-
Hey, guest user. Hope you're enjoying NeoGAF! Have you considered registering for an account? Come join us and add your take to the daily discourse.
The Math Help Thread
- Thread starter -COOLIO-
- Start date
- Status
You didn't mention, but can we assume that C is invertible in (b)? Of course, we must assume that C and D are invertible matrices in (a), or the left hand side wouldn't make sense.
If so, the two equations are due to the property (AB)^(-1) = B^(-1)A^(-1) for invertible matrices A and B. Write C = (C^(-1)) ^ (-1), and show that one side of the equation can be written as the other.
Oops, did forget to mention that. Both are n x n matrices which are invertible.
edit: Man I hate proofs. Not really sure about the first couple of steps for these (if someone could post the first 2-3 steps for each one I would be eternally grateful)
For these, always make sure to eliminate the inverse of additions. So if you have something like (A + B)^(-1), do (A + B)^(-1)*(A + B)=I. It will happen on both sides but at least it will not be the inverse of addition and so the distributive property can take effect.
Got two simple qs here, thinking about method more than anything. Finding the limit of the following two sequences.
For the first, I just divided the top and bottom by the first term of the denominator, and got (1/2n^2)/(1+2^n/8n) + 2. Is it sufficient to then say 1/(2n^2) -> 0 by the product rule, and so the first term here tends to 0(by product rule once again), and so the entire seq tends to 2?
For the second, I let the sequence (an, say) equal e^log(5n^3+6n^4)^(1/n), simplified it down, said lim an = e^lim(log 6n^4+5n^3)/n, then by L'Hopital's, this tends to 0, so an -> e^0 = 1. Is this right/is there a simpler way of looking at it? Have been away from Analysis I for a while, and turns out it makes up most of the first q for the Analysis II paper in the summer, so need to catch up on it.
This wouldn't pass as a rigorous proof, but we can eyeball the limit by considering which terms become dominant as n approaches infinity.
In (ii), the (4^n) term dominates the (2^n) term in the numerator (exponential functions always dominate polynomials, so the polynomials n^2 and n^3 have no effect relative to another), and likewise in the denominator (again, it doesn't matter that n^4 dominates n^3, because the exponentials are more dominating). The ratio of the dominant terms (2 * n^3 * 4^n) / (n^3 * 4^n) approaches 2.
In (iii), 6n^4 dominates 5n^3, so consider the limit for just (6n^4)^(1/n) = 6^(1/n) * (n^(1/n))^4. The term 6^(1/n) approaches 1 and n^(1/n) also approaches 1 (I guess we would need to know this limit).
For (a), start from the right-hand side. Let me write inv(A) instead of A^(-1) for clarity here.
C * inv(C + D) * D
= inv( inv(C) ) * inv(C + D) * inv( inv(D) ).
By the property that inv(AB) = inv(B)*inv(A) for invertible matrices, we can write the product of three inverses above as,
= inv( inv(D) * (C + D) * inv(C) )
= inv( inv(D)*C*inv(C) + inv(D)*D*inv(C) )
= inv( inv(D) + inv(C) ).
Do (b) in a similar manner. Pick one side and get to the other.
We have to be careful. We have yet to show that the equations are true in the first place, so we really shouldn't multiply both sides of the equation by (A + B) or do other operations on both sides in the hopes that that might lead to a simple tautology. Of course, we could then start from the tautology and arrive back at the equation, which would prove it, but I think it's better to work with one side only when we don't know if a given equation is true.
spelltropy
Member
Cheers for this, but can you immediately say something is dominant? (I can see it is, but not sure if in an exam I can just write it down?). For iii), things like knowing the limits of x^(1/n) and n^(1/n) are assumed, so I suppose I could come straight to the answer after getting past the first stage, but I'm not sure if the whole dominant thing (for either question) is rigorous enough... was my method incorrect?
Stumpokapow
listen to the mad man
Given the function:
3^x * x^3 * 10x^2, as x tends higher, the function most closely mirrors 3^x, rather than the x^3 or 10x^2 terms, so 3^x is dominant.
In general:
- Factorials dominate exponential functions, exponential functions with higher bases dominate exponential functions with lower bases
- Non-polynomial (exponential and factorial) functions dominate polynomial functions
- The highest order polynomial function dominates lower order polynomial functions
- Square polynomial terms dominate xlogx terms, which dominate linear terms
- Linear terms dominate log-linear terms and fractional exponent terms
- Lower log terms like loglog dominate constant terms
- Constant terms are dominated by everything
Again kgtrep notes that his/her strategy of using dominant terms is more about a quick eyeballing / sanity check than a rigorous proof, so I'd caution that as well, but dominating terms are useful in a variety of fields (computer science: algorithmic complexity, as one example) so worth thinking about.
When I looked at your work, the overall method seemed fine to me. However, I don't see how you found that (1/2n^2)/(1+2^n/8n) + 2 was equal to the first expression, and I'm not sure that you are using the term "product rule" in a way that I am familiar with (though that may be because I have been teaching a lot of basic calculus recently and the product rule for derivatives is forefront in my mind).
Edit: Yep, I just double-checked by looking at n=1 and those expressions are definitely not equal.
I have a physics test tomorrow and don't really understand how to do anything on my review. Usually I'm pretty good at reverse engineering the problems to figure them out, but I'm not really getting anywhere with it this time. My review is pretty much the same as the test, just with the numbers in the problems switched. If anyone could explain how to solve any of these problems, or just show work, I would really appreciate it.
