

If you insist on looking at the specific case of N-->oo, then:

lim[n->oo](n*1/n)

L'hopital's rule: =lim(1/1)

Therefore total distance = total distance along x-axis + total distance along y-axis = 2.

You an also use the commutative property of vector addition: A+B+C=B+A+C

problem states you move (1/n, 0) then (0, 1/n) over and over again

=(1/n,0)+(0,1/n)+(1/n,0)+(0,1/n)...

By commutative property,

=(1/n,0)+(1/n,0)+.... + (0,1/n)+(0,1/n)+...

Which means it's the same as moving directly along x-axis and then along line x=1.

Distance = 1+1=2.

And of course, beaten to the punch by bonanova. I'll add a little addition, then.

The reason it's not sqrt(2) is that you're never actually moving at a diagonal. You can consider the problem as a series of right triangles, point to point, with a=b=1/n, c=rt(2)/n. You're always moving along the two legs, not the hypotenuse, so it's the sum of all the legs.

Edited October 18, 2010 by j.green

[Guest]

even though bonnanova's explanation seems logical why is it that N * sqrt(2/N^2) not the right answer?

sqrt(2) represents the minimum travelling distance

but think about each individual step.

Each step can be divided into a component lying directly on the desired line of motion between (0,0) and (1,1) and a remaining component perpendicular to this line.

Moving strictly in the +x direction can be thought of as moving with a component on the direction (0,0) -> (1,1) as well as with a component in the direction (0,0) -> (1,-1).

This perpendicular distance is "undone" in the subsequent step as moving in the +y direction can be thought of as having the following components: (0,0) -> (1,1) and (0,0) -> (-1,1), with the second direction listed here cancelling out the perpendicular motion from the last step in the +x direction.

What does it all mean?

In the direction perpendicular to the line from (0,0) to (1,1) we are moving back and forth without any additional progress towards the goal attributable to this motion. This extra motion results in a greater travelling path, than would exist if we moved only on the line. Since this extra motion is perpendicular to the desired motion, it neither results in movement towards or away from the destination, it is just wasted distance traveled.

Does that help?

Edited October 19, 2010 by mmiguel1

[Guest]

When N is infinity

Distance traveled will still be 2 units.

However, it will take "infinity" time to reach (1,1).

_|

_|

_|

Each movement along x+ axis will be 1/N units and after "N" movements, it will be 1 unit.

Each movement along y+ axis will be 1/N units and after "N" movements, it will be 1 unit.

sum of all _ and | will be 2 units. However, you will never live enough to see it reaches (1,1) and most probably it will stay around (0.0000000000000000000001,0.0000000000000000000001) so at any finite time period, the distance traveled will be almost zero.

[Guest]

When N is infinity

Distance traveled will still be 2 units.

However, it will take "infinity" time to reach (1,1).

_|

_|

_|

Each movement along x+ axis will be 1/N units and after "N" movements, it will be 1 unit.

Each movement along y+ axis will be 1/N units and after "N" movements, it will be 1 unit.

sum of all _ and | will be 2 units. However, you will never live enough to see it reaches (1,1) and most probably it will stay around (0.0000000000000000000001,0.0000000000000000000001) so at any finite time period, the distance traveled will be almost zero.

That is not necessarily true, I never said anything about how time depends on N.

If the time it takes to perform a step is independent of N and is also positive and finite, then you are correct.

But if the time it takes to do a step is 1/N seconds, then you will reach your destination in 2 seconds.

All of integral calculus and basically all technological advancements that can be called "modern" (which I don't think would be possible without calculus) rely on the fact that the sum of an infinite number of infinitesimal values can produce a finite result. The dx is the infinitesimal in an integral case, and you are summing an infinite number of these to produce a finite value.

Edited October 20, 2010 by mmiguel1

[Guest]

you are summing an infinite number of these to produce a finite value

I agree with you, Infinite is not bound by time, space or dimensions. Thats why we cannot involve infinite in equations.

However for purpose of understanding, you divide 1 by N (infinite small units) and try to add them all, you will end up getting 1 (may be at end of times). [ 1/N * N = 1, though in pure math sense we shdnt do it]

so basically you will travel 1 unit on x and 1 unit on y (end up 2 units traveled). So let me rephrase my answer. It will cover a distance of UPTO 2 units but to me and you, it will look forever stuck at (0,0).

[bonanova]

even though bonnanova's explanation seems logical why is it that N * sqrt(2/N^2) not the right answer?

Because sqrt(2) is the length of the diagonal path from (0,0) to (1,1) and you never travel any distance along that path.

[Guest]

I agree with you, Infinite is not bound by time, space or dimensions. Thats why we cannot involve infinite in equations.

However for purpose of understanding, you divide 1 by N (infinite small units) and try to add them all, you will end up getting 1 (may be at end of times). [ 1/N * N = 1, though in pure math sense we shdnt do it]

so basically you will travel 1 unit on x and 1 unit on y (end up 2 units traveled). So let me rephrase my answer. It will cover a distance of UPTO 2 units but to me and you, it will look forever stuck at (0,0).

