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[rjc571]

Maybe one of y'all will find this easy, I'm sillily stumped.

I know it does converge (because after being fairly sure I checked on wolfram, converges to 0.514987) and I'm quite certain I just need to use comparison test. i.e. find a sequence b_n such that 0<the given sequence =< b_n and the sum of b_n from 1 to inf converges. I always have trouble picking sequences though. Am I supposed to apply the concept of telescoping series somehow maybe?

Multiply the terms of the series by (sqrt(1 + n^6) + n^3) / (sqrt(1 + n^6) + n^3) and see what happens.

 

[big ander]

Multiply the terms of the series by (sqrt(1 + n^6) + n^3) / (sqrt(1 + n^6) + n^3) and see what happens.

oh duh! I'd been trying changing the quantity within the sqrt and then factoring and a whole bunch of other messy stuff...bleh. Thank you very much. It's just been so long since I've done series convergence, at least 5 years.

 

[Tater Tot]

Help please:

simplify the expression: tan(-x)*sin(-x)+cos(-x) <--original problem

-tanx*-sinx+cosx

-(sinx/cos)*-sinx+cosx

(-sin^2x+cosx)/cosx

Then what?

 

[cpp_is_king]

Help please:

simplify the expression: tan(-x)*sin(-x)+cos(-x) <--original problem

-tanx*-sinx+cosx

-(sinx/cos)*-sinx+cosx

(-sin^2x+cosx)/cosx

Then what?

There's a couple of errors here. First of all, get rid of negative signs. No reason to leave them hanging around when they cancel each other out. negative times a negative is a positive, so just get rid of your 2 negatives:

1) tan(-x)*sin(-x)+cos(-x)
2) tan(x) sin(x) + cos(x)

Second, you've added your fractions incorrectly. You can't add 2 fractions without first finding a common denominator. In this case, the common denominator is cos x, but you didn't multiply the second term of the addition by cos x / cos x before adding.

3) sin^2(x) / cos(x) + cos^2(x) / cos(x)
4) [sin^2(x) + cos^2(x)] / cos(x)
5) 1 / cos(x)
6) sec(x)

 

[Tater Tot]

Thank you sir

 

[coldvein]

well guys .. i've got a 91% in my math class right now, which is by far the best i have ever done in math. we just started adding, subtracting, and multiplying polynomials. this is the last week of class. all that remains for me is to take the final next wednesday. i really, really, really want to do well on it and get an A in this SHIT.

 

[cpp_is_king]

well guys .. i've got a 91% in my math class right now, which is by far the best i have ever done in math. we just started adding, subtracting, and multiplying polynomials. this is the last week of class. all that remains for me is to take the final next wednesday. i really, really, really want to do well on it and get an A in this SHIT.

"shit" implies that it's a bad thing. But who are you trying to kid, we all know that math is a wonderful thing and you secretly love it.

 

[coldvein]

"shit" implies that it's a bad thing. But who are you trying to kid, we all know that math is a wonderful thing and you secretly love it.

it has actually been enjoyable. its unique. there have been many times this quarter when i felt like i'd hit a barrier, some torturous thing that i would never understand. and now those things seem easy to me. feels good. the only thing i can relate it to is learning a foreign language.

 

[cpp_is_king]

Unrelated, but in case anyone is bored, I came across this problem the other day and found it sufficiently unique and interesting that I thought someone else here might enjoy it.


You and I are playing a game where we each have $50. We begin flipping a fair coin. After each flip, if it turns up heads, you give me $1. If it turns up tails, I give you $1. The game is over when either player runs out of money.

On average, how many times should we have to flip the coin before the game is over?

 

[rjc571]

Unrelated, but in case anyone is bored, I came across this problem the other day and found it sufficiently unique and interesting that I thought someone else here might enjoy it.


You and I are playing a game where we each have $50. We begin flipping a fair coin. After each flip, if it turns up heads, you give me $1. If it turns up tails, I give you $1. The game is over when either player runs out of money.

On average, how many times should we have to flip the coin before the game is over?

