[unreality]

#0 is like a Practice Problem, the rest are actually a bit harder

(0) Take any number. Now take the square root and add 2. Do this repeatedly. What happens (no matter what you number you start with), and why?

(1) What is the value for n that gives the highest result for the nth root of n?

(2) If you pick numbers randomly from 0 to 1 (inclusive), how many picks would you expect [on average] it would take you for the total sum of the numbers you've drawn to reach or pass 1?

(3) Given the defining property of the Golden Ratio:

p² = p + 1

Prove this formula:

pⁿ = F(n)*p + F(n-1)

and this one:

pⁿ = p^n-1 + p^n-2

Note that p has the quadratic equation:

p² - p - 1 = 0

thus the values are (1+√5)/2 and (1-√5)/2, and both have the properties of p

F(n) = Fibonacci number

Fibonaccis: add the previous two: 1,1,2,3,5,8,13,21,34,55,89,etc

ie, F(7) = 13

also, negative Fibonaccis hold up to the rule that the previous two add to the next one, so a full box might be:

n:       -7  -6   -5   -4  -3   -2   -1   0  1  2  3  4  5   6    7

F(n):   -13   8   -5    3  -2   1    -1   0  1  1  2  3  5   8    13      [/codebox]

(4) extra credit for number 3: Can you create a formula based on p & n which calculates F(n) without using any other F(x) in the equation?

Edited October 14, 2008 by unreality

[andromeda]

Never mind I didn't read it carefully!

Edited October 14, 2008 by andromeda

[Prime]

Question (1) has recently been done here.

Take a derivative of the function y = x^1/x and find where it is equal to zero.

As the derivative is x^1/x*x^-2*(1 - lnx), the only way for it to be zero is lnx = 1 --> x = e.

Therefore, e^1/e is the maximum for that function.

[Guest]

0: If you go to infinity, the result will come out to 4 (or very close to 4). For numbers < 4, the square root is < 2, so adding 2 will increase the next term. However, for numbers > 4, the square root is > 2, so adding 2 will still result in a decrease of the next term. This continues until you arrive at 4, which is stable.

1: n = 3. n has to be > 1, since any root of any number > 1 is > 1. In addition, it has to be a prime number. If you take n^(1/n) and n is equal to, say a*b where a and b are prime numbers with a, b < n; then you get a^(1/n) * b^(1/n). If a is larger than b, then you end up with < a^(2/n), and if n > 2*a, then a^(2/n) < a^(1/a). The only time they are equal is when a are both 2 (powers of 2 evaluate to root 2). So looking at prime numbers, you end up with...

n = 2: 1.414; n = 3: 1.44; n = 5: 1.38; n = 7: 1.32 and on and on, and they all get smaller, so it's n = 3. (However, if this is not all integers, then it becomes more complicated)

... or you could just go with Prime's...

2: I think it's 3, since the average is <0.5 (because 0 is included) and the expected value at each time n is n*(average)

3: Okay, you got me there.

[Guest]

1) 3

2) 2 (for whole numbers, I realize now that e is the correct answer)

3) For the second equation, the p^(n-2) can be factored out to leave p^2=p+1

Edited October 15, 2008 by BigTime

[Guest]

I believe is 2. Both 0 and 1 are included, therefore, on average, 2 numbers picked would add up to exactly 1.

[Guest]

I'm not sure about the Fibonacci sequence, but it's easy to prove that it equals p^n = p^n-1 + p^n-2. Factor out a p^n-2 from each side, and you are left with (p^n-2)(p^2)=(p^n-2)(p^1+1). p^n-2 cancels, and you are left with p^2=p+1

[Prime]

The #0 task appears to be similar to what we've done in Bonanova's infinite exponential, where EventHorizon explained stable equilibrium points.

Suppose the infinite exp​ression ((a1 + 2)^1/2 +2)^1/2 + ... converges to some number x. Then after infinite steps, having got x as a result we add one more step and get the same result (x). Thus we have an equation: (x + 2)^1/2 = x.

Then x + 2 = x². Solving for positive x, we get x = 2. Checking the result: (2 + 2)^1/2 = 2.

[Guest]

Err... if I've done my math right...

