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The Math Help Thread

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F

FunkyMunkey

Banned
RJNavarrete said:
Solve the proportion.

2ep5zk0.jpg


I've never really seen a fraction written like that. wtf?
Click to expand...

Dividing is the same as being multiplied by the reciprocal, so

(2/5)*(1/y) = (1/5)*(1/11) [which is 1/55]

Divide both sides by 2/5 and you get (1/y) = (5/110) [which is (1/55)*(5/2)]

Multiply each side by y,

1 = (5/110)*y

then divide both by (5/110) [which is 1*(110/5) = y]

y = 22

edit: I over-explained it but I do that a lot so eh haha. reciprocals are invaluable
 
N

narcosis219

Member
RJNavarrete said:
Solve the proportion.

2ep5zk0.jpg


I've never really seen a fraction written like that. wtf?
Click to expand...

You can solve that by just looking at it.

Look at the numerator. It's twice as big as the one on the right. So, it follows that the denominator must be also twice as big as its counterpart on the right. This is true because they are proportional.

y = 22

Or, just solve for y lol
 
R

Rich Uncle Skeleton

Member
mcrae said:
wait, if your debt ceases to exist when you turn 65, why would you ever repay any of it?
(i obviously dont understand what you wrote)
Click to expand...

Maybe he means for there to be a minimum monthly payment, in which case I suppose it depends on what that minimum is. If you multiply it by the number of payments you must make before you turn 65, you find the least you can pay without actually paying off the loan. Then you should compare this to the least you can pay if you do pay off the loan (which will be dependent on how quickly you're able to pay it off - the faster the better. And you'd need to factor in the lost interest of paying sooner rather than later.) It's difficult without actually fixing all the variables involved, but I think this would be the general approach.

mcrae said:
can you explain more? the chances of 3 dice rolling 3 1's is 1/6*1/6*1/6= 1/216, and the chances of 3 dices rolling a 1, a 2, and a 3, are 3/6*2/6*1/6 = 1/36, or 6 times as likely
Click to expand...

I think he means fix a number of dice, say n, and then fix a multiset of digits in {1, 2, ..., 6} of cardinality i between 1 and n. For example, if you roll 5 dice what are the chances of having a pair of 2s show up (maybe he's a Yahtzee player :D ).

As the guy said, one way to go about this is to count how many rolls give the desired outcome and divide by the total number of rolls which is 6^n. Well you need to have your desired outcome appear, and then the other n-i dice are free to be whatever, so if you fixed which dice are supposed to be which (i.e. the first two dice I roll must be 2s) then there are 6^(n-i) rolls that have your desired outcome. So really the problem, as you noted, boils down to how many ways the ordering of things can be changed. First of all, you have to count how many pairs of dice there are among the 5 (i.e. you want to allow for the last two to be 2s, or the first and third, etc.). This is 5 choose 2, or more generally n choose i (look at Wikipedia's entry on binomial coefficient to see what I mean by n choose i; it's just the number of ways of choosing i things from n possible choices). Then you need to figure out how many permutations there are of your multiset. In the case 1-1-1 there's only one, but in the case 1-2-3 there are 6 (3-2-1, 2-1-3, etc.). This is a bit harder. Let k be the number of distinct digits in your desired outcome (i.e. in the case 1-1-1 there's only the digit 1; in 1-2-3 there are 3), and let m_k be the number of appearances of the k^th digit (In 1-1-1, we have m_1 = 3, and in 1-2-3 we have m_1=m_2=m_3=1). You need to know how many ways there are of choosing which slots are assigned to each of the digits. It seems to me that's

P = (i choose m_1) x (i-m_1 choose m_2) x ... x (i-m_1-...-m_(k-2) choose m_(k-1))

That is, choose where you want the m_1 appearances of your 1st digit to be, then choose among the remaining positions where you want your 2nd digit to appear, and so on. You don't have to choose where your kth digit will go since the positions of the others force it. Putting it all together I think the answer to your question is that the probability is 6^(-i) x (n choose i) x P [edit: This is false. This number is much larger than 1 for very large n. I might try to figure out how to fix it later, or maybe someone else will give a correct answer.], where P is that weird number I defined above. I have no idea if the above is right, so corrections welcome. :D
 
J

jepense

Member
Rich Uncle Skeleton said:
I think he means fix a number of dice, say n, and then fix a multiset of digits in {1, 2, ..., 6} of cardinality i between 1 and n. For example, if you roll 5 dice what are the chances of having a pair of 2s show up (maybe he's a Yahtzee player :D ).