1. Find the work of a 20 kg child running up a hill that is 50 m long and makes 20° angle with the horizontal.
2. A skier zips down a cliff that has a 300 m vertical drop. What is the velocity of the skier when it reaches the bottom of the hill? (assume no friction)
3. A parachutist with a mass of 100kg jumps out of an airplane at an altitude of 2000m. After the parachute deploys, the parachutist lands with a velocity of 10 m/s. What is the change of his energy?
4. What is the average power supplied by a 90 kg secretary running up a flight of stairs raising vertically 15 m in 8 s?
5. Water flows over a section of Niagara Falls at a rate of 3x10^6 kg/s and falls 65m. What is the power of the waterfall?
I have all of the equations in front of me and have been staring at them, but I'm feeling overwhelmed and don't know which one is for which type of problem. Again, if anyone is kind enough to help with any of these, I would be super grateful.
1. Find the work of a 20 kg child running up a hill that is 50 m long and makes 20° angle with the horizontal.
2. A skier zips down a cliff that has a 300 m vertical drop. What is the velocity of the skier when it reaches the bottom of the hill? (assume no friction)
3. A parachutist with a mass of 100kg jumps out of an airplane at an altitude of 2000m. After the parachute deploys, the parachutist lands with a velocity of 10 m/s. What is the change of his energy?
4. What is the average power supplied by a 90 kg secretary running up a flight of stairs raising vertically 15 m in 8 s?
5. Water flows over a section of Niagara Falls at a rate of 3x10^6 kg/s and falls 65m. What is the power of the waterfall?
I have all of the equations in front of me and have been staring at them, but I'm feeling overwhelmed and don't know which one is for which type of problem. Again, if anyone is kind enough to help with any of these, I would be super grateful.
Stumpokapow
listen to the mad man
It's not clear if you are in high school, where kinematics formulae are given outright, or university, where you would be expected to use elementary calculus to derive them from knowns. So I proceed with calculus, but you should still be able to follow.
Givens:
Vertical drop: This is a clue you should use the acceleration of gravity
Acceleration: 9.8 m/s^2 give or take, if you have a different constant than I remember, use the one you have.
Distance: 300m distance
Starting velocity: 0 (no given velocity)
Starting distance: 0 (no given starting point on the drop)
Looking for:
Velocity at time t, where t is the time that skiier has moved 300m
Some formula using what we already know
Acceleration ~= 9.8 m/s^2 = a
Integrate, velocity = 9.8t m/s^2 + 0m/s (starting velocity) = at + v0
Integrate, distance = 4.9t^2 m/s^2 + 0t m/s + 0m (starting distance) = at^2 + v0t + d0
(v0 and d0 here should be subscripted)
Solve for t using distance, which is a given:
300 = 4.9t^2 <-- how many seconds have elapsed when he's traveled 300 m
sqrt(300/4.9) = t <-- solve for t
t = 7.8246etc.
Plug time into velocity formula to solve for velocity:
velocity = 7.8246 s * 9.8 m/s^2 = 76.68108 m/s
76.68 m/s assuming no friction and thus no terminal velocity.
Follow this kind of structure for all of the problems, most of them are simple mathematics as long as you can use the formula. Write out your givens, write out what you're looking for. Find the series of formulae you need to move from your givens to your unknowns.
In this case:
Given: Distance, v0, d0, Acceleration
We want: velocity
Formula to move from distance, v0, d0, acceleration to velocity = use acceleration to fill out distance formula, solve for time in distance formula, use time to solve for velocity in velocity formula.
Thank you so much! I should have specified that I am in high school, so the process is a bit simpler for me. I ended up getting the correct answer with your help. In my class, I believe we didn't use kinematic equations at all for this unit, which is kind of bizarre. Many longer equations were used instead. I guess there are multiple ways to arrive at an answer though, and the kinematic equations work well.
I ended up calling a friend who took a higher level physics class than me for additional help. For that problem he said to use this equation:
PE = KE
mgh = 1/2mV^2
This is closer to the long equation we were given, but he said this excises some of the parts that are not pertinent to the more simple problems we do in my class. Both ways end up with the same answer though.
Fortunately the test isn't until the end of the day, so he's going to try to help me with the other problems before then. But if all else fails during the test, I'll just try to use kinematic equations which seem to be reliable for these kinds of problems.
Thanks again.
While conservation of energy is much easier, I feel like you should develop a feel for the kinematics before approaching that, imo.
My teacher never mentioned using kinematics once during this unit, oddly enough. She wanted us to use conservation of energy. We focused on kinematics a lot in earlier units so maybe she feels that it's time to move on. I'm not entirely sure.
I took the test earlier today and did pretty well, I think. At least for not knowing how to do anything on it at all 24 hours ago.
spelltropy
Member
eh, don't know what I did there. did it again and came out with [(2^(2-n)*n^(-1) / (n(2^(3-n)+1)] + 2 / (1+n2^(3-n)) - wolframalpha seems to agree with this one. (divided by the other term in the denominator, was slightly nicer)
By product rule I meant if an->a and bn->b then anbn->ab - i think its valid to use in this case?
If we take, for example, (2^(2-n) * (1/n)), can we instantly say this tends to 0, because of the 1/n & product rule, or is it incorrect to say that? Clearly it does (by sight), but not sure about how I'm justifying my answers. Similarly, for the question above, whilst I can appreciate that the 2^n dominates the n in the (n*2^(3-n)) term, could I justify it on paper by saying that the limit of (1/2)^n as n->infinity is 0 (a standard result on my course), and so, by the product rule, n*2^(3-n) -> 0?
