# 英国代考|MATH0020 Differential Geometry 微分几何学

Lemma 1.11.3. The following formulas hold:
\begin{aligned} \langle\overrightarrow{\boldsymbol{r}}(s+\sigma)-\overrightarrow{\boldsymbol{r}}(s), \overrightarrow{\boldsymbol{\tau}}(s)\rangle &=\sigma-\frac{k^{2}}{6} \sigma^{3}+o\left(\sigma^{3}\right), \ |\overrightarrow{\boldsymbol{h}}(s)| &=\sigma-\frac{k^{2}}{4} \sigma^{3}+o\left(\sigma^{3}\right), \ \langle\overrightarrow{\boldsymbol{\tau}}(s), \overrightarrow{\boldsymbol{\tau}}(s+\sigma)\rangle &=1-\frac{k^{2}(s)}{2} \sigma^{2}+o\left(\sigma^{2}\right), \end{aligned}
where $\overrightarrow{\boldsymbol{h}}(s)=\overrightarrow{\boldsymbol{r}}(s+\sigma)-\overrightarrow{\boldsymbol{r}}(s)$ and $\sigma>0$.
Proof. Applying Taylor’s formula and the Frenet formulas, we have

\begin{aligned} \overrightarrow{\boldsymbol{r}}(s+\sigma)-\overrightarrow{\boldsymbol{r}}(s) &=\overrightarrow{\boldsymbol{r}}(s) \sigma+\frac{1}{2} \overrightarrow{\boldsymbol{\tau}}^{\prime}(s) \sigma^{2}+\frac{1}{6} \overrightarrow{\boldsymbol{\tau}}^{\prime \prime}(s) \sigma^{3}+o\left(\sigma^{4}\right) \ &=\overrightarrow{\boldsymbol{\tau}}(s) \sigma+\frac{1}{2} k \overrightarrow{\boldsymbol{v}}(s) \sigma^{2}+\frac{1}{6}\left[k^{\prime} \overrightarrow{\boldsymbol{v}}(s)-k^{2} \overrightarrow{\boldsymbol{\tau}}(s)+\kappa \overrightarrow{\boldsymbol{\beta}}(s)\right] \sigma^{3}+o\left(\sigma^{4}\right) \end{aligned}
From this and from the properties of the Frenet frame the first formula of the lemma follows. Assume that $\vec{h}(s)=\overrightarrow{\boldsymbol{r}}(s+\sigma)-\overrightarrow{\boldsymbol{r}}(s)$. Applying the above decomposition for the vector $\vec{h}(s)$, we have
$$|\overrightarrow{\boldsymbol{h}}(s)|^{2}=\sigma^{2}+\frac{1}{4} k^{2} \sigma^{4}-\frac{1}{3} k^{2} \sigma^{4}+o\left(\sigma^{4}\right)=\sigma^{2}-\frac{1}{12} k^{2} \sigma^{4}+o\left(\sigma^{4}\right)$$
Taking the square root, we obtain the second equality of the lemma. The last equality of the lemma can be proved analogously.

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