As the guy said, one way to go about this is to count how many rolls give the desired outcome and divide by the total number of rolls which is 6^n. Well you need to have your desired outcome appear, and then the other n-i dice are free to be whatever, so if you fixed which dice are supposed to be which (i.e. the first two dice I roll must be 2s) then there are 6^(n-i) rolls that have your desired outcome. So really the problem, as you noted, boils down to how many ways the ordering of things can be changed. First of all, you have to count how many pairs of dice there are among the 5 (i.e. you want to allow for the last two to be 2s, or the first and third, etc.). This is 5 choose 2, or more generally n choose i (look at Wikipedia's entry on binomial coefficient to see what I mean by n choose i; it's just the number of ways of choosing i things from n possible choices). Then you need to figure out how many permutations there are of your multiset. In the case 1-1-1 there's only one, but in the case 1-2-3 there are 6 (3-2-1, 2-1-3, etc.). This is a bit harder. Let k be the number of distinct digits in your desired outcome (i.e. in the case 1-1-1 there's only the digit 1; in 1-2-3 there are 3), and let m_k be the number of appearances of the k^th digit (In 1-1-1, we have m_1 = 3, and in 1-2-3 we have m_1=m_2=m_3=1). You need to know how many ways there are of choosing which slots are assigned to each of the digits. It seems to me that's

P = (i choose m_1) x (i-m_1 choose m_2) x ... x (i-m_1-...-m_(k-2) choose m_(k-1))

That is, choose where you want the m_1 appearances of your 1st digit to be, then choose among the remaining positions where you want your 2nd digit to appear, and so on. You don't have to choose where your kth digit will go since the positions of the others force it. Putting it all together I think the answer to your question is that the probability is 6^(-i) x (n choose i) x P, where P is that weird number I defined above. I have no idea if the above is right, so corrections welcome. :D
Click to expand...

Won't this double count some configurations? E.g., if you want to see how many ways can one get a pair of 1s in three dice, {1, 1, a}, {1, b, 1} and {c, 1, 1} are valid for any a, b, c. But if they are a=b=c=1, you get the same configuration: {1, 1, 1}. So in fact there are only 6+5+5 valid sets, not the full 6+6+6.
 
D

Dogenzaka

Banned
._.

Math never works the way I want it to.

So I have an equation.

(1/4)^x = 2^(x+10)

Find x.

Soooo....I take the natural logarithm of both sides.

ln (1/4)^x = ln (2^(x+10))

Pull out the exponents

x ln (1/4) = x ln (2) + 10 ln (2)

So I guess I want to move all the x's to one side?

-(x ln (2)) on both sides so..

x ln (1/4) - x ln (2) = 10 ln (2)

Distribute out the x

x (ln (1/4) - ln (2)) = 10 ln (2)
x (ln (1/8)) = 10 ln (2)

x ln (1/8) = 10 ln (2)

So divide out to leave the x alone.

(x ln (1/8))/(ln (1/8)) = 10 ln (2)/ln (1/8)

x = 10 ln (2)/ln (1/8)

I get -3.333

But my answer key says the answer's -10/3...WTF

EDIT: Oh wait holy crap :lol :lol must be late for me. -10/3 = -3.3333333
 
D

Dogenzaka

Banned
Our teacher never told us how to do this...so I need help.

I have ln(5) x - ln(2x-1) = ln(4)

Normally I'd be fine with this equation but I don't know how the hell to extract that x out of the ln(2x-1) term...so I don't really know how to approach this.
 
C

close to the edge

Member
Dogenzaka said:
Our teacher never told us how to do this...so I need help.

