# Search results

1. ### Prove that T(G) is subgroup of G

1. a in T, there exists n such that a^n = e b in T, there exists m such that b^m = e (ab)^mn = (a^n)^m (b^m)^n = e^m e^n = e axiom 1 holds. 2. let e be identiy in G e^n = e ==> e in T(G) T(G) subgroup of G, it's also true that ea = ae =a for all a in G conclusion: idenity of G, e...
2. ### Prove that T(G) is subgroup of G

Homework Statement Homework Equations subgroup axioms: 1. a, b in T(G), then ab in T(G) 2. existence of identity element. 3. a in T(G), then a^-1 in T(G) The Attempt at a Solution 1. let a be in T(G), then a^n = e. let b be in T(G), then b^n = e (ab)^n = (a^n)(b^n) = (e)(e) = e axiom 1...
3. ### Quickie: vector normal to surface

My fault all along. I was trying to calculate the magnitude of the surface instead of the magnitude of the gradient of the surface. Yes i know how to calculate the magnitude of a vector thank you very much. answered.
4. ### Quickie: vector normal to surface

how does that help? i need to know how to compute the magnitude.
5. ### Quickie: vector normal to surface

nope. Take the magnitude of the llevel surface?
6. ### Quickie: vector normal to surface

Homework Statement Find vector normal to z = x^2 + y^2 - 3 at point r = (2, -1, 2) Homework Equations The Attempt at a Solution here is the markscheme. I understand how to find the gradient, but i dont understand how they calculated the magnitude. thanks
7. ### Vector space, linear transformations & subspaces

so essentially I've done the problem but in an "all in one" fashion?
8. ### Vector space, linear transformations & subspaces

Homework Statement Let V be a vector space over a field F and let L and M be two linear transformations from V to V. Show that the subset W := {x in V : L(x) = M(x)} is a subspace of V . The Attempt at a Solution I presume it's a simple question, but it's one of those where you just don't...
9. ### Simple matrix/linear algebra question, help

can you look at my edit? thanks.
10. ### Simple matrix/linear algebra question, help

I think i understand. L(1) means that we are considering the function f(x) = 1 (a straight line through y=1): L(f) = f' + f(-2)t L(1) = d(1)/dt + f(-2)t f(-2) is equal to 1 in this case, so: d(1)/dt + f(-2)t = 0 +t = t However, if we consider L(t): L(t) = t' + f(-2)t =d(t)/dt + f(-2)t...
11. ### Simple matrix/linear algebra question, help

Still doesn't make sense. L(1) = 1' + 1*(-2)t = d(1)/dt + (-2)t = 0 + (-2)t = (-2)t =/= t. what is f(-2) equal to anyway?
12. ### Simple matrix/linear algebra question, help

Homework Statement Here is the question, i know how to do part (i) but I do not understand part (ii): The Attempt at a Solution [/B] here's the solution from the marking scheme: i understand how they formed the matrix from their working out (i can se the pattern), but I do not...
13. ### Quick directional derivative question -- help please

Duh! you're awesome. Thanks.
14. ### Quick directional derivative question -- help please

Homework Statement [/B] find directional derivative at point (0,0) in direction u = (1, -1) for f(x,y) = x(1+y)^-1 The Attempt at a Solution grad f(x,y) = ( (1+y)^-1, -x(1+y)^-2 ) grad f(0,0) = (1, 0) grad f(x,y) . u = (1,0).(1,-1) = 1. seems easy but markscheme says im wromg. It says...

16. ### Lagrange multipliers, guidance needed

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17. ### Lagrange multipliers, guidance needed

ok, so then x must equal y which must equal zero. Or perhaps my calculations are wrong...
18. ### Lagrange multipliers, guidance needed

that x =y = 0, however this cannot be true due to the third equation. therefore x and y has to take multiple values that satisfy the following relationships: \frac{32}{3}xy = -24xy 1 - 3x^{2}-2y^{2}=0 going along the right lines? and thanks for the help so far.
19. ### Lagrange multipliers, guidance needed