You should be able to figure this out using a transition matrix. Here's an example on Wikipedia

The game will have 51 states:
State 1 - Both players have $50
State 2 - One player has $51 and the other has $49
State 3 - One player has $52 and the other has $48
...
State 50 - One player has $99 and the other has $1
State 51 - One player has $100 and the other has $0, and the game ends

Then the transition matrix T is as follows:
0 1 0 .... 0
.5 0 .5 0 ....0
0 .5 0 .5 0 ....0
...etc.
the 50th row is 48 zeroes followed by .5 0 .5
the 51st is 50 zeroes followed by 1

Then as per wikipedia's example, the expected length of the game is &#964;(I - T)^-1 v1, where &#964; is the initial vector (in this example, the game starts at state 1, so the initial vector is &#964; = [1 0 ... 0]), I is the identity matrix, and v1 is a column vector consisting of all ones. (Also I guess you have to delete the last row and column from T, or else I - T wouldn't be invertible.) You can just plug this in to Mathematica or something to find the answer.

 

[cpp_is_king]

That's an approach i hadnt thought of, but its possible (and more interesting) to derive a closed form expression for the answer with just pencil and paper

 

[DEAD RABBIT]

That's an approach i hadnt thought of, but its possible (and more interesting) to derive a closed form expression for the answer with just pencil and paper

I think I found the answer but I cannot claim to have found the solution by myself. Hence, spoilers:

Spoiler

I recognized the problem as a Random Walk, a subject I had studied before in university.

The (canonical) problem is phrased on wikipedia: How many 50/50 random walks N on a 1D lattice on average are necessary to be k steps to the right (or equivalently, the left)? The answer: k = N / 2 with k the average position after N walks with 50/50 left/right walk probabilities. The pen&paper derivation can be found on wiki's Random Walk page (attempting to replicate it here would be messy).

The answer to your problem for N would be 100.

 

[cpp_is_king]

Viewing it as a random walk is actually a great idea. However, something's wrong because that's not the correct answer But reading the way you've rephrased the problem, it certainly does appear to be equivalent to the problem I've stated. I'm not sure if the issue is with the statement of the problem (seems unlikely) or the derivation of the answer, but I don't really know much about theory of random walks, so I guess I'll have to go browse the Wikipedia page and see what their derivation looks like.

The easiest way to see that the answer isn't correct is if you consider the case of k=1. According to the formula k = N/2, then N = 2, but actually N=1 because no matter what the outcome of the first coin toss is, the game is over immediately.

I do like that idea though, it seems like it should work in theory.

 

[Tater Tot]

Hi again, just need quick help on how to find the inverse of a function.

f(x) = x/squareroot(x+4)

x=y/squareroot(y+4)

x*squareroot(y+4)=y

then i got stuck

 

[cpp_is_king]

Hi again, just need quick help on how to find the inverse of a function.

f(x) = x/squareroot(x+4)

x=y/squareroot(y+4)

x*squareroot(y+4)=y

then i got stuck

x^2 (y+4) = y^2

x^2 y+ 4x^2 = y^2

y^2 - x^2 y - 4x^2 = 0

now use quadratic formula to solve for y

 

[Kenka]

GAF, there is an upcoming draw to determine who will face who in a fairly known soccer competition here in Europe. Teams in the column on the left can draw teams listed on the upper line in the table of probabilities hereunder:

Now my question is: what's the probability of having at least one of the following draws?

Valencia - Manchester
Valencia - Juventus
Valencia - Schalke
Madrid - Manchester
Madrid - Juventus
Madrid - Schalke

Thank you much !

 

[Kenka]

Quick computation, out of shape when it comes to probabilities though, but at first glance:
Chance that one of the Valencia matches happens: 0.6205 (i.e. 62.05%)
Chance that none of the Valencia matches happen: 0.3795 (i.e. 37.95%)
Chance that none of the Valencia matches happen but one of the Madrid matches: 0.6205*0.3795=0.2355 (i.e. 23.55%)

Total chance=85.6%

There's a chance I missed something. It's been a while since I did probability theory. When's the drawing ?

Dec, 20th. I think you did something a bit shady. Did you substract 23.55 from 100 to reach 85.6? Also, you multiplied together two mutually exclusive events on line 4.

OK, probability of a Valencia game: ~60%
Probability of a Madrid game: 60% also.
Probability that both a Valencia and a Madrid game happen: 0.6*0.6= 0.36
Probability that a Madrid game happens but not a Valencia game (and vice versa): 0.6*0.4= 0.24
Probability that none happens: 0.4*0.4= 0.16 = 16%

The last one is the most interesting. It means that there is a 84% chance at least one of these games will happen.

Wohoo, I did it alone!

 

[Kenka]

Thank you! This looks complicated, I'd better plunge back into my books to refresh my poor brain.