I have ln(5) x - ln(2x-1) = ln(4)

Normally I'd be fine with this equation but I don't know how the hell to extract that x out of the ln(2x-1) term...so I don't really know how to approach this.
Click to expand...
The inverse function for the natural logarithm is e^x and because every part of the equation has the ln, they all cancel each other out, which leaves you with a normal equation.
 
D

Dogenzaka

Banned
close to the edge said:
The inverse function for the natural logarithm is e^x and because every part of the equation has the ln, they all cancel each other out, which leaves you with a normal equation.
Click to expand...

Wait...so are you telling me ANYTIME I have an equation with ALL ln(blahblah), I can cancel the ln's out and just work with what's in the parenthesis?

ln(4) + ln(2x+3) - ln(4-3x) = ln(15) would for example...become...

4 + (2x+3) - (4-3x) = 15?
 
H

harriet the spy

Member
Dogenzaka said:
Our teacher never told us how to do this...so I need help.

I have ln(5) x - ln(2x-1) = ln(4)

Normally I'd be fine with this equation but I don't know how the hell to extract that x out of the ln(2x-1) term...so I don't really know how to approach this.
Click to expand...

ln(5) x or ln(5x)?
 
H

harriet the spy

Member
Dogenzaka said:
Wait...so are you telling me ANYTIME I have an equation with ALL ln(blahblah), I can cancel the ln's out and just work with what's in the parenthesis?

ln(4) + ln(2x+3) - ln(4-3x) = ln(15) would for example...become...

4 + (2x+3) - (4-3x) = 15?
Click to expand...
No no no!

ln(a*b)=ln(a)+ln(b)

so ln(4)+ln(2x+3)=ln(8x+12)

In the example above, you would write

ln(4*(2x+3))=ln(15*(4-3x))
and since ln is one-to one
this implies
4*(2x+3)=15*(4-3x)
 
D

Dogenzaka

Banned
harriet the spy said:
No no no!

ln(a*b)=ln(a)+ln(b)

so ln(4)+ln(2x+3)=ln(8x+12)

In the example above, you would write

ln(4*(2x+3))=ln(15*(4-3x))
and since ln is one-to one
this implies
4*(2x+3)=15*(4-3x)
Click to expand...

So....you're saying....

ln(5) x - ln(2x-1) = ln(4)

uhhh....but I can't just cancel everything out because there's that lonely annoying x right there that messes with the logarithmic terms.

Unless you're saying I can just do

x ln(10x-5) = ln(4)

Which becomes x e^(ln(10x-5)) = e^(ln(4))

Which then cancels out and becomes x(10x-5) = 4

Which becomes 10x^2-5x = 4?

Which then becomes a quadratic equation at 10x^2-5x-4 = 0?
 
P

phantom13

Neo Member
Dogenzaka said:
So....you're saying....

ln(5) x - ln(2x-1) = ln(4)

uhhh....but I can't just cancel everything out because there's that lonely annoying x right there that messes with the logarithmic terms.

Unless you're saying I can just do

x ln(10x-5) = ln(4)
Click to expand...

Unfortunately, no you cant. xln(5) is actually equal to ln(5^x) so you'd get ln(5^x(2x-1)) = ln(4).
 
S

SDMz

Neo Member
An easy way to remember how to work with logs (well it's the way I remembered it)

Since e^(a + b) = e^a * e^b holds it would make sense for the following to hold

ln(e^a * e^b) = ln( e ^(a + b)) = a + b = ln (e^a) + ln(e^b)

With this you can remember that a multiplication inside a logarithm translates to the sum of the logarithm of the products.

