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The Math Help Thread
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Hypertrooper
Member
Hey guys!
Does anyone have a good site with exercises for linear algebra? The best would be if they are sorted in categories like linear functions, groups, vector space etc., so I can test my knowledge. Solutions would be great, but at the end the site does not really need ones.
I already got a bunch of exams, but I want to do exercises to specific subjects, so I get to know the proposition and theorems much better.
Does anyone have a good site with exercises for linear algebra? The best would be if they are sorted in categories like linear functions, groups, vector space etc., so I can test my knowledge. Solutions would be great, but at the end the site does not really need ones.
I already got a bunch of exams, but I want to do exercises to specific subjects, so I get to know the proposition and theorems much better.
I just searched for "linear algebra problems" and this came up on the top of the list:
hans.math.upenn.edu/~kazdan/504/la.pdf
There are two hundred problems or so, with no solutions (you have us), but they seem to cover all that you are looking for.
spelltropy
Member
Quick multivariable calculus q here: have a function f(x,y) = x^2+y^2-2x-4y. Trying to find the set of points at which the direction of steepest descent of the function is (1/sqrt2, -1/sqrt2)... I found the negative of the gradient vector and then that = (1/sqrt2, -1/sqrt2) - however, I only get one value of x and y, so not sure what I'm doing wrong? (the question asks for the set of points)
boviscopophobic
Member
They're asking for the gradient vector to be in a certain direction, which is not the same as the gradient vector being exactly equal to the specified direction vector -- there could be an additional scalar factor involved.
spelltropy
Member
so any multiple of the (x,y) i found? cheers for the help
boviscopophobic
Member
Not necessarily a multiple of the particular vector you found. More like, you want all points (x,y) such that the negative of the gradient vector there is some positive scalar multiple of (1/sqrt(2), -1/sqrt(2)).
Hypertrooper
Member
Hey guys! I need help with this task:
My initial thought is to put the coefficient in a matrix and try to calculate the dim in that way:
Which would get me dim(V*)=2.
My initial thought is to put the coefficient in a matrix and try to calculate the dim in that way:
Which would get me dim(V*)=2.
I think your approach and answer are correct. To be rigorous, this is how I thought about the problem:
Let e1*, e2*, e3*, e4* form the "standard" basis for V*, the space of linear functionals from V = R^4 to the field R. The basis functions are defined by,
e1*(v) = v1,
e2*(v) = v2,
e3*(v) = v3,
e4*(v) = v4.
In other words, e_i* returns the i-th component of v (which is in terms of the standard basis for V = R^4). Clearly (maybe?), V* has a dimension of 4, the same as V.
Now consider a subspace of V*, formed by taking linear combinations of f, g, and h. What is the dimension of this subspace?
Well, in terms of the four basis for V*, we can think of f, g, h as vectors in R^4 (I'll write these in capital letters to distinguish them from the functionals f, g, h), i.e.
F = [5, 9, 59, 85]^T,
G = [5, 9, 0, 85]^T,
H = [0, 0, 135, 0]^T.
Since we have three vectors, we know that the dimension of the subspace is at most 3. But we also already know how to determine how many of these vectors are linearly independent: consider the equation a*F + b*G + c*H = 0 (here, 0 is the zero vector in R^4, but more precisely, we are considering the zero functional in V*). Solve for a, b, c like you did.
uncleniccius
Member
OK, Google doesn't seem to help (I tried and people dispute it) so I'll ask here.
Tanx=Sinx/Cosx
My textbook has a written answer which suggests Tan^2x=sin^2x/Cos^2x
Is this true?
Tanx=Sinx/Cosx
My textbook has a written answer which suggests Tan^2x=sin^2x/Cos^2x
Is this true?
Yes, the second equation is just the first one squared. They are both true.
uncleniccius
Member
Thanks!
It doesn't work for the other equations though, does it? Cosecx doesn't equal 1+cotx?
No, that one isn't true. But I think I see what is confusing you. Note that (1+cot x)^2 is equal to 1+2cot x+cot^2 x and not just 1+cot^2 x.
uncleniccius
Member
That makes sense, I wasn't thinking about it in that way.
Identities, along with everything mathematical seem to be difficult to get used to, but get much easier with time.
