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...
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...
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.
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
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...
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...
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...
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...
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.
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...
\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...
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...
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.
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...
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.
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.
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.
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...
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...