Okay, I see what you're saying about the product rule now. If you want to apply that, you need to make sure that both of the individual limits exist. So 2^(2-n)*(1/n) -> 0 since both 2^(2-n)=4/2^n -> 0 and 1/n -> 0. You can't use the same reasoning on n*2^(3-n) since the sequence n (or 8n, depending on how you break it up) does not converge. In this case, L'Hopital's rule is probably the easiest way to go.
The obvious example to see why you need both sequences to converge is n*1/n. Clearly this sequence converges to 1 even though 1/n -> 0.
To avoid trying to type a properly formatted matrix, I'll call the right-hand side of the equation B. To solve for A, just invert the operations on the left hand side like you would in any algebra problem. Start by taking the inverse of both sides to get
A^T-2I = B^(-1).
There are different methods for calculating B^(-1), but I'm guessing that you know a formula that finds the inverse of any 2 x 2 matrix. Then add 2I to both sides to get
A^T = B^(-1) + 2I,
where adding 2I just means adding 2 to each of the diagonal entries of B^(-1). Finally take the transpose to find
A = (B^(-1)+2I)^T,
where transposing a 2 x 2 matrix just means exchanging the non-diagonal entries.
Ahhhh, ok. That's super simple lol. I have two more questions
Same wording as the first you helped me with(sorry for poor formatting).
([1 0]A}^(-1)=[1 0]
([2 1] }blank [2 2]
Second question is how do I determine if a vector is in the row space of a matrix?
We can interpret a matrix-vector multiplication A*x as a linear combination of the columns of A, where the weights come from the entries of x. In other words, if A is a m x n matrix and x is a vector of size n,
A * x = x1 * a1 + x2 * a2 + ... + xn * an.
The row space of A is (literally) the space of all vectors of size n that can be formed by taking a linear combination of the rows of A. But the rows of A are the same as the columns of A^T!
Hence, to find if a vector b of size n is in the row space of A, we can solve the equation A^T * x = b and see if there is a solution. The solution x (of size m) would tell us how to take the linear combination of the rows of A to get b.
So like, I've been using Khanacademy for a month or two now, starting from scratch as I've always been horrible at maths and have been generally pretty stoked about the progress I made in the beginning up until now.
But heading into the Algebra 1 course, I've started to encounter a lot of exercises involving manipulating formulas like in ( FV=PV(1+r)^n ) where you should find r.
In some cases I can kind of work out a solution but in this particular case I have no idea where to start.
Aside from that I can't grasp how to solve and work with units and unit word problems. I'm sitting here watching hours and hours of videos but I just can't understand what's going on.
Especially when you're supposed to start dividing different values and use fractions.
I understand how to calculate fractions and cross-multiply but having a word problem and seeing two fractions next to one another is a totally different thing.
Other than that I haven't encountered anything horribly difficult but there's something incredibly difficult about this particular part and I don't intend to leave any gaps.
​​
If anyone could give some pointers I'd be extremely grateful..
God knows why I can learn any language in the history of the world without problems but can't for the heck of it be "good" at maths.
But heading into the Algebra 1 course, I've started to encounter a lot of exercises involving manipulating formulas like in ( FV=PV(1+r)^n ) where you should find r.
In some cases I can kind of work out a solution but in this particular case I have no idea where to start.
Aside from that I can't grasp how to solve and work with units and unit word problems. I'm sitting here watching hours and hours of videos but I just can't understand what's going on.
Especially when you're supposed to start dividing different values and use fractions.
I understand how to calculate fractions and cross-multiply but having a word problem and seeing two fractions next to one another is a totally different thing.
Other than that I haven't encountered anything horribly difficult but there's something incredibly difficult about this particular part and I don't intend to leave any gaps.
​​
If anyone could give some pointers I'd be extremely grateful..
God knows why I can learn any language in the history of the world without problems but can't for the heck of it be "good" at maths.
franbatista123
Member
It's easy, just calculate g(3) and then if the result is, for example, 5, do f(5).
thesaucetastic
Unconfirmed Member
This is a formula for compound interest, right? Where FV = final value, and PV = present value?
So to get r, we've got to work on isolating it from the rest of the equation. First, we get rid of PV by dividing both sides:
FV/PV = PV/PV * (1+r)^n
FV/PV = (1 + r)^n
Now we've got to get rid of the exponent. Do you know about fractional exponents? Like how you can take the square root of something by raising it to the 1/2th power? Same thing here, except you're raising each side to the 1/n th power.
(FV/PV)^(1/n) = (1 + r)^[
(1/n)](FV/PV)^(1/n) = 1 + r
Then you subtract 1 from both sides and you're done! All you have to do now is to plug in the numbers given to you for FV, PV, and n.
(FV/PV)^(1/n) - 1 = r
For word problems, maybe try finding one you're struggling with and posting it here? Personally, I can't tell you much specific if I don't know the context.
Here's a question I'm having trouble with:
The probability of getting to the i-th block is (1/2)^(i -1), but this doesn't take into account the time you waste getting halfway there and then returning back to the starting point.
Right now you're assuming the cab doesn't take any time to take you back, but for extra points you can assume it takes c*n time, where c is a constant of your choice.
Man I hate statistics.
boviscopophobic
Member
The key observation here is that after you've first returned to the bar, everything about the problem is exactly the same as before, except that you wasted some number of steps beforehand. Let's label the blocks 0, 1, ..., n, with n being home and 0 being the bar. Then if you start at 0, the time T taken to first return to 0 (the "hitting time") can be 2, 3, 4, .., n, or infinity (meaning you never returned to the bar because you arrived at home). If E is the expected time to get home starting from the bar, then the expected total time to get home given that T = k < infinity is equal to E + k (can you supply the argument why?)Here's a question I'm having trouble with:
The probability of getting to the i-th block is (1/2)^(i -1), but this doesn't take into account the time you waste getting halfway there and then returning back to the starting point.
