Homework Statement
I'm trying to show that every affine function f can be expressed as:
f(x) = Ax + b
where b is a constant vector, and A a linear transformation.
Here an "affine" function is one defined as possessing the property:
f(\alpha x + \beta y) = \alpha f(x) + \beta f(y)...
Homework Statement
Is the set of all continuous functions (defined on say, the interval (a,b) of the real line) a vector space?
Homework Equations
None.
The Attempt at a Solution
I'm inclined to say "yes", since if I have two continuous functions, say, f and g, then their sum f+g...
This is the quick and, in some ways, "simple" way to get the answer.
But I'm also thinking that most people don't see normed inner product spaces until they get to a course on real analysis, which might be a bit ahead of where the OP is at.
But yes, you've basically laid out the right...
Yes, correct. With inner products, we have:
v \cdot w = |v| |w| \cos \theta
where theta is the angle between the two vectors.
In your original problem statement, you also indicated that:
v \cdot w = \left(-\frac{1}{2}\right) |v| |w|
Now, as a next step, using these two facts, are...
Geometrically, the lengths of the vectors are involved in the dot product, but so is the angle between the two vectors.
Do you know what this relationship is?
The radius limits are correct.
We start at 0, and move out to a radius of 2.
For the angle limits, consider the following:
Start at the origin (0,0,0) and move up the z-axis to 2: (0,0,2).
Now, move the radius vector "down" (i.e, rotate it), until instead of pointing "up", it's now...
Your answer is close to what I get, but I would get different limits of integration.
For the transformation, I would use:
x = r \sin \theta \cos \phi
y = r \sin \theta \sin \phi
z = r\cos \theta
If you work out the "integrand", I get:
r^3 \cos \theta
but then we're...
Are you able to find the curl of this vector field?
In general, if the curl of a vector field is zero, that means that (a) it's the gradient of a scalar (potential) function; and (b) the line integral of scalar potential function is path-independent.
Remember too, area must have units of "area".. that is to say "length*length"..
So "dA" must have units of length*length.
The unit of r is "length"
The unit of dr is "length"
The unit of dtheta is none, or dimensionless...
You can't have dA = r^2drd\theta, b/c then you'd be left w/ units of...
Think about how "r" changes as "r" changes (I know that sounds silly)...
And think about how "theta" changes as "theta" changes.
So first think about r.. that's the simpler one.
Suppose we go out to a distance "r" (in polar coordinates) from the center of origin. Then we go just a bit...
I know you've already "solved" that part, but personally I don't like thinking of exactness in terms of "P's" and "Q's", unless those things have been carefully defined.
What you're really trying to do w/ exactness suppose you have a solution to the D.E. given by:
u(x,y) = c
Then you...
You'll end up w/ a double integral.. Something that looks like:
A = \int \int dA(r,\theta)
What is the differential element for the radius?
What is the differential element for the angle?
I don't want to give the whole answer away, but let's say that (this is somewhat sloppy...
OK, so suppose we have a set H such that given x,y \in H, we know that \alpha x + (1-\alpha)y \in H for all alpha.
The object is to prove that H is a hyperplane.
Note that this equation can be rewritten as:
y + \alpha(x-y) \in H
So if we can prove that the difference between two...
Homework Statement
I'm working on a problem involving hyperplanes and factor spaces. It involves a bit of setup. I'll describe first the definitions. Suppose you have a vector space K, of dimension n. Suppose you have a linear subspace L of K. Choose a vector x_0 \in K, then the hyperplane H...
Yes, essentially.
You mentioned in the OP something about a "rank-nullity" theorem.
I'm not sure exactly what you meant about "rank-nullity" theorem, but yes, the rank of a matrix is the number of linearly independent columns (or rows), and so if the number of rows is k, then you must...
I wouldn't even mess around w/ an augmented matrix, or trying to actually, "physically" reduce the matrix into some kind of row-echelon form.
Just consider the linear system:
A \cdot x = 0
A is a k-by-n matrix.. that is, it has k rows and n columns.
x is an n-dimensional vector.
0...
For the sake of illustration, suppose we are working in V = R^2, D=2, and suppose you choose S=3 vectors. You desire to prove that these three vectors must be linearly dependent in R^2.
Form a linear combination of these vectors. You desire to verify whether or not
a_1v_1 + a_2v_2 +...
Good point, but that may be ambiguity more to my use of the word "polynomial"...
What I mean is simply powers of x raised to some real number r_i. After all, x^{-2} and x^{-3} are linearly independent also, yes?
True, but we could just multiply by x^{r_1+1} to get back to the induction...
Homework Statement
Given a set of polynomials in x:
x^{r_1}, x^{r_2},...,x^{r_n}
where r_i \neq r_j for all i \neq j (in other words, the powers are distinct), where the functions are defined on an interval (a,b) where 0 < a < x < b (specifically, x \neq 0), I'd like to show that this...
Yes, so according to Mathematica the form works out to:
\int \frac{dx}{x\sqrt{x-a}} = \frac{2}{\sqrt{a}}\tan^{-1}\left(\sqrt{\frac{x-a}{a}}\right)
which is fine. I think that answers my question.
I have another related question though. This whole question/thread is taken from a...
My question is not so much about how to do the change of variables..
That I'm pretty sure I can do.
My question is more that, once I do the change of variables, I get an expression like:
\left(\frac{dr}{d\theta}\right)^2 = \frac{r^2(r-a)}{a}
or, perhaps "simplifying"...
Homework Statement
From the relation:
A(x^2+y^2) -2Bxy + C =0
derive the differential equation:
\frac{dx}{\sqrt{x^2-c^2}} + \frac{dy}{\sqrt{y^2-c^2}} = 0
where c^2 = AC(B^2-A^2)
The Attempt at a Solution
I'm able to (more or less) do the derivation, but I think the correct...
Homework Statement
Make the following change of variables:
x = r \cos \theta
y = r \sin \theta
and integrate the following equation:
(xy'-y)^2 = a(1+y'^2)\sqrt{x^2+y^2}
The Attempt at a Solution
First it's worth noting that the equation x^2+y^2=a^2 (even without changing...