triplestation
Member
Oh wow, I can't believe I didn't see that lol
I'll get to fixing this right now, thanks a bunch!
Note to self: (2x)^2 =/= 2x^2
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The Math Help Thread
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I'll get to fixing this right now, thanks a bunch!
Note to self: (2x)^2 =/= 2x^2
GiantEnemyGoomba
Member
We're currently doing partial fractions. By looking up the answer to this problem and working backwards, I was able to figure out the two fractions that I should've been able to find, but I can't seem to figure out an intuitive way I was supposed to solve the problem in the first place without knowing the answer.
It's late so sorry if none of that made sense lol.
We're currently doing partial fractions. By looking up the answer to this problem and working backwards, I was able to figure out the two fractions that I should've been able to find, but I can't seem to figure out an intuitive way I was supposed to solve the problem in the first place without knowing the answer.
It's late so sorry if none of that made sense lol.
Substitution integral with u=e^x and dx/du= e^(-x)= 1/u
triplestation
Member
EDIT: fixed!
Stumpokapow
listen to the mad man
look at the lines between du and dx; where did the 3 go?
du = -sin(3x) . 3 dx
dx = - du/(3 . sin(3x))
when you remove the term, remove -1/3, not -3
du = -sin(3x) . 3 dx
dx = - du/(3 . sin(3x))
when you remove the term, remove -1/3, not -3
Stumpokapow
listen to the mad man
Hey Stumpokapow, can you talk Evilore into implementing Mathjax? Thx!
I don't think it's going to happen to be 100% honest, so you're probably better off coming up with a good workflow to render LaTeX to a PNG.
triplestation
Member
A Human Becoming
More than a Member
Looking at this data, is geometric mean probability essentially the confidence level? I really need to get a book on statistics. It's been too long since I took the course in college.
Stumpokapow
listen to the mad man
The geometric mean probability is derived from the likelihood function, and is a measure of certainty about a person's score, but is not something like a confidence interval (you cannot add or remove this from the estimate to get a sense of where the person could be). NOMINATE is kinda weird (and not an introductory topic!) and so I don't have a great interpretation for you as to what to do with it.
I will say that that's kinda a weird question to be asking. Would it help if I did a walkthrough of how NOMINATE works? The site doesn't do a good job of explaining and Howard and Keith's book is not free and the earlier papers on it are not super accessible either and you can sort of figure out whether it helps you understand what you're looking at, how to interpret it, whether or not to believe it, and whether the GMP matters for you.
DW-Nominate is an effort to assign spatial positions in one or several dimensions to legislators based on roll-call votes. It is based on spatial modelling assumptions.
The idea is that imagine you, I, Bernie Sanders, Hillary Clinton, and Donald Trump are all voters in the legislature. Imagine for a second that there are no parties, and also imagine that you don't know the labels liberal or conservative or where we're from or anything about what we're voting on. In right life, we might say maybe Trump is to Hillary's "right" politically, Clinton to your right, you to my right, and me to Bernie's right (just assume for the sake of argument that this is true), so the lineup would look like: Bernie / Me / You / Hillary / Trump How would we come up with this? Okay, one way to do it is to look at how we vote. Imagine we have the following votes:
Death Penalty for Everyone: Donald Trump votes yes, everyone else votes no
Lower capital gains taxes: Trump and Hillary vote yes, everyone else votes no
Nationalize the airline industry: Bernie votes yes, everyone else votes no
Puppies for all: Everyone votes yes
America should change its flag to the Canadian flag: Everyone votes no
Compel federal employees to be a part of the union: Bernie and I vote yes, you Hillary and Trump vote no.
In the real worlds, we have hundreds and thousands of votes and hundreds of legislators. But just in our own little world, could you organize all of us from "left to right"? Yeah, I think you could. You'd intuit that people who are more politically similar will vote together more often. So if we have lots of votes where it's 4-1 and Trump is the 1, then we can imagine Trump is off to one side. If we have 3-2 votes and you're always on the 3 side, then you're probably in the middle? We could line up all of the people, and then make lines to "cut" between people people to represent the votes, and we can do this in a way that one side of the line is yes, one side of the line is no, and everyone is on the right side of the votes.
So, imagine we did our lineup like this:
Bernie Hillary Trump Me You
Let's look at our votes
Death Penalty:
Bernie Hillary | Trump You Me <-- this looks like we're voting with Trump
Bernie Hillary Trump | You Me <-- this looks like Bernie and Hillary are voting with Trump
| Bernie Hillary Trump You Me <-- this looks like we all agree
All of these possible arrangements have "errors". The cut point cannot possibly explain the vote with this arrangement.
