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I will reproduce the statement and part of the proof of Theorem 10.28 from Lee's book. Note that $I(V, W)$ below is the index form, but this is not really relevant for my question. Theorem 10.28. Let $(M, g)$ be a Riemannian manifold. Suppose $\gamma: [a, b] \to M$ is a unit-speed geodesic segment without interior conjugate points. If $V$ is any proper normal piecewise smooth vector field along $\gamma$, then $I(V, V) \geq 0$, with equality if and only if $V$ is a Jacobi field. In particular, if $\gamma(b)$ is not conjugate to $\gamma(a)$ along $\gamma$, then $I(V, V) \gt 0$ unless $V = 0$. First the proof reparametrizes so that we work over $[0, b]$ and define $p = \gamma(0)$. Then we choose an orthonormal basis $(w_1, \ldots, w_n)$ for $T_pM$ such that $w_1 = \gamma'(0)$. Then we define $J_\alpha$ for $\alpha = 2, \ldots, n$ as the unique normal Jacobi fields along $\gamma$ satisfying $J_\alpha(0) = 0$ and $D_t J_\alpha(0) = w_\alpha$. Because there are no interior conjugate points, we can show that $(J_\alpha(t))$ forms a basis for the orthogonal complement of $\gamma'(t)$ in $T_{\gamma(t)}M$ for each $t \in (0, b)$. Then for $t \in (0, b)$ we can write $$V(t) = v^\alpha(t) J_\alpha(t)$$ for piecewise smooth functions $v^\alpha: (0, b) \to \mathbb{R}$. Now the proof says that the $v^\alpha$ can be extended to piecewise smooth functions on $[0, b]$. I understand the proof that the $v^\alpha$ can be extended on $[0, b)$, and I also understand how to prove that the $v^\alpha$ can be extended on $[0, b]$ if $J_\alpha(b) = 0$ for all $\alpha$ (in both cases we work in normal coordinates and use the explicit form of the Jacobi fields in normal coordinates and Taylor's theorem). Part where I am stuck Now the proof states (If $J_\alpha(b) = 0$, the argument is the same as for $t = 0$, while if not, it is even easier.) But I'm confused how to prove this in the case where $J_\alpha(b) \neq 0$ for some $\alpha$. My difficulty is that I can't rule out a case such as $J_2(b) \neq 0$, $J_3(b) \neq 0$ and $v^2(t) = \frac{A}{b - t}$, $v^3(t) = \frac{B}{b - t}$ for some nonzero constants $A, B$ such that $AJ_2(t) + BJ_3(t) = \mathcal{O}((b - t)^2)$ as $t \to b^-$ (it is possible that $J_2(b)$ and $J_3(b)$ are linearly dependent, since $\gamma(b)$ can be a conjugate point). Then we have $$V(t) = \frac{A}{b - t} J_2(t) + \frac{B}{b - t} J_3(t)$$ and $V(t) \to 0$ as $t \to b^-$ so $V$ could still be piecewise smooth on $[0, b]$ even though the component functions are not (so the proof must show that the Jacobi fields are such that this doesn't happen). How do I prove $v^\alpha$ has a piecewise smooth extension on $[0, b]$ in the case where $J_\alpha(t) \neq 0$ for some $\alpha$?

It is well known that $$\int e^x \mathrm{d}x=e^x+C$$ Is there a way to prove this is true without using $\int e^x \mathrm{d}x = e^x+C$, differentiating both sides of course work, but I want to look for ways that involve integration techniques, for example using integration by parts. Here is what I was able to do, the integral can be written as $$\int e^x \mathrm{d}x=\int 1 e^x\mathrm{d}x$$ By integration by parts, we have $$\int 1e^x\mathrm{d}x = 1\cdot e^x - \int e^x\mathrm{d} 1 = e^x +C$$ However here I still used the fact that $\int e^x \mathrm{d}x = e^x+C$, is there anyway to get around this?

I am currently reading "functions of one complex variable". In chapter 3 page 54 there is a section which I don't understand, and I am seeking alternative explanations. Let $G_1, G_2$ be open connected sets; to try to find an analytic function $\Gamma$ such that $\Gamma(G_1) = G_2$, we try to map both $G_1$ and $G_2$ onto the open unit disk. If this can be done, $\Gamma$ can be obtained by taking the composition of one function with the inverse of the other. As an example, let $G$ be the open set inside two circles $ \Gamma_1$ and $\Gamma_2$, intersecting at points $a$ and $b$ $a \neq b$. Let $L$ be the line passing through $a$ and $b$, and give $L$ the orientation $(\infty, a, b)$. Then $T_z = (z, \infty, a, b)$ maps $ L $ onto the real axis $(T_\infty = 1 , T_a = 0, T_b = \infty)$. Since $T$ must map circles onto circles, $T$ maps $\Gamma_1$ and $\Gamma_2$ onto circles through $0$ and $\infty$. That is, $T(\Gamma_1)$ and $T(\Gamma_2) $ are straight lines. By the use of orientation, we have that $T(G) = \{ w : \epsilon < \arg(w) < \epsilon' \}$ for some $ \epsilon' > 0 $, or the complement of some such closed sector. By the use of an appropriate power of $ z $ and possibly a rotation, we can map this wedge onto the right half-plane. Now, composing with the map $(z - i)(z + 1)^{-1}$ gives a map of $G$ onto $D = \{ z : |z| < 1 \}$. There are a lot of leaps in logic in this paragraph. I will try and spell out the ones I don't understand: $T$ maps $L$ onto the real axis. I assume this is because anything on $L$ is technically co-circular (because a line with infinity is homeomorphic to a circle?), and thus the cross product is real. By the use of an appropriate power of $ z $ and possibly a rotation, we can map this wedge onto the right half-plane. Now, composing with the map $(z - i)(z + 1)^{-1}$ gives a map of $G$ onto $D = \{ z : |z| < 1 \}$. I don't understand any of this really, could someone explain what is going on here, possibly with an example?

Matt Parker has asserted multiple times that the Thompson problem is unsolved. This asks for an algorithm -- explicit in how specified by Smalle -- which finds the minimum energy configuration of n points on a sphere under electrostatic repulsion. Surely that can't be right. Maybe it's not a very practical algorithm but can't we just shove it into the decision procedure for Th(R) to approximate a minimal solution to any desired accuracy? The function to be minimized is just a sum over 1/distance. Indeed, if you knew how many configs achieved the global minimum you could just brute force search and bound (since on any closed set of domain excluding equal vertexs all partials will be bounded so you can just find as many minimums as exist and make mesh fine enough and bound everything else to be above them). Am I being dumb, are Wikipedia and Parker just wrong or is this another one of those cases where non-logicians are implicitly demanding a solution in familiar functions not arbitrary computable approximations but not saying so?

I need to prove that for two concentric circles $D\subset E$, for $T$ a Möbius transformation which does not turn the circles inside out (i.e.$T(D)\subset T(E))$ and $\infty$ being outside $T(E)$: $\frac{r(T(E))}{r(T(D))}\geq\frac{r(E)}{r(D)}>1$. I have already shown that $r(T(D))=\frac{r}{\vert(u^2-1)(1-r^2)+1\vert}$ for $T$ being a special stretch map. Now I need the argument to generalise this result. Can anyone give me a hint? The task is found in the book Indras pearls by David Mumford, Caroline Series and David Wright on p. 94 as project 3.5 if anybody is interested in this. Many thanks in advance!