Fine example of explanation, I hope I'll understand it tonight after some education about the topic.

 

[Charmicarmicat]

I was wondering if anyone could help a complete maths novice with something I read in a book on relativity.

So the first part was: (cT)^2 = 1^2 + (vT)^2

I can follow this ok, that's not the problem, what threw me is they then say that this can be solved with equation T^2 = 1 / (c^2 – v^2), but they don't show the process so I don't know how they got there! To actually quote the book "this isn't a book on maths, so you'll just have to trust that we got it right"!!

I'm sure this is really basic stuff and I'm just a dunce, but if anyone could briefly explain, or even direct me to some decent tutorials on this kind of thing online that would be awesome.

 

[Leezard]

I was wondering if anyone could help a complete maths novice with something I read in a book on relativity.

So the first part was: (cT)^2 = 1^2 + (vT)^2

I can follow this ok, that's not the problem, what threw me is they then say that this can be solved with equation T^2 = 1 / (c^2 – v^2), but they don't show the process so I don't know how they got there! To actually quote the book "this isn't a book on maths, so you'll just have to trust that we got it right"!!

I'm sure this is really basic stuff and I'm just a dunce, but if anyone could briefly explain, or even direct me to some decent tutorials on this kind of thing online that would be awesome.

It's just two steps:
1 = (cT)^2 - (vT)^2 = T^2 *(c^2 - v^2)
=> T^2 = 1/(c^2 - v^2)

 

[bwtw]

GAF, there is an upcoming draw to determine who will face who in a fairly known soccer competition here in Europe. Teams in the column on the left can draw teams listed on the upper line in the table of probabilities hereunder:


Now my question is: what's the probability of having at least one of the following draws?

Valencia - Manchester
Valencia - Juventus
Valencia - Schalke
Madrid - Manchester
Madrid - Juventus
Madrid - Schalke

Thank you much !

Just wondering, how did you get the %s for each team drawing X given the complication of not being able to draw a team from your own country?

 

[xJavonta]

Reading this thread makes me realize how bad I am at math, holy shit.

 

[DEAD RABBIT]

Viewing it as a random walk is actually a great idea. However, something's wrong because that's not the correct answer But reading the way you've rephrased the problem, it certainly does appear to be equivalent to the problem I've stated. I'm not sure if the issue is with the statement of the problem (seems unlikely) or the derivation of the answer, but I don't really know much about theory of random walks, so I guess I'll have to go browse the Wikipedia page and see what their derivation looks like.

The easiest way to see that the answer isn't correct is if you consider the case of k=1. According to the formula k = N/2, then N = 2, but actually N=1 because no matter what the outcome of the first coin toss is, the game is over immediately.

I do like that idea though, it seems like it should work in theory.

Damn, of course, stupid of me that I didn't try out the trivial cases (the solution looked so nice, I thought it had to be correct). But your example tells me exactly where the problem is. There is a cut-off or game-over scenario in the problem statement that is not incorporated in the Random Walk model that I used.

For instance, imagine 75 units of money being exchanged. In the space of all permutations, there are permutations where the first 50 units go to either player one/left or player two/right with still 25 money exchanges (or walks) left. But the game is over (the other player is out of money), the permutation ends here. However, for a Random Walk the permutation simply continues until all N walks are done. Hence it is missing the game-over feature.

With the RW equation that I used, you average the walk position with the entire width of a Pascal triangle at the (N+1)th row multiplied by the probability and the difference between the walks to the left and right (the row index of the triangle). This row represents the 'final state' of the game. But because the Pascal triangle gets cut off at 50 walks, it gives you two problems.

The first is that the Pascal's triangle entries change because the entries beyond the boundary (beyond 49) are not added to the entries of the next row within the boundary. Try for instance cutting off a Pascal triangle at 2. From row 3 and beyond, you end up summing quantities that are multiples of 3. At higher cut-offs, the progressions look more complicated. Seems like a lot of hassle to solve in a general way. The second problem is that you also need to average over the left & right boundaries of the Pascal triangle, because they are part of the final state of the game. The first 49 coin tosses you can ignore (the first 50 rows of the Pascal triangle), because they're not part of the final state. The remaining rows however are important.

I found out that there are several names to your problem. The most common one seems to be Gambler's ruin or a version of it. In physics, it's apparently called the Random Walk with Absorbing Walls problem.

No luck finding/understanding the solution yet, but my quest for one had been interesting to say the least.

