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## Question

#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 = pn-1 + pn-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 by unreality

## Recommended Posts

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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.

Still, I don't understand the cube analogy. Or the formula for the probabilities for the consecutive steps.

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Since Archimedes pre-dates calculus
Area of a circle A= pi * r2

and

Volume of a sphere V= (4/3) * pi * r3

so did you solve it?

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so did you solve it?

I know the answer Archimedes gave. And there are ways to arrive at his answer that require some assumptions. I have never come across a non-calculus proof however, as it would probably be very lengthy. For your "tricylinder" intersection volume, I have seen a non-calculus determination for it, but it relies on the formula for the "bicylinder".

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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.

1. Proof that every point lying on the surface that passes through (1,0,0), (0,1,0), and (0,0,1) of the unit-cube has its xyz coordinates adding up exactly to 1. (That shouldn't be all that difficult.)

2. Even more interesting, how do you derive that series consisting of number of trials times their respective probabilities is represented by Taylors's polynomial?

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

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

Excuse me but did u see my post #13?!

I gave a complete answer to (3)

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Excuse me but did u see my post #13?!

I gave a complete answer to (3)

The induction proof looks simple and sound.

Here is the same, slightly re-shuffled, which may or may not make it more clear.

Check that the rule works for n=2:

p2 = p*F2 + F1

Since both F1 and F2 are equal to 1, we get:

p2 = p + 1, which is true by definition of p.

Now suppose, the rule holds for some number n:

pn = pFn + Fn-1

Then multiplying both sides by p, we have:

p*pn = p2Fn + p*Fn-1

Substituting p2 in the right side by p+1 (in accordance with "golden rule"):

p*pn = (p+1)*Fn + p*Fn-1

pn+1 = Fn + p*Fn + p*Fn-1

pn+1 = Fn + p*(Fn + Fn-1)

Noting that Fn + Fn-1 is Fn+1 by definition of Fibonachi series, we have:

pn+1 = p*Fn+1 + Fn

And so the rule holds for n+1

QED

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1. Proof that every point lying on the surface that passes through (1,0,0), (0,1,0), and (0,0,1) of the unit-cube has its xyz coordinates adding up exactly to 1. (That shouldn't be all that difficult.)

2. Even more interesting, how do you derive that series consisting of number of trials times their respective probabilities is represented by Taylors's polynomial?

I think Prime forgot her coffee this morning... XP

1. x+y+z=1 is the equation of the plane passing through (1,0,0), (0,1,0), and (0,0,1).

2. Derivation? The Taylor series expansion of e^x is Sum{x^n/n!} which is simply Sum{1/n!} when x=1. That "happens to be" the series I obtained for the expectation.

I say "happens to be" in quotes, because there is something that bothers me about this solution. How many times have you seen a solution of "e" or "e^x" by sheer coincidence? This makes me wonder, is the expectation also e^x for values of x other than 1? It appears to be true for 0<x<1, which in turn suggests there ought to be a simpler solution to this problem that does not involve the Taylor series. Something along the lines of:

E(x) = a + \int_0^x{E(t)dt}

where E is the expected number of random numbers that can be drawn before their sum > x (or t). I played around a bit, but the closest I got was E(x) = 1 + \int_0^x{E(x-t)dt} (where t is the first number drawn), but that doesn't quite work. Maybe someone with better math skills wants to have a go at it...

D.

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I think Prime forgot her coffee this morning... XP

1. x+y+z=1 is the equation of the plane passing through (1,0,0), (0,1,0), and (0,0,1).

2. Derivation? The Taylor series expansion of e^x is Sum{x^n/n!} which is simply Sum{1/n!} when x=1. That "happens to be" the series I obtained for the expectation.

I say "happens to be" in quotes, because there is something that bothers me about this solution. How many times have you seen a solution of "e" or "e^x" by sheer coincidence? This makes me wonder, is the expectation also e^x for values of x other than 1? It appears to be true for 0<x<1, which in turn suggests there ought to be a simpler solution to this problem that does not involve the Taylor series. Something along the lines of:

E(x) = a + \int_0^x{E(t)dt}

where E is the expected number of random numbers that can be drawn before their sum > x (or t). I played around a bit, but the closest I got was E(x) = 1 + \int_0^x{E(x-t)dt} (where t is the first number drawn), but that doesn't quite work. Maybe someone with better math skills wants to have a go at it...

D.

