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The problem is that the Lipschitz constant L must satisfy L \geq \sup\left\{\left|\frac{f(x) - f(y)}{x - y}\right| : (x,y) \in \Omega^2, x \neq y\right\}. (If f is differentiable, then we have \lim_{x\to y} \left| \frac{f(x) - f(y)}{x - y}\right| = |f'(y)| and we would also require L \geq \sup...

Depending on the calculator, it might be that the sequence of keys [3] [-] [x2] is interpreted as (-3)^2 not -(3^2).

\tau is regarded as a parameter of the IVP (2,3), so d/dt here means \partial/\partial t. We know that both \bar{E} and 0 are n - 1 times differentiable with respect to t: 0 trivially, and \bar E because it is the solution of the IVP (2,3), so its first n - 1 derivatives with respect to t exist...

You can't put \alpha = -|m| for m \in \mathbb{Z} directly into the definition J_\alpha(x) = \sum_{n=0}^\infty \frac{(-1)^n}{n!\Gamma(n + \alpha + 1)}\left(\frac{x}{2}\right)^{2n+\alpha} but that doesn't mean that J_{-m} is not defined. Frobenius's Method for the Bessel equation xy'' +xy'...

There is only one complex infinity: It is the single point you must add to the complex plane to make it toplogically conjugate to a sphere. In the real line we treat +\infty and -\infty as being distinct because the real line is ordered. The complex plane is not.

Use the binomial expansion: (1 + x)^{\alpha} = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!}x^2 + \dots + \frac{\alpha(\alpha - 1) \cdots (\alpha - n + 1)}{n!}x^n + \dots,\qquad |x| < 1. For small |x|, f(x) = (1 - x)^{1/2} can be expanded as is. A Taylor series of x^{1/2} about 1 should also...

I think you should include the constraint explicitly in the Lagrangian: \mathcal{L} = \tfrac12m(\dot s^2 + s^2 \dot \varphi^2) - mgs\sin \varphi + m\lambda(\varphi - \omega t). How do you justify the last line? You correctly found the derivatives of the Lagrangian, but you didn't put them...

\frac{\frac{a}{a+1} + \frac{1}{2a+1}}{1 - \frac{a}{a+1}\frac{1}{2a+1}} = \frac{\frac{a}{a+1} + \frac{1}{2a+1}}{1 - \frac{a}{a+1}\frac{1}{2a+1}} \cdot \frac{(a+1)(2a+1)}{(a+1)(2a+1)} = \frac{a(2a+1) + (a+1)}{(2a+1)(a+1) - a} = \frac{ 2a^2 + 2a + 1}{2a^2 + 2a + 1} = 1.

To guarantee boundedness of the derivatives we require that they be continuous (hence C^1) on a compact domain. This is the generalisation of the theorem that a function continuous on a closed, bounded interval is bounded, but a function continuous on an unclosed bounded interval may not be...

A basic upper bound for the value of an integral of F: [a,b] \to \mathbb{R} is \int_a^b F(s)\,ds \leq (b- a)\sup_{s \in [a,b]} F(s). By extension, given I = \int_0^1 \sum_{i=1}^n\|\mathbf{F}_i(t,s)\||x_i - y_i|\,ds we can obtain \begin{split} I &\leq \sum_{i=1}^n \sup_s \|\mathbf{F}_i\||x_i -...

"All roots of natural numbers" might also include, for example, i, -1 and -i as fourth roots of 1, so its not that simple. Even if we restrict attention to real roots, we should also include -1.

I think from (\log |\sin x|)' = t the next step is \log |\sin x| = \log |A| + \tfrac12 t^2 and hence |\sin x| = |A|e^{\frac12 t^2}. It follows from this that \sin x and A have the same sign, so we can drop the absolute value signs.

It's not sufficient to note the existence of a fixed point of the iteration. The fixed point might be unstable, or if it is stable then the initial value might not be in its domain of stability. Those things need to be checked. In this case, you can show that for any x \geq 0, \left|\sqrt{2x}...

To continue my earlier post, from \cosh v_0 = \frac{1 + \epsilon^2}{1 - \epsilon^2} we can obtain \begin{split} \epsilon^2 &= \frac{\cosh v_0 - 1}{\cosh v_0 + 1} \\ &= \frac{2\sinh^2(v_0/2)}{2\cosh^2(v_0/2)} \\ &= \tanh^2(v_0/2)\end{split} so that \frac{2\epsilon^2}{(1 - \epsilon^2)^2}\ln...

We have \int_0^\infty \frac{u}{(u + \epsilon^2)(1 + u)^2}\,du = \int_0^\infty \frac{\epsilon^2}{(1 - \epsilon)^2}\left(\frac{1}{1 + u} - \frac{1}{u + \epsilon^2}\right) + \frac{1}{1 - \epsilon^2}\frac{1}{(1 + u)^2}\,du. There is in fact no difficulty with \begin{split} \int_0^\infty \frac{1}{u...