 

[cpp_is_king]

Damn, of course, stupid of me that I didn't try out the trivial cases (the solution looked so nice, I thought it had to be correct). But your example tells me exactly where the problem is. There is a cut-off or game-over scenario in the problem statement that is not incorporated in the Random Walk model that I used.

For instance, imagine 75 units of money being exchanged. In the space of all permutations, there are permutations where the first 50 units go to either player one/left or player two/right with still 25 money exchanges (or walks) left. But the game is over (the other player is out of money), the permutation ends here. However, for a Random Walk the permutation simply continues until all N walks are done. Hence it is missing the game-over feature.

With the RW equation that I used, you average the walk position with the entire width of a Pascal triangle at the (N+1)th row multiplied by the probability and the difference between the walks to the left and right (the row index of the triangle). This row represents the 'final state' of the game. But because the Pascal triangle gets cut off at 50 walks, it gives you two problems.

The first is that the Pascal's triangle entries change because the entries beyond the boundary (beyond 49) are not added to the entries of the next row within the boundary. Try for instance cutting off a Pascal triangle at 2. From row 3 and beyond, you end up summing quantities that are multiples of 3. At higher cut-offs, the progressions look more complicated. Seems like a lot of hassle to solve in a general way. The second problem is that you also need to average over the left & right boundaries of the Pascal triangle, because they are part of the final state of the game. The first 49 coin tosses you can ignore (the first 50 rows of the Pascal triangle), because they're not part of the final state. The remaining rows however are important.

I found out that there are several names to your problem. The most common one seems to be Gambler's ruin or a version of it. In physics, it's apparently called the Random Walk with Absorbing Walls problem.

No luck finding/understanding the solution yet, but my quest for one had been interesting to say the least.

Glad someone else is enjoying it

Try this:

Suppose N(a, b) is the expected number of turns required for the game to end if the players have a and b dollars respectively, so that youre trying to find the value of N(50, 50).

1) What is the base case? (What are the conditions on a, b for the game to be over). Express it mathematically

2) Can you write down a recurrence expressing N(x, x) in terms of something else? the recurrence should contains only a single recurring call to N. If it doesnt, think for a minute until you get an idea

Spoiler

3) Continue in this fashion trying to succesively get N(x, x) in terms of a single other occurrence of N until you notice a pattern. For a super simple example, consider the recurrence:

f_n = f_{n-1} + 2

then you can write:

f_{n-1} = f_{n-2} + 2

And back-substituting you find that:

f_n = f_{n-2} + 4
= f_{n-3} + 6
= f_{n-4} + 8
...
= f{n-k} + 2k

See if you can do that here.

 

[Espresso]

Can someone who's knowledgable about statistics help me out here?

The useful life of a computer terminal at a university computer center is known to be normally distributed with a mean of 3.25
years and a standard deviation of .5 years.

a) Historically 22% of the terminals have had a useful life less than the manufacturer&#8217;s advertised life. What is the manufacturer&#8217;s advertised life for the computer terminals?
b) What is the probability that a computer terminal will have a useful life of at least 3 but less than 4 years?



Edit 2: Does anyone know of a good online resource for statistics dummies?

 

[Partial Gamification]

Can someone who's knowledgable about statistics help me out here?

The useful life of a computer terminal at a university computer center is known to be normally distributed with a mean of 3.25
years and a standard deviation of .5 years.

a) Historically 22% of the terminals have had a useful life less than the manufacturer&#8217;s advertised life. What is the manufacturer&#8217;s advertised life for the computer terminals?
b) What is the probability that a computer terminal will have a useful life of at least 3 but less than 4 years?



Edit 2: Does anyone know of a good online resource for statistics dummies?

a) Given the Normal Distribution (its a common stats curve):

you can see how the standard deviations "chop-up" the area under the curve into regions that have consistent measure (edit: from Nromal Curve to Normal Curve). Note the area under this curve is 100%, or one as it does not reach the horizontal axis but comes infintesimally close. Anyway, that is just gibberish backstory but I think a visual is always good to see what you are working with; and as a further side, I think the Normal Distribution came out of the Guiness Brewery as a paper authored by "Student."

I don't know how you are approaching this (or how you are expected to approach this). I had a Z-Table and looked up the value and guesstimated three years, using the equation: z-score = ( X - mean) / standardDeviation.

b) For this, you need the | Prob(4yrs)-Prob(3yrs) | , the above equation can be used to solve for these values.

