Permanently A
Junior Member
Oh yeah, it is. Missed that part. Thanks!
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The Math Help Thread
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Oh yeah, it is. Missed that part. Thanks!
boviscopophobic
Member
F is defined piecewise, but in your code for definition of ax, you've omitted the pieces x < 0 and x > 3. For the questions shown, your definition happens to give the right answer (because all values of x that are asked about are between 0 and 3), but you'd get the wrong answer for other values of x. You could try something like
ax <- function(x) { ifelse(x < 0, 0, ifelse( x<= 3, x^3/27, 1 ) ) }
although there are other ways of doing this as well.
Hypertrooper
Member
Hey guys,
I have to proof "2^n+1 >= n^2+n+2" with mathematical Induction.
I do the base case with n = 1.
2^1+1 = 2^2 = 4 = 1^2 + 1 + 2. True.
Now I am doing the induction step: P
=> P(n+1)
2^(n+1)+1 >= (n+1)^2 + (n+1) + 2
2^n+2 >= n^2+2n+1+n+3
2^n+2 >= n^2+3n+4
2^n+2 can be written as 2^(n+1)*2. We proved with the base case that 2^n+1 is bigger or equal to n^2*+n+2, so we insert it for 2^n+1 which gets us to this:
(n^2+n+2)*2 = 2*n^2 + 2n + 4.
2*n^2 + 2n + 4 > n^2+3n+4. q.e.d. (I personally think it's clear that 2*n^2 is bigger than n^2. 2*n^2 grows faster than n^2 does. And we can ignore the rest (2n+4 or 3n+4))
But it can't be that easy, can it? I do not want a solution, just a hint.
Edit.: Further thoughts:
Because we have ">=", we are allowed to do:
2*n^2+2n+4 >= n^2+3n+4 |-n^2 |-2n |-4
n^2 >= n | /n
n >= 1
q.e.d.
I have to proof "2^n+1 >= n^2+n+2" with mathematical Induction.
I do the base case with n = 1.
2^1+1 = 2^2 = 4 = 1^2 + 1 + 2. True.
Now I am doing the induction step: P
=> P(n+1)2^(n+1)+1 >= (n+1)^2 + (n+1) + 2
2^n+2 >= n^2+2n+1+n+3
2^n+2 >= n^2+3n+4
2^n+2 can be written as 2^(n+1)*2. We proved with the base case that 2^n+1 is bigger or equal to n^2*+n+2, so we insert it for 2^n+1 which gets us to this:
(n^2+n+2)*2 = 2*n^2 + 2n + 4.
2*n^2 + 2n + 4 > n^2+3n+4. q.e.d. (I personally think it's clear that 2*n^2 is bigger than n^2. 2*n^2 grows faster than n^2 does. And we can ignore the rest (2n+4 or 3n+4))
But it can't be that easy, can it? I do not want a solution, just a hint.
Edit.: Further thoughts:
Because we have ">=", we are allowed to do:
2*n^2+2n+4 >= n^2+3n+4 |-n^2 |-2n |-4
n^2 >= n | /n
n >= 1
q.e.d.
Charmicarmicat
Banned
I have a sort of math question, about whether I should continue to study it. I'm basically a mature student who's doing a part time degree in astronomy/physics. However the math module I'm currently doing is making my head spin, and I know it's only basic stuff so far. At present I'm covering negative and fractional indices, I can follow my text, but once I get onto questions I just get lost. I know things are going to get exponentially harder as I go, is it worth me continuing if I'm already struggling on simple stuff? It's a horrible decision as physics is my passion, but I'm not sure I have the smarts to follow it to degree level 
I have a sort of math question, about whether I should continue to study it. I'm basically a mature student who's doing a part time degree in astronomy/physics. However the math module I'm currently doing is making my head spin, and I know it's only basic stuff so far. At present I'm covering negative and fractional indices, I can follow my text, but once I get onto questions I just get lost. I know things are going to get exponentially harder as I go, is it worth me continuing if I'm already struggling on simple stuff? It's a horrible decision as physics is my passion, but I'm not sure I have the smarts to follow it to degree level
Math is a skill that you need to practice and develop. Being able to read and understand the text is great, but you do need to struggle with those problems to build up your ability. If you find that you are getting completely lost on your assigned problems, then maybe you need to find some easier problems to try first, and work your way up to the ones that are giving you trouble. It might take some extra effort on your part, but don't let an early setback stop you from pursuing what you love.