Susbstituting q = \Lambda x = \Lambda \sqrt{u} gives \begin{split} \int_0^\infty \frac{q^3}{(q^2 + m^2)\left(1 + \frac{q^2}{\Lambda^2}\right)^2}\,dq &= \Lambda^2 \int_0^\infty \frac{x^3}{(x^2 + \epsilon^2)(1 + x^2)^2}\,dx \\ &= \frac{\Lambda^2}{2} \int_0^\infty \frac{u}{(u + \epsilon^2)(1 +...

You know that (2 + \sin (x/2))^2 > 0 for any x. So y can only be negative when \cos(x/2) < 0, which is not the case for x \in [0, \pi].

\left| a_n - \frac{\pi}{12}\right| = \int_0^{2-\sqrt{3}} \frac{x^{4n}}{1 + x^2}\,dx < \int_0^{2-\sqrt{3}} x^{4n}\,dx = \frac{(2 - \sqrt{3})^{4n+1}}{4n+1} is a tighter bound; depending on what you're doing convergence as (4n+1)^{-1} may not be adequate.

If x + y = 2a and xy = a^2 then x and y are the roots of \begin{split} 0 &= (z - x)(z - y) \\ &= z^2 - (x + y)z + xy \\ &= z^2 - 2az + a^2 \\ &= (z - a)^2.\end{split} Thus (x,y) = (a,a) is the only possibility.

Are you sure about that? What you have written for the numerator is 1 - x^{4n} = (1 - x^{2n})(1 + x^{2n}). Did you mean (1 - x^4)^n? If so, you end up with a polynomial in x, which you can integrate analytically: \begin{split} a_n &= \int_0^{2 - \sqrt{3}} (1 - x^4)^{n-1}(1 - x^2)\,dx \\ &=...

You can't integrate \int (t - 2)^2\sqrt{t}\,dt = \int t^{5/2} - 4t^{3/2} + 4t^{1/2}\,dt?

Sadly, I suspect the answer is that it is because she was female.

Did you consider t = x + 2, dx = dt?

You can almost always rewrite \sum_{k=-\infty}^{\infty} a_k = a_0 + \sum_{k=1}^\infty (a_k + a_{-k}). To calculate the derivative, I would abstract the details until they are needed. So start with F(t) = \left(\sum_{k=-\infty}^{\infty} f_k(t)\right)^n and proceed to F'(t) =...

I don't agree with your result for v. Your equation for v' can be reduced to \begin{split} 0 &= xy_1v'' + (2xy_1' - y_1)v' \\ &= \frac{x^2}{y_1} \left( \frac{y_1^2}x v'' + \left( \frac{2y_1y_1'}{x} - \frac{y_1^2}{x^2}\right)v'\right) \\ &= \frac{x^2}{y_1} \frac{d}{dx}\left( \frac{y_1^2...

Easier is \begin{split} \int_0^{2\pi}\int_0^{2\pi} \cos(\phi - \phi')\,d\phi\,d\phi' &= \int_0^{2\pi}\left[ \sin(\phi-\phi') \right]_{\phi=0}^{\phi = 2\pi}\,d\phi' \\ &= \int_0^{2\pi}\sin(2\pi - \phi') - \sin (-\phi')\,d\phi' \\ &= \int_0^{2\pi}0\,d\phi' \\ &= 0.\end{split}

In terms of plane polar coordinates (r,\phi) and (r,\phi') you have \theta = \phi - \phi'. To integrate over a circle in the (r',\phi') you must not only integrate with respect to r' from 0 to R, but also with respect to \phi' between 0 and 2\pi. The integral of \cos(\phi - \phi') over a...

I assume by A(x) you mean \operatorname{Ai}(x) or vice-versa.

In principle, you can use the same technique: define the discrete fourier transforms of both variables, \begin{split} \hat\nu(\zeta, t) &= \sum_{n=-\infty}^\infty \nu_n(t)e^{in\zeta} \\ \hat u(\zeta, t) &= \sum_{n=-\infty}^\infty u_n(t)e^{in\zeta}\end{split} Then taking the DFT of your PDEs you...

I'd just like to note that, in the proposed solution by taking logs before differentiating, one should first simplify \ln(x^5) = 5 \ln x and \ln(\sqrt{x^2 + 2}) = \frac12\ln(x^2 + 2) before taking the derivative, thereby saving an application of the chain rule.

It is fairly easy to show from the definition of the limit of a sequence and the definition of a sequence diverging to +\infty that \lim_{n \to \infty} |b_n| = \infty\quad\Rightarrow\quad\lim_{n \to \infty} \frac{1}{|b_n|} = 0 and \lim_{n \to \infty} |b_n| = 0 \quad\Rightarrow\quad \lim_{n \to...

Set u = x + x^{-1}. Then x^2 + x^{-2} = u^2 - 2 and x^3 + x^{-3} = u^3 - 3u so that u^3 + u^2 - 2u - 30 = (u- 3)(u^2 + 4u - 10) = 0. Then completing the square in x gives (2x - u)^2 = u^2 - 4 and the choice u = 3 leads directly to (2x - 3)^2 = 5. The other roots u = -2 \pm \sqrt{14} lead to...