Right now you're assuming the cab doesn't take any time to take you back, but for extra points you can assume it takes c*n time, where c is a constant of your choice.
Man I hate statistics.
Now by enumerating the probabilities of the various hitting times, you can write down and solve an equation for E. I get E = 2^n + 2^(n-1) - 2.
Edit: Note that I'm assuming that the cab doesn't take any travel time, but that movements are "synchronized" hourly. So at t=0 you start at the bar, at t=1 you've made it a block forward. At that point you either call a cab or walk forward another block, meaning that at t=2, you are at the bar (with probability 1/2) or at block 2 (with probability 1/2). It is not entirely clear to me if this is how the problem wording is intended to be interpreted. Regardless, the idea of the analysis is similar even if you change the motion model a bit.
This is a formula for compound interest, right? Where FV = final value, and PV = present value?
So to get r, we've got to work on isolating it from the rest of the equation. First, we get rid of PV by dividing both sides:
FV/PV = PV/PV * (1+r)^n
FV/PV = (1 + r)^n
Now we've got to get rid of the exponent. Do you know about fractional exponents? Like how you can take the square root of something by raising it to the 1/2th power? Same thing here, except you're raising each side to the 1/n th power.
(FV/PV)^(1/n) = (1 + r)^[
(FV/PV)^(1/n) = 1 + r
Then you subtract 1 from both sides and you're done! All you have to do now is to plug in the numbers given to you for FV, PV, and n.
(FV/PV)^(1/n) - 1 = r
For word problems, maybe try finding one you're struggling with and posting it here? Personally, I can't tell you much specific if I don't know the context.
Here's a question I'm having trouble with:
The probability of getting to the i-th block is (1/2)^(i -1), but this doesn't take into account the time you waste getting halfway there and then returning back to the starting point.
Right now you're assuming the cab doesn't take any time to take you back, but for extra points you can assume it takes c*n time, where c is a constant of your choice.
Man I hate statistics.
Thanks for the help with the problem, gonna look up fractional exponents some more; guess it just gets very confusing for me when they mix it up with letters.
And about my last question; take this problem for example:
"A group of 6 farmers harvested 27 square kilometers of farmland over the harvest season.
Each farmer harvested 0.6 square kilometers per day. Working together, they harvested 52,000 bushels of soybeans per day.
- How many bushels of soybeans did the farmers harvest over the whole season?"
I feel like what I'm missing here is not how to solve a problem but not knowing the logical reason behind why the problem is being solved in a certain way?
Easy example to prove my point, I had problems with division for the longest time ( still not perfect ) because all I'd ever been taught was that if you divide X by Y you get C.
Before starting to draw figures to graphically see why that was so, it just wouldn't stick.
Maybe the problems I'm stating here are laughably easy to most people but we all have our difficulties I guess.
Stumpokapow
listen to the mad man
Here's a question I'm having trouble with:
The probability of getting to the i-th block is (1/2)^(i -1), but this doesn't take into account the time you waste getting halfway there and then returning back to the starting point.
Right now you're assuming the cab doesn't take any time to take you back, but for extra points you can assume it takes c*n time, where c is a constant of your choice.
Man I hate statistics.
One strategy might be to solve for n=1 (special case because of the "guaranteed block"), n=2, n=n+1 and then generalize visually. Treat the solution as a sum from 0 to infinity of p(i) t(i). p(i) is probability of taking i cabs, while t(i) is the cost of taking that many cabs. Here's my guess:
Say the man has to walk only one block. How long will it take him to get home?
1 hour.
Say the man has to walk only two blocks (the guaranteed block, and then one random block). How long will it take him to get home?
1. Say he takes no cabs, then it will take him 1 hour + 1 hour = 2 hours to get home (p=0.5)
2. Say he takes one cab, then it will take him 1 Hour + (cab time) + 1 hour + 1 hour = 3 hours + (cab time) to get home (p=0.25)
3. Say he takes two cabs, then it will take him 1 hour + (cab time) + 1 hour + (cab time) + 1 hour + 1 hour = 4 hours + 2*(cab time) to get home (p=0.12)
... (he might take an infinite number of cabs, so collapse this into a sum)
Sigma(i=0 to infinity) p(i) t(i)
p(i) = 0.5^(i+1)
t(i) = 2 + i + ci where c = cab time
Sigma(i=0 to infinity) (0.5^(i+1)) (2 + i + ci)
(i is number of cabs)
Say needs to walk a third, how much time does this add marginally?
Well, say he takes 0 cabs in the final block (p=0.5) then he pays 1 extra hour for his third marginal block. Say he takes 1 cab in the final block (p=0.25) then he pays c + E[f(n-1)] + 1. Say he takes 2 cabs in the final block (p=0.125), then he pays 2c + 2E[f(n-1)] + 1. So total marginal cost of the nth block is: sum(i=0 to infinity) p(i) t(i) = sum(i=0 to infinity) (0.5^(i+1)) 1 + ci + E[f(n-1)]i.