Thanks again for your help
Hypertrooper
Member
That's not what I exactly thought, but I knew that the elements of the dual space are the linear factors and that's where I got the idea. Thanks for explaining it in detail. Very appreciated.
The next task I have a problem with is following exercise:
That's how I tried it:
The correct solution is:
No problem, you got me to learn dual basis again, so...
For your second question, I'm not sure what a descriptive matrix/display matrix is. Can you tell us the definition? I'll think about what to do.
If anyone knows what to do, please feel free to answer in place of me.
All right, I have something, Hypertrooper. The question isn't hard to solve, but is confusing cause of that term descriptive/display (I'm sure this is not standard) and the ambiguous definitions of v1*, v2*. I think you have to assume that v1*, v2* mean the dual basis of v1, v2, and not something else, i.e.
<v1, v1*> := v1*(v1) = 1,
<v2, v1*> := v1*(v2) = 0,
<v1, v2*> := v2*(v1) = 0,
<v2, v2*> := v2*(v2) = 1.
Note that I introduced the angular brackets (they often indicate some kind of bilinear pairing, and are called a duality pairing). Also, <v1, v1*> = <v1, f(v1)>, etc.
The question is just asking you to write a matrix representation of the linear map f. We know that, given a linear map f and some choice of basis representing the vector spaces that are the domain and the range of f, we can write down a matrix M (an array of numbers) to represent the map f and its linearity.
----
Let me first illustrate what to do with a simple example. For a tensor T that maps from R^n to R^m (take the standard basis for R^n and R^m), the matrix M is our usual m x n matrix, and it satisfies the following relation:
T(x) = M*x, for all x in R^n.
This means, the j-th column of the matrix M (call it Mj) is defined to be the action of T on the j-th unit basis vector in R^n.
Mj := M*ej = T(ej).
But Mj is an element in R^m, so it can be uniquely written as a linear combination of the unit basis vectors in R^m. This allows us to define the (i, j)-th entry of M (call it Mij) as the coefficient in front of the i-th basis vector for Mj.
Not only that, because R^m is an inner product space (using the inner product <x, y> = x^T * y), we can more conveniently define Mij as,
Mij := <ei, M*ej> = <ei, T(ej)>.
----
Same thing applies to our problem. f is a linear map from V to V*, and we have already chosen the standard basis to be the basis for V and V*. Since V and V* both have a dimension of 2, M will be a 2 x 2 matrix.
Note that, in terms of v1 = [1, -1]^T and v2 = [-1, 2]^T, the standard basis of V can be written as,
e1 = [1, 0]^T = 2*v1 + v2,
e2 = [0, 1]^T = v1 + v2.
Hence, the entries of M are,
M11 := <e1, M*e1>
= <e1, f(e1)>
= <2*v1 + v2, f(2*v1 + v2)>
= 4*<v1, f(v1)> + 2*<v1, f(v2)> + 2*<v2, f(v1)> + <v2, f(v2)>
= 4(1) + 2(0) + 2(0) + 1(1)
= 5,
M12 := <e1, M*e2>
= <2*v1 + v2, f(v1 + v2)>
= 2*<v1, f(v1)> + 2*<v1, f(v2)> + <v2, f(v1)> + <v2, f(v2)>
= 2(1) + 2(0) + 1(0) + 1(1)
= 3,
M21 := <e2, M*e1>
= <v1 + v2, f(2*v1 + v2)>
= 2*<v1, f(v1)> + <v1, f(v2)> + 2*<v2, f(v1)> + <v2, f(v2)>
= 2(1) + 1(0) + 2(0) + 1(1)
= 3,
M22 := <e2, M*e2>
= <v1 + v2, f(v1 + v2)>
= <v1, f(v1)> + <v1, f(v2)> + <v2, f(v1)> + <v2, f(v2)>
= 1(1) + 1(0) + 1(0) + 1(1)
= 2.
Note, we used the bilinearity of the duality pairing and the linearity of f in order to "expand the terms."
<v1, v1*> := v1*(v1) = 1,
<v2, v1*> := v1*(v2) = 0,
<v1, v2*> := v2*(v1) = 0,
<v2, v2*> := v2*(v2) = 1.