Shouldn't the negatives look like this?

n: -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
F(n): 13 -8 5 -3 2 -1 1 0 1 1 2 3 5 8 13

Working backwards from the positives: 2-1=1; 1-1=0; 1-0=1; 0-1=-1. All the signs from there on are flipped.

What? You thought I was actually going to answer the question?

[Prime]

For the question (2) the average number of picks of random numbers to add up to 1 or more, the intuitive answer 2 is wrong.

Think of it this way, the probability to pick exactly 1.0 out infinite number of points is zero for any finite number of attempts. Thus, for the purpose of average, getting 1 with just one pick does not come into play. The minimum number of picks that would count in calculation of average is 2. And then there are cases when it takes more than 2 picks to add up to 1. So the average must be greater than 2.

Here is my reasoning, which may or may not lead to a correct result in this case. (It's a matter of probability).

Once again the formula for the average is

2*p₂ + 3*p₃ + 4*p₄ + 5*p₅ + … where each p_n is the probability to add up to 1 or more exactly after n picks.

The average value for the first pick is 0.5. If the second pick value is between 0.5 and 1.0 – the job is done in two steps. The probability of such second pick is 1/2 (because it has to fall within 1/2 of the total interval). Thus the first term of the infinite series above: 2*(1/2).

The probability for the second pick to fall within 0 to 0.5 interval thus not making total in 2 picks is also 1/2. When it does fall within that interval, on average it cuts it in half. Thereafter to make the total in 3 picks, the third pick must fall within the average of 3/4 of the total length and not to make total -- within 1/4 of the total length. Thus the second term of the series is 3*(1/2 * 3/4).

Likewise, the third term is 4*(1/2 * 1/4 * 7/8), and so on.

The general formula for each term of the infinite series: n*(2^n-1 - 1)/2^n(n-1)/2 where n represents the number of picks.

I am not sure how to solve it at the moment, but it adds up in spreadsheet to just over 2.64.

Edited October 15, 2008 by Prime

[Guest]

n being a whole number, I guess it is n=2

[bonanova]

e.

It would have to be shown by calculation, however.

Simulations give too wide a variance to remove doubt.

[Guest]

#3 can be prooved by recursion

F(n) = F(n-1) + F(n-2) by definition

P^2 = p + 1 = p.F(2) + F(1)

suppose this is true until p^(n-1) = p.F(n-1) + F(n-2)

And let us demonstrate it for P^n:

P^n = p.p^(n-1) = p^2.F(n-1) + p.F(n-2)

substituting F(n-2) by F(n) - F(n-1) in the above equation we get:

p^n = p.F(n) + (p^2-p).F(n-1)

but p^2 = p + 1 thus p^2 - p = 1

and hence we get: p^n = p.F(n) + F(n-1)

Same reasoning applies to the second formula where we suppose that:

p^(n-1) = p^(n-2) + p^(n-3)

and hence p^n = p.p^(n-1) = p^(n-1) + p^(n-2)

[Guest]

For the question (2) the average number of picks of random numbers to add up to 1 or more, the intuitive answer 2 is wrong.

Think of it this way, the probability to pick exactly 1.0 out infinite number of points is zero for any finite number of attempts. Thus, for the purpose of average, getting 1 with just one pick does not come into play. The minimum number of picks that would count in calculation of average is 2. And then there are cases when it takes more than 2 picks to add up to 1. So the average must be greater than 2.

Here is my reasoning, which may or may not lead to a correct result in this case. (It's a matter of probability).

Once again the formula for the average is

2*p₂ + 3*p₃ + 4*p₄ + 5*p₅ + … where each p_n is the probability to add up to 1 or more exactly after n picks.

The average value for the first pick is 0.5. If the second pick value is between 0.5 and 1.0 – the job is done in two steps. The probability of such second pick is 1/2 (because it has to fall within 1/2 of the total interval). Thus the first term of the infinite series above: 2*(1/2).

The probability for the second pick to fall within 0 to 0.5 interval thus not making total in 2 picks is also 1/2. When it does fall within that interval, on average it cuts it in half. Thereafter to make the total in 3 picks, the third pick must fall within the average of 3/4 of the total length and not to make total -- within 1/4 of the total length. Thus the second term of the series is 3*(1/2 * 3/4).