Here is a basic "handbook"
and an awesome "cookbook"

edit: note my graph is not drawn well. Here is the first link's bit on Normal Distributions

 

[youngwerther]

I need some help with chemistry, specifically the Clausius-Clapeyron equation, because I don't know how to use it.

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 degrees Celsius, and 393.0 Torr at 45.0 degrees Celsius. Calculate the molar heat of vaporization for this liquid (kJ/mol).

I don't know what to do with logs.

The equation is

ln P1/P2 = delta-Hvap/R (1/T2 - 1/T1)

Conceptually, I know to convert the temperatures to Kelvin so the units cancel with R, the gas constant (8.314 J/K mol), and then plug and chug, but I don't know what to do with the log. Anyone help?

 

[Enkidu]

I need some help with chemistry, specifically the Clausius-Clapeyron equation, because I don't know how to use it.

A certain liquid has a vapor pressure of 92.0 Torr at 23.0 degrees Celsius, and 393.0 Torr at 45.0 degrees Celsius. Calculate the molar heat of vaporization for this liquid (kJ/mol).

I don't know what to do with logs.

The equation is

ln P1/P2 = delta-Hvap/R (1/T2 - 1/T1)

Conceptually, I know to convert the temperatures to Kelvin so the units cancel with R, the gas constant (8.314 J/K mol), and then plug and chug, but I don't know what to do with the log. Anyone help?

I am not at all familiar with this equation, but you have both P1 and P2, right? So why not just plug them in and calculate the logarithm? It doesn't seem like you need to do anything mathematically complicated here.

 

[youngwerther]

I don't know anything about logarithms.
Technically I don't have the proper math prerequisites to be in this class but I got in anyways.
It hasn't been a problem until now, I just don't know how to use this one equation.

I also realize that this is probably something incredibly simple and once someone shows me what to do I'll slap my forehead, but until then I'm totally lost at the mysterious little letters "ln".

 

[Enkidu]

I don't know anything about logarithms.
Technically I don't have the proper math prerequisites to be in this class but I got in anyways.
It hasn't been a problem until now, I just don't know how to use this one equation.

I also realize that this is probably something incredibly simple and once someone shows me what to do I'll slap my forehead, but until then I'm totally lost at the mysterious little letters "ln".

Do you have access to a calculator? Generally you don't calculate logarithms by hand, you either use a calculator or in some cases a table. If you don't have the natural logarithm on your calculator you can get it by doing log10(x)/log10(e).

 

[Partial Gamification]

I don't know anything about logarithms.
Technically I don't have the proper math prerequisites to be in this class but I got in anyways.
It hasn't been a problem until now, I just don't know how to use this one equation.

I also realize that this is probably something incredibly simple and once someone shows me what to do I'll slap my forehead, but until then I'm totally lost at the mysterious little letters "ln".

Just to add on the topic of logarithms, it derives from Latin; coined by Scottish mathematician John Napier (1550-1617), lit. "ratio-number," from Gk. logos "proportion, ratio, word:" logos+ arithmos ":number." It is a function, incidentally inverse to the exponential function.

Without deriving the value, consider the number e (and that below is not an inverse but just the function re-written) inv(ln(x)) = e^x (if that makes sense- small error below)

Think of it like scientific notation, in that it is just a shorthand for large and small numbers (but remember it is a function, and the sci-note is more an operation multiuplying by powers of ten). Logarithmic scales are used on slide rules for calculation by measurement. People, pre-modern computers, would look-up values in log tables -books filled with logs. Wolfram has a decent site for topics such as this.

 

[Lonely1]

I dunno if this is the correct place to ask this since is more of a CS question. But I consider CS a sub-branch of Logic, so lets go!

As I understand, swarm intelligence metaheuristic algorithms doesn't have a "halting" condition, right? Although, traditionally it's said that such algorithms halt when they achieve convergence, verifying Convergence is undecidable.

Does adding additional a "halting" condition is "cheating"? AKA, is no longer the same system as defined in a Computing Network framework (Gershenson, 2010)? I believe so, but I could be wrong. What does GAF thinks?

Which begs my last question, Elementary Cellular Automaton don't have halting condition either. How does Cook goes around this issue in order to prove that the rule 110 is Turing Complete? I gave a quick read to his proof but didn't found the answer. Sadly, I don't have the time atm for an in-depth review of the article.

Maybe GAF knows the answer. If not, writing this helped me to sort my ideas.

