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[survivor]

Yeah, that's correct. Recall though that an equivalence relation has to contain (x, x) for any x in X (i.e. we must have x~x for all x). So for { (x, f(x)) | x in X } to be an equivalence relation, f(x)=x always. In other words, the only f that gives an equivalence relation is the identity function (and the corresponding partition consists of singleton sets).

Ah I see. So for my last example I ended up with { (a,a), (b,b), (c,c), (d,d) }. I kinda guessed it wasn't equivalent, but I see that it is now.

Thanks.

 

[mcrae]

nvm!

 

[PistolGrip]

Hey GAF, try an help me with this problem:

I want to have a function that with one parameter I can change the curvature but in the end it should reach the same point.

At first I thought y = a * sqrt(X) with a being the parameter I manipulate but that doesn't work as the curve will reach a separate point at the end. Then I figure taking arcs of ellipses would do the trick but I also want the curvature to be 0 in which case its just a straight line. I have seen this in photoshop where I can change the curvature of text but I am having a hard time figuring out how to do is as a function.

 

[luoapp]

Hey GAF, try an help me with this problem:

I want to have a function that with one parameter I can change the curvature but in the end it should reach the same point.

[/IMG]http://i51.tinypic.com/2mean3t.jpg[/IMG]

At first I thought y = a * sqrt(X) with a being the parameter I manipulate but that doesn't work as the curve will reach a separate point at the end. Then I figure taking arcs of ellipses would do the trick but I also want the curvature to be 0 in which case its just a straight line. I have seen this in photoshop where I can change the curvature of text but I am having a hard time figuring out how to do is as a function.

Polynomial interpolation

 

[PistolGrip]

Thank you sir. Looks like this is what I was looking for.

 

[CoffeeJanitor]

This seems really elementary but I'm having trouble with my solution. I was just wondering if someone could quickly check this systems of equations problem that I have -- I have been doing the problem for over an hour now and have been having issues, as Wolfram has been giving me some weird ass solutions--I don't know whether or not its just Wolfram's odd algorithm or if I'm actually right. I just want someone to double check this so I can maybe not go insane.

Anyways here's the problem:

Find an expression for all of the solutions of the given system:

x + 4y - 3z + 2w = 0

3y + 5z - 4w = 0

 

[Zeona]

This seems really elementary but I'm having trouble with my solution. I was just wondering if someone could quickly check this systems of equations problem that I have -- I have been doing the problem for over an hour now and have been having issues, as Wolfram has been giving me some weird ass solutions--I don't know whether or not its just Wolfram's odd algorithm or if I'm actually right. I just want someone to double check this so I can maybe not go insane.

Anyways here's the problem:

Find an expression for all of the solutions of the given system:

x + 4y - 3z + 2w = 0

3y + 5z - 4w = 0

( x | y | 2x + 11y | 2.5x + 14.5y )

 

[CoffeeJanitor]

( x | y | 2x + 11y | 2.5x + 14.5y )

Sorry...Could you explain your answer? I think they wanted it in the form x = Ay + Bz, z = Cy + Dw, etc., but I could be wrong.

Thanks.

 

[bachikarn]

Sorry...Could you explain your answer? I think they wanted it in the form x = Ay + Bz, z = Cy + Dw, etc., but I could be wrong.

Thanks.

He just means: (x,y,z,w) = ( x, y, 2x + 11y , 2.5x + 14.5y )

Which is the same as z = 2x + 11y, w = 2.5x + 14.5y

You can do some algebra to his solution if you want it to be in the form of x = Ay + Bz, z = Cy + Dw.

 

[CoffeeJanitor]

Alright that makes sense. Thanks!

 

[mcrae]

hey guys, how do i convert y=3.6(1.29)^x to the form y=ae^cx (where e is a constant) the answer is y=3.6(e)^0.2546x, but im not sure how to get it. something to do with logarithms i think?

or s=10^(1.71-1.30t) to the form s=ab^t

totally lost

 

[Feep]

hey guys, how do i convert y=3.6(1.20)^x to the form y=ae^cx (where e is a constant)

or s=10^(1.71-1.30t) to the form s=ab^t

totally lost

The first one...3.6(1.20)^x = 3.6(e^ln(1.20))^x = 3.6(e^0.182)^t = 3.6e^0.182t

The second one...10^(1.71 - 1.30t) = 10^1.71 * 10^-1.30t = 51.29 * (10^-1.30)^t = 51.29 * 0.05^t

Just some tricky exponent rule questions

 

[mcrae]

The first one...3.6(1.20)^x = 3.6(e^ln(1.20))^x = 3.6(e^0.182)^t = 3.6e^0.182t

The second one...10^(1.71 - 1.30t) = 10^1.71 * 10^-1.30t = 51.29 * (10^-1.30)^t = 51.29 * 0.05^t

Just some tricky exponent rule questions

havent touched this stuff in a year, my mind was completely wiped... but it makes sense after reviewing a bit now, thanks a bunch feep!

