I have posted this question in mathSE a few weeks ago (and proposed a bounty) but so far got no response.

In the book Riemanian geometry (by do-Carmo), the following result is proved (Corollary 2.9 page 115):

$\text{(1) } \|J(t) \|^2=t^2-\frac 13kt^4+R(t)$ where $(\frac{R(t)}{t^4}) \stackrel{t\rightarrow 0}{\longrightarrow} 0$.

($J(t)$ is a Jacobi field, $k$ is the sectional curvature of the relevant plane)

**An immediate corollary (2.10 in the book) is:**

$\text{(2) } \|J(t) \|=t-\frac 16kt^3+\tilde R(t)$ where $(\frac{\tilde R(t)}{t^3}) \stackrel{t\rightarrow 0}{\longrightarrow} 0$

**Question:**

In the case $k=0$ (I only assume $K_p(\sigma)=0$, i.e the sectional curvature at a given point w.r.t a given plane is zero), how can we determine if the geodesics spread faster or slower compared to the Euclidean case? (Then $ \|J(t) \|=t$).

**In other words, what governs the next term of the taylor expansion?**

(Perhaps the values of the sectional curvature in nearby points or its derivatives?)

The proof of $\text{(1)}$ was via calculation of the derivatives of $f(t)=\|J(t)\|^2 $ (using the metricity of the connection). It turns out that: $$f(0)=f'(0)=f^{(3)}(0)=0, f''(0)=2 ,f^{(4)}(0)= -8k.$$

**I am asking what is $f^{(5)}(0)$?**

If $J(t)=\exp_p(tv(s))$, $v(0)=v,v'(0)=w, \sigma = \operatorname{span} \{v,w\}\subseteq T_pM $, then: $J(0)=0,J'(0)=w$.

By Jacobi's equation $J''(0)=J^{(2)}(0)=0$ as well. An easy calculation shows that:

$$ f^{(5)}(0)= 20 \langle J^{(2)}(0),J^{(3)}(0) \rangle + 10 \langle J^{'}(0),J^{(4)}(0) \rangle + 2 \langle J(0),J^{(5)}(0) \rangle = 10 \langle w,J^{(4)}(0) \rangle$$

**So, the question amounts to calculating $J^{(4)}(0)$.**

I have tried to differentiate the Jacobi equation twice, but couldn't obtain anything useful. (do-Carmo differentiated it once, in order to calculate $J^{(3)}(0)=-R(v,w)v$).