 

[Partial Gamification]

I dunno if this is the correct place to ask this since is more of a CS question. But I consider CS a sub-branch of Logic, so lets go!

As I understand, swarm intelligence metaheuristic algorithms doesn't have a "halting" condition, right? Although, traditionally it's said that such algorithms halt when they achieve convergence, verifying Convergence is undecidable.

Does adding additional a "halting" condition is "cheating"? AKA, is no longer the same system as defined in a Computing Network framework (Gershenson, 2010)? I believe so, but I could be wrong. What does GAF thinks?

Which begs my last question, Elementary Cellular Automaton don't have halting condition either. How does Cook goes around this issue in order to prove that the rule 110 is Turing Complete? I gave a quick read to his proof but didn't found the answer. Sadly, I don't have the time atm for an in-depth review of the article.

Maybe GAF knows the answer. If not, writing this helped me to sort my ideas.

I found this interesting and beyond my full comprehension but these are the bits I geleened, thanks for the cause to look into it! Consider this at least a bump for the question, I'd be interested to know more.

From Page 6 :Gershenson

ACO and PSO have been used mainly for optimization. This explains why their f is the minimum (best) of the solutions found. In contrast, ANNs have been used to solve many dierent tasks, e.g. classication, generalization, recognition, error correction, and time sequence retention. Still, all of the architectures can be described as computing a function f in a distributed fashion. This is because they require the interaction of nodes to produce f.

It is interesting to note that, even when ACO and PSO are inspired by swarming systems, algorithms of ANN and ACO are more similar between themselves than with PSO, in the sense that they update edges, while PSO algorithms update nodes. However, the models can be extended from networks to hypernetworks (Johnson, 2010), where there is a duality between nodes and edges, i.e. one can exchange nodes and edges while preserving the functionality of the hypernetwork. In this case, PSO particles can be described as hyperedges, and their interactions as nodes. Then, the PSO algorithm a would update hyperedges.

From pages 69-71, another text* talks of examining single particles in PSO optimizaiton. Showing convergence is usually difficult. "From a statistical point of view, each particle in a PSO forms a Markov Chain. ...a push for theoretical adavncemetn on the understanding of the interactions of multiple Markov Chains." There is sample MATLab pseudocode also that employs a while(criterion) loop and given that the celular automata are a discrete set counted over time, as generations or levels, it seems likely that it wouldn't be cheating if convergence was shown at each node rather than throughout the system; but this is where I am not completely certain. These seem like "best-fit algorithms" that are degined to be halted but this might only address the slow-timescale.

Gershenson again:

It should be noted that multiple dynamical scales are an important feature to enable adaptation (Holland, 1975, 1995). A system can function at a \fast" scale, while adaptation can work at a \slow" scale. When the situation of the system changes, adaptation can change the function of the system to cope with the new situation.

*Xin-She Yang Nature-Inspired Metaheuristic Algorithms: Second Edition

 

[youngwerther]

Where am I going wrong here?

A certain substance has a heat of vaporization of 48.23 kJ/mol. At what Kelvin temperature will the vapor pressure be 7.00 times higher than it was at 291 K?

Ok, so I use the Clausius-Clapeyron equation.

ln(P2/P1) = DHvap/8.314(1/T1-1/T2)

Since the gas constant is in J/Kxmol, I convert the heat of vaporization from kJ to J to get 48230 J/mol.
Since the vapor pressure is 7 times higher, that means P2/P1 is 7.00
And I'm looking for a temperature, T2.

So I have

ln(7.00)=48230/8.314(1/291 - 1/T2)

After the algebra I get -265.13 K. That's wrong. What have I done wrong?

 

[cpp_is_king]

Where am I going wrong here?

A certain substance has a heat of vaporization of 48.23 kJ/mol. At what Kelvin temperature will the vapor pressure be 7.00 times higher than it was at 291 K?

Ok, so I use the Clausius-Clapeyron equation.

ln(P2/P1) = DHvap/8.314(1/T1-1/T2)

Since the gas constant is in J/Kxmol, I convert the heat of vaporization from kJ to J to get 48230 J/mol.
Since the vapor pressure is 7 times higher, that means P2/P1 is 7.00
And I'm looking for a temperature, T2.

So I have

ln(7.00)=48230/8.314(1/291 - 1/T2)

After the algebra I get -265.13 K. That's wrong. What have I done wrong?

is the correct answer T2=322.478? Thats what i get when i do the algebra. if thats correct, at least it means youve done everything right up until the last step

 

[Something Wicked]

Where am I going wrong here?