I take it, you cannot solve those problems.

1. Whatever math book you are looking at to get your answers, may not show the proof/derivation of "x+y+z=1 is the equation of the plane passing through (1,0,0), (0,1,0), and (0,0,1)", because it is an 8-th grade school problem.

I'm just curious, can you manage to solve it on your own?

2. I didn't ask you why Taylor polynomial approximates e. I asked, how do you see that Taylor polynomial describes the average we are to find in this problem?

I am beginnig to suspect, that you are borrowing "your solutions" from some reference, while not understanding clearly the meaning thereof.

You also show very poor deductive skills by refering to Prime as "her".

If you feel the 2 questions I posted here are below your station, I invite you to solve my restatement of another Unreality's probability topics.

Since I constructed that problem on my own, there is a possibility that it can't be googled up, or found in a book. I think the first of the questions there is relatively easy.

Edited by Prime
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I take it, you cannot solve those problems.

1. Whatever math book you are looking at to get your answers, may not show the proof/derivation of "x+y+z=1 is the equation of the plane passing through (1,0,0), (0,1,0), and (0,0,1)", because it is an 8-th grade school problem.

I'm just curious, can you manage to solve it on your own?

2. I didn't ask you why Taylor polynomial approximates e. I asked, how do you see that Taylor polynomial describes the average we are to find in this problem?

I am beginnig to suspect, that you are borrowing "your solutions" from some reference, while not understanding clearly the meaning thereof.

You also show very poor deductive skills by refering to Prime as "her".

If you feel the 2 questions I posted here are below your station, I invite you to solve my restatement of another Unreality's probability topics.

Since I constructed that problem on my own, there is a possibility that it can't be googled up, or found in a book. I think the first of the questions there is relatively easy.

Furrfu.

For your future reference, I do not waste my time responding to childish and unprovoked personal attacks.

Good day.

D.

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1. Whatever math book you are looking at to get your answers, may not show the proof/derivation of "x+y+z=1 is the equation of the plane passing through (1,0,0), (0,1,0), and (0,0,1)", because it is an 8-th grade school problem.

I'm just curious, can you manage to solve it on your own?

A plane is uniquely determined by 3 points.

(1,0,0) is a solution to x+y+z=1

(0,1,0) is a solution to x+y+z=1

(0,0,1) is a solution to x+y+z=1

Therefore, the equation x+y+z=1 describes the plane passing through (1,0,0), (0,1,0), and (0,0,1)

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A plane is uniquely determined by 3 points.

(1,0,0) is a solution to x+y+z=1

(0,1,0) is a solution to x+y+z=1

(0,0,1) is a solution to x+y+z=1

Therefore, the equation x+y+z=1 describes the plane passing through (1,0,0), (0,1,0), and (0,0,1)

That's too simple, although hard to argue with. This proof involves guessing the equation. Consider your 2-d analogy where the points are described by y = 1 - x function. If I gave you a pair of points (x1, y1) and (x2, y2), then you'd have to go on solving a system of equations to tell me what is the equation for the straight line passing through those two points. For example, given (.2, .8) and (.5, .5), after solving it for a straight line, you'd also need to notice that line's funny property of x+y=1.

"A plaine is uniquely determined by 3 points" is one essential part of it. Another, that an equation of type x+y+z=a describes a straight plane. There is also a somewhat more involved geometrical proof.

But that aside, I don't see the problem (2) the average number of random picks as being solved here. There was a hypothesis that it can be expressed as Taylor series approximating e, but no reasoning to support that.

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Gentlemen, let's please keep it civil

mojail: If I read it correctly, your proof is based on this:

""suppose this is true until""

You just 'supposed' something was true, and then didn't go back and back it up later (again, if I read correctly, sorry if I'm mistaken)

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Gentlemen, let's please keep it civil

mojail: If I read it correctly, your proof is based on this:

""suppose this is true until""

You just 'supposed' something was true, and then didn't go back and back it up later (again, if I read correctly, sorry if I'm mistaken)

1). We are civil. That's just our brand of civility.

2). Mojail's proof is called a method of mathematical induction proof.

Mojail first demonstrated that the theorem holds for one specific case.

Then made a supposition that the theorem holds for some case n-1. That supposition is valid, because of the preceding demonstration.

Then proved that the theorem also true for n. Since n-1 is an arbitrary number, the proof is valid.