 

[mcrae]

but maybe i lied, haha. how do you establish that in general x = e^ln(x) ?

 

[Feep]

but maybe i lied, haha. how do you establish that in general x = e^ln(x) ?

Those functions are exact opposites of each other; they cancel each other out. ln is natural logarithm, log base e.

ln x = y ---> e^y = x

So if you put ln e^x = ?, e^? = e^x, therefore ? = x.

 

[mcrae]

ok, thanks.

having some trouble figuring out the lines of best fit for sets of data that are best modeled non-linearly. the equations we've been given work to find b in the form y=bx+a, but i cant find the proper value for a unless its already a linear equation.

near the bottom i've used a=(averageY)-b*(averageX) to try and find a, and i get -1.19877, even though the trendline in excel says it should be 0.0633. anyone know what im doing wrong?

edit: so i figured im supposed to be using ln, not log. fuck.

 

[cpp_is_king]

Rule of thmb that is true slightly less than 100% of the time: never use log

 

[TimesEunuch]

Hey GAF, try an help me with this problem:

I want to have a function that with one parameter I can change the curvature but in the end it should reach the same point.

At first I thought y = a * sqrt(X) with a being the parameter I manipulate but that doesn't work as the curve will reach a separate point at the end. Then I figure taking arcs of ellipses would do the trick but I also want the curvature to be 0 in which case its just a straight line. I have seen this in photoshop where I can change the curvature of text but I am having a hard time figuring out how to do is as a function.

There are literally an infinite number of ways to do this. You would need to be more specific about what properties you require of the curvature / limit as the parameter goes to infinity / etc. For example, here is a plot of x = y exp(-t(1-y)) for t=0, 1, 10, 100, 1000, t being the parameter (the x and y axis are reversed in that plot).

As an aside, your picture reminded me of the front cover of "Real Analysis" by Frank Morgan.

 

[Mudkips]

Y = red banana.

 

[Therion]

Hey GAF, try an help me with this problem:

I want to have a function that with one parameter I can change the curvature but in the end it should reach the same point.

At first I thought y = a * sqrt(X) with a being the parameter I manipulate but that doesn't work as the curve will reach a separate point at the end. Then I figure taking arcs of ellipses would do the trick but I also want the curvature to be 0 in which case its just a straight line. I have seen this in photoshop where I can change the curvature of text but I am having a hard time figuring out how to do is as a function.

The simplest function that comes to mind is y=x^a. For a>0 the graph will always contain (0,0) and (1,1). If you want to change the end point to something else, it's simple to tweak. It can get as curvy as you might want (by adjusting a) while staying in the bounding rectangle specified by the end points.

 

[PistolGrip]

There are literally an infinite number of ways to do this. You would need to be more specific about what properties you require of the curvature / limit as the parameter goes to infinity / etc. For example, here is a plot of x = y exp(-t(1-y)) for t=0, 1, 10, 100, 1000, t being the parameter (the x and y axis are reversed in that plot).

As an aside, your picture reminded me of the front cover of "Real Analysis" by Frank Morgan.

Thanks. Polynomial interpolation was the right answer. Found a nice website to generate these functions for me.

http://www.math.ucla.edu/~ronmiech/Interpolation/HTMDOCS/Introduction/Interpolation_Applet.htm

 

[TimesEunuch]

The simplest function that comes to mind is y=x^a.

Why did I miss this example? Seriously, sometimes I think I'm coming down with early-onset Alzheimer's.

Y = red banana.

I approve.

 

[mcrae]

who likes algebra?
A=B/C
H=F/B
G=B/(C+F)

how do i find G in terms of A and H, or A in terms of G and H?

 

[hemtae]

who likes algebra?
A=B/C
H=F/B
G=B/(C+F)

how do i find G in terms of A and H, or A in terms of G and H?

