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April 28[edit]

Differential Equation[edit]

I attempted to work out the position, velocity, and acceleration as functions of time for two electrons repelling themselves, but I got stuck when it boiled down to z/x(t)^2=a(t), where z is a constant, x(t) is position, and a(t) is acceleration of x(t). How would this be solved? or are there any resources that I could use? 72.197.202.36 (talk) 00:54, 28 April 2008 (UTC)[reply]

OK...

x''=zx^{-2}

x''x'=z{\dfrac {x'}{x^{2}}}.

Integrating,

{\dfrac {1}{2}}(x')^{2}={\dfrac {-z}{2x}}

x'={\sqrt {\dfrac {-z}{x}}}

Which looks much more solvable. -mattbuck (Talk) 02:30, 28 April 2008 (UTC)[reply]

Set z=1 for simplification. That amounts to chosing a unit of time. Note that the energy, kinetic + potential,

E={\dfrac {1}{2}}(x')^{2}+x^{-1}

is a constant of motion

E'=x'x''-x^{-2}x'=x'(x''-x^{-2})=0\,

Then you are left with a differential equation of the first order

{\frac {dx}{dt}}=x'={\sqrt {2(E-x^{-1})}}\,

Separate variables and integrate

t=f(x)=\int {\frac {dx}{\sqrt {2(E-x^{-1})}}}\,

Solve for x

x=f^{-1}(t)\,

Bo Jacoby (talk) 06:35, 28 April 2008 (UTC).[reply]

Interesting. mattbuck's answer misses a constant of integration but results in a dirt-simple solution that you might guess if you simply stare at the original equation for long enough. Bo's answer leads to a truly horrible, ugly solution set that depends very strongly on the constants of integration, but I guess must be the full general solution. Ouch. --Prestidigitator (talk) 18:35, 28 April 2008 (UTC)[reply]

Note that z>0 for repulsive forces. Mattbuck's case, E=0, leads to

t=\int {\frac {dx}{\sqrt {-2x^{-1}}}}={\frac {\sqrt {2}}{3i}}{\sqrt {x^{3}}}+t_{0}\,

which is mathematically correct but has no physical interpretation because the time coordinate cannot be imaginary. So the energy must be positive.

Using the distance at which the velocity is zero as unit of length

x=E^{-1}\cdot y

you get the constant outside the integral

t=\int {\frac {dx}{\sqrt {2(E-x^{-1})}}}={\frac {1}{\sqrt {2}}}\int {\frac {d(E^{-1}\cdot y)}{\sqrt {E-(E^{-1}\cdot y)^{-1}}}}={\frac {1}{\sqrt {2E^{3}}}}\int {\frac {dy}{\sqrt {1-y^{-1}}}}\,

Bo Jacoby (talk) 10:27, 29 April 2008 (UTC).[reply]

Sequences[edit]

I can't find the link between these sequences :(

1. 20, 18, 16, 14, .., 14, 16, 18, 20. I don't get it, is it 12?

2. 2, 3, 5, 6, .., 9, 11, 12, 14. Does the sequence go + 1 + 2? If it does, the missing numbr is 8, isn't it?

3. 2, 3, 5, .., 17, 33, 65. In this one do you double it and subtract 1? If so its 9 isnt it?

Thanks guys --Hadseys 10:32, 28 April 2008 (UTC)[reply]

One good way to deal with this type of sequence might be to compute differences, differences of differences, ..., until you spot a pattern.

For example, for the first

20, 18, 16, 14, .., 14, 16, 18, 20

-2 -2 -2 ? ? 2 2 2

So you could expect the sequence of differences to be -2 -2 -2 -2 2 2 2 2, and then the missing number is 12.

In such a simple case it might be easier to spot the pattern directly though.

In the second case, the differences are indeed 1 and 2 alternating, so the missing number should be 8.

In the third case, the differences are 1, 2, ?, ?, 16, 32. You might recognize the powers of two here, and in this case adding the powers of two is the same as doubling and subtracting one. So yes, 9 apparently. -- Xedi (talk) 11:38, 28 April 2008 (UTC)[reply]

A good way to judge whether a rule you find for a sequence makes sense is to compare its complexity with the amount of data you have. The more data, the more complicated you can afford the expression to be. For example, the given entries for sequence 3 can be perfectly fitted by

a_{n}={\frac {3n^{5}-37n^{4}+217n^{3}-527n^{2}+764n+60}{240}}

, which would make the missing entry 9.05. However, the complexity of this rule is not justified by the relatively little data we have. The rule

a_{n}=2^{n-1}+1

is remarkably simpler and thus a much better candidate. Similarly,

a_{n}=12+2|n-5|\;\!

and

a_{n}={\frac {6n+1-(-1)^{n}}{4}}

are adequate for expressing the first and second sequences.