1/\log(1 + t) doesn't have a Taylor series about t = 0, because the function is not defined (and therefore not differentiable) there: \log(1 + 0) = \log 1 = 0. It does have a Laurent series about t = 0.

Note that as P and Q are self-adoint on a finite-dimensional complex inner product space (V, \langle \cdot, \cdot \rangle), their eigenvalues are real and they each have a basis of orthogonal eigenvectors. By definition, a linear map L: V \to V is positive definite if and only if \langle Lx ,x...

You have \mathbf{B} = \mathbf{b}(\mathbf{x})t where \nabla \cdot \mathbf{b} = 0. Hence in the absence of sources, (\nabla \times \mathbf{b})t = \frac{1}{c^2} \frac{\partial \mathbf{E}}{\partial t} so that \mathbf{E} = \frac{c^2t^2}{2} (\nabla \times \mathbf{b}) + \mathbf{E}_0(\mathbf{x}). From...

In this case, \frac{\partial f}{\partial y} means differentiation of f with respect to its first argument, and \frac{\partial f}{\partial y'} means differentiation of f with respect to its second argument, in both cases with the other argument held constant. You are differentiating not f, but...

If f_K(x) = \sum_{k=1}^K \frac{\arctan(kx)}{k^2} then the series expansion f_K(x) = \sum_{k=1}^K \sum_{n=0}^\infty \frac{(-1)^nk^{2n-1}x^{2n+1}}{2n+1} is only valid for x\in \bigcap_{k=1}^K \{t \in \mathbb{C} : k|t| < 1\} = \{ t \in \mathbb{C}: |t| < K^{-1}\} so that the radius of convergence...

Another approach is proof by contradiction: Assume that f'(0) exists and see what follows.

We know that f(0) = 0, so it suffices to show that \lim_{x \to 0} \frac{f(x)}{x} = \lim_{x \to 0} \lim_{K \to \infty} \sum_{k = 1}^{K} \frac{\arctan(kx)}{kx} \frac 1k does not exist. That \lim_{y \to 0} \frac{y}{\tan(y)} = 1 might be useful.

Assuming your logs are to base 2 then this is one of the possible solutions. It is convenient to use logs to base 2; then the left hand side is \log_{x/8}(x^2/4) = 2\frac{ \log_2(x/4)}{\log_2(x/8)} = \frac{2 (\log_2 x - 1)}{\log_2 x - 3} and the right hand side is 7 + \log_2(8/x) = 10 -...

Let f_i(x) = (1 - x)^i - (1 -2x)^i with x \in [0,1]. This is continuous with f_i(0) = 0 for all i \geq 1, so the problem comes for large k.

See post #3.

I'm not sure that's correct, or at least the notation is confusing. This is incorrect. \epsilon here is playing the role of x and v is playing the role of a, but the role of f is played by (v \mapsto L(v^2)) rather than L. So we should find \begin{split} L(v^2 + 2\epsilon v + \epsilon^2) &=...

Quicler is, for positive integers n and m, 12n - 20m = 4(3n - 5m) = 0\quad\Leftrightarrow\quad (n,m) = (5k,3k), k \in \mathbb{N} and the order of 12 in \mathbb{Z}_{20} is found by taking k = 1.

Your error is in stating that \vec \omega \times \vec r = r \omega \sin \theta. That deals with the magnitude of the cross-product, but not its direction; that must also depend on \phi. EDIT: Even the magnitude is only correct when \sin \theta \geq 0; outside of this range you must multiply by...

Your integral is f(r)r^2 \int_0^{\pi} \sin \theta \int_0^{2\pi} (\vec \omega \times \vec r)\,d\phi\,d\theta. Without loss of generality, you can assume \hat\omega = \omega \hat z. Then \vec \omega \times \vec r = \omega(x \hat y - y \hat x). Integrating either x = r \cos \phi \sin \theta or y =...

What do we know about T_{\mathrm{air}}? Have you tried a Laplace transform in time? That gives \kappa(p\hat T - T_0) = \frac{\partial^2 \hat T}{\partial x^2} where \kappa= \rho c_p subject to \hat T(0,p) = \frac{T_s}{p}, \quad \left.-k\frac{\partial \hat T}{\partial x}\right|_{x=L} = h(\hat...

The OP did state "While trying to solve problem in Hydrodynamic stability ...". However, since the OP hasn't told us which problem they are trying to solve or how they derived this system, we can but speculate. I did misinterpret the situation in my earlier post. The system of PDEs looks like...

This is a linearised stability problem, so you are looking for normal modes. Your assumption, in view of the boundary conditions, is that (\theta, \psi) = (\Theta e^{k_nx}\sin( n\pi y), \Psi e^{k_nx}\sin (n \pi y)) for constant (\Theta, \Psi) and positive integer n. If k_n has strictly...