Adding 2\pi to the argument of a complex number does not change its real or imaginary parts, but does add 2\pi i to its logarithm.
This is why we need to choose an interval of width 2\pi in which we regard the argument as lying. The standard branch uses [-\pi, \pi). The reason for this is that...
In that example, the system will look in the current directory before looking in /bin or /usr/bin. It will therefore run the dodgy ./ls rather than the authentic /bin/ls. And the malicious user can modify ./ls so it doesn't list itself when imitating the output of /bin/ls.
This can be...
B^* and Q^* relate to the same network, G^*, as stated in the first line of the proof. The definitions of B and Q are not given in this extract - presumably they are defined earlier in the text - but the notation strongly suggests that B and Q both relate to the single network G.
The length of a curve should be the same, whether you start measuring it from the left end (dx/dt > 0) or the right end (dx/dt < 0). Some details of the derivation will change if dx/dt < 0; in particular the assumption that f(\alpha) = a < b = f(\beta) must be replaced by f(\alpha) = b > a =...
The second equation in (1.18) follows from (1.17) by first applying it to \mathbf{A} \times (\mathbf{B} \times \mathbf{E}) and then setting \mathbf{E} = \mathbf{C} \times \mathbf{D}.
For the first, you can use the cyclic nature of the scalar triple product A \cdot (B \times C) = B \cdot (C...
Why did you assume that
\sqrt{(i- 1)^2} = i - 1 rather than the correct (in the principal branch)
\sqrt{(i-1)^2} = \sqrt{-2i} = \sqrt{2} e^{-i\pi/4} = 1 - i?
Note that
g\sin 2\theta = \frac{160}{56^2}\,\mathrm{m}\,\mathrm{s}^{-2} = \frac{5}{98}\,\mathrm{m}\,\mathrm{s}^{-2}. Using g = 9.8\,\mathrm{m}\,\mathrm{s}^{-2} then gives \sin 2\theta = \frac12, and \sin 30^{\circ} = \frac12 is one of the results which you should know.
The difficulty here is to compute (adequate approximations to) 9^{1/7} and 14^{8/7} = 14 \cdot 14^{1/7} using "a calculator that can only add, subtract, multiply, divide".
Your notation is confusing. You don't mean y \in Az = 0 etc.; you mean y \in \{ z \i n\mathbb{Z}^m: Az = 0 \}, but you can just write "Let y \in \Lambda^{\perp}(A). Then Ay = 0."
The central point is that if Ay = 0 then we can write Ay = AUU^{-1}y so that U^{-1} y \in \Lambda^{\perp}(AU); hence...
r = r_0 + 0.9t is only valid if dr/dt is constant.
Why did you assume that dr/dt was constant? The question only tells you that dr/dt = 0.900\,\mathrm{cm}/\mathrm{s} when r = 6.50\,\mathrm{cm}.
There are two options:
1) You can place the origin of the cylindrical coordinates at the origin.
(2) You can place the origin of the cylindrical coordinates at the centre of the circle x^2 + (y-1)^2 = 1.
The first of these leads to a simple integral; the other does not.
This is a consequence of the definition of the Darboux integral.
The upper and lower sums of a real-valued function with respect to a partition are real numbers by closure of addition and multiplication. The lower integral, as the supremum of the lower sums, is a real number by the least upper...
This example is complicated by the fact that x_\xi does not exist at \xi = 0. In this case, a characteristic in \xi > 0 intersects the characteristic x = 0 when
t = \frac{2\xi}{2 - \xi}. This first happens at
\max \left\{ 0, \inf_{\xi \geq 0} \frac{2\xi}{2 - \xi} \right\} = 0.
If the curve has an equation of the form y = f(x) then for each x there exists at most one y such that (x,y) is on the curve. It is given that (2,10) is on the curve, so (2,19) is not on the curve and Erika is wrong. Alternatively, it may be that the assumption that the curve has an equation of...
A shock forms when neighbouring charcteristics first intersect. Characteristics corresponding to neighbouring values of \xi will intersect when (x(\xi,t),t) = (x(\xi + \delta \xi, t), t) which to first order in \delta \xi requires \frac{\partial x}{\partial \xi} = 0.
I think it's continuity of exp we need, since we're not pulling the limit inside the log.
\lim_{x \to a} f(x) = \lim_{x \to a} \exp(\ln f(x)) = \exp(\lim_{x \to a} \ln f(x)).
You cna get the Laurent series for \operatorname{sech}(\pi z) about i/2 by setting t = z - \frac i2 and then
\begin{split}
\frac{1}{\cosh(i\pi/2 + \pi t)} &= \frac{1}{\cosh(i\pi/2)\cosh \pi t + \sinh(i\pi/2) \sinh \pi t} \\
&= \frac{1}{i\sinh \pi t} \\
&= \frac{1}{i\pi t} \left( 1 -...
You seem confused. There is no arbitrary constant in a definite integral, which is what M(R) = 2\pi\int_0^R r\rho(r)\,dr is, and a basic result of integration theory (Riemann or Lebesgue) is that for any function f, \int_a^a f(x)\,dx = 0.