Edit: I should mention that I think your feelings are pretty common among people learning math. If you're a mature student doing a part-time degree, maybe you don't see as many people struggling around you as if you were a traditional student. Just be aware that most people do go through the same thing.
Charmicarmicat
Banned
Cool, thanks for the positive reply, what I needed to hear! I'll keep on struggling and hopefully it will start to click somewhere down the line.
Also, you have most excellent Devy avatar
How am I supposed to know what this phase shift is?
-1.1? -1.15? -1.2? wtf.
By looking at the wave, you know everything but the phase shift.
Starting with the generic equation, F(x) = Acos(Bx C) + D
A: amplitude is A
B: period is (2π
/|B|C: phase shift is C/B
D: vertical shift is D
Plug in what you know, and solve for what you don't. Hope that helps a little, I can't write out a solution right now.
By looking at the wave, you know everything but the phase shift.
Starting with the generic equation, F(x) = Acos(Bx – C) + D
A: amplitude is A
B: period is (2π
C: phase shift is C/B
D: vertical shift is D
Plug in what you know, and solve for what you don't. Hope that helps a little, I can't write out a solution right now.
I dunno, I came up with pi/2, but that's wrong evidently :/ Came to that conclusion because for 9sin(x) to = 9, sin x = 1, sin pi/2 = 1, tada, but wrong
Edit: Well, on webassign, it said the formula y = 9sin((2pi/5)x + pi/2) was correct, but it says the phase shift is not -pi/2. Is such a thing even possible? Oh, I guess it would be -5pi/4?
Imagine g(x)=g
for x!=y, then using certain theorem (I'm sure you know which), you can prove existence of a point z between x and y such that g'(z)=0. Imagine there's an additional w (for simplicity assume w isn't between x and y) such that g(x)=g(w). What can you conclude?Try viewing it that way.
Biggest-Geek-Ever
Member
For the Mathematica sleuths out there, I'm trying to use Lagrange multipliers to find the absolute extrema of the fuction f(x,y, z) = ye^(x-z) with the constraints 9x^2+4y^2+36z^2=36, xy+yz = 1. Setting up the system of equations is easy, but I haven't been able to get Mathematica to solve the system, numerically or otherwise. It will just think for hours until it eventually crashes.
The FindMaximum and NMinimize commands will actually spit out the absolute extrema, but I would like to know if there's a way to get Mathematica to solve the system. Any ideas?
If you were able, it would have been easier to express the graph as a cosine function. Since there is a peak at x = 0, you won't even have to worry about finding a phase shift.
The FindMaximum and NMinimize commands will actually spit out the absolute extrema, but I would like to know if there's a way to get Mathematica to solve the system. Any ideas?
Close. As SummitAve mentioned, your phase shift is C/B. If C = -pi/2 and B = 2pi/5, then your phase shift is -pi/2/(2pi/5) = -pi/2 x 5/2pi = 5/4.I dunno, I came up with pi/2, but that's wrong evidently :/ Came to that conclusion because for 9sin(x) to = 9, sin x = 1, sin pi/2 = 1, tada, but wrong
Edit: Well, on webassign, it said the formula y = 9sin((2pi/5)x + pi/2) was correct, but it says the phase shift is not -pi/2. Is such a thing even possible? Oh, I guess it would be -5pi/4?
If you were able, it would have been easier to express the graph as a cosine function. Since there is a peak at x = 0, you won't even have to worry about finding a phase shift.
FUBAR McDangles
Member
Alright bros I have a few questions and would appreciate any help:
I have no idea what to do for that.
And the other is this one:
I am really struggling with this one. I figured it would go 2((x^2)+8) (2x), with the (x-8) just being negated to 0, which would simplify to 4x((x^2)+8) but that isn't right. I don't know what to do.
Any help?
I have no idea what to do for that.