Likewise, the third term is 4*(1/2 * 1/4 * 7/8), and so on.

The general formula for each term of the infinite series: n*(2^n-1 - 1)/2^n(n-1)/2 where n represents the number of picks.

I am not sure how to solve it at the moment, but it adds up in spreadsheet to just over 2.64.

Hmmm...I did a quick simulation in Excel and it came out to 2 almost exactly, so I am not sure if my simulation was done incorrectly or the answer is indeed very close to 2

[Guest]

e.

It would have to be shown by calculation, however.

Simulations give too wide a variance to remove doubt.

Suspicion confirmed.

Obviously, n must be at least 1. The probability that two random numbers do not add up to at least 1 is 1/2. Think of throwing a dart at the unit square. The points for which x1+x2<1 lie below the line passing through (0,1) and (1,0), or half the unit square. For n=3, it becomes the probability of throwing a dart below a line passing through (0,a) and (a,0), where x1, the first number chosen, is 1-a. Again, it is the area of the triangle compared to the unit square, or a^2/2. However, to obtain the total probability, we must integrate for all possible values of x1:

P(3) = \int_0^1{x^2/2 dx} = x^3/(2*3) = 1/6. (sorry, no eqn. editor...)

The probabilities for n=4, 5, etc.... can be found recursively to obtain P(n) = 1/n!

The expected value of n is given by:

E(n) = \sum_1^\infty{n * P(n)} = \sum_1^\infty{n/n!} = \sum_1^\infty{1/(n-1)!} = 1+1+1/2+1/6+1/24 + ...

which is the Taylor series expansion of e^x at x=1.

[Prime]

Hmmm...I did a quick simulation in Excel and it came out to 2 almost exactly, so I am not sure if my simulation was done incorrectly or the answer is indeed very close to 2

I wrote a quick simulation in Excel and got an average result of 2.7283 over just several thousand trials.

Suspicion confirmed.

Obviously, n must be at least 1. The probability that two random numbers do not add up to at least 1 is 1/2. Think of throwing a dart at the unit square. The points for which x1+x2<1 lie below the line passing through (0,1) and (1,0), or half the unit square. For n=3, it becomes the probability of throwing a dart below a line passing through (0,a) and (a,0), where x1, the first number chosen, is 1-a. Again, it is the area of the triangle compared to the unit square, or a^2/2. However, to obtain the total probability, we must integrate for all possible values of x1:

P(3) = \int_0^1{x^2/2 dx} = x^3/(2*3) = 1/6. (sorry, no eqn. editor...)

The probabilities for n=4, 5, etc.... can be found recursively to obtain P(n) = 1/n!

The expected value of n is given by:

E(n) = \sum_1^\infty{n * P(n)} = \sum_1^\infty{n/n!} = \sum_1^\infty{1/(n-1)!} = 1+1+1/2+1/6+1/24 + ...

which is the Taylor series expansion of e^x at x=1.

I didn't get where factorial came from. Also, I cannot agree with your calculation of the probability to make 1 with three picks -- P(3) = 1/6.

Your own estimate of the probability to get 1 from 2 steps is 1/2. So the probability to get 1 on the 3rd try is just 1/3 of the remaining 1/2? That does not make sence. The individual probability for consecutive steps to get 1 should increase. My calculation: after not getting 1 from the first 2 picks the probability to get it n the 3rd pick is 3/4, thus overall probability to get it from 3 picks 1/2 * 3/4 = 3/8.

[Guest]

I wrote a quick simulation in Excel and got an average result of 2.7283 over just several thousand trials.

I didn't get where factorial came from. Also, I cannot agree with your calculation of the probability to make 1 with three picks -- P(3) = 1/6.

Your own estimate of the probability to get 1 from 2 steps is 1/2. So the probability to get 1 on the 3rd try is just 1/3 of the remaining 1/2? That does not make sence. The individual probability for consecutive steps to get 1 should increase. My calculation: after not getting 1 from the first 2 picks the probability to get it n the 3rd pick is 3/4, thus overall probability to get it from 3 picks 1/2 * 3/4 = 3/8.

One way to visualize it for n=3 is to picture the probability as the volume of the unit cube under the plane passing through (0,0,1), (0,1,0), and (1,0,0), which is 1/6. A bit trickier for n=4...