Thus the final cost should be:
f(1) = 1
f
= f(n-1) + sum(i=0 to infinity) (0.5^(i+1)) (1 + ci + E[f(n-1)]i)For c=0 (no cab cost):
f(1) = 1
f(2) = 1 + 2 = 3
f(3) = 3 + 4 = 7
f(4) = 7 + 8 = 15
f(5) = 15 + 16 = 31
f(n>0) = (2^n) - 1
For c=1 (cab costs same as walking a block):
f(1) = 1
f(2) = 1 + 3 = 4
f(3) = 4 + 6 = 10
f(4) = 10 + 12 = 22
f(5) = 22 + 24 = 46
etc.
Hopefully at least some of this is something useful to go on. Did not check for algebraic or sequence calculation errors use this as logical scaffolding, nothing else.
52,000 bushels/day * number of days in a season = number of bushels per season.
How many days did it take them?
0.6 sq km a day / farmer * 6 farmers = 3.6 sq km a day
27 square kilometers / season / 3.6 sq km a day = 7.5 days to get through 27 kilometers (ie to get through a season)
52,000 * 7.5 = 390,000
thesaucetastic
Unconfirmed Member
Thanks for the help, guys. I don't think I completely get it yet, but you've given me a start at least. Here are the hints my prof gave me:
X_i = # of hours needed to get home if he is i blocks away from home, i=0,1,...,n.
Hint: there is a simple set of recurrence relations you can write down for E[X_i]
By enumerating the probabilities, you mean kind of what stump was doing?
X_i = # of hours needed to get home if he is i blocks away from home, i=0,1,...,n.
Hint: there is a simple set of recurrence relations you can write down for E[X_i]
So k is the time to your first return, right? And is E going straight to the house from the bar without any returns (i.e., thing I calculated before)? So k is just accounting for the wasted steps you took before resetting. I assume that's the argument?
By enumerating the probabilities, you mean kind of what stump was doing?
This is basically how I was interpreting the problem, but I realized it might be limiting once you try factoring in cab time (i.e., things wouldn't be synchronized on the hour anymore, because the cab takes time c*i if you made it i blocks before returning).
You know, I was trying to do exactly this, but I didn't recognize they collapsed into sums like that. Man I feel dumb.
Stumpokapow
listen to the mad man
Note that in my interpretation of the problem, say the guy is at block 2 at time t0 and takes a cab, the cab teleports him to the bar, and then one hour later (time t0+1) he's at block 1 (guaranteed block). If the idea is that the cab takes him to the bar and he immediately moves to block 1 (time t0), then that's going to be a different calculation.
Likewise, I assume that if the cab takes time, he is at the bar at time t0 + c, and at block 1 (guaranteed block) at time t0+1+c--so either c is an integer or if c is a fraction, that it does not disrupt the flow of time otherwise (ie there is no "wait until the top of the hour" penalty, he just goes to his next "turn")
Likewise, I assume that if the cab takes time, he is at the bar at time t0 + c, and at block 1 (guaranteed block) at time t0+1+c--so either c is an integer or if c is a fraction, that it does not disrupt the flow of time otherwise (ie there is no "wait until the top of the hour" penalty, he just goes to his next "turn")
boviscopophobic
Member
No, E is the expected time including all possible false starts. Anyway, stump's method is maybe a more elementary way to think about this particular problem -- I'm thinking about it in terms of Markov chains.
The trouble is, if you don't discretize to "every hour on the hour", then there isn't really enough information in the problem to specify how you behave in continuous time. You have a 1/2 probability of turning back "every hour", but it doesn't say anything about (e.g.) when within that hour you turn back, or how you make forward progress during that hour -- and of course, both of those make a difference if you want to specify nonzero travel time for the cab.
thesaucetastic
Unconfirmed Member
Yeah, I asked my professor about it being a Markov chain, but he said not to think of it in terms of that, so I'm guessing he's looking for a more elementary approach.
But yeah, it's not very specific about the time frame. I'm just going to assume it's an hour from the point of time you arrive at the bar. So if you got to the bar at 1:05, you'd be at block 1 at 2:05.
spelltropy
Member
i was giving this differential geometry question a go last night, but kept getting 0 as my answer using the transformation, so was wondering if someone could check. I thought it might be the fact that i had been looking at geometry questions for a good 10/11 hours, but I just tried it again and still got 0:
Evaluating the double integral of xy dx dy over the square omega with vertices (0,0), (1,1), (2,0), (1,-1) using the transformation x=(u+v)/2, y=(u-v)/2, and directly.
I found the jacobian to be -1/2, and the u,v limits both to be 0 and 2. For the "directly" part I used x,y limits of (0,2) and (-1,1) respectively, not sure if I'm beating around the wrong bush here, but the answer came out as 2.
Also, if you're faced with a question such as "Parametrise this shape in three different ways", what kind of process do you go through to figure them out. Eg, a 'simple' one that cropped up was a circle with centre (1,2) and radius 3 - I got (1+3cost,2+3sint), not sure what else?
Evaluating the double integral of xy dx dy over the square omega with vertices (0,0), (1,1), (2,0), (1,-1) using the transformation x=(u+v)/2, y=(u-v)/2, and directly.
I found the jacobian to be -1/2, and the u,v limits both to be 0 and 2. For the "directly" part I used x,y limits of (0,2) and (-1,1) respectively, not sure if I'm beating around the wrong bush here, but the answer came out as 2.
Also, if you're faced with a question such as "Parametrise this shape in three different ways", what kind of process do you go through to figure them out. Eg, a 'simple' one that cropped up was a circle with centre (1,2) and radius 3 - I got (1+3cost,2+3sint), not sure what else?
boviscopophobic
Member
The region of integration is symmetric about the x axis, and the integrand has odd symmetry about the x axis, so the answer is 0. For the direct approach, your region of integration is incorrect, because it describes a square with sides parallel to the x and y axes, rather than rotated 45 degrees as in the problem statement. However, even with the incorrect region you should still get 0, so you should double check your work.
