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On a horse with no name?
A very short quiz
- Thread starter noway2
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I messed up #2 in my head because I got the 0.2 minutes right, but then I treated that as 20 sec (doofus move). If I had just thought of it as 1/5 I would have done better.
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OP
WRT the third one, this will probably flake out more people out as it's an obscure section of mathematics, but I'll try to put it simply.
In this example, we're talking about the number of lily pads at a point in time. Furthermore were modeling the situation such that time occurs in discrete intervals of number of days. At a particular point in time, or day, there are X lily pads and the time is measured in integer values. In other words, it's 1, 2, 3, etc not 1.5.
So at one point in time, let's call the days or time 'k" for simplicity we have X lily pads. At the next time, advancing one step in time or k+1 we have twice as many lily pads, or in other words we have 2X lily pads as we did at time k.
This lets us write the difference equation, y(k+1) = 2*y(k) where y is the number of lily pads (I changed to y from X because y is typically the dependent variable in algebra.
Now here's the trick. Going from time k to k+1 is an advance in time. Let's define this operation as an "advance" and call it the letter p. Therefore to advance in time, we operate on the function or multiply it by p. So now, p*y(k) = 2*y(k). In this particular system, there is no "forcing" function that impacts the number of lily pads and instead it's purely a function of the previous value. Therefore we only have what is called a complementary solution that can be found by solving the homogenous equation or p*y(k) - 2*y(k) = 0. Notice all we did is an algebraic subtraction. We can rewrites this by factoring out the y(k) as y(k)(p-2)=0. This equation had two solutions, either y(k) is zero or p=2. Either term would make the expression true, equaling zero). The solution of zero lily pads, y(k) =0 is called the trivial solution, therefore the solution we want is p=2. Notice here that that we've done is treat the time advance operation as an algebraic entity, which as Oliver Heavyside proved a long time ago, is a valid process.
Given the roots of the characteristic (homogenous equation) we know the solution takes the form of (root)^k or in this case 2^k or 2 raised to the k power. The actual generic solution is y(k) = A*2^k where A is a multiplicative constant determined by the initial conditions or in this case we know that at day 48 there is 100 percent coverage, let's call this 1.0. So now we can solve for A as saying y(48), which is 1 = A*2^48 or A = 1/(2^48). With this solution we can find the percent coverage at any day k.
There is a simpler way to look at it. Let's say that at day 0, the initial condition,there is 1 lily pad. Therefore 1 lily pad = A* 2^0. Since 2 raised to the zero power is 1 (anything to the zero power is identically one) 1=A*1 or A=1. Therefore the number of lily pads is y(k) =1*2^k.
Now we have the solution for all time, k or in table form
Time : lily pads
0 : 1
1 : 2
2 : 4
3 : 8
And so on. Notice that at each preceding time period we have half the number of lily pads. Therefore if at time k=48 we have 100% coverage we will half half the coverage one delay prior or at t=47.
The numbers work out for the other value of A and you will get .5 for k=47 or half coverage, but the constant is messy.
So there is you discrete time calculus lesson for today. This is what popped into my head on the way home yesterday.
Next time, we will solve this system with a forcing function that each day frog turds with seeds cause there to be an addition of an extra lily pad or y(k+1) = 2*y(k) + 1 which will have both a complementary solution and a particular solution with the general solution being the sum of the two.
In this example, we're talking about the number of lily pads at a point in time. Furthermore were modeling the situation such that time occurs in discrete intervals of number of days. At a particular point in time, or day, there are X lily pads and the time is measured in integer values. In other words, it's 1, 2, 3, etc not 1.5.
So at one point in time, let's call the days or time 'k" for simplicity we have X lily pads. At the next time, advancing one step in time or k+1 we have twice as many lily pads, or in other words we have 2X lily pads as we did at time k.
This lets us write the difference equation, y(k+1) = 2*y(k) where y is the number of lily pads (I changed to y from X because y is typically the dependent variable in algebra.
Now here's the trick. Going from time k to k+1 is an advance in time. Let's define this operation as an "advance" and call it the letter p. Therefore to advance in time, we operate on the function or multiply it by p. So now, p*y(k) = 2*y(k). In this particular system, there is no "forcing" function that impacts the number of lily pads and instead it's purely a function of the previous value. Therefore we only have what is called a complementary solution that can be found by solving the homogenous equation or p*y(k) - 2*y(k) = 0. Notice all we did is an algebraic subtraction. We can rewrites this by factoring out the y(k) as y(k)(p-2)=0. This equation had two solutions, either y(k) is zero or p=2. Either term would make the expression true, equaling zero). The solution of zero lily pads, y(k) =0 is called the trivial solution, therefore the solution we want is p=2. Notice here that that we've done is treat the time advance operation as an algebraic entity, which as Oliver Heavyside proved a long time ago, is a valid process.
