Cross posted from my question:
This is a formula/theorem (written below) that I found mentioned in a medical imaging paper, without any proof or any further detail. I'd appreciate if anyone could please provide either the detail or the sketch to prove it.
Let $c=\exp_p(tV)$ be a geodesic on a Riemannian manifold $M$. Let $W(t)$ be a parallel vector field $c$ with $W(0)=W$. Consider the Jacobi field $J(t)=(D\exp_p)_{tV} ( {tW} )$.
Then: $\lim_{t\to 0}|\frac{W(t)-\frac{J(t)}{t}}{t}|=0.$
For constant curvature spaceforms, when I take normal Jacobi fields the formula holds good. But I'm not sure how to prove it for general manifolds.
P.S. in the paper that I mentioned, the formula is mentioned somewhat heuristically, but this is my interpretation of their words.
EDIT I: I can also see that if we take compare the norms of $W(t)=W$ and norms of $J(t)$, which is $|J(t)|=t|W|+O(t^3)$ (See Do Carmo, P. 115), then clearly,
$lim_{t\to 0}\frac{|W(t|)-\frac{|J(t)|}{|t|}}{|t|}=0.$
But still it doesn't prove the result, but shows that the claim could be correct.
EDIT II:I think I've got an idea, it's working so far, but not fully yet. I considered the function $f(t):=\langle tW(t)-J(t),tW(t)-J(t)\rangle =||tW(t)-J(t)||^2$ and showed so far that $f(0)=f'(0)=f''(0)=0$. So if I do some careful work on the fourth derivative $f''''(0)$ now, hopefully I'll have it!
(1) Consider Chapter 2 exercise 2 (57p.) in the book "Riemannian Geometry" by Do Carmo: So $$ \bigg(W-\frac{J}{t}\bigg)' = \lim_t \frac{ P^{-1}_{0,t} \bigg(W-\frac{J}{t} \bigg) + \bigg(W-\frac{J}{t}\bigg)(0) }{t} = \lim_t \frac{W-\frac{J}{t} }{t} $$ where $P_{0,t} : T_{p} M \rightarrow T_{c(t)}M$ is a parallel transport.
Here we used : $P_{0,t}^{-1} $ goes to $id$ if $t$ goes to $0$. And $$ J(0)=0,\ J'(0)=W(0) \Rightarrow J = tW(t) + o(t^1) $$
(2) And $$ \bigg(W-\frac{J}{t}\bigg)' = \frac{J't-J}{t^2} $$
Let $X:=J't-J$ so that $$ X(0)=0,\ X'(0)= (J't-J)' = J'' t =-tR(c',Y)c'=0$$ $$ X''(0)=J'''t+J''=-R(c',Y)c'=0$$
Hence $$ X(t)= \sum_{i=3}^n \frac{t^i}{i!} E_i(t)+ o(t^n) $$ for some parallel vector fields $E_i$ s.t. $E_i(0) =\nabla_{c'}^{(i)} X(0)$. Hence $\lim_t \frac{X}{t^2} =0$.