Another useful tool with logarithms comes from the equality (e^a)^b = e^(ab). Then it would make sense for the following to hold:

ln((e^a)^b) = ln(e^(ab)) = ab = b* ln(e^a)

Here you see if you take something to the power b inside the logarithm it's the same as multiplying the logarithm with b.

another thing of note is the effect of minus signs:

ln( x) - ln(y) = ln(x) + ln(y^(-1)) = ln(x* y^(-1)) = ln(x/y)

So as phantom13 noted we get x*ln(5) - ln(2x-1) = ln(5^x / (2x -1)) and you'll need to solve

e^(x*ln(5)/(2x-1)) = 5^x/(2x-1) = 4
 
F

Fuzzery

Member
God dammit, I just read this as the "Meth Help Thread," clicked on it, and was thinking of something to post. Then I saw the posts :lol
 
M

Minamu

Member
Rich Uncle Skeleton said:
Maybe he means for there to be a minimum monthly payment, in which case I suppose it depends on what that minimum is. If you multiply it by the number of payments you must make before you turn 65, you find the least you can pay without actually paying off the loan. Then you should compare this to the least you can pay if you do pay off the loan (which will be dependent on how quickly you're able to pay it off - the faster the better. And you'd need to factor in the lost interest of paying sooner rather than later.) It's difficult without actually fixing all the variables involved, but I think this would be the general approach.
Click to expand...
Sounds about right :) I suppose I'd need to know some numbers to get it working. Just looking for some general formula. I don't know what's stopping one from not paying a single cent but I don't think the loaner would be very thrilled :lol Getting additional bank loans etc would be difficult.
 
J

jepense

Member
SDMz said:
So as phantom13 noted we get x*ln(5) - ln(2x-1) = ln(5^x / (2x -1)) and you'll need to solve

e^(x*ln(5)/(2x-1)) = 5^x/(2x-1) = 4
Click to expand...
In general, if you have equations combining exponentials, logarithms and algebraic terms (of the type e^x = -x or 10 ln x = x), they are transcendental and the solutions cannot be written explicitly.
If you want x, find it numerically.
 
R

Rich Uncle Skeleton

Member
jepense said:
Won't this double count some configurations? E.g., if you want to see how many ways can one get a pair of 1s in three dice, {1, 1, a}, {1, b, 1} and {c, 1, 1} are valid for any a, b, c. But if they are a=b=c=1, you get the same configuration: {1, 1, 1}. So in fact there are only 6+5+5 valid sets, not the full 6+6+6.
Click to expand...

Definitely. Realized this last night. So it ends up giving probability much larger than 1. :lol I'm not sure how to fix it.

Edit: OK, let's see if this works. One way to handle the 1-1-1 case is to see how many rolls there are with exactly two occurrences of 1 (15) and add it to the number of rolls with three 1s (1). More generally, if k and m_i are as I first defined them, let l_i be between m_i and n and consider the number N(l_1,...,l_k) of rolls with exactly l_i occurrences of the i^th digit. I claim that, letting L = l_1 + ... + l_k and P(l_1,...l_k) be the number of permutations of a multiset with l_i occurrences of its i^th member, this is

N(l_1,...,l_k) = (n choose L) P(l_1,...,l_k) (6-k)^(n-L).

Then if we sum up N(l_1,...,l_k) over all k-tuples with l_i between m_i and n and L less than or equal to n, we should get what we want. Call it N. Then the desired probability is 6^(-n)N. Of course, this formula is not very friendly-looking; even if it's correct, there's probably a nicer answer. :D
 
A

Aeris130

Member
Not really sure if I'm understanding what l_i is correctly. If I'm trying to roll [1,4,4,5] in any order:

m_1=1, m_2=2, m_3=1

m_1=1, l_2=2, l_3=2, l_4=1, L=6

? Am I missing something?
 
R

Rich Uncle Skeleton

Member
Aeris130 said:
Not really sure if I'm understanding what l_i is correctly. If I'm trying to roll [1,4,4,5] in any order:

m_1=1, m_2=2, m_3=1

m_1=1, l_2=2, l_3=2, l_4=1, L=6

? Am I missing something?
Click to expand...