Using the first equation, you get that C*A = B, so substituting that into the last equation you get G = (C*A)/(C+F). Using the second equation you get B*H = F so you can also substitute that into the third equation getting (C*A)/(C+B*H). Substituting in for the B you get (C*A)/(C+C*A*H). Factoring out the bottom gives you (C*A)/(C*(1+A*H)). You can cancel out the C for a final answer of A/(1+A*H).

For your second equation, you should solve the last equation for C. Which should give you C=(B-F*G)/G. Substituting that into the first equation you get A = (B*G)/(B-F*G). Solving the second equation for F gives you F = B*H. Substituting that into the first equation gives you A = (B*G)/(B-B*H*G). Factoring out the bottom gives you (B*G)/(B*(1-H*G)). Cancelling out the B on the top and bottom gives you the final answer of A = G/(1-H*G)

 

[mcrae]

For your second equation, you should solve the last equation for C. Which should give you C=(B-F*G)/G. Substituting that into the first equation you get A = (B*G)/(B-F*G). Solving the second equation for F gives you F = B*H. Substituting that into the first equation gives you A = (B*G)/(B-B*H*G). Factoring out the bottom gives you (B*G)/(B*(1-H*G)). Cancelling out the B on the top and bottom gives you the final answer of A = G/(1-H*G)

im on my 5th page of rough work here, and for some reason just didnt clue in at all to the way you were supposed to do that. it was the same way when i took an introductory logic course a year back, just don't intuitively get it.

using the same above relationships, and D=(B+F)/(C+F), how do i get A in terms of H and D?
is it D/(1+H-DH) ?

 

[stevematic]

dR/dt=kR(50-R), when t=0, R=5

show that R=50/1+(9e^(-50kt))

Can anyone please show me how to get to that?

 

[hemtae]

dR/dt=kR(50-R), when t=0, R=5

show that R=50/1+(9e^(-50kt))

Can anyone please show me how to get to that?

You need to rearrange the equation so you have dR/(R(50-R)) = kdt. Then you use some partial fractions magic so you can integrate the left side. I get (1/(50R))+(1/(50-R))dR = kdt.
Then you integrate both sides, plug in the conditions it gives you to find the constant of integration, then do some law of exponents algebra magic to get the final answer.

If you need help at one of the magic steps, I'll do my best to explain further lol

im on my 5th page of rough work here, and for some reason just didnt clue in at all to the way you were supposed to do that. it was the same way when i took an introductory logic course a year back, just don't intuitively get it.

using the same above relationships, and D=(B+F)/(C+F), how do i get A in terms of H and D?
is it D/(1+H-DH) ?

The only reason I knew to do that was because I worked myself in a circle doing it what I thought was the more straightforward way. There may be a more elegant solution to reaching the same answer.

And yes that's what I get as well

 

[mcrae]

The only reason I knew to do that was because I worked myself in a circle doing it what I thought was the more straightforward way. There may be a more elegant solution to reaching the same answer.

And yes that's what I get as well

i appreciate the help

 

[stevematic]

You need to rearrange the equation so you have dR/(R(50-R)) = kdt. Then you use some partial fractions magic so you can integrate the left side. I get (1/(50R))+(1/(50-R))dR = kdt.
Then you integrate both sides, plug in the conditions it gives you to find the constant of integration, then do some law of exponents algebra magic to get the final answer.

If you need help at one of the magic steps, I'll do my best to explain further lol

Thanks for that.

I get to log(5)/50 - log(45) = kdt and I'm stuck, lol.

 

[-COOLIO-]

quick statistics question.

if the probability of a is 0.04 and the probability of b is 0.03, how do you find the conditional probability?

normally its P(A|B) = P(A intersect B)/P(B)

but how do i get the intersect with straight percentages?

 

[Therion]

quick statistics question.

if the probability of a is 0.04 and the probability of b is 0.03, how do you find the conditional probability?

normally its P(A|B) = P(A intersect B)/P(B)

but how do i get the intersect with straight percentages?

I'm no stats guy, but I don't think there is enough information to solve this. You need to know some relation between A and B. I mean, if they are totally independent, then P(A|B) = 0.04. If they are mutually exclusive, then P(A|B)=0. If B only happens as a consequence of A, then P(A|B)=1. Or it could be anything in between.

 

[cpp_is_king]

I'm no stats guy, but I don't think there is enough information to solve this. You need to know some relation between A and B. I mean, if they are totally independent, then P(A|B) = 0.04. If they are mutually exclusive, then P(A|B)=0. If B only happens as a consequence of A, then P(A|B)=1. Or it could be anything in between.