One trivial way to get alternate parameterizations is to alter the "speed". In your parameterization, the complete curve is described when t runs from 0 to 2pi. If you replace t with a different monotone function that also ranges from 0 to 2pi, then you can trace out the same curve with a different velocity profile. Depending on the problem, there may also be more natural, less trivial ways to get a different parameterization, but I'm not getting any brilliant ideas at the moment.
I'm not entirely sure that I am understanding your first question correctly, but those four vertices are (x,y)-coordinates, right? If so, the bounds of integration for x should depend on y since the lines bounding the region are not parallel to the axes.
As far as the second question, I will assume that a "shape" is always some kind of closed curve. If so, then any parametrization must produce the same points, so the only things you can really change are the speed and direction at which you traverse the curve, or the starting/ending point. So for your example, you might have the parametrizations:
(1+3cos t, 2+3sin t) where t is in [0, 2pi)
(1+3cos t, 2-3sin t) where t is in [0, 2pi)
(1+3cos t^2, 2+3sin t^2) where t is in [0, sqrt(2pi))
(1+3cos (t+pi/2), 2+3sin (t+pi/2)) where t is in [0, 2pi)
The first is your original parametrization, the second traverses the curve in the opposite direction, the third traverses the curve at a non-constant speed, and the fourth has a different starting point.
spelltropy
Member
About the first: yep, the vertices are in (x,y) coordinates. To convert them (for completing the question using transformations/Jacobian), I figured the region consists of four lines, equations of which are given by y=x, y=-x, y=x-2, y=x-2. Putting x=(u+v)/2 and y=(u-v)/2 into these, I got u=0, v=0, u=2, v=2, so they were my new bounds for the integral. Is this wrong?
For "directly" solving the question, how can I represent the x bound in terms of y? I see where you're coming from - do i have to split the integral into two? one for the line where y=x, then another for y=2-x?
Second question: thats exactly what I was trying to think of - I guess i just need to spend some more time looking at curves and how they change when t is replaced with t^2 and such. Cheers!
You and me both buddy
Linear Algebra:
My teacher said that in the Matrix Equation Ax = b, the # columns in the matrix A have to equal the # rows in the vector x, in order for Ax to be equal to b. This makes sense, but thinking about it geometrically, it seems like if A and x are mismatched in the way stated above, you should still be able to find a solution in some cases. Let's pretend this is in 3-space with x,y, and z axis'.
In this case the resultant vector b isn't using the 3rd dimension z. Couldn't you just consider the vector x to be a vector in 3-space with the 0 vector as it's z component? That would make the equation solvable, and give a free variable for the 3rd column of A. In the case where we make the free variable equal to 0, the solution should exist, I think, as a 2-dimensional vector defined in 3-space.
If none of this makes any sense, I'm sorry, I'm on like 2 hours of sleep.
My teacher said that in the Matrix Equation Ax = b, the # columns in the matrix A have to equal the # rows in the vector x, in order for Ax to be equal to b. This makes sense, but thinking about it geometrically, it seems like if A and x are mismatched in the way stated above, you should still be able to find a solution in some cases. Let's pretend this is in 3-space with x,y, and z axis'.
In this case the resultant vector b isn't using the 3rd dimension z. Couldn't you just consider the vector x to be a vector in 3-space with the 0 vector as it's z component? That would make the equation solvable, and give a free variable for the 3rd column of A. In the case where we make the free variable equal to 0, the solution should exist, I think, as a 2-dimensional vector defined in 3-space.
If none of this makes any sense, I'm sorry, I'm on like 2 hours of sleep.
Stumpokapow
listen to the mad man
Trying to understand. I don't think you're not making sense, but I am not understanding what you're arguing.
The matrices presented are non-conformable under multiplication. So it's not clear how you would process the multiplication. You appear to be arguing that we should treat an n x m dimensional matrix (in this case n x 1) as an n* x m matrix padded with 0s to arbitrary size in order to make it conformable to anything left-multiplying by it? So then if the left matrix was a 3x8000 matrix, you would treat this vector as an 8000x1 vector with 0s throughout except the first two?
In this case B_(1,1) (the field marked m in your example) would be equal to ak + bl + c*(???? what would go here in your hypothetical)? You appear to be treating it as a 0? So then the c and f terms are wholly irrelevant in this case; they play no role in the resulting vector B. Rather than pretending that the vector you call X has an invisible 0, why not simply left-multiply X with the two left columns of A as appropriate? I would ask then what the substantive meaning of c and f actually are? Like, filling in values for the left matrix and the middle vector, what do you think the right vector is telling you in your scenario?
My linear algebra is pretty weak though.
The matrices presented are non-conformable under multiplication. So it's not clear how you would process the multiplication. You appear to be arguing that we should treat an n x m dimensional matrix (in this case n x 1) as an n* x m matrix padded with 0s to arbitrary size in order to make it conformable to anything left-multiplying by it? So then if the left matrix was a 3x8000 matrix, you would treat this vector as an 8000x1 vector with 0s throughout except the first two?
In this case B_(1,1) (the field marked m in your example) would be equal to ak + bl + c*(???? what would go here in your hypothetical)? You appear to be treating it as a 0? So then the c and f terms are wholly irrelevant in this case; they play no role in the resulting vector B. Rather than pretending that the vector you call X has an invisible 0, why not simply left-multiply X with the two left columns of A as appropriate? I would ask then what the substantive meaning of c and f actually are? Like, filling in values for the left matrix and the middle vector, what do you think the right vector is telling you in your scenario?