Sorry for the confusion. You're right that m_1 = 1, m_2 = 2, and m_3 = 1, which is to say we need at least one 1, two 4s, and one 5. It's possible however that we'll have extra 1s, 4s, or 5s. As jepense pointed out, we have to be careful about double counting these, which is what I hope summing over the k-tuples (l_1,...,l_k) helps with. What we want to do is first consider the rolls where there is exactly one 1, two 4s, and one 5. If there are say 6 dice total, the other two dice are allowed to be anything other than 1, 4, or 5. And there are 12 ways of rearranging 1, 4, 4, 5 (six choices for where the two 4s go, and then two orderings on 1 and 5 for each of these), so there are 12 x 3^2 such rolls. Now do the same sort of reasoning but when there are exactly two 1s, two 4s, and one 5. Then we'll have handled l_1 = 1, l_2 = 2, l_3 = 1 and l_1 = 2, l_2 = 2, l_3 = 1. If we keep going and do this for ever triple (l_1, l_2, l_3) with l_1 at least 1, l_2 at least 2, l_3 at least 1 (so we have at least as many 1s, 4s, and 5s as we need), and their sum is not more than n (we better not have more occurrences of 1s, 4s, and 5s than we have dice), and sum up all the valid rolls we find, we'll see how many rolls there are that work (hopefully without double counting any). Dividing this with the total number of rolls, 6^6, and we get the probability.
 
R

Rich Uncle Skeleton

Member
LethaL ImpuLse said:
Can anyone help me find the derivatives of these functions?

1. y = sin^3(pix-2)

2. y = (sinx)^x
Click to expand...

I take it you've learned the chain rule recently. What you want to do is write the given functions as compositions of functions whose derivatives you know, i.e. define functions f and g (or f, g, and h) such that y = f(g(x)) (or y = f(g(h(x))) ). There is a bit of an art to this. One way to think about it is suppose I asked you to evaluate the function at say x = 0; what would you have to do first? Whatever you do first is your innermost function, just as if you plugged x = 0 into f(g(h(x))), the first thing you'd have to do is evaluate h(0). So for example, in 1, the first thing you'd do is find pi(0) - 2, so maybe you should choose h(x) = pix - 2. What should g and f be? Keep reasoning the same way. Once you have them written as compositions you can start applying the chain rule.
 
Q

quilter46761

Member
LethaL ImpuLse said:
Can anyone help me find the derivatives of these functions?

1. y = sin^3(pix-2)

2. y = (sinx)^x
Click to expand...


Building, think of these as 'stuff' problems.

derivative of (sinx)^x = derivative of ((stuff)^x) * (derivative of stuff)
so, derivative of (sinx)^x = ((sinx)^x * ln sinx * 1) * (cosx)
derivative of a power function * derivative of sinx

When you have multiple steps, the trick I learned was to think of the multiple steps as vegetables.

derivative of (sin (pi*x - 2))^3 = derivative of (( potatoes )^3) * derivative of (sin (tomatoes)) * derivative of tomatoes

where potatoes are (sin (pi*x - 2)) and tomatoes are (pi*x - 2)
don't mutilate the vegetables until they're ready

derivative of (sin (pi*x - 2))^3 = (3 (sin (pi*x - 2))2) * (cos (pi*x - 2)) * (pi - 0 )
simplifies to (6pi(sin(pi*x -2)^2)(cos(pi*x -2)))

Dunno if this helped at all. just a way to keep all the chain rule parts in order.
 
A

Aeris130

Member
Rich Uncle Skeleton said:
Sorry for the confusion. You're right that m_1 = 1, m_2 = 2, and m_3 = 1, which is to say we need at least one 1, two 4s, and one 5. It's possible however that we'll have extra 1s, 4s, or 5s. As jepense pointed out, we have to be careful about double counting these, which is what I hope summing over the k-tuples (l_1,...,l_k) helps with. What we want to do is first consider the rolls where there is exactly one 1, two 4s, and one 5. If there are say 6 dice total, the other two dice are allowed to be anything other than 1, 4, or 5. And there are 12 ways of rearranging 1, 4, 4, 5 (six choices for where the two 4s go, and then two orderings on 1 and 5 for each of these), so there are 12 x 3^2 such rolls. Now do the same sort of reasoning but when there are exactly two 1s, two 4s, and one 5. Then we'll have handled l_1 = 1, l_2 = 2, l_3 = 1 and l_1 = 2, l_2 = 2, l_3 = 1. If we keep going and do this for ever triple (l_1, l_2, l_3) with l_1 at least 1, l_2 at least 2, l_3 at least 1 (so we have at least as many 1s, 4s, and 5s as we need), and their sum is not more than n (we better not have more occurrences of 1s, 4s, and 5s than we have dice), and sum up all the valid rolls we find, we'll see how many rolls there are that work (hopefully without double counting any). Dividing this with the total number of rolls, 6^6, and we get the probability.
Click to expand...