Some related reading is Occam's razor and overfitting. -- Meni Rosenfeld (talk) 12:21, 28 April 2008 (UTC)[reply]

How did you come up with those latter two? Did you just see it, or was there a method you used? GromXXVII (talk) 22:43, 28 April 2008 (UTC)[reply]

For the first sequence - I know that it is symmetric around 5 (hence

|n-5|

), increases by 2 for every step (hence the 2) and starts at 12 (hence the 12). For the second - I know it increases linearly with additive fluctuations due to parity, hence it is a linear combination of 1, n and

(-1)^{n}\;\!

; the rest of the work was done by Mathematica. -- Meni Rosenfeld (talk) 23:24, 28 April 2008 (UTC)[reply]

Water tank[edit]

I need help with the following questions:

A certain tank holds 20 gallons of water. When a tap over it is turned on water flows from it into the tank at the rate of 4 gallons a minute. When a plug in the bottom of the tank is pulled out water drains away from the tank at the rate of 6 gallons a minute.

1. With the tank empty and the plugin, the tap is turned on. How many minutes will it take to fill the tank?

I've worked the answer out to be 5 minutes because 20 ÷ 4 = 5 minutes

2. With the tank full, the tap is turned on and the plugged pulled out. How many minutes will it take to empty the tank?

I dont know the method to work this out

Well, every minute that goes by, the amount is increased by 4 and decreased by 6, so in essence, the net loss is 2 gallons every minute. So if it's full, it should take 10 minutes to empty. Ziggy Saw dust 15:24, 29 April 2008 (UTC)[reply]

3. With the tank empty and the plug in, the tap is turned on for 3 minutes. It is then turned off and the plug pulled out. How many minutes will it take to empty the tank?

I worked this out to be 2 because at a rate of 4 galons a minute, 4 × 3 = 12 gallons in. At an expulsion rate of 6 gallons a minute, the answer is 12 ÷ 6 = 2 minutes.

4. With the tank full, the plug is pulled out and the tap turned on. After 4 minutes the tap is turned off with the plug left out. How many more minutes will it take to empty the tank.

The tap is on for 4 minutes so 4 × 4 = 16 gallons put in. So 16 ÷ 6 is how long it would take to drain the take. Am i right?

Well, with tap on + plug out it loses 2 gallons per minute, so in 4 minutes it would lose 8 gallons, and after the tap was turned off it would lose 6 per minute, and 12/2=6, so... 6 minutes. You went about it a pretty bizarre way, but you got the right answer. Ziggy Saw dust 15:24, 29 April 2008 (UTC)[reply]

5. The tank is full. The plug is pulled out for 2 minutes. The tap is then turned on for 3 minutes with the plug still out. How many gallons will be in the tank at the end of this time.

If the plug is pulled out for 2 minutes, the tank has lost 12 gallons. If it is switched on for 3 minutes, it will have gained the 12 gallons, so would the answer be 20 gallons?

You're right on the first part, but when the tap is turned on, the plug is still out, so actually it will be losing 2 gallons per minute, so 3 mins = 6. 12-6=6, so 6 gallons. Ziggy Saw dust 15:24, 29 April 2008 (UTC)[reply]

6. The tank is empty with the plug in. The tap is turned on for 4 minutes. The plug is then pulled out with the tap left on for another 6 minutes. How many gallons will be in the tank at the end of this time.