It will cover a distance of UPTO 2 units but to me and you, it will look forever stuck at (0,0).

Well... in my last post, I wasn't talking about distance at all, just time. And actually I was saying it would travel the whole way in just 2 seconds, right before our eyes. This is of course assuming that the time it takes to do a step in the +x or the +y direction is 1/N seconds, which was not stated in the initial problem, but proposed only for the sake of argument in my last post.

Basically I'm saying that you cannot assume anything about how long it will take to get to (1,1) because time was never mentioned in the original post. To demonstrate this, I say, that ok, we all know that the steps are getting smaller as N gets bigger. What if also, the time it takes to do a step also gets smaller by the same amount as the distance? In that case, the object of interest would be moving at a stable average speed.

For example, let's start with N=10

The first step in the +x direction travels 0.1 meters and takes 0.1 seconds.

The +y direction step travels 0.1 meters and takes 0.1 seconds.

The speed of the object here is 0.1 meters / 0.1 seconds = 1 meter/second

Let's make N bigger, like N = 10000

The first step in the +x direction travels 0.0001 meters and takes 0.0001 seconds.

The +y direction step travels 0.0001 meters and takes 0.0001 seconds.

The speed of the object here is 0.0001 meters/0.0001 seconds = 1 meter/second

Notice that the speed did not change, even though N got a thousand times bigger!

In this scenario, the speed is independent of N. If N were to go to infinity, the speed of the dot would still be 1 meter/second.

It was already proved that the distance traveled is 2 meters, so the time taken would be 2 meters / 1 meter/second = 2 seconds

This is just a counter example for your claim that the dot would take an infinite amount of time to move. It was never actually specified how time depends on N, so in fact we can't really say anything about how long it will take.

[Guest]

Well... in my last post, I wasn't talking about distance at all, just time. And actually I was saying it would travel the whole way in just 2 seconds, right before our eyes. This is of course assuming that the time it takes to do a step in the +x or the +y direction is 1/N seconds, which was not stated in the initial problem, but proposed only for the sake of argument in my last post.

Basically I'm saying that you cannot assume anything about how long it will take to get to (1,1) because time was never mentioned in the original post. To demonstrate this, I say, that ok, we all know that the steps are getting smaller as N gets bigger. What if also, the time it takes to do a step also gets smaller by the same amount as the distance? In that case, the object of interest would be moving at a stable average speed.

For example, let's start with N=10

The first step in the +x direction travels 0.1 meters and takes 0.1 seconds.

The +y direction step travels 0.1 meters and takes 0.1 seconds.

The speed of the object here is 0.1 meters / 0.1 seconds = 1 meter/second

Let's make N bigger, like N = 10000

The first step in the +x direction travels 0.0001 meters and takes 0.0001 seconds.

The +y direction step travels 0.0001 meters and takes 0.0001 seconds.

The speed of the object here is 0.0001 meters/0.0001 seconds = 1 meter/second

Notice that the speed did not change, even though N got a thousand times bigger!

In this scenario, the speed is independent of N. If N were to go to infinity, the speed of the dot would still be 1 meter/second.

It was already proved that the distance traveled is 2 meters, so the time taken would be 2 meters / 1 meter/second = 2 seconds

This is just a counter example for your claim that the dot would take an infinite amount of time to move. It was never actually specified how time depends on N, so in fact we can't really say anything about how long it will take.

This reminds me of Zeno's paradox sort of.

Think about though, how many points exist on the number line between 0 and 1?

If you are an object moving on the number line from 0 to 1 in 1 second, how many points do you cross?

An infinite number of points!

But if you think about it, how many time points are there between elapsed time = 0 seconds to elapsed time = 1 second?

Also an infinite number!

As an object moving from x=0 to x=1 in the time t=0 to t=1, you are traversing an infinite number of points in space and an infinite number of points in time.

Although they are both infinite, there are just as many time points traversed as space points.

The ratio in this case is 1.

This is aspect of calculus. The derivative is one infinitesimal value divided by another infinitesimal value.

Even though these values are infathomably small, they still have a finite ratio. The space vs time case above dealt with infinities rather than infinitesimals, but the reciprocal of an infinitesimal is an infinity.

df/dx, the derivative of f with respect to f has infinitesimals df and dx.

but

df/dx = (1/dx) / (1/df)

(1/dx) and (1/df) are infinities.

And these infinities have a finite ratio, namely, the derivative.

So to sum up some stuff,

(1) an infinite sum of infinitesimals can yield finite values

and

(2) ratios of infinities or infinitesimals can yield finite values

(1) Is the principle behind integral calculus

(2) is the principle behind differential calculus

Together, these simple ideas are responsible for the modern age (in my opinion, yeah yeah, i suppose there was some other important stuff too)

Edited October 20, 2010 by mmiguel1

[Guest]

Surely if you are only ever travelling in the x direction or the y direction, no matter how many times you turn from one direction to the other the distance you travel is X + Y. So 2