My linear algebra is pretty weak though.
It sounds like you're understanding me correctly. I'm just envisioning a 3-dimensional space, but any solutions would exist in a 2-d plane. The table in front of me has a 2-d surface but it's in 3-dimensional space. Any vectors that existed on the surface of the table wouldn't have vertical components, (or the coefficent on the vertical components would be zero, depending on how things are defined).
Is there a way to describe a 2-d vector within a higher dimensional space in this way?
I'm not sure what the c and f would represent, but I'm not sure what coefficients in the row containing a free variable represent for ANY system, really. I'm only in the middle of the first chapter of our book, maybe some of this stuff gets answered later.
DEAD RABBIT
Member
Linear Algebra:
My teacher said that in the Matrix Equation Ax = b, the # columns in the matrix A have to equal the # rows in the vector x, in order for Ax to be equal to b. This makes sense, but thinking about it geometrically, it seems like if A and x are mismatched in the way stated above, you should still be able to find a solution in some cases. Let's pretend this is in 3-space with x,y, and z axis'.
In this case the resultant vector b isn't using the 3rd dimension z. Couldn't you just consider the vector x to be a vector in 3-space with the 0 vector as it's z component? That would make the equation solvable, and give a free variable for the 3rd column of A. In the case where we make the free variable equal to 0, the solution should exist, I think, as a 2-dimensional vector defined in 3-space.
If none of this makes any sense, I'm sorry, I'm on like 2 hours of sleep.
Well, you can invert your view of the problem and consider first your equation as two constrained equations and as long as #columns matrix = #rows vector, the matrix product operation is applicable.
Whether it is solvable depends on what your free variables are, I assume k and l. If this 'third dimension' variable (lets say q) is fixed, then the equations are solvable. It is equivalent to subtracting constant values q * c and q * f from m and n respectively.
How to interpret this geometrically is a different story.
On the one hand, considering Euclidean space, what you are essentially doing is masquerading a translation (which is computed by vector addition/subtraction) as a matrix operation. This confuses the geometric meaning, because matrices are usually used for rotating and scaling. Your parameter space is 3D (k,l,q), but your geometric space is still 2D (k,l). You can show this to yourself by rewriting your '3D' matrix product as a sum of a 2D matrix product with a 2D vector. In geometric terms, you multiply your input vector (k,l) with matrix (a,b,d,e) and then add a vector to q*(c,f) to it, where k,l are geometric variables and q is a scaling variable for your translation vector.
On the other hand, in more abstract mathematics they intentionally handle matrices and vectors like you have. For instance in the study of conic sections (parabola, hyperbola, conics) they use homogeneous vectors as a basis to define the implicit equation that defines the Euclidean shape.
The matrix equation in your earlier post is not well-defined, but I think you are on track of something that is of interest in academia right now. It's called compressive sensing (CS).
Traditional linear algebra says that the equation Ax = b has infinitely many solutions if A has fewer rows than columns--really, if the size of b, the observation/output, is less than the size of x, the source/input. We say that the equation is underdetermined.
CS tells us that, no, if we know a priori that x is sparse (has a lot of zeros, and in essence, lives in a lower dimension), then we can still get a unique solution by solving the l1-minimization problem x* = argmin ||Ax - b||_{1}, and that l1-minimization works better than any other lp-minimization with p > 1.
It's easy to explain why l1 works pictorially. As shown below, the equation Ax = b forms a hyperplane--a line in 2D, a plane in 3D, etc. We can think of the lp-minimization problem as drawing a sequence of lp-"balls" until it hits the hyperplane. (A ball is the set of all vectors whose size, measured in the lp-norm, is at most the ball's radius. Note that we really should be starting out with a ball of infinite size and shrinking it down just until it no longer intersects the hyperplane---the opposite of what I've drawn below.)
We see that using the l1-norm, it's very likely that the minimizer x* will land on one of the axes and result in having a lot of zero entries. With the l2-norm (and same for all other lp-norms for p > 1), this is not the case.
That's the most I can explain without embarrassing myself cause I don't understand the theory too much, haha. As you can see, CS is a multi-disciplinary subject that combines linear algebra, convex optimization, and probability theory, so I find that quite interesting. Perhaps you do too and will go on learning more about it.
Edit: My bad, I had made a blunder of thinking about the overconstrained system, not the underdetermined system.
This is to settle an argument between 2 of my friends. What is the probability of rolling at least 2 sixes on 3 dice? Friend A says it is approximately 16/216(7.41%) due to binomial distribution. Friend B says it is 1/36(2.78%) due to only needing the probability of getting sixes on 2 dice. They need an arbiter on this, and I'm choosing GAF.
Wouldn't it be 5/216? Total combination is 6*6*6=216. 1/6 for dice 1, 1/6 for dice 2, 5/6 for dice 3. Edit: Never mind, got confused.
Binomial distribution. Friend B is wrong because while you only need to get sixes on two dice, having a third dice increases your chances of getting at least two sixes.
Expanding on this: dice 1 and 2 gets six, dice 3 has 5 choices without 6. Same thing if you think of combination 1 and 3, and 2 and 3. That is 5*3=15. Finally, all 3 dices are 6. So the total number of choices for at least 2 dices with 6 is 16. The total number of combination is 216. Therefore 16/216.
Stumpokapow
listen to the mad man
2 sixes or at least 2 sixes?