A certain substance has a heat of vaporization of 48.23 kJ/mol. At what Kelvin temperature will the vapor pressure be 7.00 times higher than it was at 291 K?

Ok, so I use the Clausius-Clapeyron equation.

ln(P2/P1) = DHvap/8.314(1/T1-1/T2)

Since the gas constant is in J/Kxmol, I convert the heat of vaporization from kJ to J to get 48230 J/mol.
Since the vapor pressure is 7 times higher, that means P2/P1 is 7.00
And I'm looking for a temperature, T2.

So I have

ln(7.00)=48230/8.314(1/291 - 1/T2)

After the algebra I get -265.13 K. That's wrong. What have I done wrong?

The equation is setup correctly; you goofed on the algebra.

 

[youngwerther]

I wonder where I messed up, I did it twice that way and once with the equation set up a different way.
Did I do the ln wrong? I did the ln of 7 on my calculator and got 1.95.

edit - cpp, yes your answer is correct.

 

[cpp_is_king]

I wonder where I messed up, I did it twice that way and once with the equation set up a different way.
Did I do the ln wrong? I did the ln of 7 on my calculator and got 1.95.

edit - cpp, yes your answer is correct.

Your calculation of ln(7) was correct.

ln(7.00)=48230/8.314(1/291 - 1/T2)

1.946 = 5801.06(1/291 - 1/T2)

0.000335441 = 1/291 - 1/T2

1/T2 = 1/291 - 0.000335441

1/T2 = 0.00310099

T2 = 322.478

 

[DEAD RABBIT]

Glad someone else is enjoying it

Try this:

Suppose N(a, b) is the expected number of turns required for the game to end if the players have a and b dollars respectively, so that youre trying to find the value of N(50, 50).

1) What is the base case? (What are the conditions on a, b for the game to be over). Express it mathematically

2) Can you write down a recurrence expressing N(x, x) in terms of something else? the recurrence should contains only a single recurring call to N. If it doesnt, think for a minute until you get an idea

Spoiler

3) Continue in this fashion trying to succesively get N(x, x) in terms of a single other occurrence of N until you notice a pattern. For a super simple example, consider the recurrence:

f_n = f_{n-1} + 2

then you can write:

f_{n-1} = f_{n-2} + 2

And back-substituting you find that:

f_n = f_{n-2} + 4
= f_{n-3} + 6
= f_{n-4} + 8
...
= f{n-k} + 2k

See if you can do that here.

Solved it. Yeah... the RW idea was nice, but your method is much cleaner.

Relations:
N(a,b) = N(b,a)
N(a,0) = 0

First recursion:
N(x,x) = 0.5(N(x+1,x-1) + 1) + 0.5(N(x-1,x+1) + 1)
= N(x+1,x-1) + 1

The trick was to add 1 to the recursion terms. That was the one thing I was missing. Once I realized the mathematical inconsistency of leaving out the addition of this 1, it was pretty trivial solving the rest.

Second recursion
N(x,x) = 0.5(N(x+2,x-2) + 1) + 0.5(N(x,x) + 1) + 1
-> N(x,x) = N(x+2,x-2) + 1 + 1 + 2 = N(x+2,x-2) + 4
Third recursion.
N(x,x) = 0.5(N(x+3,x-3) + 1) + 0.5(N(x+1,x-1) + 1) + 4
-> N(x,x) = N(x+3,x-3) + 1 + 8 = N(x+2,x-2) + 9

The progression 1,4,9,... is quadratic:
-> N(x,x) = N(x+k,x-k) + k^2

For the end-game scenario:
x = k
N(x,x) = N(2x,0) + x^2 = x^2

For x = 50, the answer is 2500.

 

[CHUNKYBOWSER]

I'm having a little trouble with a Calculus problem... Just differentiating [sqrt(x)]^x. Any help?

Some of the possible answers include ln if that helps at all.

 

[DEAD RABBIT]

I'm having a little trouble with a Calculus problem... Just differentiating [sqrt(x)]^x. Any help?

Some of the possible answers include ln if that helps at all.

You have to rewrite sqrt(x) into an exponent first

Spoiler

like so: [sqrt(x)]^x = [exp(ln(x)/2)]^x = exp(xln(x)/2)

so that you can apply the chain & product rules. You can't apply those rules directly on x^x-like equations.