Note that I introduced the angular brackets (they often indicate some kind of bilinear pairing, and are called a duality pairing). Also, <v1, v1*> = <v1, f(v1)>, etc.
The question is just asking you to write a matrix representation of the linear map f. We know that, given a linear map f and some choice of basis representing the vector spaces that are the domain and the range of f, we can write down a matrix M (an array of numbers) to represent the map f and its linearity.
----
Let me first illustrate what to do with a simple example. For a tensor T that maps from R^n to R^m (take the standard basis for R^n and R^m), the matrix M is our usual m x n matrix, and it satisfies the following relation:
T(x) = M*x, for all x in R^n.
This means, the j-th column of the matrix M (call it Mj) is defined to be the action of T on the j-th unit basis vector in R^n.
Mj := M*ej = T(ej).
But Mj is an element in R^m, so it can be uniquely written as a linear combination of the unit basis vectors in R^m. This allows us to define the (i, j)-th entry of M (call it Mij) as the coefficient in front of the i-th basis vector for Mj.
Not only that, because R^m is an inner product space (using the inner product <x, y> = x^T * y), we can more conveniently define Mij as,
Mij := <ei, M*ej> = <ei, T(ej)>.
----
Same thing applies to our problem. f is a linear map from V to V*, and we have already chosen the standard basis to be the basis for V and V*. Since V and V* both have a dimension of 2, M will be a 2 x 2 matrix.
Note that, in terms of v1 = [1, -1]^T and v2 = [-1, 2]^T, the standard basis of V can be written as,
e1 = [1, 0]^T = 2*v1 + v2,
e2 = [0, 1]^T = v1 + v2.
Hence, the entries of M are,
M11 := <e1, M*e1>
= <e1, f(e1)>
= <2*v1 + v2, f(2*v1 + v2)>
= 4*<v1, f(v1)> + 2*<v1, f(v2)> + 2*<v2, f(v1)> + <v2, f(v2)>
= 4(1) + 2(0) + 2(0) + 1(1)
= 5,
M12 := <e1, M*e2>
= <2*v1 + v2, f(v1 + v2)>
= 2*<v1, f(v1)> + 2*<v1, f(v2)> + <v2, f(v1)> + <v2, f(v2)>
= 2(1) + 2(0) + 1(0) + 1(1)
= 3,
M21 := <e2, M*e1>
= <v1 + v2, f(2*v1 + v2)>
= 2*<v1, f(v1)> + <v1, f(v2)> + 2*<v2, f(v1)> + <v2, f(v2)>
= 2(1) + 1(0) + 2(0) + 1(1)
= 3,
M22 := <e2, M*e2>
= <v1 + v2, f(v1 + v2)>
= <v1, f(v1)> + <v1, f(v2)> + <v2, f(v1)> + <v2, f(v2)>
= 1(1) + 1(0) + 1(0) + 1(1)
= 2.
Note, we used the bilinearity of the duality pairing and the linearity of f in order to "expand the terms."
themagicalkitsune
Member
This problem has been bugging me for a couple of days (ordinary diff. eqs):
Prove that y'(x) = (y(x))^(1/3) + x, y(1) = 0 [x >= 1, y >= 0] has a unique solution.
Picard's theorem of existence and uniqueness doesn't apply here.
I have a kinda dumb argument that I'm not sure will get me anywhere: y' is always > 0 so any solutions are 1-to-1 and increasing. If they start at the same value, they must be the same solution. I have a feeling though that this argument is weak and that one can easily construct a counterexample.
Prove that y'(x) = (y(x))^(1/3) + x, y(1) = 0 [x >= 1, y >= 0] has a unique solution.
Picard's theorem of existence and uniqueness doesn't apply here.
I have a kinda dumb argument that I'm not sure will get me anywhere: y' is always > 0 so any solutions are 1-to-1 and increasing. If they start at the same value, they must be the same solution. I have a feeling though that this argument is weak and that one can easily construct a counterexample.