But if we did our lineup like this:
Bernie Me You Hillary Trump
Then our cut line works fine:
Bernie Me You Hillary | Trump <-- everyone is correctly classified
(Trump Hillary You Me Bernie would also work, there's nothing magic about the left)
Okay, but it gets more complicated. Not everything is one dimensional. Bernie Sanders, for example, despite obviously being on the left of the spectrum on all those issues, isn't for gun control. So if we have a gun control vote, and Trump and Bernie vote no, and everyone else votes yes, what happens to our perfect little ordering? No matter how we line people and votes up, you can't draw a line that put Trump and Bernie together and the rest of us apart. So we have an "error" in our line-up.
Bernie | Me You Hillary Trump <-- this predicts that Trump votes with us
Bernie Me You Hillary | Trump <--- this predicts that Bernie votes with us
Sometimes this might happen because there's an issue that doesn't follow conventional ideology--like imagine a vote that's "The West Coast is the Best Coast", and all the west coast people regardless of ideology vote yes, and all the east coast people vote no. Another example of this was true during the civil rights era. Southern democrats typically voted yes on New Deal stuff like Northern democrats, but they voted no on Civil Rights stuff, even when Republicans and Northern Democrats voted yes. So maybe instead of putting us in a line, you put us all in a two-dimensional grid. Now, instead of the dividing us using vertical lines for votes, you can use lines at any angle. A vertical line would imply only the first dimension spatial position matters for the vote, a horizontal line would imply that liberals and conservatives unite but everyone is divided along this second dimension. It's not even guaranteed that the first dimension means liberal/conservative, but since liberal/conservative is the dominant social division in legislators in the US, that's what people typically call NOMINATE's first dimension. The process works without knowing anything about the content of the votes though, so there's no intrinsic reason why liberal/conservative is that first dimension. Maybe it isn't, exactly?
Howard and Keith set out to figure out how complicated American politics was in terms of spatial dimensions. By definition the more dimensions you add, the more errors you can reduce. But some errors are actual errors (maybe Trump agrees to vote a certain way because Bernie is out sick and Bernie will cover it later--so it looks like they each vote very strangely once). So the question becomes how many dimensions you add? Well, most of the variation in how people vote can be explained along a single spatial dimension. In some periods in history, like civil rights, a second dimension is handy. Two dimensions explain almost all the variation. So the weekly DW-NOMINATE numbers are reported in two dimensions.
NOMINATE works using a very very complicated method of joint scaling and error reduction. This is quite complicated, quite computationally complex. In Keith and Howard's original book on nominate, they have a chapter with a toy example like the one I just mentioned. But the thing you need to know as a USER is that NOMINATE works by trying out orders of spatial position in order to minimize the "error" that happens from them. The more you can reduce the error, the better. It turns out, though, that how you actually measure errors is very difficult, and how you measure improvement or reduction in errors is also very difficult.
This is made complicated that NOMINATE is trying to work across both chambers (the House and Senate) that don't have the exact same votes (homework question: How do they do this? It's pretty cool when you figure it out), and across time. How do we tell if Democrats are further to the left today than 30 years ago? It's difficult, because it's not like we just have the same votes every year. NOMINATE does it using some very sneaky assumptions about stability over time. It's also tricky because some people have two years of sporadic attendance and a few votes, while others have 40 year career patterns of voting.
Almost there. Remember how I said calculating error is hard? Here's an example: In the real US senate, there are 100 senators. Which is worse, an error that says Bernie Sanders is a hard-right Republican, or an error that says that a Republican from New Hampshire is slightly more liberal than they are (or some other moderate being slightly on the wrong side of the cut)? Some people say the two are the same error, but actually the first one is much worse. It's a BIG miss. So that's where probabilistic elements come into play. As a result, they calculate likelihood functions (if you study more stats you'll get to these in an MLE class or in machine learning) to make it clear how certain they are about their classification of people. And now we're back to the geometric mean probability. You can't just take the probability or the likelihood function output and add them to the coordinates they give you to get an answer as to what it is 100%.
Stumpokapow
listen to the mad man
Found my PDF copy of Ideology and Congress so you can see how it's calculated, although I don't think this excerpt explains "what it does" very well:
A Human Becoming
More than a Member
Thanks for that very thorough explanation Stump. I think I have a better grasp of it now, but I'll read it over again and also check the Wikipedia entry. I did look out for the book, but no libraries within 50 miles carry it. I came across this data reading this FiveThirtyEight article talking about the two dimensions. I'm trying to decipher what interesting data can be extracted.