What about combinations that contins some additional 1/4/5, as well as some other number? Are they covered in 12 x 3^2 or "(l_1, l_2, l_3) with l_1 at least 1, l_2 at least 2, l_3 at least 1"?
 
C

cwmartin

Member
ili0926 said:
picture_2.png


Can anyone help me with part c? Specifically whether or not the result should be surprising/obvious?
Click to expand...

Man i dont like the way your book words problems, like at all. That being said I could help with a and b, but i learn the envelope method this week, maybe I can help then. love econ
 
R

Rich Uncle Skeleton

Member
Aeris130 said:
What about combinations that contins some additional 1/4/5, as well as some other number? Are they covered in 12 x 3^2 or "(l_1, l_2, l_3) with l_1 at least 1, l_2 at least 2, l_3 at least 1"?
Click to expand...

Everything is covered by one of those (l_1,l_2,l_3)s. 12 x 3^2 corresponds to (1,2,1), except I forgot a factor. It should be 15 x 12 x 3^2, because we need to choose which 4 dice we're using for the 1s, 4s, and 5s (15 choices), then how we want the 1s, 4s, and 5s to be arranged (12), then what the other two dice are (3^2). In the notation I used before, I'm saying N(1,2,1) = 15 x 12 x 3^2. That's the number of rolls with exactly one 1, two 4s, and one 5. If you have one extra 4, it's covered by N(1,3,1). One extra 5 is N(1,2,2). An extra 1 and an extra 5 is N(2,2,2). Some of these allow for there to be dice that aren't 1, 4, or 5. Others don't. For example, (1,2,2) means one 1, two 4s, and two 5s. You have one die left over, which could be 2, 3, or 6. (2,2,2) on the other hand uses all 6 dice, so there isn't anything that's not 1, 4, or 5. This is why in the formula for N(l_1,l_2,...,l_k), I multiplied by (6-k)^(3-L). 3-L of the dice are not one of our desired numbers, so they're allowed be chosen from all 6-k other faces on the die.
 
W

wizword

Banned
Discrete mathmatics 2 proofs. I am fucked. This is where I get the majority of my test points off because of these stupid proofs.
Prove that m equivalent n (mod p) then m^2 equivalent n^2 (mod p)

Let P greater than or equal to 2. Define the mapping θ: Z(p) implies [Z]p by θ(m) = [m]p
 
J

jepense

Member
wizword said:
Discrete mathmatics 2 proofs. I am fucked. This is where I get the majority of my test points off because of these stupid proofs.
Prove that m equivalent n (mod p) then m^2 equivalent n^2 (mod p)

Let P greater than or equal to 2. Define the mapping θ: Z(p) implies [Z]p by θ(m) = [m]p
Click to expand...
Simple definitions:
m = n (mod p) ->
m = kp + n, for some k ->
m^2 = k^2p^2 + 2kpn + n^2 = (k^2p+2kn)p + n^2 ->
m^2 = n^2 (mod p) (since there is an l so that m^2 = lp + n^2)

All the proofs for simple modular arithmetic go this way.

I'm not sure what the notation in the second example means.
 
Â

∀ Narayan

Member
Rich Uncle Skeleton said:
So you want an equivalent equation with only x and y occurring? Well, x = rcos(theta), right? And r is the square root of (x^2 + y^2).
Click to expand...
Heh. Yeah, but that isn't the answer. I just got to work and did the problem in my head. I got the answer.

r+rcosθ=3 -> r=3-rcosθ -> sqrt(x^2+y^2)=3-x -> x^2+y^2=(3-x)(3-x) -> x^2-y^2=9-3x-3x+x^2 -> y^2=-6x+9

Guess I just had a brain fart, or something. Really not a difficult problem at all.
 