The mass of the disc 0 \leq r < R is M(R) = 2\pi\rho_0 \int_0^R re^{-r/h}\,dr. It follows immediately that M(0) = 0. Since
\frac{d}{dr} re^{-r/h} = (1 - rh^{-1})e^{-r/h} we have
re^{-r/h} = he^{-r/h} - h\frac{d}{dr} re^{-r/h} and thus
\int_0^R re^{-r/h}\,dr = h^2(1 - e^{-R/h}) - hRe^{-R/h}.
The relevant categorisation is as elliptic (Poisson), parabolic (diffusion) or hyperbolic (wave).
The equation for displacement of an elastic medium has at leading order two time derivatives on the left and two space derivatives on the right, both with positive coefficieints; we have therefore...
I'm not sure why you need to do that. The characteristics are curves in the (x,t) plane, or more properly the half-plane t > 0. If you're having difficulty expressing them in the form t(x), then use x(t), which you already have.
You can see that the characteristics are x = \frac14t^2 - \frac12t...
Part II (Chapters 8 to 12) covers everything relevant to differential and integral calculus in \mathbb{R}^3 (partial derivatives; div, curl, grad; line integrals, surface integrals, volume integrals; Stokes' theorem and the divergence theorem), which is what an introductory multivariate calculus...
To get to this point, you expanded a determinant by its final column. What happens if you reverse that process after calculating the dot product?
Hint: Write \mathbf{w}_n = D_1\mathbf{e}_1 + \dots + D_{n}\mathbf{e}_{n}.
How would you indicate the derivative of an arbitrary function? It's not "unknown", in the sense that it must be specified before the PDE can be solved rather than being obtained as part of the solution, but nonetheless it is not possible to give its derivative in terms of t before it is specified.
(Please copy your code and paste it between tags; don't just post a screenshot. It is better to call your variable tau rather than use a unicode symbol.)
(t > t_0).all is a method; its boolean value is True. Thus your function will only ever return the unmodified Vt. You need to add () after...
As I stated above, every (n - 1) dimensional subspace is \{ \mathbf{x} \in \mathbb{R}^n: \mathbf{x} \cdot \mathbf{a} = x_1a_1 + \cdots + x_na_n = 0\} for some \mathbf{a} \neq 0. This is analagous to a plane in \mathbb{R}^3. In higher dimensions one might more properly call it a hyperplane.
An (n-1) dimensional subspace in \mathbb{R}^n is analagous to a plane in \mathbb{R}^3: they are given by an equation of the form \mathbf{x} \cdot \mathbf{n} = 0.
Consider:
1. If \alpha : V \to W is a linear map and B = \{b_i\} is a basis for V, then \alpha(B) spans \alpha(V). This follows from linearity of \alpha: If v \in V then v = \sum_i a_ib_i and
\alpha(v) = \alpha\left(\sum _ia_i b_i\right) = \sum_i a_i \alpha(b_i).
2. The ith column of a...
The matrix representation of a linear map depends on the choice of basis. In your example of polynomials of degree at most 2, taking the basis as the first three Chebyshev polynomials will give a different representation of the derivative than taking \{1, x, x^2\}. Or given any three distinct...
If your function is of the form g(x) + h(y), the partial derivative with respect to one variable will never depend on the other; if you want that then you need a more general form.
I'm not sure this is the best approach to this problem; could you please post the exact problem statement?
You should have
\ddot x = -\sum_{n=1}^\infty n^2\omega^2 (a_n \cos (n\omega t) + b_n \sin (n\omega t)).
I think the idea is that x(t) = a_1 \cos \omega t + \epsilon x_p(t) so that...
You are told to calculate the fields between the plates. Start with \mathbf{E}. You are asked to assume that the plates are infinite. Do you know how to find the field between a pair of infinite flat plates a fixed distance apart with a prescribed potential difference between them?
You're told how \mathbf{P} relates to \mathbf{E}; the question states
\mathbf{P} = \epsilon_0(\chi_1 + \chi_3E^2)\mathbf{E}. The definition
\mathbf{D} = \epsilon_0\mathbf{E} + \mathbf{P} does not assume a linear relationship between \mathbf{P} and \mathbf{E}.
Set u = y' + 2y. Then
\begin{split}
u' - u &= (y'' + 2y') - (y' + 2y) \\ &= y'' + y' - 2y \\&= x^2. \end{split} This gives you two first order linear ODEs to solve.
(This works because \lambda^2 + \lambda - 2 = (\lambda - 1)(\lambda + 2).)
If you do use induction, then your base case is to show that the result holds for n = 0 and n = 1, and your inductive hypothesis is that if the result holds for n and n-1 then it holds also for n + 1. That will invole some manipulation of binomial coefficient identities.
However, there may be...
You have
h(x,t) = \sum_{n=0}^\infty \frac{P_n(x)}{n!}t^n and you wnat to show show that h(x,t) = A(t)e^{xt} for a specific A(t).
You found after applying the recurrence relation that
\frac{\partial h}{\partial x} = \sum_{n=1}^\infty \frac{P_{n-1}(x)}{(n-1)!}t^n. But the right hand side is...
Given x \in \mathbb{R}, x^2 is uniquely determined: the laws of arithmetic would not hold if there were two or more possible values for x^2. But given c > 0 there are always two numbers such that x^2 = c, because the laws of arithmetic require (-x)^2 = x^2.
By convention, for c \geq 0 we define...