Consider the process \dot{Z_t}=a_t-\theta*a_{t-1} where \theta=0.8, a_t are iid N(0,1), and \dot{Z_t}=Z_t - \mu.
(a) Find the theoretical ACF (autocorrelation function) \rho_k for k=1,2,3,4.
(b) FInd the theoretical PACF (partial autocorrelation function) \phi_{kk} for k=1,2,3,4.
(a) Find the theoretical ACF (autocorrelation function) \rho_k for k=1,2,3,4.
(b) FInd the theoretical PACF (partial autocorrelation function) \phi_{kk} for k=1,2,3,4.
themagicalkitsune
Member
If you only have to work with the series expansion of e^x and log^x, how can you prove e.g that log(e^x) = x? For some reason induction won't work as the series are infinite.
cpp_is_king
Member
The obvious way. Write the series expansion of log(x) first. Then look for where x appears, and replace it with the series expansion of e^x. Now think real hard until you see it
Edit: Actually I'm wrong. I worked it out in my head but but when I try it on paper it doesn't work out as nicely as I thought it did.
kingwingin
Member
need help with a finding the inverse of a function.
y = 8 - 3x
I keep getting -y = (8-x)/3 and the back of the book says y = (8-x)/3.
how do they get rid of the minus symbol on the y
y = 8 - 3x
I keep getting -y = (8-x)/3 and the back of the book says y = (8-x)/3.
how do they get rid of the minus symbol on the y
cpp_is_king
Member
Book is correct, can you show your work?
Stumpokapow
listen to the mad man
How to find the inverse of a function:
1) Switch x and y
2) Solve for y
1) Switch x and y
y = 8 - 3x -> x = 8 - 3y
2) Solve for y:
Add 3y to both sides
x + 3y = 8 - 3y + 3y
Simplify
x + 3y = 8
Subtract x from both sides
x + 3y - x = 8 - x
Simplify
3y = 8 - x
Divide both sides by 3
3y/3 = (8-x)/3
Simplify
y = (8-x)/3
kingwingin
Member
I did the below except moved the 8 over first instead of the 3y
that makes sense. so when moving an equation around like that why does it screw up when I move the 8 first?. is there an order I should follow for future reference? non of the other questions gave me problems except this one. been studying functions all night and I was killing myself on this one
Stumpokapow
listen to the mad man
The order doesn't matter. Any operation you do to both sides of the equation doesn't change the result. You just probably forgot that -3y means -3y and not 3y at some point when you were working on it.
y = 8 - 3x
Switch x and y
x = 8 - 3y
Subtract 8 from both sides
x - 8 = 8 - 3y - 8
Simplify
x - 8 = -3y
Divide both sides by -3
(x-8)/(-3) = -3y/-3
Simplify
y = -(x-8)/3
Simplify again
y = (8-x)/3
spelltropy
Member
hi, i'm doing a question regarding minimal polynomials and the jordan canonical form of a matrix. basically, i found the minimal polynomial fine using the method of finding the characteristic then testing if the matrix (say A) satisfies any reducible version of c(A) (probably the wrong wording, but i hope you can see what i'm getting at). This also seems to be the method every set of notes/examples i've found online uses.
However, our lecturer did it a bit different, and didn't explain it very well - he basically took each of the basis vectors (e1, e2, e3), then multiplied A by them repeatedly until he found a multiple (say A^3e1) which was linearly dependent on the e1, Ae1, A^2e1. He then formed some matrix (this is the part where I have no clue what's going on), say B, and took the determinant of (B-xI). He then did the same for the other basis vectors, so he had three expressions, at which point he found the LCM of the three to give the minimal polynomial.
I can see how this might be faster if the matrix has a very long characteristic polynomial (i think), but I'm really not sure how the matrix was made.
Could anyone explain what's going on? Perhaps do an example? Thanks
However, our lecturer did it a bit different, and didn't explain it very well - he basically took each of the basis vectors (e1, e2, e3), then multiplied A by them repeatedly until he found a multiple (say A^3e1) which was linearly dependent on the e1, Ae1, A^2e1. He then formed some matrix (this is the part where I have no clue what's going on), say B, and took the determinant of (B-xI). He then did the same for the other basis vectors, so he had three expressions, at which point he found the LCM of the three to give the minimal polynomial.