If it's turned on for 4 minutes the tank gains 16 gallons. And that's all i understand

The 4 minutes with the 4g/m from the tap is 16 gallons, and then the plug is taken out, so net loss per minute is 2 gallons, so 2x6=12, and 16-12=4, so 4 gallons. Ziggy Saw dust 15:24, 29 April 2008 (UTC)[reply]

Could somebody help me with these please, thanks --217.171.129.79 (talk) 12:39, 28 April 2008 (UTC)[reply]

You seem to be having trouble with all the questions (to whit 2, 4, 5 and 6) that involve the tap being on and the plug out at the same time. Under these circumstances, four gallons of water flow into the tank every minute and 6 gallons flow out. So what is the overall change in the amount of water every minute? Algebraist 13:08, 28 April 2008 (UTC)[reply]

2 gallons? —Preceding unsigned comment added by 217.171.129.79 (talk) 13:15, 28 April 2008 (UTC)[reply]

If it is 2 gallons, is Q2. 10 minutes? Q4 6 minutes? Q5 14 gallons? And the last one 0 gallons? --217.171.129.79 (talk) 13:19, 28 April 2008 (UTC)[reply]

Yes, total flow = what flows in - what flows out, so here you get that the total flow is 4-6 = -2 gallons per minute (the minus sign shows that water is flowing out). So you got Q2 right.

For Q4, there are 20 gallons initially, and for the first four minutes, 2 gallons are flowing out per minute. After these first four minutes, how many gallons are left ? Then water flows out at 6 gallons per minute.

So four minutes of water flowing in fills it by 2 gallons, so there are 8 gallons in the tank. So the answer is 8 ÷ 6? So 1½? —Preceding unsigned comment added by 217.171.129.79 (talk) 13:39, 28 April 2008 (UTC)[reply]
- Not exactly. You have 20 gallons at the beginning, and 2 are flowing out per minute, so you get 20 - 4*2 = 12 gallons left after four minutes. Then water continues flowing out at 6 gallons per minute.

So it's 2 minutes?

Yes.

For Q5, start at 20 gallons, it flows out at a rate of 6 gallons per minute for 2 minutes, then flows out at 2 gallons per minute for 3 minutes. Can you work out how much is left ?

2 gallons?

Yes.

For Q6, after 4 minutes there are 16 gallons as you said. After that, water is flowing out at a rate of 2 gallons per minute for six minutes. Can you see how many gallons will be left ?

4 gallons?

Yes.

Hope that helps -- Xedi (talk) 13:35, 28 April 2008 (UTC)[reply]

The crumby thing about this problem is that it that the wording is ambiguous. Neither of these interpretations would be unreasonable:

A certain tank holds 20 gallons of water. When a tap over it is turned on water flows from it into the tank at the rate of 4 gallons a minute. When [the tap is off and] a plug in the bottom of the tank is pulled out water drains away from the tank at the rate of 6 gallons a minute.
A certain tank holds 20 gallons of water. When a tap over it is turned on water flows from it into the tank at the rate of 4 gallons a minute. When [the tap is on and] a plug in the bottom of the tank is [also] pulled out water [is lost] from the tank at the rate of 6 gallons a minute.

(Equivalently, one could interpret "drains away from the tank at the rate of" as either the flow out of the open plug, or the net rate at which the total volume of water changes.) The first interpretation is the one that is assumed above and is probably what the author meant. But the second interpretation could also be obtained logically from the narrative flow of the problem statement and would result in completely different answers. --Prestidigitator (talk) 18:56, 28 April 2008 (UTC)[reply]

And since this is a math rather than physics problem we don't have to take into account that the flow rate will decrease with decreasing head, so the 6gpm number would only be true if the tank was completely full. Loren.wilton (talk) 08:15, 29 April 2008 (UTC)[reply]

Positive=Negative???????[edit]

s=1+2+4+8+16.... (s-1)=2+4+8+16.... 2s=2+4+8+16.... Thus 2s=(s-1)

     s=-1

Is this possible? Why? —Preceding unsigned comment added by 202.179.74.246 (talk) 13:55, 28 April 2008 (UTC)[reply]

WHAAOE: See 1 + 2 + 4 + 8 + · · ·. In brief, if you want to give this sum a finite value, then yes, -1 is the sensible one to choose. It's probably more natural to say that the sum is either infinite or undefined, however. Algebraist 14:28, 28 April 2008 (UTC)[reply]

[edit conflict] We have an article about this: 1 + 2 + 4 + 8 + · · ·. The short answer: The mistake is in the first statement, "s=1+2+4+8+16....". In the real numbers, with the usual definition of summation, the sum does not converge and thus you cannot assign a real value to it. There are alternative settings in which the sum is indeed equal to -1, but those diverge remarkably from your everyday arithmetic. -- Meni Rosenfeld (talk) 14:32, 28 April 2008 (UTC)[reply]