1. \frac{\partial L}{\partial x}= (-64x(1+4x^{2}+4y^{2})^{-3}) - 6x\lambda = 0 2. \frac{\partial L}{\partial y}= (-64y(1+4x^{2}+4y^{2})^{-3}) - 4y\lambda = 0 3. \frac{\partial L}{\partial \lambda}= 1 - 3x^{2}-2y^{2}=0 if you get an equation for lambda from the first and second equation...
20. ### Lagrange multipliers, guidance needed

\frac{\partial L}{\partial x}= (-64x(1+4x^{2}+4y^{2})^{-3}) - 6x\lambda = 0 \frac{\partial L}{\partial y}= (-64y(1+4x^{2}+4y^{2})^{-3}) - 4y\lambda = 0 \frac{\partial L}{\partial \lambda}= 1 - 3x^{2}-2y^{2}=0 this leads to a contradiction though, using the first and second equation we get...
21. ### Lagrange multipliers, guidance needed

ok, then the Lagrange equation we are interested in is: L(x,y, \lambda)=\frac{4}{(1+4x^{2}+4y^{2})^2}- \lambda(3x^{2} + 2y^{2}-1) what do you mean by form of k_f(x,y)? could you come up with a mini-alternative example to what you mean, I just don't see how you can simplify it, unless this has...
22. ### Lagrange multipliers, guidance needed

my k_f(x,y) is: \frac{4}{(1+4x^{2}+4y^{2})^2} How would you simplify this/the problem? I tried simplifying with wolfram alpha, it doesn't return anything useful. my k_g(x,y) is: \frac{24}{(1+36x^{2}+16y^{2})^2} please, someone help. It's a bit of an emergency.
23. ### Lagrange multipliers, guidance needed

Homework Statement f(x,y) is function who's mixed 2nd order PDE's are equal. consider k_f: determine the points on the graph of the parabloid f(x,y) = x^2 + y^2 above the ellipse 3x^2 + 2y^2 = 1 at which k_f is maximised and minimised. The Attempt at a Solution is this the...
24. ### Partial derivatives

yes I know. I fixed it. I rewrote f'(u) as partial f/partial u and I treated it as multiplication with the partial derivatives. You should end up with two different partials at the bottom. if you plug it in I don't know how to show LHS = RHS.
25. ### Partial derivatives

I realized my mistake. I rewrote f'(u) as partial f/partial u, and I saw my mistake when taking the 2nd derivative (it was more clear). However, it STILL doesn't work.
26. ### Partial derivatives

\frac{\partial w}{\partial t}=\frac{\partial w}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial w}{\partial v}\frac{\partial v}{\partial t}. \frac{\partial w}{\partial u} = f'(u). \frac{\partial w}{\partial v} = g'(v). \frac{\partial u}{\partial t}=-a. \frac{\partial v}{\partial...
27. ### Partial derivatives

I tried caltulating the individual components and I did what titas.b said to do. It didn't workout when I tried to show each side is equal. for example, how would you do partial w/ partial u? is that equal to f'(u)? maybe my individual compenents are incorrect.
28. ### Partial derivatives

Homework Statement let w(u,v) = f(u) + g(v) u(x,t) = x - at v(x,t) = x + at show that: \frac{\partial ^{2}w}{\partial t^{2}} = a^{2}\frac{\partial ^{2}w}{\partial x^{2}} The Attempt at a Solution w(x-at, x+at) = f(x-at) + g(x+at) \frac{\partial }{\partial t}(\frac{\partial...
29. ### Field axioms, subspaces

show that W is a subspace of the vector space V (and V is a vector space over the Field F = {0, 1}). 1.x = x since 1 is in the field and x is in W (and W is a subspace of V) then we can use the scalar multiplication identity from V. Ergo, scalar multiplication holds for this subspace. Proof...
30. ### Field axioms, subspaces

I think I've found a more robust proof for scalar multiplication: 0, 1 in F_2 x in W consider all possible combinations: 0x, 0x + 1x, 1x + 1x, 0x + 0x, 1x + 1x + 1x 0x = 0_{W} 0x + 1x = 0_{W} + x = x 1x + 1x = (1 + 1)x = 0x = 0_{W} 0x + 0x = 0_{W} 1x + 1x + 1x = (1 + 1)x + 1x = 0x + 1x =...