Hypertrooper
Member
Woah! This helped a lot! I tried to find a similar exercise in the home assignment, but all they asked was what I calculated. This is going to be a great help for tomorrow! (If I get such a exercise in the exam)
For today and my last biggest problem is following:
My way to tackle this is to transform the matrix in a echelon form:
(I would do Ax=0 first, so I have the solution space if the assignment is asking me to find every solution which is not the case here)
If I did transform Ax=b in echelon form, I would easily see which a I have to take for a non-solution to this linear function. But if you tried it to transform in E_3, I am getting to some nasty calculation which seems wrong to me. There should be a easier way.
Biggest-Geek-Ever
Member
Recall the Invertible Matrix Theorem states that any square matrix that is invertible is row equivalent to the identity matrix. There's a property of determinants that ties into invertible matrixes that should give you the solution to this problem.Woah! This helped a lot! I tried to find a similar exercise in the home assignment, but all they asked was what I calculated. This is going to be a great help for tomorrow! (If I get such a exercise in the exam)
For today and my last biggest problem is following:
My way to tackle this is to transform the matrix in a echelon form:
(I would do Ax=0 first, so I have the solution space if the assignment is asking me to find every solution which is not the case here)
If I did transform Ax=b in echelon form, I would easily see which a I have to take for a non-solution to this linear function. But if you tried it to transform in E_3, I am getting to some nasty calculation which seems wrong to me. There should be a easier way.
I agree. A zero determinant means either there is no solution or infinitely many solutions. Once you find the values of a for which the determinant is zero, go back to the matrix equation, and do the row reduction for each a (there is only one candidate here, a = -324) to see whether there is no solution or there are infinitely many.
I also tried considering the real and imaginary parts of each equation. This results in a 6 x 6 system (with real coefficients and real unknowns), but of course, requires more work in row reduction.
I'm curious, what class are you taking that makes you solve these problems? It surely can't be an undergraduate one.
Hypertrooper
Member
It is not. I am studying Computer science at German university with Math as my minor. Unfortunately I had been lazy the whole semester and did the minimum work. Well. I am getting my revenge now.
That's quite a program they have there. Here in the US, |a| = 324 would never happen--maybe 4 at most, with 0 or 1 very likely. Best wishes on your exam.
Sorry, I don't know enough to prove it rigorously, but have you considered the equation as a fixed point problem (y' - x)^3 = y? It's unfortunate that there aren't many options for proving existence and uniqueness for nonlinear equations.
Well, it must at least exist because you can put it as:
y=int y^(1/3) - x dx
And the integral is possible because since y' exists, y must be differentiable under the interval, and conversely integrable.
You have to be careful. While we can write the equation y(x) = int_{1}^{x} y(t)^(1/3) - t dt, we don't know yet if this equation has any solution. This is how the proof of Picard's theorem goes, under a strong assumption of Lipschitz continuity; if I'm not mistaken, OP is saying that the assumption is not true here, so we cannot use Picard's theorem.
I see, oh well. Unfortunately I am a physics major, so real analysis isn't an area I have dabble much in.
Haha, I don't either. I work with finite elements, so all my solutions to differential equations are weak solutions.
loaf of bread
Member
What is the best online resource for Discrete Math help
So, method of successive approximation.
You have y'=f(t,y) and y(0)=0, the initial value which you can later correct by change in variable.
Suppose there is a function y=p(t) that satisfies initial value problem.
Basically, you have the integral: p(t)=int_0^t f(s, p(s)) ds
Choose an initial function p_0(t)=0 and replace p(s)=p_0(s)
After the integral you have p_1. Replace it once again and repeat process until p_k(t) = p_k+1(t), basically proof by induction, which makes p_k(t) a solution.
Then you show the sequence converge and that the function satisfy the integral.
Finally prove uniqueness by assuming two different solutions and then subtracting them:
|p1(t)-p2(t)| <= A*(int_0^t p1(s)-p2(2) ds)
In the end, p1(t)=p2(t)
You have y'=f(t,y) and y(0)=0, the initial value which you can later correct by change in variable.
Suppose there is a function y=p(t) that satisfies initial value problem.
Basically, you have the integral: p(t)=int_0^t f(s, p(s)) ds
Choose an initial function p_0(t)=0 and replace p(s)=p_0(s)
After the integral you have p_1. Replace it once again and repeat process until p_k(t) = p_k+1(t), basically proof by induction, which makes p_k(t) a solution.