[Prime]

One way to visualize it for n=3 is to picture the probability as the volume of the unit cube under the plane passing through (0,0,1), (0,1,0), and (1,0,0), which is 1/6. A bit trickier for n=4...

How and why would that correspond to the probability in question?

Consider two probabilities:

1). You have just picked 1 random number and it is less than 1. What is the probability that your very next pick will get the total to 1 or over?

2). You have just picked 10 random numbers and their total is still less than 1. What is the probability that your next (11th) pick will take total to 1 or over?

Shouldn't probability (2) be greater than probability (1)? Wouldn't it be very likely that when you haven't made the total after 10 picks, you'd be somewhere very close?

According to your estimates the probability to make total with the second pick (after the first pick didn't make it) is 1/2. Whereas, the probability to make total on the third pick (after previous 2 picks didn't make it) is 1/3. That is the more picks you made, less likely the next pick is to make the total. That just cannot be.

[Prime]

Hmmm...I did a quick simulation in Excel and it came out to 2 almost exactly, so I am not sure if my simulation was done incorrectly or the answer is indeed very close to 2

Here is my simulation program:

Sub randpick()

Randomize

picksum = 0#

totpicks = 0#

For i = 1 To 1000

    picksum = Rnd

    totpicks = totpicks + 1

    Do While picksum < 1#

        picksum = picksum + Rnd

        totpicks = totpicks + 1

    Loop

Next i

ActiveCell.Value = totpicks / 1000

End Sub[/codebox]

[Guest]

According to your estimates the probability to make total with the second pick (after the first pick didn't make it) is 1/2. Whereas, the probability to make total on the third pick (after previous 2 picks didn't make it) is 1/3. That is the more picks you made, less likely the next pick is to make the total. That just cannot be.

Uh... no. 1/6 is the probability of choosing 3 random numbers that do NOT sum to more than 1. So, 2/3 of the time that you choose 2 numbers that still do not add up to 1, you will exceed it on the third.

And, they are not estimates. Those are exact probabilities.

[Guest]

well today I took the PSAT and one of the questions were very similar to this problem

Given the defining property of the Golden Ratio:

p2 = p + 1

Prove this formula:

pn = F(n)*p + F(n-1)

[Guest]

How and why would that correspond to the probability in question?

Pick a point (x,y,z) in the unit cube. If x+y+z<1, the point you have picked lies between the origin, and the plane passing through (1,0,0), (0,1,0), and (0,0,1). If the probability distribution is uniform, then the probability of picking a point in the region where x+y+z<1 is the volume of that region.

[unreality]

(0) is 4

(1) and (2) are both e

I haven't seen much work on (3) or (4) though... most of the people that did (3) just did the pⁿ =

p^n-1 +p^n-2 proof, not the pⁿ = F(n)*p + F(n-1)

anyway, good job on #2, that one's a toughie

[Guest]

Pick a point (x,y,z) in the unit cube. If x+y+z<1, the point you have picked lies between the origin, and the plane passing through (1,0,0), (0,1,0), and (0,0,1). If the probability distribution is uniform, then the probability of picking a point in the region where x+y+z<1 is the volume of that region.

It's easiest to start by thinking of the 2D case. If x+y<1, then you can represent that as a point in the unit square below the line y=-x+1, which is just a diagonal going across the square. If the probability distribution is uniform, then the probability of picking a point in the region where x+y<1 is the area of that region.

From there, it's pretty simple to generalize to the 3D case as d3k3 did. As he said, though, it gets harder for 4 and above.

[unreality]

I'm adding two problems These ones are as old as Archimedes

(5) Two cylinders of unit radius intersect at 90° ... what is the volume common to both cylinders?

You can use calculus OR basic geometry

(6) Much harder, what is the volume common to THREE intersection cylinders? (all at 90°). You'll probably need calculus for this one...

[Prof. Templeton]

I'm adding two problems These ones are as old as Archimedes

(5) Two cylinders of unit radius intersect at 90° ... what is the volume common to both cylinders?

You can use calculus OR basic geometry

Since Archimedes pre-dates calculus

Area of a circle A= pi * r²

and

Volume of a sphere V= (4/3) * pi * r³