For a gaussian function $\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-mean)^2}{2\sigma^2}}$, for which we introduce $\max=\frac{1}{\sigma\sqrt{2\pi}}$, the exact value at $mean+1\sigma$ is $\max/e^{1/2}$. Also, the exact value of the logarithm ($\ln$) of this function at $mean+1\sigma$ is $\ln{\max}-1/2$.
If we do a Taylor development of this gaussian function at second order, we obtain that at the position of the mean + $1\ \sigma$, the approximate value of the gaussian is max/2. This is rather far from the correct value.
If we do a Taylor development of the logarithm of this gaussian function at second order, we deduce that at the position of the mean + $1\ \sigma$, the approximate value of the gaussian is $\ln{\max}-1/2$, so we find exactly the correct value.
Why do we find exactly the correct value at $mean+1\sigma$ when using the Taylor development of the ln of the gaussian, but not if we do the Taylor development of the gaussian itself ?
Consider the infinite sum $$f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}=\sum_{n=0}^\infty (-1)^n\frac{ 2^{-n-\frac{1}{2}} \sigma ^{-2 n-1}}{\sqrt{\pi } n!}(x-\mu)^{2n}$$ $$f(\mu+\sigma)=\sum_{n=0}^\infty (-1)^n \frac{2^{-n-\frac{1}{2}}}{\sqrt{\pi } \sigma n!}=\color{red}{\frac{1}{\sqrt{2 e \pi } \sigma }}$$
$$g(x)=\log(f(x))=\log \left(\frac{1}{\sqrt{2 \pi } \sigma }\right)-\frac{(x-\mu )^2}{2 \sigma^2}$$ $$g(\mu+\sigma)=\log \left(\frac{1}{\sigma }\right)-\frac{1}{2}-\frac{1}{2} \log (2 \pi )$$ $$e^{g(\mu+\sigma)}=\color{red}{\frac{1}{\sqrt{2 e \pi } \sigma }}$$