S

Starfish_Oxide

Member
Can GAF's almighty mathematical ability help me how do do this... got an exam on Friday and dont have a clue how to do it




ok update i can do the first to parts but cant get the last 2..help please
 
Â

∀ Narayan

Member
New problem. I'm having trouble illustrating this problem, and I think my lack of proper illustration contributes to the fact that I can't seem to get the right answer.

An airplane has an airspeed of 150km/h. It is to make a flight in a direction of 70 degrees while there is a 25km/h wind from 340 degrees. What will the airplane's actual heading be?

I can do the math, I'm just having trouble visualizing it. The section I'm working on deals with vectors and applications.
 
F

Feep

Banned
An interesting problem here, GAF.

• There are n cards in a box numbered 1 through n. We draw cards
successively and at random with replacement. If the ith draw is the
card numbered i, we say that a match has occurred.
a) What is the probability of at least one match in n trials?
b) What happens if n increases without bound?
Click to expand...

I'm fairly sure a) is 1 - ((n - 1) / n) ^ n. Correct me if I'm wrong.

b), however, poses a problem. It asks what a) is as n approaches infinity. From my calculator, the graph appears to be asymptotic, honing in on a value around 63.2%. Does anyone know how to prove this number mathematically? I feel like this limit problem may involve Taylor expansions, or something, and I do NOT remember how to do those.
 
J

jepense

Member
Starfish_Oxide said:
ok update i can do the first to parts but cant get the last 2..help please
Click to expand...
A steady state is a vector s = [x,y] so that s = Ts. That is, if you apply the transition to the steady state, you don't actually change the distribution.

Feep said:
I'm fairly sure a) is 1 - ((n - 1) / n) ^ n. Correct me if I'm wrong.

b), however, poses a problem. It asks what a) is as n approaches infinity. From my calculator, the graph appears to be asymptotic, honing in on a value around 63.2%. Does anyone know how to prove this number mathematically?
Click to expand...
Lim (1-1/n)^n = 1/e.
 
F

Feep

Banned
jepense said:
The number e is DEFINED as the limit e = lim_n (1+1/n)^n. So, e^x = lim_n (1+1/n)^xn = lim_m (1+x/m)^m (m=xn), and 1/e = e^-1 = lim_m (1-1/m)^m.
Click to expand...
Heh. Guess I didn't know the formal definition of e. Thanks for the help! :D
 
S

Starfish_Oxide

Member
jepense said:
A steady state is a vector s = [x,y] so that s = Ts. That is, if you apply the transition to the steady state, you don't actually change the distribution.
Click to expand...

Thanks, but i that the definition of a steady state vector? and would this be correct?

so vector v=Tv v=[1,2]

[1,2]=[.8,.1][1,2]
........ [.2,.9]

[1,2]=[1,2]
 
J

jepense

Member
Starfish_Oxide said:
Thanks, but i that the definition of a steady state vector? and would this be correct?

so vector v=Tv v=[1,2]

[1,2]=[.8,.1][1,2]
........ [.2,.9]

[1,2]=[1,2]
Click to expand...
"Steady state" usually means a state that evolves dynamically (people jump between R and D) but whose macroscopic features stay the same over time (total number of people in R and D stay the same). Since the system here evolves according to v_(n+1) = T v_n, and steadyness means v_(n+1) = v_n (=v), we must have v = T v in the steady state. That is, v is an eigenvector corresponding to the eigenvalue 1, which you should've find in (b).

And no, that is not correct according to the problem. Your matrix corresponds to 0.2 switching from first to second and 0.1 switching from second to first. (And [1,2] does not satisfy x+y = 1, you'd need to divide by 1+2.)
 
S

Starfish_Oxide

Member
Thanks for your help with the last one Jepense but ive got another one



Its part B i can graph it and all that it is just finding the area of the enclosed area and also the volume generated by the solid revolving around the x-axis
 
J

jepense

Member
Starfish_Oxide said:
Its part B i can graph it and all that it is just finding the area of the enclosed area and also the volume generated by the solid revolving around the x-axis
Click to expand...
B is an integration exercise. The area enclosed by functions a(x) > b(x) is the integral of (a(x)-b(x))dx from x0 to x1, where x0 and x1 are the intersection points of the functions ( a(x0) = b(x0) ).