Given the roots of the characteristic (homogenous equation) we know the solution takes the form of (root)^k or in this case 2^k or 2 raised to the k power. The actual generic solution is y(k) = A*2^k where A is a multiplicative constant determined by the initial conditions or in this case we know that at day 48 there is 100 percent coverage, let's call this 1.0. So now we can solve for A as saying y(48), which is 1 = A*2^48 or A = 1/(2^48). With this solution we can find the percent coverage at any day k.
There is a simpler way to look at it. Let's say that at day 0, the initial condition,there is 1 lily pad. Therefore 1 lily pad = A* 2^0. Since 2 raised to the zero power is 1 (anything to the zero power is identically one) 1=A*1 or A=1. Therefore the number of lily pads is y(k) =1*2^k.
Now we have the solution for all time, k or in table form
Time : lily pads
0 : 1
1 : 2
2 : 4
3 : 8
And so on. Notice that at each preceding time period we have half the number of lily pads. Therefore if at time k=48 we have 100% coverage we will half half the coverage one delay prior or at t=47.
The numbers work out for the other value of A and you will get .5 for k=47 or half coverage, but the constant is messy.
So there is you discrete time calculus lesson for today. This is what popped into my head on the way home yesterday.
Next time, we will solve this system with a forcing function that each day frog turds with seeds cause there to be an addition of an extra lily pad or y(k+1) = 2*y(k) + 1 which will have both a complementary solution and a particular solution with the general solution being the sum of the two.
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BigWaylon
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Mine both got 1/3...3/3
Gonna test the wife and son (10) tomorrow. Pretty good chance he scores higher.. Math is not her strong suit...
Both said $.10 for #1
Both got #2 correct
He said 23 and she said 24 for #3
Jerk... You made me spit my coffee this morn.
We've had an interesting time trying to hire people from Oxford, your wit indicates you possibly moved there.
Heh. Ok, since I caused you to waste some nectar of the gods, I'll confess that I do indeed now have an Oxford address. Only been here 4 months now, but I've gotten a handle on sitting out on the front porch and counting cars go by.
OP
I said the quiz was short, not an explaination of the solution. The sad / scary thing is that the 3rd problem kept bugging me all day and then it hit me on the way home as far as the theory behind it. I've always found signal processing and the math behind it to be fascinating. It was a subject that gave me a lot of trouble in school and I didn't really get it until years later when I started studying it on my own.
I used to have an office with a grease board and would write simple equations or circuits and solve them on it. My coworkers would look at it and call me an a hole.
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But the obvious answer to #3 is "the day before 48, when there is half as much coverage, one.doubling earlier". That isn't even a trick question. If you found a theory behind that, more power to you!
The solutions are all short too :-/
It's an exercise in ignoring the obvious but wrong answer to figure out the simple correct answer, although I know exponentials from binary and computers so well that 24 is a nonsense answer and didn't even register. So is 47 though. 2^48 of anything is bigger than a pond! Or rather 2^-48 of the pond on the first (zeroth really) day is one tiny lily pad.
OP
No, they're not trick questions. They should be intuitively obvious, but that's the thing, they're designed such that the intuitively obvious answer is wrong. What's fascinating is that such a small percentage of the population gets them right, I had a suspicion, which I stated in the first post, that the results would be better here. If the poll is to be believed this group does better than the MIT students who were an outlier.
Same here. Working with computers for decades made it intuitive to throw out the 24. It should have been a no brainer but it wasn't either, which is curious. Intuitively I recognized it as an exponential sequence, but sonething in the working of it kept bugging me, until it finally hit me while driving. While I calculated the answer with logs, there was still something about it driving me nuts. I've given the quiz to others and they've said the ball and bat one continues to irritate them. What's fascinating about that question is the study said that if they used similar questions people got them correct a lot more and there is sonething about the $1 and 10 cents that triggers and immediate response.
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All three. No real math needed. Like someone said above, mostly intuition.
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I really like it & there's great hunting at local farms if you get a few good 'ole boy connections.
I've a few customers in the area, and will be there next week trying to get some more.