 

[cpp_is_king]

Another way is to write y = x^(x/2)

then taking the natural log of both sides to get rid of the exponent:

ln = (x/2) * ln(x)

Now differentiate both sides. Note that when you differentiate the left side it qualifies as "implicit differentiation", assuming you've learned that. And you use the product rule on the other side

y' / y = (x/2) * (1/x) + ln(x)/2 = 1/2 + ln(x)/2

Now substitute back in the original value of y for the left hand side where you have 1/y.


y'/(x^(x/2)) = (1 + ln(x))/2

y' = (1 + ln(x)) x^(x/2) / 2

 

[cpp_is_king]

For the end-game scenario:
x = k
N(x,x) = N(2x,0) + x^2 = x^2

For x = 50, the answer is 250.

Good job! I should point out though that 50^2 is 2500, not 250 But you had the formula right, so full credit

So with that out of the way, I have to make a confession. The Wikipedia article for Random Walk actually says that the expected translation distance after n steps is in fact proportional to Sqrt[n], not n/2. So I think you're original idea was actually fine, and I'm not quite sure where the n/2 thing came from. But at least now you know how to derive that number Plus I just thought the derivation was kind of interesting, since you don't usually think about recurrences in multiple variables.

 

[AgentChris]

Jenny buys bags of peanut butter cups and brings them to school to sell. She charges a dime for one pbc but sometimes she's willing to accept a nickel. She sold 35 pbc's for a total of $2.85.

I know the answer, I'm not sure how to put it algebraically; 13 @ a nickel and 22 @ a dime.

In each group there are either 1 high school student and 3 nursery school kids or 2 high school students and 4 nursery school kids. There are a total of 23 high school students and 61 nursery kids in all. How many groups are there in all?

Same thing, Not too sure how to put it algebraically; 15 groups of 1HS & 3K, 4 groups of 2HS & 4K.
15(3)=45
(8/2)4=16
16+45=61

 

[Kazerei]

x = number of pbcs sold for a dime
y = number of pbcs sold for a nickel

x + y = 35
0.1x + 0.05y = 2.85

a = number of groups with 1 student and 3 kids
b = number of groups with 2 students and 4 kids

a + 2b = 23
3a + 4b = 61

 

[AgentChris]

x = number of pbcs sold for a dime
y = number of pbcs sold for a nickel

x + y = 35
0.1x + 0.05y = 2.85

a = number of groups with 1 student and 3 kids
b = number of groups with 2 students and 4 kids

a + 2b = 23
3a + 4b = 61

Thanks a lot!

 

[DEAD RABBIT]

Good job! I should point out though that 50^2 is 2500, not 250 But you had the formula right, so full credit

So with that out of the way, I have to make a confession. The Wikipedia article for Random Walk actually says that the expected translation distance after n steps is in fact proportional to Sqrt[n], not n/2. So I think you're original idea was actually fine, and I'm not quite sure where the n/2 thing came from. But at least now you know how to derive that number Plus I just thought the derivation was kind of interesting, since you don't usually think about recurrences in multiple variables.

The expectation value of <k> = N/2 is mentioned in a collapsed section on the wiki page: "Derivation of \sqrt{n} dispersion proportionality". I interpreted it incorrectly. The way they count the total length of the steps (they count from the left, not the middle) is basically equivalent to saying that <k> = 0, which in retrospect is totally obvious for a 50-50 random walk (it's also mentioned previously as Sum(E(Z_i)) = 0).

For a pure random walk, there are no boundaries and it can easily make more than 2k steps left or right within k^2 expected steps, provided that k is large enough. Thus it is summing over more walks than we want. For this reason, I am not sure if you can say the calculations are the equivalent.

 

[TheFatOne]

I'm stuck on critical points in Calc1 GAF. I understand critical points, but I don't know how to solve for the critical points in a trig function. For instance

f(x) = sin(x)cos(x)
f'(x) = cos^2(x)-sin^2(x) = cos(2x)


I know that I have to set cos(2x)= 0, but I don't know how to solve it at this point. It's the only thing I don't understand so far in this chapter in my Calc class.

 

[cpp_is_king]

When does cos(x) = 0? Its one of the first things you learn about trig functions

 

[TheFatOne]

When does cos(x) = 0? Its one of the first things you learn about trig functions

Fuuuu I completely forgot about the unit circle. Was getting so damn nervous because of my Calc test tomorrow. Thanks for the reminder.