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In order to follow a branch of solutions to an implicit equation on the manifold $M = \lbrace x \in \mathbb R^n, \|x\|_p = 1 \rbrace $, I'm interested in the following problem. Given two points $a$ and $b$ in $M$, is there a natural way to extrapolate them to a third point $c$, as in the flat case $c = 2b - a$ ?

In particular, I'm trying the following strategy. Find v such that $\exp_b(-v) = a$, and compute $c$ as $\exp_b(v) = a$, where $\exp_b$ is the exponential map on $T_bM$. This should be a well-posed problem for $b$ and $a$ close enough, but my knowledge of differential geometry is very limited, and I do not know if it possible to solve this problem in closed form for general $p$. I expect it only makes sense for $p \geq 2$, but I'd be happy to be proved wrong. I have solved the case $p=2$ using the formula from (it reduces to simple geometry)

If the inverse problem (logarithm map) is too hard to solve, I'm still interested in the direct problem (exponential map), because if I have an explicit derivative of the curve available, I can use that as $v$. Of course, a trivial strategy is to extrapolate linearly the curve and then project back, but I feel that's clumsy.

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