A brief diversion on to Taylor series expansions – partly based on the Wikipedia article on Taylor’s theorem. I have been working on this for a few days now (Fields medal always going to elude me, I’m afraid!) – and still not fully worked it all out, but it is close enough for me to post it here and hope that someone might explain the bits I have missed…

We have a function which we want to approximate at a point by a Taylor expansion:

Simplifying this to the first few terms:

with < >** should be **

It is fairly obvious. I hope, why the *zeroth* term is and the first term is : just the starting point and the distance along the tangent . Then becomes the correction for the ‘error’ this crude approximation will have. (The first term here is the *linear approximation*.)

Essentially, we can generate a Taylor expansion (or in my case something that starts to look very similar) with repeated application of L’Hospital’s rule. Here goes…

Applying L’Hospital’s rule

Starting to look like a Taylor expansion now…

**Update**: Professor Rubin comments (I have moved this up here both because he knows what he is talking about and because I can get the LaTeX to work):

In the second formula, I’m pretty sure you want rather than . Unfortunately, I think you went off the rails around “Applying L’Hospital’s rule”: F’(x) – P’(x) would be the , which (assuming continuity of ) would be , not . If you go back to the first line after “Here goes…” and differentiate (we’ll assume is arbitrarily smooth), you get .

## 4 responses to “Taylor series expansions and L’Hospital’s rule”

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In the second formula, I’m pretty sure you want \(x \rightarrow a\) rather than \(a \rightarrow x\). Unfortunately, I think you went off the rails around “Applying L’Hospital’s rule”: F'(x) – P'(x) would be the \(\lim_{x \rightarrow a} \theta(x)\), which (assuming continuity of \(\theta\)) would be \(\theta(a)\), not \(\theta(x)\). If you go back to the first line after “Here goes…” and differentiate (we’ll assume \(\theta\) is arbitrarily smooth), you get \(F'(x) – P'(x) = \theta(x) + \theta'(x)(x-a)\).

Isn’t that what I have anyway? I know that means also, which appears wrong!!

[…] Paul A. Rubin, of Michigan State University (to whom I now owe a whole night of drinks), pulled me up about my use of L’Hospital’s […]