Usually these types of questions state that a and b are independent events. In this case P(a intersect B) = 0.3 x 0.4

I assume the op just forgot to include that

 

[krameriffic]

Except many times you will need to do integration by parts like 5 times to get a function you can integrate....

I was never taught the shortcuts for these like with regular derivatives so I don't know the single rule. In fact I haven't been taught anything. My teacher is from China and whispers in broken English to the board and writes incredibly small. I am fucked for Calc 2...

This is true. Typically, they teach you to do something called tabular integration or the tabular method. You can look it up. It's basically a shorthand method for piecing together what repeated integrations by parts would give you.

Thanks for that.

I get to log(5)/50 - log(45) = kdt and I'm stuck, lol.

You appear to have performed the integration over R, but you still have a dt in there. Integrating kdt over t should just give you kt, in which you just plug in your condition for t.

 

[Takao]

Data Permutations pains:

A Canadian postal code consists of 6 characters of 3 letters alternating with 3 digits. An example is M4N 0R3. How many possible postal codes exist under this system if you cannot have repeated numbers or letters?

If you're allowed repetition it goes:

(Letters)
26P3= 26! (divided by) 23! = 15600

(Numbers)
10P3= 10! (divided by) 7! = 720

15600 x 720 = 11 232 000

But I don't know how to account for not allowing repetition.

Question 2: Determine the number of ways that the 12 members of the boys' baseball team can be lined up if Joe, Tanner, and Josh must all be together in that order.

I did: 12P3 = 12! (divided by) 9!= 1320

Correct?

Question 3: State the number of ways that the 9 members of the debating club can be lined up for a picture if Fraiser must be on the far left and Samantha and Charlotte must be together in that order from left to right.

My answer = 12 possibilities ...

 

[Therion]

If you're allowed repetition it goes:

(Letters)
26P3= 26! (divided by) 23! = 15600

(Numbers)
10P3= 10! (divided by) 7! = 720

15600 x 720 = 11 232 000

But I don't know how to account for not allowing repetition.

You're actually not allowing repetition in this answer. 26P3 = 26*25*24. You are not allowing the second letter to repeat the first, or the third to repeat the first or second. If repetition is allowed, this would be 26^3.

For question 2, you could consider the trio as a single boy, since no matter how you move them around, they can't change order. So then you would just need the number of ways to arrange 10 people.

You could do question 3 in a similar way, counting the pair as a single person and ignoring the person who stays in a fixed position, as he cannot affect the number of arrangements.

 

[nincompoop]

Usually these types of questions state that a and b are independent events. In this case P(a intersect B) = 0.3 x 0.4

I assume the op just forgot to include that

If A and B are independent events then P(A | B) is just P(A), so I assume the poster didn't forget that.

 

[Takao]

You're actually not allowing repetition in this answer. 26P3 = 26*25*24. You are not allowing the second letter to repeat the first, or the third to repeat the first or second. If repetition is allowed, this would be 26^3.

For question 2, you could consider the trio as a single boy, since no matter how you move them around, they can't change order. So then you would just need the number of ways to arrange 10 people.

You could do question 3 in a similar way, counting the pair as a single person and ignoring the person who stays in a fixed position, as he cannot affect the number of arrangements.

Other people have told me it would be 25P3 and 9P3 to account for no repetition. Have I been had?

Why is it 10 though? If we count the trio as a single group wouldn't that be 11?

For the third question, it's 7! and that makes sense as Fraiser is 1 group, and Samantha/Charlotte are another (bringing the total to 9 groups/students).

 

[Therion]

Other people have told me it would be 25P3 and 9P3 to account for no repetition. Have I been had?

It should be 26P3 and 10P3 for no repetition. I'm not sure why their numbers are one less.

Why is it 10 though? If we count the trio as a single group wouldn't that be 11?

Think of it this way. If you have twelve people, which we'll name simply:

A B C D E F G H I J K L

and we group 3 together, say A, B, and C, we get this:

(ABC) D E F G H I J K L

Ten "people" to arrange.

For the third question, it's 7! and that makes sense as Fraiser is 1 group, and Samantha/Charlotte are another (bringing the total to 9 groups/students).

This is 7!, but Frasier doesn't count as a group at all. We take him out of the picture because no matter how the rest are arranged, he'll just be stuck on the end. So that leaves 8 to arrange, but two get stuck together like in the last problem. Therefore we have 7 "people" to arrange, plus a Frasier on the end that we don't count.