And the other is this one:
I am really struggling with this one. I figured it would go 2((x^2)+8) (2x), with the (x-8) just being negated to 0, which would simplify to 4x((x^2)+8) but that isn't right. I don't know what to do.
Any help?
Biggest-Geek-Ever
Member
For the first one, the question is asking you to use differentials to estimate ΔV, which is the maximum amount a calculated volume using V(r) will be off from the actual amount. In other words:
ΔV ≈ dV = dV/dr * dr
dV/dr is the derivative of the volume function with respect to r, and dr is the maximum error in your radius. From there it should just be finding your derivative and plugging in values.
For the second question, it looks like it's just brute forcing with the chain rule. Let (x^2+8)^2 be your f and (x-8) be your g, so the first derivative will be f'g + g'f.
FUBAR McDangles
Member
Where does pi go when computing the derivative? Is the derivative of the function 4(pi)r^2 ? So then I take the estimated radius to put in at r, then multiply that all by .003, right?
Second question makes sense now. Brain fart. I'll work through that in a bit. Thanks for all this, by the way.
EDIT: Alright I'm struggling with the second one now.
So f= ((x^2)+8)^2 , g=(x-8)
f'g= 4x((x^2)+8) (x-8)
g'f= ((1)((x^2)+8)^2)
How do I simplify the f'g? Does it just make it all (4x^2)((x^2)+8) - 32x((x^2)+8)?
Biggest-Geek-Ever
Member
You've got it exactly right, since pi is a constant.
And you are welcome.
And you are welcome.
Pi is just a constant, dV/dr is just 4pi(r)^2 like you said.
FUBAR McDangles
Member
You've got it exactly right, since pi is a constant.
And you are welcome.
Had to edit in my conundrum on the second question, if you could look at that please.
Biggest-Geek-Ever
Member
I would multiply 4x and (x-8) together to get 4x^2-32, simplifying f'g to (4x^2-32x)(x^2+8) and use the product rule on that. The derivative of that should be (8x-32)(x^2+8) + (4x^2-32x)(2x), and then tack on the derivative of g'f to get your second derivative of h(x).
Alternatively, you could actually fully expand out h(x) at the start. It might get slightly messy, but you'll be left with a polynomial, and taking the second derivative of that should be simple.
FUBAR McDangles
Member
Thanks for this. Helps me a lot.
By the expanding out h(x) at the start, do you mean doing this:
((x^2)+8)^2 (x-8) -> (FOIL the first part, then the second)
= (x^5)+(16x^3)+(64x) - (8x^4)-(128x^2)-512
And so on from there?
As messy as that is, it would probably be easier for me haha.
EDIT: Hell yeah I got it right. Definitely should have been able to figure it out myself but I will take all the help I can get. Thanks a bunch for all the help!
Biggest-Geek-Ever
Member
Your second term should be 16x^3, but otherwise that's correct.
FUBAR McDangles
Member
Yeah I caught it when I went through again. Like I said, thank you very much for all of the help. I really appreciate it.
Biggest-Geek-Ever
Member
When a hyperbola has a center at zero with a foci on the x-axis, you are probably aware the formula is:
x^2/a^2-y^2/b^2 = 1
What the problem means by "write the formula for x^2" is to rearrange the equation so that x^2 is on the left and everything else is on the right. So:
x^2/a^2 = 1+y^2/b^2
x^2 = a^2(1+y^2/b^2)
Keep in mind that c^2 = a^2 + b^2, where c is the x-coordinate of the focus and a is the x-coordinate of the vertex. Hopefully you can see what needs to be done from there.
x^2/a^2-y^2/b^2 = 1
What the problem means by "write the formula for x^2" is to rearrange the equation so that x^2 is on the left and everything else is on the right. So:
x^2/a^2 = 1+y^2/b^2
x^2 = a^2(1+y^2/b^2)
Keep in mind that c^2 = a^2 + b^2, where c is the x-coordinate of the focus and a is the x-coordinate of the vertex. Hopefully you can see what needs to be done from there.
Got it! Thanks, dude. We've been working on parabolas and ellipses as well, so I'm used to "a" being the largest distance between points, with "c" being the smaller one. Need to remember that with hyperbolas it's the other way around.