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Gentlemen, let's please keep it civil

mojail: If I read it correctly, your proof is based on this:

""suppose this is true until""

You just 'supposed' something was true, and then didn't go back and back it up later (again, if I read correctly, sorry if I'm mistaken)

Hi unreality,

First, plz let me apologize if i was rude in my last message, didn't mean it to sound like this,

Second, you are right, i started my proof with "suppose this is true until", but this is a well known method of proving properties related to mathematical series...

In fact, when u need to prove a formula related to a mathematical series, u start by verifying that it works for the first order, then u suppose it is true until a certain order (n-1) and then u use that to verify that it holds for the order n...

Maybe i am not explaining it well, but i am confident that my proof is strong and its not just some suppositions!

Any other opinions on this matter?

Again i am sorry if i was blunt in my previous post

Cheers...

Edit: Prime i just read ur reply, thank you for making things clear

Edited by mojail
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Yes but for the second:

You "supposed" this is true:

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

Then obviously it's going to be true for p^n = p^(n-1) + p^(n-2), just multiply both sides by p. But you gave no reasoning for the original supposition. It's like saying:

pdog = pdog-1 + pdog-2

That works, so plug 'n' in for 'dog' and voila!

I just don't see the logic, sorry

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for the first proof:

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

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

Great so far!

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

Okay, I see this and assume you will later prove it

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)

You didn't prove it. You used backwards reasoning. Clearly if this is true:

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

Then this is true:

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

But the fact is, you can't get this:

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

From the special case of:

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

Directly, at least. F(2) and F(1) are both 1, so you could use those in everything

for example

g(x) = x2

g(2) = 1 * 22

g(2) = F(2) * 22

"suppose this is true until n"

g(n) = F(n) * n2

Which is totally not true

edit: I hope I'm not coming across as mean!! I'm sure you know more math than me, I'm just having a hard time understanding this concept ;D

Edited by unreality
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Yes but for the second:

You "supposed" this is true:

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

Then obviously it's going to be true for p^n = p^(n-1) + p^(n-2), just multiply both sides by p. But you gave no reasoning for the original supposition. It's like saying:

I just don't see the logic, sorry

You know what? I am sorry I really don't have neither the time nor the will to explain more, but it is very logical!

And u are asking us to be civil and then your replies aren't as civil as you are requesting!!!

So I am going to drop this topic...

Good luck figuring out the logic!

Edit: I suggest you go back to your high school courses for clarifications!

Edited by mojail
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Whoa, did you see my edit?

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Whoa, did you see my edit?

well yes i saw u edit!

OK i am going to explain it rapidly once more...

When u prove that the property is good for the first order, then if u can prove that it is true for the order n by supposing it is for the order (n-1) then u have proved that the property works for all the orders!

I am sorry maybe i am not being clear, but this method of induction is really well known!

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Um sorry i just saw ur last edit!

Again i know i got carried away, but ur post was a little provocative to me

no hard feelings

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for example

g(x) = x2

g(2) = 1 * 22

g(2) = F(2) * 22

"suppose this is true until n"

g(n) = F(n) * n2

Which is totally not true

yes u are right, and it is not true because u can not prove that it works for order (n+1)

In other terms, u can suppose that: g(n) = F(n) * n2

but can u prove from this equation that: g(n+1) = F(n+1) * (n+1)2 ?

In this case no, this is why the supposition is not true...

I hope this would clarify things up

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Ohhhh I see. Thanks! Wow I feel like an idiot now! Sorry for doubting you

Edited by unreality
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...

Okay, I see this and assume you will later prove it

...

There is no need for latter proof. It was already demonstrated by example. This is how mathematical induction proof works.

1). Demonstate, the rule works for some specific case using specific numbers. (That was done here with n=2)

2). Assume there exists such n for which the rule holds. (Perfectly legal assumption in view of the preceding demonstration, that already proved that assumption.)

3). Show mathematically that assuming the rule is true for some n, it will also hold for n+1. (And that's the end of it.)

Here is a popular example of the inductive proof:

Prove that all men are bald

1). Suppose a guy had just one hair. Is he bald? Of course, he is!

2). Regard a guy who has n hair, while still being bald.

3). Now say, that guy grows just one additional hair and has n+1 hairs. Is he no longer bald? No way! One more hair won't do it.

QED

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Yeah I understand it now Thanks guys

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Ohhhh I see. Thanks! Wow I feel like an idiot now! Sorry for doubting you

no problem glad everything is clear now

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