The volume generated by revolving the curves is the integral of pi( a(x)^2-b(x)^2 )dx from x0 to x1. The idea is to split the volume to thin cylindrical pieces with an area pi a^2 - pi b^2 and height dx.
 
C

close to the edge

Member
Hey, need help real quick: How do I find out how many numbers smaller than 1000 are not divisible by 3? I know that 1000/3 gives me the quantity of numbers under 1000 that are divisible by 3, I just need to find out the opposite. Is it 1000-(1000/3) in this case? I'm sure it's simple and obvious but my brain doesn't seem to want to do math anymore today, so yeah.

edit: Okay, just asked a friend and he did the same thing I did (1000-333).
 
Q

quilter46761

Member
close to the edge said:
Hey, need help real quick: How do I find out how many numbers smaller than 1000 are not divisible by 3? I know that 1000/3 gives me the quantity of numbers under 1000 that are divisible by 3, I just need to find out the opposite. Is it 1000-(1000/3) in this case? I'm sure it's simple and obvious but my brain doesn't seem to want to do math anymore today, so yeah.

edit: Okay, just asked a friend and he did the same thing I did (1000-333).
Click to expand...


Yeah, its as simple as you thought it was.

And, I was figuring out how to make gifs tonight, so I figured this thread would be mostly like to appreciate...

output.gif
 
R

Rolio

Member
solve by making an appropriate substitution:

2x^(2/3) + 7x^(1/3) - 15 = 0

I know that x^(2/3) = cube root of x^2 (or is it square root of x^3), but I cannot think of how to proceed. I see that the denominators of the exponents are the same. Hmm.
 
B

bachikarn

Member
Rolio said:
solve by making an appropriate substitution:

2x^(2/3) + 7x^(1/3) - 15 = 0

I know that x^(2/3) = cube root of x^2 (or is it square root of x^3), but I cannot think of how to proceed. I see that the denominators of the exponents are the same. Hmm.
Click to expand...

Let y = x^1/3. So your equation is now:

2y^2 + 7y - 15 = 0.

Solve that equation for y (quadratic formula or factoring). Whatever you get, set it equal to x^1/3, and solve for x.


Ventrue said:
So these questions are largely to do with generating functions.

1. I've got (what I believe to be) the closed form for the generating function of the Lucas series: L(z) = (2-z)/(1-z-z^2)

I'm trying to get the sum Sn of the first n Lucas numbers. S(z), the generating function for the sum, is L(z)/(1-z), correct? So that's S(z) = (2-z)/(1-z-z^2)(1-z).

However, I need an explicit formula from that. Yet I don't see how to get it in a form like 1/(1-bz)^k or anything like that. Is my work up to this point wrong, or is my approach to finding an explicit formula wrong?

2. Trying to find an explicit formula for (note: [this means subtext])

a[n] = -(L[0]a[n-1] + L[1]a[n-2]...+L[n-1]a[0]), a[0] = 1. I should be able to do most of it, but I'm at a loss at to how to begin: how do I find a[n]'s generating function?

3. Trying to find a sum for sequences like:

a[n] = (nC0)3^0 + (nC1)3^1... + (nCn)3^n.
a[n] = 1 + sqrt(2) +...sqrt(n).

Again, I just don't know where to start. They're not arithmetic or geometric series. I feel I need a generating function in here somewhere to sum them, but I don't know how to find it.
Click to expand...

I don't know anything about this stuff, but sounds like you are in a pretty high level math class. You might want to try http://www.physicsforums.com/ or http://www.artofproblemsolving.com/Forum/viewforum.php?f=218 if no one here can help you.
 
V

Ventrue

Member
bachikarn said:
I don't know anything about this stuff, but sounds like you are in a pretty high level math class.
Click to expand...
It's not too high level, I don't think. I've seen people here solve harder stuff (AFAIK).
 
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