Biggest-Geek-Ever
Member
First, I would rewrite it as
(y^2+4y) +(-3x^2+6x) = 11 → (y^2+4y) - 3(x^2-2x) = 11
Now, you want to complete the square for y and x. Doing so for both should be straight forward enough, making sure you keep in mind the -3 that is multiplying the x's. Finally, simplify the expression into a hyperbolic form you are familiar with.
Edit: Subtracting the 11 to the left side is an extraneous step, so I just edited to exclude that.
(y^2+4y) +(-3x^2+6x) = 11 → (y^2+4y) - 3(x^2-2x) = 11
Now, you want to complete the square for y and x. Doing so for both should be straight forward enough, making sure you keep in mind the -3 that is multiplying the x's. Finally, simplify the expression into a hyperbolic form you are familiar with.
Edit: Subtracting the 11 to the left side is an extraneous step, so I just edited to exclude that.
Ugh, I'm awful at completing squares. How do I do that? y^2 + 4y + 4 = 11?
Biggest-Geek-Ever
Member
To help with that, I'll look at a general equation of the form x^2+bx = c. To complete the square, you take one half of the coefficient b and square, so equivalently (1/2b)^2 = 1/4*b^2. You now add this to both sides of the equation:
x^2+bx+1/4b^2 = c +1/4b^2
The left side can now be factored into (x+1/2b)^2
With your problem, if we apply this to the y's first, you will get:
(y^2+4y+4) -3(x^2-2x) = 11 + 4
(y+2)^2 - 3(x^2-2x) = 15
You might be seeing where this is going now. Try and work out completing the square for the x's.
Damn it, I'm really sorry but I'm still completely lost. So I start with -3(x^2 - 2x) = 0?
I'm taking -3x^2 + 6x, 1/2 of 6 = 3, so I'm getting (x+3)^2? I have no idea what I'm doing. :/
Biggest-Geek-Ever
Member
When you factor out the -3, you get -3(x^2-2x). 1/2 squared of -2 is +1. So you're function will now be:
(y+2)^2 - 3(x^2-2x+1) = 15 - 3 (This is negative since you have to multiply the +1 by the -3 outside the paranthesis)
Now factor the x term:
(y+2)^2 - 3(x-1)^2 = 12
Divide by 12:
((y+2)^2)/12 -3((x-1)^2)/12 = 1
((y+2)^2)/12 -((x-1)^2)/4 = 12
This is the equation of a hyperbola with the center (1,-2). To find the foci, solve c^2 = a^2 + b^2 for c, with a^2 being 12 and b^2 being 4:
c^2 = 12 + 4 = 16
c = ±√16 = ±4
To find the foci, you take your center and add c to the y-coordinate. So the foci are then (1,-2±4) = (1,2) and (1,-6). So the answer looks to be choice 3.
Sorry to blow through that, but I'm heading off to bed and didn't want to leave you hanging. If you're confused, there's some great youtube videos out there that go through similar problems. I think this is a goodie: http://www.youtube.com/watch?v=TFTw5s1wsWM
So I am trying to work out some made up problems dealing with a camera in graphics/linear algebra and I've gotten stuck on a particular part. So I'm trying to take a volume and normalize it around the origin. There's two different volumes I'm dealing with and each has separate equations for normalizing. One volume is orthogonal, so I only have to translate it and then scale it. The other is a frustum so I have to shear it if it isn't symmetric, scale it and then use what's called perspective normalization. The issue I'm coming across is that there are 6 variables, left, right, up, down, near, far. and that near and far just don't seem to add up correctly and I'm trying to figure out why.
for reference
http://www.gomezconsultants.com/CSE5280/Viewing/ViewVolume1.gif
http://www.gomezconsultants.com/CSE5280/Viewing/ViewVolume2.gif
For the frustum volume we have the shearing matrix
For the orthogonal volume we have the translation matrix(just look at the second matrix. The 2 is supposed to be a z)
So I'm having issues with the sign for near and far. It seems that if I set them to be positive then it works for the frustum but if I set them negative it works for the orthogonal so I'm going kind of crazy not understanding what I'm doing wrong. It's gotta be something extremely stupid that I'm not seeing.