It is true that the sequence is not convergent, but it is not true that we cannot assign a value to it. Many mathematicians don't, but some do. Feel free to put 1+2+4+8+...=−1. But some statements regarding finite sums and convergent series ceases to hold true when you do so. A finite sum of positive numbers is positive, but an infinite sum of positive numbers may be negative, 1+2+4+8+...=−1. A finite sum of integers is an integer, but an infinite sum of integers may be an noninteger: 1+3+9+27+...=−1/2, because the sum S solves the equation S=1+3S. You may insert or delete parentheses in a finite sum, but not in an infinite sum: 1−1+1−1+1−1+1−1+...=1/2 because the sum S solves the equation S=1−S, but (1−1)+(1−1)+(1−1)+(1−1)+...=0. Bo Jacoby (talk) 15:46, 28 April 2008 (UTC).[reply]

You can assign those values, but it's not particularly meaningful to do so when working in the standard real numbers. You have to define an alternative metric (or, equivalently, an alternative absolute value) to make the sequences converge. If the sequence doesn't converge, you run into all kinds of problems when you try and manipulate it. --Tango (talk) 16:58, 28 April 2008 (UTC)[reply]

Does it have to come from changing the metric? Is analytic continuation equivalent to that in some way? Black Carrot (talk) 18:40, 28 April 2008 (UTC)[reply]

Well "converges to" means "gets close to" - as far as I can tell, the only way to change whether or not something converges is to change the meaning of "close" which corresponds to changing the metric. That's just from an intuitive perspective, not a rigorous one, obviously, so it's possible there's some technicality I've missed that may be significant. --Tango (talk) 12:51, 29 April 2008 (UTC)[reply]

Completely off topic: even if convergence was the only way of doing things, convergence in general does not require a metric: a topology suffices. Algebraist 16:12, 29 April 2008 (UTC)[reply]

True, you could get rid of the metric completely, rather than just changing it, and replace it with an arbitrary topology. If you want to leave the metric as it is, though, your definition of convergence is fully determined (unless I'm misunderstanding something). --Tango (talk) 16:18, 29 April 2008 (UTC)[reply]

Convergence is not the only way of assigning a value to a series. One does not run into 'all kinds of problems'. The series 1+2+4+...=−1 is used in the standard computer representation of integers: minus one is represented in binary by all ones. It makes sense. Theretical physicists, troubled by divergent series in their perturbation calculations, might be comforted by knowing that some divergent series do define values rather than being senselessly infinite. Bo Jacoby (talk) 09:09, 29 April 2008 (UTC).[reply]

Um, physics doesn't know anything about binary representations. Radix-dependent methods of assigning values to sums of divergent series aren't likely to find much applicability in physics. --Trovatore (talk) 18:45, 29 April 2008 (UTC)[reply]

What Bo has given are two distinct arguments in favor of choosing 1+2+4+...=-1. The first has to do with binary representations of integers in digital computing; The second has nothing to do with binary representations or radices, only with the fact that it "works" in some physical contexts. Apparently other divergent sums, such as 1 + 2 + 3 + 4 + · · ·=-1/12, are applicable in physics. -- Meni Rosenfeld (talk) 19:35, 29 April 2008 (UTC)[reply]

I followed the link to John Baez's page, and it appears that what they're talking about is not really 1+2+3+4+..., but rather ζ(−1). By the tame you get to the representation 1+2+3+4+..., the information that this is supposed to be interpreted as a value of the zeta function has simply been lost. The same for the other supposed sums represented here. The values given are context-dependent, and the contexts are not all the same; contrary to Bo's assertion, you will definitely get into all kinds of problems if you mix and match them. --Trovatore (talk) 19:50, 29 April 2008 (UTC)[reply]

One does run into problems with non-convergent sequences. You've mentioned that the addition is no longer associative. I consider addition not being associative to be a problem. --Tango (talk) 20:01, 29 April 2008 (UTC)[reply]

True, but that's not really a problem for the series we were originally talking about (1 + 2 + 4 + 8 + · · ·), since it converges p-adically and p-adic addition most certainly is associative (and even commutative). —Ilmari Karonen (talk) 12:10, 30 April 2008 (UTC)[reply]

Series converges in p-adic metric when p=2, but not for general p, surely ? Gandalf61 (talk) 12:41, 30 April 2008 (UTC)[reply]

Right, my mistake. I should've just said "converges 2-adically". —Ilmari Karonen (talk) 17:06, 30 April 2008 (UTC)[reply]