Then you show the sequence converge and that the function satisfy the integral.
Finally prove uniqueness by assuming two different solutions and then subtracting them:
|p1(t)-p2(t)| <= A*(int_0^t p1(s)-p2(2) ds)
In the end, p1(t)=p2(t)
Right, the existence proof in Picard's theorem concerns exactly this. The integral equation is an example of fixed point problems, which take the form of L
= y for some operator L. For such problems, we can make an initial guess y0, define the next iterate (guess) to be y_{k + 1} := L(y_{k}), and keep repeating this to see if y_{k} converges (in some norm) to something--call it y*. Then, y* satisfies the equation L = y, and we have found a solution.Picard's theorem says if L is Lipschitz continuous in y, then the sequence {y1, y2, ...} that we get from the integral equation will converge. In our problem, L = L(y, x) = y^(1/3) + x, and we can show that y^(1/3) is not Lipschitz continuous in y (mainly because the derivative of y^(1/3) is not bounded at y = 0). So we have to try something else to prove existence and uniqueness.
Right, the existence proof in Picard's theorem concerns exactly this. The integral equation is an example of fixed point problems, which take the form of L
In our problem, L = L(y, x) = y^(1/3) + x, and we can show that y^(1/3) is not Lipschitz continuous in y (mainly because the derivative of y^(1/3) is not bounded at y = 0). So we are forced to be more clever in proving existence and uniqueness.
Oh so it's the same thing. I thought that because it was a section in nonlinear 1st order that it would be different. Also because I don't know Lipschitz continuity. ^_^
themagicalkitsune
Member
^ Yes that is Picard's method. However I got a nice message from a gaffer which builds up on the fact that solutions are 1-1 and hence invertible: rewrite the equation in terms of x and notice that now the relevant function satisfies Lipschitz continuity for the x - variable (as well as the other requirements to apply Picard). The solution is unique - reverting back, we get a unique solution y. Thanks for everyone's input - I'm still in the very beginning of learning diff eqs.
and notice that now the relevant function satisfies Lipschitz continuity for the x - variable (as well as the other requirements to apply Picard). The solution is unique - reverting back, we get a unique solution y. Thanks for everyone's input - I'm still in the very beginning of learning diff eqs.boviscopophobic
Member
Geometric way: the constraint x+y+z = 12 defines a plane, which we intersect with the first octant. The sum of squares x^2+y^2+z^2 corresponds to the squared radius of a sphere centered at the origin and passing through (x,y,z). Obviously, to minimize this sum, we should choose the sphere to be just big enough to intersect the constraint surface. Also obviously, this will happen when the sphere is tangent to the plane. Then symmetry implies that x = y = z = 4.
"Calculus" way, which is more generally applicable even to problems without obvious geometric structure: use Lagrange multipliers.
spelltropy
Member
does anyone have a good resource for learning java ridiculously quickly? i've been shafted by the module registration system at my uni and now have to do the programming module that I was planning to deregister (and hence didn't go to lectures for it for nearly 8 weeks...) and now I have an assignment worth 30% of the grade for it due on Sunday, so naturally i'm panicking a bit!
I think you had meant (s*50*10^-6) * Vo, i.e. solve
((Vo - Vi)/20,000) + (Vo/10000) + (Vo/20000)*s = 0.
Since we know that we want to write Vo/Vi = (...), let's divide both sides of the equation above by Vi, assuming that Vi does not equal to 0.
We get,
1/20000 * (Vo/Vi) - 1/20000 * (1) + 1/10000 * (Vo/Vi) + s/(20000) * (Vo/Vi) = 0.
By the distributive property, this equation becomes,
(1/20000 + 1/10000 + s/20000) * (Vo/Vi) = 1/20000.
For convenience, let's go ahead and multiply both sides by 20000. Then,
(1 + 2 + s) * (Vo/Vi) = 1.
Hence, assuming s in your application never equals -3,
Vo/Vi = 1/(3 + s).
Just got this question in calc 2 and I have no idea how to do it:
My first instinct is to just solve for y and then take the derivative of that, but that feels kind of wacky and doesn't necessarily line up with what we went over today (Inverse functions of the natural log, e^x, etc). Any help?