So the whole purpose of shearing is to make sure the volume is symmetric about the x and y. if it isnt then every vector needs to be multiplied with the about matrix. For example, lets say our window was top = 1, bottom = -1, left = -2, right = 0 and n and f = whatever. The idea would be so after the shear the point (0, 0, z) would be sheared to (1, 0, z) This means the right side has moved over 1, so the left side should move to (-1, 0, z). So using the equation above we know that x' = z*(l+r/(2n)) = z*(-2/(2n)) = z*(-1/n). Another thing that we know is that z will be negative since cameras point in the negative direction ,so lets say z = -1. x' = (1/n) Now that I have this I have to assume that n has to be positive, and that n is just a distance, which I've read in a few places that they are. Thats great and all, but then I try to apply this to the translation matrix for orthogonal translation and it seems that n and f need to be negative.
Lets take a 2x2x2 that is from the original simply translated 2 units down the negative z direction. That means there's a vector on the cube at (0,0,-1) which is the center of the closest face. Now the second equation is supposed to translate the origin of the cube to the origin of the camera. Since I'm translating I need to use 4 components since translations are linear. So this vector will only change in the z direction and the equation becomes z' = -1*1 + 1*-(f+n)/2 = -1 + -(f+n)/2. So I know my answer needs to be (1, 0, 0), yet the only way for that to happen is if f and n are negative, yet in the previous example f and n are positive even though they're in the same direction. This is really confusing me and I don't know what I'm doing wrong. There's something with the signs that isn't quite working out. It might just be that it's late...
At the same time I'm seeing this stuff for ortho...
So maybe the equations wrong...idk That one won't work either, as it would be -1*-1 + (3+1)/2 = 3 and not 1....bah
And another
Why are all of these different....The last one posted is what MSDN is using so I would assume it's correct, but my answers are still wrong. If I had a cube that is 3 units away on the -z axis (so it goes from -3 to -5 on the z) and is a 2x2x2 then the new z would be z' = -2/(f-n)*z + (f+n)/(f-n) = -2/(2)*-3 - 8/2 = 3-4 = -1. This should be positive 1, as the vector I chose was 0, 0, 3 which is the closest face, and when scaled/translated it should be over the origin...
for reference
http://www.gomezconsultants.com/CSE5280/Viewing/ViewVolume1.gif
http://www.gomezconsultants.com/CSE5280/Viewing/ViewVolume2.gif
For the frustum volume we have the shearing matrix
For the orthogonal volume we have the translation matrix(just look at the second matrix. The 2 is supposed to be a z)
So I'm having issues with the sign for near and far. It seems that if I set them to be positive then it works for the frustum but if I set them negative it works for the orthogonal so I'm going kind of crazy not understanding what I'm doing wrong. It's gotta be something extremely stupid that I'm not seeing.
So the whole purpose of shearing is to make sure the volume is symmetric about the x and y. if it isnt then every vector needs to be multiplied with the about matrix. For example, lets say our window was top = 1, bottom = -1, left = -2, right = 0 and n and f = whatever. The idea would be so after the shear the point (0, 0, z) would be sheared to (1, 0, z) This means the right side has moved over 1, so the left side should move to (-1, 0, z). So using the equation above we know that x' = z*(l+r/(2n)) = z*(-2/(2n)) = z*(-1/n). Another thing that we know is that z will be negative since cameras point in the negative direction ,so lets say z = -1. x' = (1/n) Now that I have this I have to assume that n has to be positive, and that n is just a distance, which I've read in a few places that they are. Thats great and all, but then I try to apply this to the translation matrix for orthogonal translation and it seems that n and f need to be negative.
Lets take a 2x2x2 that is from the original simply translated 2 units down the negative z direction. That means there's a vector on the cube at (0,0,-1) which is the center of the closest face. Now the second equation is supposed to translate the origin of the cube to the origin of the camera. Since I'm translating I need to use 4 components since translations are linear. So this vector will only change in the z direction and the equation becomes z' = -1*1 + 1*-(f+n)/2 = -1 + -(f+n)/2. So I know my answer needs to be (1, 0, 0), yet the only way for that to happen is if f and n are negative, yet in the previous example f and n are positive even though they're in the same direction. This is really confusing me and I don't know what I'm doing wrong. There's something with the signs that isn't quite working out. It might just be that it's late...
