# Linear approximation and differentiability

If a function ${f\colon \mathbb R\rightarrow \mathbb R}$ is differentiable at ${a\in \mathbb R}$, then it admits good linear approximation at small scales. Precisely: for every ${\epsilon>0}$ there is ${\delta>0}$ and a linear function ${\ell(x)}$ such that ${|f(x)-\ell(x)|<\epsilon \,\delta}$ for all ${|x|<\delta}$. Having ${\delta}$ multiplied by ${\epsilon}$ means that the deviation from linearity is small compared to the (already small) scale ${\delta}$ on which the function is considered.

For example, this is a linear approximation to ${f(x)=e^x}$ near ${0}$ at scale ${\delta=0.1}$.

As is done on this graph, we can always take ${\ell}$ to be the secant line to the graph of ${f}$ based on the endpoints of the interval of consideration. This is because if ${L}$ is another line for which ${|f(x)-L(x)|<\epsilon \,\delta}$ holds, then ${|\ell-L|\le \epsilon \,\delta}$ at the endpoints, and therefore on all of the interval (the function ${x\mapsto |\ell(x)-L(x)|}$ is convex).

Here is a non-differentiable function that obviously fails the linear approximation property at ${0}$.

(By the way, this post is mostly about me trying out SageMathCloud.) A nice thing about ${f(x)=x\sin \log |x|}$ is self-similarity: ${f(rx)=rf(x)}$ with the similarity factor ${r=e^{2\pi}}$. This implies that no matter how far we zoom in on the graph at ${x=0}$, the graph will not get any closer to linear.

I like ${x\sin \log |x|}$ more than its famous, but not self-similar, cousin ${x\sin(1/x)}$, pictured below.

Interestingly, linear approximation property does not imply differentiability. The function ${f(x)=x\sin \sqrt{-\log|x|}}$ has this property at ${0}$, but it lacks derivative there since ${f(x)/x}$ does not have a limit as ${x\rightarrow 0}$. Here is how it looks.

Let’s look at the scale ${\delta=0.1}$

and compare to the scale ${\delta=0.001}$

Well, that was disappointing. Let’s use math instead. Fix ${\epsilon>0}$ and consider the function ${\phi(\delta)=\sqrt{-\log \delta}-\sqrt{-\log (\epsilon \delta)}}$. Rewriting it as

$\displaystyle \frac{\log \epsilon}{\sqrt{-\log \delta}+\sqrt{-\log (\epsilon \delta)}}$

shows ${\phi(\delta)\rightarrow 0}$ as ${\delta\rightarrow 0}$. Choose ${\delta}$ so that ${|\phi(\delta)|<\epsilon}$ and define ${\ell(x)=x\sqrt{-\log \delta}}$. Then for ${\epsilon \,\delta\le |x|< \delta}$ we have ${|f(x)-\ell(x)|\le \epsilon |x|<\epsilon\,\delta}$, and for ${|x|<\epsilon \delta}$ the trivial bound ${|f(x)-\ell(x)|\le |f(x)|+|\ell(x)|}$ suffices.

Thus, ${f}$ can be well approximated by linear functions near ${0}$; it’s just that the linear function has to depend on the scale on which approximation is made: its slope ${\sqrt{-\log \delta}}$ does not have a limit as ${\delta\to0}$.

The linear approximation property does not become apparent until extremely small scales. Here is ${\delta = 10^{-30}}$.

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