 

[cashman]

Alright gaf, I'm having a tough time wrapping my head around this Calc II problem

1) Prove that for any curve w(t) that stays on the some sphere centered at the origin, i.e ||w(t)|| = b for some constant b, one must have w(t) orthogonal to w'(t) for all t (Hint: use the dot product/Leibniz rule for dot products to differentiate w(t) * w'(t) = b^2

Then use 1) to show that for any curve r(t) the unit tangent T(t) and principal unit normal N(t) are orthogonal for corresponding t's.

 

[cpp_is_king]

Alright gaf, I'm having a tough time wrapping my head around this Calc II problem

1) Prove that for any curve w(t) that stays on the some sphere centered at the origin, i.e ||w(t)|| = b for some constant b, one must have w(t) orthogonal to w'(t) for all t (Hint: use the dot product/Leibniz rule for dot products to differentiate w(t) * w'(t) = b^2

Then use 1) to show that for any curve r(t) the unit tangent T(t) and principal unit normal N(t) are orthogonal for corresponding t's.

||w(t)||^2 = w . w = b^2

2 w . w' = 0

W . W' = 0

Thats the definition of orthogonal.

Can you take it feom there?

 

[cashman]

||w(t)||^2 = w . w = b^2

2 w . w' = 0

W . W' = 0

Thats the definition of orthogonal.

Can you take it feom there?

Yep I got it. Thank You, greatly appreciated!

 

[The Technomancer]

Alright, here's a general-ish question, some complex arithmetic for my EE course that's driving me nuts.

(V - 20i)/5 + V/20i + (V+10)/-10i = 0

I'm just not sure how to solve for V here

 

[QuantumBro]

Alright, here's a general-ish question, some complex arithmetic for my EE course that's driving me nuts.

(V - 20i)/5 + V/20i + (V+10)/-10i = 0

I'm just not sure how to solve for V here

Can you multiply everything by 20i?

 

[CRS]

Technomancer, you can simply further (dividing by 5 and -10i), move everything that has a "V" onto the other side of the equation and then distribute the V out. Solve for V.

 

[MrBig]

More of a physics question but I need help on this.

The question was something along the lines of pushing on a book with 75 newtons of force at a 25* angle, the coefficient of static friction was .76, and I believe you had to find the acceleration.

Another question involved turning a brick on its smaller side and telling whether that had the same or smaller coefficient of friction.

I have no idea how I missed this lesson on friction out of everything else.

 

[Slizz]

Anyone want to take a stab at this for me pllllz?

What is the expected value of the proposition?

 

[cpp_is_king]

Anyone want to take a stab at this for me pllllz?

What is the expected value of the proposition?

He has a free throw rate of 226/372, or .60753. The chances of him making the next 2 in a row are thus (.60753)(.60753) = .36909. Thus, you will make $16 with probability .36909, and you will lose $11 with probability 1 - .36909.

(16)(.36909) + (-11)(1 - .36909) = 5.90542 - 6.94002 = -$1.03.

Often you're dealing with problems where events are completely independent of one another, i.e. the outcome of one event does not affect the outcome of another event. But in this case, there is a level of skill involved, and his existing free throw rate is indicative of his skill level. So it's safe to assume that he will continue with roughly .60753 free throw accuracy.

 

[Slizz]

He has a free throw rate of 226/372, or .60753. The chances of him making the next 2 in a row are thus (.60753)(.60753) = .36909. Thus, you will make $16 with probability .36909, and you will lose $11 with probability 1 - .36909.

(16)(.36909) + (-11)(1 - .36909) = 5.90542 - 6.94002 = -$1.03.

Often you're dealing with problems where events are completely independent of one another, i.e. the outcome of one event does not affect the outcome of another event. But in this case, there is a level of skill involved, and his existing free throw rate is indicative of his skill level. So it's safe to assume that he will continue with roughly .60753 free throw accuracy.

Thanks a bunch. I get stuck on stupid parts of the problem for some reason. Like I understand the formula in the end for expected value, just little things getting there messes me up.

 

[Necromanti]

So, I'm asking this for a friend... He wants to know how you'd isolate the variable k in this equation:

B = [1-(1-(p/(k/n))^2]^k

While this is just for fun (I think), any help would be appreciated.

 

[alveus]

So, I'm asking this for a friend... He wants to know how you'd isolate the variable k in this equation:
While this is just for fun (I think), any help would be appreciated.

The parentheses don't match up in this equation. There needs to be another ) somewhere.