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The terms $x+\dfrac1x$ call for a change of variable with an exponential $x=e^t$ to yield $2\cosh(t)$. Hence, $$\cosh(t)=-\frac12.$$ As the RHS is negative, the substitution must involve complex numbers and $x=e^{it}$. Then $$\cos(t)=-\frac12,$$ $$t=\pm\frac{2\pi}3$$ and $$x=e^{\pm i\frac{2\pi}3}.$$

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Hint: \large \begin{align}f(t\zeta_1, t\zeta_2) &= \left(t\zeta_2 - t^2\zeta_1^2\right)\left(t\zeta_2 - 2t^2\zeta_1^2\right) \\ \\ &=t^2\left(\zeta_2 - t\zeta_1^2\right)\left(\zeta_2 - 2t\zeta_1^2\right) \end{align}

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Here is another solution. Consider the function $f(x) = m((x+A)\cap A)$. \begin{align} |f(x+\Delta x)-f(x)|&=\left|m((x+\Delta x+A)\cap A)-m((x+A)\cap A)\right| \\ &=\left|\int_{A\cap (A+x+\Delta x)}1 dy-\int_{A\cap (A+x)}1dy\right| \\ &= \left|\int_A 1_{(A+x+\Delta x)}dy-\int_A 1_{(A+x)}dy\right| \\ &= \left|\int_A (1_{A}(y-x-\Delta ...

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Clearly for any $x$ and Lebesgue measurable set $A$, $m(x+A)=m(A)$. If $x\notin B=A-A$, then $$(A+x)\cap A=\emptyset \hspace{4 mm} \text{i.e.}\hspace{4 mm} m((A+x)\cup A)=2m(A)$$ We can suppose that $m(A)>0$ and $A$ compact, as we can limit it to a compact subset of finite measure. Suppose it's not true. Then there is a sequence $x_n\to0$ such that ...

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Let $f = 1_A$ $g = 1_A$ take $h=f * g$ note that $$\|f*g\|_1 = \|f\|_1 \|g\|_1.$$ Since $\text{supp} h \subset A+A = B$, $h$ is continuous and $\|h\|_1>0$ there is a $x \in \Bbb{R}$ such that $h(x)>0$. Use the continuity of $h$ to conclude that $h(y)>0$ for every $y \in (x - \delta, x + \delta)$. So $(x-\delta,x + \delta) \subset B$

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First you'll want to show that for $g: \mathbb{R} \to \mathbb{R}$ $$\mathscr{F} \left ( \frac{ d g }{dx} \right ) = i \xi \mathscr{F}( g )$$ where the Fourier transform takes $\mathscr{F} : u(x) \to \hat u( \xi)$. Apply this to $\Delta u$ now to see $$\mathscr{F} ( \Delta u ) = - || \xi ||^2 \hat u$$ Since you'll take this as a spacial transform (not ...

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By the form of the equation, $x$ and $\dfrac1x$ are both solutions, and their product is $p:=1$. Their sum is drawn from the equation, $s:=x+\dfrac1x=-1$. From the sum and the product, we derive the difference, by $$d^2=s^2-4p=-3.$$ Then $$x=\frac{s\pm d}2=\frac{-1\pm i\sqrt3}2.$$

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Let me expand my comment into a proper answer. The map $Y(t)$ is a Jacobi field along the geodesic $\gamma(t)=\exp(tV)$ for any choices of $V$ and $W$. If you want to normalize your geodesics to have unit speed, then you want $\|V\|=1$ to make $Y$ into a Jacobi field along a geodesic. A Jacobi field along the geodesic in direction $V$ that vanish at $p$ is ...

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You have that $$d(x,z)^2 = \langle x-z, x-z \rangle = \langle x-y+y-z, x-y+y-z \rangle$$ $$= \langle x-y, x-y \rangle + 2 \langle x-y, y-z \rangle + \langle y-z, y-z \rangle$$ $$= d(x,y)^2 + 2 \langle x-y, y-z \rangle + d(y,z)^2$$ Then by Cauchy-Schwarz, $$\leq d(x,y)^2 + 2 d(x,y)d(y,z) + d(y,z)^2$$ $$\leq \left( d(x,y) + d(y,z) \right)^2$$

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