Annoying maths puzzle.
- Thread starter Redingold
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Here's a math problem for you:
Say you have an infinite plane which is moving along a single axis with a speed given by the equation u=U*cos(wt) where U and w are constants. What is the velocity distribution of the semi-infinite fluid above this plane? Assume constant properites. I'll post the solution later if anybody is interested.
Say you have an infinite plane which is moving along a single axis with a speed given by the equation u=U*cos(wt) where U and w are constants. What is the velocity distribution of the semi-infinite fluid above this plane? Assume constant properites. I'll post the solution later if anybody is interested.
aah, didnt think of adding a curved lineRedingold said:
clever
this one's slightly simplerRedingold said:
you just add a diagonal line to the plus to change the + to a 4
then you get IX4II = 411
or 1 x 411 = 411
they annoy me so much though
Ya, I get it now. the thing's supposed to be in parametric terms. Gotcha.Kubanator said:
My bad. But then again, I think the problem is not exactly written right. For example, I didn't exactly understand the term step, and I immediately took it as meaning that you could move in another axis at any given moment, even though we were supposed to be moving in polar coordinates only(rotations with respect to the Z axis).
But I still think that believing you can answer the question only in terms of rotations is wrong. You need to specify where you go with those steps, don't you?
Yeah, and it will get annoying with one step, then half a step, and then a quarter of a step. but then again, I guess if there was just one step, then that's it, I believe (cosKubanator said:
-1,sin), but then again, you have to account for those extra bitty steps, ugh...
Bleh...ILPPendant said:
Tired. Have not slept for 28 hours. Will put out a correct solution tomorrow.
The quick way of doing it is to find the sum to N and then take the limit as N tends to infinity.Gruthar said:
Naturally this only works if a sequence is convergent. Which I believe those are.
I think the answer is ((4 - 2 cos)/(5-4 cos), 2 sin/(5 - 4 cos)
Here's how you arrive at this answer: I'll write the point (x,y) in the plane as the complex number x + i y. We'll use Euler's formula, e^(it) = cos(t) + i sin(t).
The distance you travel in the kth step is (cos(kn) + i sin(kn))/2^k = w^k, where w = e^(in)/2. So after k steps, your position is 1 + w + w^2 + ... + w^k. So your limiting position is the infinite sum 1 + w + w^2 + w^3 + .... This is a geometric series whose sum is 1/(1-w), which simplifies to the above answer.
(My first post on The Escapist!)
)/(5-4 cos), 2 sin/(5 - 4 cos)Here's how you arrive at this answer: I'll write the point (x,y) in the plane as the complex number x + i y. We'll use Euler's formula, e^(it) = cos(t) + i sin(t).
The distance you travel in the kth step is (cos(kn) + i sin(kn))/2^k = w^k, where w = e^(in)/2. So after k steps, your position is 1 + w + w^2 + ... + w^k. So your limiting position is the infinite sum 1 + w + w^2 + w^3 + .... This is a geometric series whose sum is 1/(1-w), which simplifies to the above answer.
(My first post on The Escapist!)
I like this.Chief Oppupu said:I think the answer is ((4 - 2 cos
Here's how you arrive at this answer: I'll write the point (x,y) in the plane as the complex number x + i y. We'll use Euler's formula, e^(it) = cos(t) + i sin(t).
The distance you travel in the kth step is (cos(kn) + i sin(kn))/2^k = w^k, where w = e^(in)/2. So after k steps, your position is 1 + w + w^2 + ... + w^k. So your limiting position is the infinite sum 1 + w + w^2 + w^3 + .... This is a geometric series whose sum is 1/(1-w), which simplifies to the above answer.
(My first post on The Escapist!)
Since you're new you probably don't realise this but you can use {sup}{/sup} to put things in superscript rather than filling the lines with circumflexes. It just looks a little neater.
Well, that's my starting point, so I can only explain how I came up with those equations. To clarify, n is the degree constant, and x is the nth term. 2^-x is the distance traveled at each iteration, and xn is the angle of travel at each iteration. To break those up into Cartesian coordinates, I just used the usual distance*cos(theta), distance*sin(theta)ILPPendant said:
form. The final point is just a sum of all the terms.
To double check my equations, I plugged in the first few terms and compared them to what I worked out doing the first few terms manually. Initially, x = 0, so cos(0)/(2^0) = 1 and sin(0)/(2^0) = 0. So the first point is (1,0).
Next, x = 1, so cos
/(2^1) = cos/2 and sin/(2^1) = sin/2Being a sum, the next point is 1 + cos
/2, sin/2At x = 2, you get cos(2n)/(2^2) and sin(2n)/(2^2)
for the point 1 + cos
/2 + cos(2n)/4, sin/2 + sin(2n)/4And so on. I think I have an answer, but it looks nasty. Will post in a few.