I can see how this might be faster if the matrix has a very long characteristic polynomial (i think), but I'm really not sure how the matrix was made.
Could anyone explain what's going on? Perhaps do an example? Thanks
Try sin(x+π/4)=sinxcos(π/4)+cosxsin(π/4) and you're done. You can find cosx easily.
My brain is fried at the moment. How do you arrive to that answer? I know it must be an identity... Probably.
Yeah, it's an identity. The first one:
http://www.sosmath.com/trig/Trig5/trig5/img6.gif
For cosx, you just need this first one:
http://www.sosmath.com/trig/Trig5/trig5/img2.gif
spelltropy, maybe it's just me on Sunday morning, but I don't quite follow your explanation. Can you give us more specifics about what your lecturer did (what is c(A) and the matrix B), replicate his/her example?
Majestad, it is an identity. We can show that sin(A + B) = sin(A) * cos(B) + cos(A) * sin(B), for any two real values A and B.
spelltropy
Member
Sorry yeah, read over it and its a bit badly explained. C(A) is the characteristic polynomial of A. He formed the matrix B by finding the coefficients a,b,c in A^3e1 = cA^2e1+bAe1+ae1 and placing them on the right side of the new matrix B. He then had two random 1s placed (not sure where they came from).
Heres the matrix -
and this is the matrix B he calculated:
He then said do the same for e2,e3,e4 and work out the characteristic of each of these matrices. The matrix B is apparently the matrix wrt (e1, Ae1, A^2e1)
Thanks for the help kgtrep and Qurupeke.
Why is (π/4) replaced into the Y value?
Also, I get confused with these A < X < B problems, this part specifically. What does that exatly mean? I understand some number is less than X, and X is less than this other value, but I don't know where this fits into the solving part of the problem.
If you are only given that sin x=1/3, then there are two possible values for cos x. The fact that x is between 0 and pi/2, i.e. x is in the first quadrant, means that cos x must be positive.
boviscopophobic
Member
Sorry yeah, read over it and its a bit badly explained. C(A) is the characteristic polynomial of A. He formed the matrix B by finding the coefficients a,b,c in A^3e1 = cA^2e1+bAe1+ae1 and placing them on the right side of the new matrix B. He then had two random 1s placed (not sure where they came from).
Heres the matrix -
and this is the matrix B he calculated:
He then said do the same for e2,e3,e4 and work out the characteristic of each of these matrices. The matrix B is apparently the matrix wrt (e1, Ae1, A^2e1)
I started writing something out, then discovered it was a lot of work, so instead I'll just point you to Wikipedia. In particular, look at the section on Computation. In your case, T is the linear transformation that has matrix A with respect to the standard basis. There is a T-invariant subspace with basis vectors {e1, A*e1, A^2 * e1}, and B is the matrix of the restriction of T to that subspace. We can then compute the minimal polynomial of B, which is denoted by mu_{T,e1}(t) in the discussion.
Sorry yeah, read over it and its a bit badly explained. C(A) is the characteristic polynomial of A. He formed the matrix B by finding the coefficients a,b,c in A^3e1 = cA^2e1+bAe1+ae1 and placing them on the right side of the new matrix B. He then had two random 1s placed (not sure where they came from).
Heres the matrix -
and this is the matrix B he calculated:
He then said do the same for e2,e3,e4 and work out the characteristic of each of these matrices. The matrix B is apparently the matrix wrt (e1, Ae1, A^2e1)
boviscopophobic's link to the wiki page allowed me to get the gist, but to be honest, I'm not sure how your lecturer got the matrix B, why he looked at det(B - xI), and other technical things going on in the wiki page.
The explanation below is my interpretation of what's going on, so it may not be correct and you should tread it carefully.
----
The idea is simple. Given a matrix A, find the minimal polynomial p(x) such that p(A) = 0, the zero matrix. (Recall that the minimal polynomial has a leading coefficient of 1 and has the smallest degree such that p(A) = 0.)
Now, by definition of the zero matrix,
p(A) = 0.
<=>
p(A) * x = 0 * x = 0 (the zero vector), for all x in R^4.
<=>
We satisfy all four equations:
p(A) * e1 = 0 (the zero vector),
p(A) * e2 = 0,
p(A) * e3 = 0,
p(A) * e4 = 0.