Biggest-Geek-Ever
Member
I doubt there's any way to solve explicitly for y in terms of x, but that's ok with implicit differentiation (and really, hence the name
). What you want to do is take the derivative of both sides, and whenever you differentiate with respect to y, include a multiplication by dy/dx or y', the derivative you are solving for. Once this is done, you can explicitly solve for the derivative, but it will be in terms of x and y.For your problem, you would use the product on the left side, and before differentiating the right, use the property of logariths to turn it into a sum. After differentiating, you should have:
(8x^7)e^-5y + (x^8)(-5e^-5y) y' = 1/x + (1/y) y'
Solve for y' now and you're good.
math wizards, i need help counting fish eggs.
i know i should have either measured the weight of a known number of eggs and the total amount of eggs OR counted how many eggs for a given volume then determined the final volume, but all i did was get a full volume (2,720 mL), and an average egg diameter (4 mm).
i used the volume of a sphere to get the volume of the average egg (33.5 mm^3), converted that to ml (0.034 mL), and found how many times that would go into my total volume of 2,720 mL to arrive at approximately 80,000 eggs.... but of course this wouldnt account for the significant amount of empty space between each egg in the graduated cylinder.
is there any math tricks to try and determine how much that empty space would account for of the total 2,720 mL?
i know i should have either measured the weight of a known number of eggs and the total amount of eggs OR counted how many eggs for a given volume then determined the final volume, but all i did was get a full volume (2,720 mL), and an average egg diameter (4 mm).
i used the volume of a sphere to get the volume of the average egg (33.5 mm^3), converted that to ml (0.034 mL), and found how many times that would go into my total volume of 2,720 mL to arrive at approximately 80,000 eggs.... but of course this wouldnt account for the significant amount of empty space between each egg in the graduated cylinder.
is there any math tricks to try and determine how much that empty space would account for of the total 2,720 mL?
You will probably want to look up random close pack theory and HCP packing in terms of hard spheres.
The maximum density that hard spheres can have over a specific volume is 74%. This can be calculated by looking at crystal lattice structures and analyzing the FCC/HCP crystal structure, but to save on calculations, this is basically pi/(3*2^(1/2) = 0.74.
Now, in actual experimental conditions, this becomes closer to 64% (or lower) if the packing is random and not well ordered (depending on the conditions of randomness).
Stumpokapow
listen to the mad man
So, this is the class of problems you're talking about:
http://en.wikipedia.org/wiki/Sphere_packing
It depends on how you packed the eggs. Given the size and the fact that you don't mention how you packed it, we can assume that this is a random packing:
http://en.wikipedia.org/wiki/Random_close_pack
It also depends on how compact the eggs are. But assuming you've compacted them as well as might be expected, you would expect to have filled 64% of the total volume. So just naively treat your container like it's 1750 mL and you should have your answer: ~50,000 eggs.
This is subject to significant error.
(Also, are the eggs actually round, or just round-ish?)
not compacted, other then by the weight of the eggs on top of them. the eggs are pretty darn round, but fish eggs are squishy, so i imagine the lower eggs are less round then the upper ones. i suppose i'll have to wait till they can be handled a bit and do some proper measuring to get a good idea, but 50,000 sounds great!
or, go out and get some more and do it right this time =)
thanks.
sorry i missed this earlier. thank you as well. would it be safe to assume a range between 74 and 64%? if the eggs are squishy, would that skew more towards, or even beyond, the 74% end?
Stumpokapow
listen to the mad man
The figures he's giving you are that optimum packing of spheres gets you 74% volume usage, random packing of spheres gets you 64% usage. You clearly didn't manually pack the eggs, so random packing is the better model. To the extent that squishiness allows you to pile more eggs in there than you'd be able to if the eggs were firmer, you will get gains in efficiency. I think if an accurate figure is important to you, you don't have all the information you need, but if a ballpark figure is important to you, then you can get it.
ballpark is fine at the point, i'll stick with the 64%
quick update (as if anyone really cares), i found a table in the literature which had eggs/qt for a given diameter. this chart gave me virtually the same number as the 36% empty space predicted from the sphere packing wiki. so, thanks again. fairly confident with the number now.
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