Binomial RV:
1/6 probability of success (p=1/6)
3 trials (n=3)
2 successes or 3 successes (i=2 or i=3)
P(X=2) = (3 choose 2) (1/6)^2 (5/6) = 0.069444
P(X=3) = (3 choose 3) (1/6)^3 = 0.004629
At least two sixes: 0.074 (7.41%)
Exactly two sixes: 0.069444 (6.94%)
Friend B is not correct, because he fails to take into account the fact that the two sixes might be on the first two dice, the first and third dice, or the last two dice, or that they might occur on all three dice.
MoG EclipsE
Member
Hey guys, I am in serious need of help. I have no damn clue how to do this problem, and I have a final in this class in a couple of days. halp This is the last question on my last assignment, any help would be greatly appreciated.
All temperatures are in degrees celcius, not kelvin, that is a mistake. Not that it really matters though
Also note: the initial temperature of the 3m x 3m plate is 10 degrees and the temp. around the plate is given:
Thank you in advance for any help!
Edit: Small mistake, the equation should read:
This is the last question on my last assignment, any help would be greatly appreciated.All temperatures are in degrees celcius, not kelvin, that is a mistake. Not that it really matters though
Also note: the initial temperature of the 3m x 3m plate is 10 degrees and the temp. around the plate is given:
Thank you in advance for any help!
Edit: Small mistake, the equation should read:
Hey guys, I am in serious need of help. I have no damn clue how to do this problem, and I have a final in this class in a couple of days. halp
All temperatures are in degrees celcius, not kelvin, that is a mistake. Not that it really matters though
Also note: the initial temperature of the 3m x 3m plate is 10 degrees and the temp. around the plate is given:
Thank you in advance for any help!
Edit: Small mistake, the equation should read:
Since we are to use first-order forward integration in time, we have,
We know T(x, y; t = 0), as it's given by the initial condition. In particular, the initial temperature is 10 degrees Celsius at the four points.
All that remains is to evaluate the time derivative (for convenience, I'll drop the argument t = 0) while using the central finite difference scheme to approximate the Laplacian operator. Recall the 5-point stencil:
I've written the formula above in the usual notation; it looks complicated, but it's really not. The (x_i, y_j) is the point at which we want to evaluate the Laplacian. Central scheme says we can make a second-order approximation using 5 nearby points including itself. The -1 in the index means look to the left/below of this point, while +1 means look to the right/above of this point. h is the stencil size--assumed to be the same in x and y directions--and is equal to 1 m in your problem, so I'll omit this number below. (Make sure that the units are consistent for real problems.)
For example, according to the scheme, the Laplacian at point #1 is,
20 + 10 + 10 + 30 - 4(10) = 30,
while at point #2, it is,
10 + 50 + 10 + 30 - 4(10) = 60.
Now you have all the information needed to approximate the time derivative at t = 0 at the four points:
You would repeat the procedure above, and use the information from t = 0.1 to approximate the temperature at t = 0.2 at point #3. The problem doesn't explicitly say this, but the temperature on the boundary is fixed in time, so we do not need to update the temperature for the boundary points at t = 0.1.
I'm having a problem with my discrete math homework,
I have to prove this:
I know that x ( x + 1 )... (x + n ) = x^(n+1 upperscore) and that factorial n can be represented as x^(x_) or 1^(x upperscore)
The previous exercise made me prove this:
and apparently it is required for this part. (I did this last demonstration fine)
Also, I don't know how to search the definitions in english, how is this called?
n^(4_) = n * (n-1) * (n-2) * (n-3)
n^(4 upperscore) = n * (n+1) * (n+2) * (n+3)
in spanish it is called caida and levante, but I CAN'T FIND ANYTHING ABOUT THIS ONLINE
I have to prove this:
I know that x ( x + 1 )... (x + n ) = x^(n+1 upperscore) and that factorial n can be represented as x^(x_) or 1^(x upperscore)
The previous exercise made me prove this:
and apparently it is required for this part. (I did this last demonstration fine)
Also, I don't know how to search the definitions in english, how is this called?
n^(4_) = n * (n-1) * (n-2) * (n-3)
n^(4 upperscore) = n * (n+1) * (n+2) * (n+3)
in spanish it is called caida and levante, but I CAN'T FIND ANYTHING ABOUT THIS ONLINE
I'm having a problem with my discrete math homework,
I have to prove this:
I know that x ( x + 1 )... (x + n ) = x^(n+1 upperscore) and that factorial n can be represented as x^(x_) or 1^(x upperscore)
The previous exercise made me prove this:
and apparently it is required for this part. (I did this last demonstration fine)
Also, I don't know how to search the definitions in english, how is this called?
n^(4_) = n * (n-1) * (n-2) * (n-3)
n^(4 upperscore) = n * (n+1) * (n+2) * (n+3)
in spanish it is called caida and levante, but I CAN'T FIND ANYTHING ABOUT THIS ONLINE
I don't think we use those notations in the US. Pretty cool names, though. The fall and the lift, heh. The caida matches the permutation operator, where
P(n, k) := n! / (n - k)! =
* (n - 1) * ... * (n - k + 1),and we can get the levante by shifting the index and considering
P(n + k - 1, k) = (n + k - 1) * (n + k - 2) * ... *
.But I'm not quite sure what you mean by Delta^n and (x - 1)^(-1,underscore). Do you mean, apply the Laplacian operator (second derivative with respect to x) n times to the function (x - 1)^(-1,underscore)? Can you tell us how we evaluate the caida for a negative power?
I suspect that the answer to your problem lies with the Leibniz rule, but I'm having trouble seeing what the functions f and g should be, and how we are to use your previous problem.
- Status