So suppose that the minimal polynomial p(x) takes the form p(x) = a0 + a1 * x + a2 * x^2 + a3 * x^3 + a4 * x^4 for some coefficients a0, ..., a4. This means,
p(A) = a0 * I + a1 * A + a2 * A^2 + a3 * A^3 + a4 * A^4.
Note, because A is a 4 x 4 matrix, we know that the degree of the minimal polynomial p(x) cannot be greater than 4. However, it is possible that the degree is less than 4 (this turns out to be the case here), which is why your lecturer wanted to look at A^k * e1 starting from the lowest power k = 0, 1, ...
Using Matlab, I found that
I * e1 = [1; 0; 0; 0]
A * e1 = [3; -1; 0; 1]
A^2 * e1 = [8; -4; 0; 3]
A^3 * e1 = [20; -12; 0; 7]
Notice that the third entry has been always 0, so A^3 * e1 must be linearly dependent upon the first three, and we don't need to calculate A^4 * e1.
We indeed find that,
-4 * (I * e1) + 8 * (A * e1) - 5 * (A^2 * e1) + (A^3 * e1) = 0,
so we guess the minimal polynomial to be the cubic polynomial,
p(x) = -4 + 8x - 5x^2 + x^3.
To show that this is actually the case, we can compute p(A) and show that this is the zero matrix. This would imply that the remaining three equations p(A) * e2 = 0, p(A) * e3 = 0, and p(A) * e4 = 0 are true.
One thing I'm not sure about is, if we get a cubic polynomial from looking at A^k * e1, whether it is possible that the minimal polynomial has an even lower degree, had we looked at A^k * e2, A^k * e3, or A^k * e4 first.
If so, I don't think that we should check whether the equation p(A) = 0 is true after finding a candidate polynomial, and say that we're done if the equation is true.
If so, I don't think that we should check whether the equation p(A) = 0 is true after finding a candidate polynomial, and say that we're done if the equation is true.
themagicalkitsune
Member
Anyone know what ring R((X)) is in algebraic terms? For example R[X] is the polynomial ring and R[[X]] is the formal power series ring. Has anyone seen this notation anywhere?
spelltropy
Member
Thanks for the help, kgtrep and boviscophobic. I sort of picked up on what was going on in the example in the lecture notes (where he didn't actually make the weird B matrix..) and managed to do the 3 examined questions using it.
I've also been having trouble with a more straightforward question. Basically, there are 2 bases ei ={1, x, x^2} and ei' = {(1+2x), (1-x)^2, (1-x)^2}. I found the matrices for the linear maps ei -> ei' and the reverse. The next question asked to write the polynomial 3-x+x^2 as a linear combination of e1', e2', e3' - I managed to get an answer which works (-3/2, 2, 5/2 respectively), but I'm not sure how exactly I got it, and how to write it out. Any suggestions?
I've also been having trouble with a more straightforward question. Basically, there are 2 bases ei ={1, x, x^2} and ei' = {(1+2x), (1-x)^2, (1-x)^2}. I found the matrices for the linear maps ei -> ei' and the reverse. The next question asked to write the polynomial 3-x+x^2 as a linear combination of e1', e2', e3' - I managed to get an answer which works (-3/2, 2, 5/2 respectively), but I'm not sure how exactly I got it, and how to write it out. Any suggestions?
You have a redundant element in the second basis. Can you tell us the correct basis?
spelltropy
Member
Eh sorry, being a bit dumb today. Second basis is (1+x^2), (1-x)^2, 1+2x
No problem. I'd approach it by viewing polynomials of degree 2 as vectors in R^3, and taking the dot product to find the linear map (I find this trick very natural and easy to remember).
We first note that, in terms of the standard basis {e1 = 1, e2 = x, e3 = x^2}, we can write,
e1 = [1; 0; 0] (since e1 = 1 * e1 + 0 * e2 + 0 * e3, etc.)
e2 = [0; 1; 0]
e3 = [0; 0; 1]
and write the basis {e1' = 1 + x^2, e2' = 1 + 2x, e3' = 1 - 2x + x^2} as,
e1' = [1; 0; 1]
e2' = [1; 2; 0]
e3' = [1; -2; 1].
Now, we can write a quadratic polynomial p as a linear combination of either basis:
(p =) p1 * e1 + p2 * e2 + p3 * e3 = p1' * e1' + p2' * e2' + p3' * e3'.
The trick is to take the dot product of both sides with respect to e1 to find that,
p1 = 1 * p1' + 1 * p2' + 1 * p3'.
