thanks for the help.
I don't know how to evaluate
\frac{d \log(u)}{d \log(x)}
I understand that I would be differentiating \log(u) as a function of \log(x), but I don't see how that is a function of \log(x).
thanks for the help.
OK, that was productive. Here's what I did:
v := \frac{y^{\prime}}{x} \Longrightarrow v^{\prime} = \frac{x y^{\prime \prime} - y^{\prime}}{x^2} \Longrightarrow x^2 v^{\prime} + y^{\prime} = x y^{\prime \prime}
Substituting,
x^2 v^{\prime} + y^{\prime} =...
EDIT: my problem is solved, thank you to those who helped
Homework Statement
Solve:
x y^{\prime \prime} = y^{\prime} \log (\frac{y^{\prime}}{x})
Note: This is the first part of an undergraduate applications course in differential equations. We were taught to solve second order...
Thanks for the reply.
Alright... I think I need to work on my intuition for this stuff to better understand your help.
Anyway, I think I've got something decent, and I'm done with it. For anyone looking at this thread for help, I found this link helpful...
Thanks for the response.
"If you sketch your curve and thing of things flowing left to right, they would be flowing in the direction of increasing r from your curve."
I don't understand what you are saying here. Could you please explain more explicitly? I am thinking you are talking about...
Homework Statement
Compute the flux of \overrightarrow{F}(x,y) = (-y,x) from left to right across the curve that is the image of the path \overrightarrow{\gamma} : [0, \pi /2] \rightarrow \mathbb{R}^2, t \mapsto (t\cos(t), t\sin(t)).
A (2-space) graph was actually given, and the problem...
Assuming we are dealing in the realm of undergraduate linear algebra...
Continuing on what WannabeNewton said, think about why \mathbb{R} is a subspace of itself, and then consider some nonempty set that is not \mathbb{R} or the set consisting of just 0.
Recall that a nonempty set is a...
The horizontal and vertical line test is an intuitive way to see if a graph belongs to an injective function. To be clear, if the graph belongs to a function, and a horizontal line drawn anywhere on the graph means that the line will only intersect the graph at most once, then the function is...
thanks for the reply.
I still find the concept odd, but it is a bit more clear, thanks. I think I just need time for this to settle in my mind, maybe.
A few questions, though.
One, when we are referring to the eigenvectors as e_1 and e_2, are we referring to them as [1,0] and [0,1], the...
Do you mean letting y=5? If so, then yes, you get \pm \sqrt{-1}, which is not a real number, and therefore there does not exist any x in the domain of f such that f(x)=5. Consequently, the range of f does not include 5, for example. That is enough to show that surjectivity is not held by f, as...
What is your reasoning? Substitute some values for x into f and see what you get, or what you can't get. Does your answer change? Even better, use graphing software to visualize f, and then everything should be pretty clear. (Example, search "wolframalpha" into a search engine and type "f(x) = 4...
Your notation for the function definition isn't correct, I think.
I believe you meant f : \mathbb{R} \rightarrow \mathbb{R}, x \in \mathbb{R} \mapsto 4 - x^2.
Here, \mathbb{R} is both the domain and codomain of f.
Surjectivity is the property that the image of the domain of f, which is...
Homework Statement
Use Lagrange multipliers to find the eigenvalues and eigenvectors of the matrix
A=\begin{bmatrix}2 & 4\\4 & 8\end{bmatrix}
Homework Equations
...
The Attempt at a Solution
The book deals with this as an exercise. From what I understand, it says to consider...
Alright, after probably too much work, I think I've gotten it worked out with delta-epsilon proof.
The following is the tactic I used.
Just to give clarity, we want to show that, if, for any real number \varepsilon > 0, there exists a real number \delta > 0 such that 0 < ||(x,y) - (0, 0)||...
You might have made a common mistake, but I might be wrong.
You want to find a \delta so that you can get from 0 < x^2 + y^2 < \delta (*) to
| \frac{x^2 \sin^2(y)}{x^2 + 2y^2} - 0| < \epsilon (**)
by good old, cold, mathematical logic and stuff (it would be good to state beforehand that your...
If I understand correctly, since our conclusion is a "result 1 OR result 2" conclusion, and it appears we can only have one or the other, but not both, i.e. the "OR" as an "exclusive-or," then we must show both that result 1 can occur while result 2 does not occur, and, likewise, that result 2...
haha, yeah, definitely. Yeah, I remember finding the chain rule a little weird. Just sit on it for a bit, and don't worry if it doesn't make "perfect sense" immediately.
Alright, so, I gather the theorem is, "For any x \in \mathbb{R}, \left\{m, n\right\} \subseteq \mathbb{N}, if both 0 \leq x \leq 1 and m \geq n, then x^m \leq x^n."
I'd start by breaking it into cases.
First, consider when x=0. Then, clearly we have our conclusion.
Similar, if we consider...
I believe the contrapositive here would be what you guessed it to be. As others have said, I think its easier to prove directly. For fun, let's prove the contrapositive, worded better for proving:
"For any vector u, in some vector space V, and any scalars a,b, in some field F, consider the...
A good way to do this is to break it down so its easier to apply the chain rule for derivation.
I'll assume we are considering the function f(x)=\sin((\pi x)^2).
To break this down, define the function S(x)=x^2.
Also, define the function g(x)=\pi x.
(Note that these x parameters in...
I'll assume were working with x \in \mathbb{R}.
Just to be clear of the definition of the floor function, it is the function \left\lfloor \right\rfloor : \mathbb{R} \longrightarrow \mathbb{Z}, with the mapping x \in \mathbb{R} \longmapsto \mathrm{max} \left\{ y \in \mathbb{Z} : y \leq x...
Homework Statement
Let z \in \mathbb{C}. Prove that z^{1/n} can be expressed geometrically as n equally spaced points on the circle x^2 + y^2 = |z|^2, where |z|=|a+bi|=\sqrt{a^2 + b^2}, the modulus of z.
Homework Equations
//
The Attempt at a Solution
My problem is that I am...
Thanks for the reply, HallsofIvy.
Oh dear, I see. Hopefully this will be my last change of answer:
f(x,y) = (1 - \frac{1}{2}x^2) (1 - xy) = 1 - xy - \frac{1}{2}x^2
Do I now have the correct idea? D=
Thanks for the reply, Zondrina.
Ohhh, so that is what is meant by "n-th degree!"
Alright, then I would have
f(x,y) = (1 - \frac{1}{2}x^2) (1 - xy + x^2y^2 - x^3y^3)
so
f(x,y) = 1 - xy + x^2y^2 -x^3y^3 - \frac{1}{2}x^2 + \frac{1}{2}x^3y - \frac{1}{2}x^4y^2 + \frac{1}{2}x^5y^3...
Homework Statement
Find the third degree Taylor polynomial about the origin of
f(x,y) = \frac{\cos(x)}{1+xy}
Homework Equations
The Attempt at a Solution
From my ventures on the Internet, this is my attempt:
I see that
\cos(x) = 1 - \frac{1}{2}x^2 + \frac{1}{4!}x^4 - \cdots...