At the same time I'm seeing this stuff for ortho...
So maybe the equations wrong...idk That one won't work either, as it would be -1*-1 + (3+1)/2 = 3 and not 1....bah
And another
Why are all of these different....The last one posted is what MSDN is using so I would assume it's correct, but my answers are still wrong. If I had a cube that is 3 units away on the -z axis (so it goes from -3 to -5 on the z) and is a 2x2x2 then the new z would be z' = -2/(f-n)*z + (f+n)/(f-n) = -2/(2)*-3 - 8/2 = 3-4 = -1. This should be positive 1, as the vector I chose was 0, 0, 3 which is the closest face, and when scaled/translated it should be over the origin...
Thanks a ton, man. Got it.
Thank you so much for the help
GiantEnemyGoomba
Member
I have a question about a problem I ran across in my Calc 1 class:
Find an expression for the nth derivative of 1/(2x+1)
I know the answer is (-1)^n(n!)(2^n)(2x+1)^-(n+1), but I was wondering if there's a more intuitive way of finding the answer rather than just relying on pattern recognition. Any ideas?
Find an expression for the nth derivative of 1/(2x+1)
I know the answer is (-1)^n(n!)(2^n)(2x+1)^-(n+1), but I was wondering if there's a more intuitive way of finding the answer rather than just relying on pattern recognition. Any ideas?
Biggest-Geek-Ever
Member
As far as I know, unfortunately, it is just a matter of pattern recognition. You've probably already worked out a example that shows the nth derivative of x^n is n!, and that's certainly easy enough to grasp without grinding out derivatives. But this only applies if n is a nonnegative integer. If n is a negative integer, things don't turn out so friendly, as the factorial of such a number is more or less undefined.
GiantEnemyGoomba
Member
Bleh, yeah that unfortunately seems to be the consensus. It's an interesting problem to do, I'm just hoping it doesn't end up on a test. Thanks for your input though!
I have a graph of a cosine graph that shows (0, -3) is a low point. i.e it's increasing right after. It shows that MAYBE at x = pi, y = 0. Doesn't say for sure, looks like it. How the fuck am I supposed to figure out the period of this graph?
amplitude = 3, and the graph is reflected, so -3 = -3 cos bx. x = 0 at the only given point, so it's impossible to solve for b with what's fucking given.
amplitude = 3, and the graph is reflected, so -3 = -3 cos bx. x = 0 at the only given point, so it's impossible to solve for b with what's fucking given.
I am not able to understand the question completely. Do you need to find the period or b? At least one of the value should be given to calculate the other. Anyway, cos x has a period of 2π, whereas cos bx will have a period of 2π/b If period is given numerically, and you have to solve for b, then equate the two equations and find b. If b is given just as a random constant, and you just have to state period then it will be 2π/b
I have to find the period and b actually. And the only thing given about the graph was (0, -3) is a trough/low point. And the maybe (pi, 0) was a point. Which, as it turns out according to the book is correct. For the answer it lists 4pi/3 as the period, 3/2 = b. And lists the equation as y = -3 cos 3/2x.
I don't know how I was supposed to derive this information without more than just 0, -3 as a low point.
Missed this info. So now put (pi, 0) in the equation, you'll get b x pi = 0 This means that b can be 1/2 or 3/2 in the domain of x belonging to [0, 2pi]
1/2 will give you a period of 4pi whereas 3/2 will give you a period of 4pi/3. Since 4pi/3 is the shorter period, b will be 3/2.
So b = 3/2 and period = 4pi/3
Orbis Tabula
Member
I know it's a while later, and you probably know you're right, but I got the test back and yeah, you got it. Luckily, the professor decided not to grade the question because too many students had trouble with the wording and were unable to solve it the way he wanted. I would have been fine taking the loss on that problem, but I'm not gonna complain if he decides to ignore it. Just means I ended up with a 98.95. Which is nice, but...Fuck professors who take off one twentieth of a point at a time. Just give me the 100, man!
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