Similarly, take the dot product with respect to e2 and e3 to get two more equations:
p2 = 0 * p1' + 2 * p2' - 2 * p3',
p3 = 1 * p1' + 0 * p2' + 1 * p3'.
Since p is arbitrary, we have a matrix equation that relates the coefficients {p1, p2, p3} in one basis to the coefficients {p1', p2', p3'} in another basis:
[p1; p2; p3] = [1, 1, 1; 0, 2, -2; 1, 0, 1] * [p1'; p2'; p3'].
We can set p1 = 3, p2 = -1, p3 = 1, and solve the matrix equation to find that p1' = -3/2, p2' = 2, p3' = 5/2.
boviscopophobic
Member
It's the ring of formal Laurent series (formal power series allowing for finitely many terms with negative exponents). If R is a field then R((X)) is the same as the field of fractions of R[[X]], as suggested by the notation.
themagicalkitsune
Member
Ah I see. It did come up on a discussion of the Laurent series and I jotted it down in my notes, but I didn't define it and going over them now left me confused. Thanks!
themagicalkitsune
Member
Do you guys know of any good books for an introduction to ring theory? I'm sort of going off by my notes and mathworld and would like something more concrete.
Pardon the double post.
Pardon the double post.
Xan Kriegor
Banned
Hey guys, I have a debate going on during this party (I know I know). It's about probability
So you have 7 numbers in double:
1234567
1234567
One guy draws a numbers and that number goes out of the set.
What is the probability that a second guy draws the same number as the first one? Remember, the number goes out of the set after it's drawn by the first guy. Thanks a million!
(I know I know). It's about probability So you have 7 numbers in double:
1234567
1234567
One guy draws a numbers and that number goes out of the set.
What is the probability that a second guy draws the same number as the first one? Remember, the number goes out of the set after it's drawn by the first guy. Thanks a million!
Hey guys, I have a debate going on during this party
So you have 7 numbers in double:
1234567
1234567
One guy draws a numbers and that number goes out of the set.
What is the probability that a second guy draws the same number as the first one? Remember, the number goes out of the set after it's drawn by the first guy. Thanks a million!
Only one second guy, right? Then it's simply 1/13, there's one out of thirteen that is the same number as the first one.
Stumpokapow
listen to the mad man
Imagine the numbers are on slips of paper in a bag. When someone draws a slip of paper, the paper is removed from the bag.
I draw a 1.
Your turn. How many slips are left in the bag? That's your denominator. How many slips say 1 in the bag? That's your numerator.
I draw a 1.
Your turn. How many slips are left in the bag? That's your denominator. How many slips say 1 in the bag? That's your numerator.
Xan Kriegor
Banned
Thank you so much. I thought it was 1/13 through my common sense.. But when it comes to statistics, my common sense has been proven to be wrong before.
FortuneFaded
Member
If you want some similar, more interesting problems. Look at these:
https://en.wikipedia.org/wiki/Monty_Hall_problem
https://en.wikipedia.org/wiki/Birthday_problem
https://en.wikipedia.org/wiki/Barber_paradox
https://en.wikipedia.org/wiki/Harmonic_series_%28mathematics%29
(Not a probability problem but people would be suprised what 1/2 + 1/3 + 1/4... adds up to)
spelltropy
Member
Sorry for taking so long to reply (i thought i already had!), but this is perfect, thanks a lot.
i've got a small question about vector analysis - I have a plane P = {(x,y,z): y=z>=0}. I'm trying to parametrise a semicircle which lies on this plane. Am i right in thinking this plane is the one midway between the y and z planes? If there's a semicircle of radius 2 in it, is the parameterization r(t)=(2cost,2sint, z(t))? How do I work out z(t)? I'm being slow after being away from this stuff for 5 months
I'm not sure what you mean by the y and z planes, but the definition of P tells you that any point in the plane must have the same y and z coordinates. So your parametrization is forced to look like (a(t), b(t), b(t)). Are you familiar with any parametrizations of the sphere? You could think of this problem as asking you to find the intersection of the sphere of radius 2 with the plane P and work from there.Sorry for taking so long to reply (i thought i already had!), but this is perfect, thanks a lot.
i've got a small question about vector analysis - I have a plane P = {(x,y,z): y=z>=0}. I'm trying to parametrise a semicircle which lies on this plane. Am i right in thinking this plane is the one midway between the y and z planes? If there's a semicircle of radius 2 in it, is the parameterization r(t)=(2cost,2sint, z(t))? How do I work out z(t)? I'm being slow after being away from this stuff for 5 months
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