Taylor expansion of smooth functions on manifolds

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arrigo.b...@gmail.com

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Apr 5, 2022, 1:40:36 PM4/5/22
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I wonder if Manopt could be helpful to solve this problem. Let's sat we have a smooth function f() defined on a manifold M. What I am looking for is an algorithm able to compute a low order polynomial approximation of f() around some point X \in M, essentially a (multivariate) truncated Taylor expansion. I think that Manopt has all the machinery needed to do this but I could not find any literature on this problem.

Ronny Bergmann

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Apr 5, 2022, 4:07:26 PM4/5/22
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Hi,
this sounds like a very interesting topic, though I think there is a challenge: A “first order polynomial” i.e. geodesic-like (more complicated probably in the multivariate case, but I have an idea) is something I can easily imagine. How would you define higher order ones? I think this might not have a unique answer and then the question is which one to choose to Taylor.

arrigo.b...@gmail.com

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Apr 6, 2022, 1:18:22 PM4/6/22
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I don't think I would ever need to go to higher than second order.

Nicolas Boumal

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Apr 7, 2022, 2:40:54 AM4/7/22
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Hello, 

I'm assuming your function f is real valued. 

One often useful notion of Taylor expansion then is to consider f(R_x(v)) for small v in the tangent space at x, where R_x maps that tangent space to M: it's a retraction, such as the exponential map for example.

If the retraction is second order (which is the case for the exponential map) we can write:

f(R_x(v)) = f(x) + <grad f(x), v> + (1/2) <Hess f(x)[v], v> + O(||v||³)

Manopt can help you with computing the Riemannian gradient and Hessian for many manifolds. 

Is this the type of Taylor expansion you had in mind? 

I explain much more about them in my book which you can download from my webpage: nicolasboumal.net/book

Best, 
Nicolas 

arrigo.b...@gmail.com

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May 16, 2022, 2:43:48 PM5/16/22
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Hi Nicolas, sorry for my late reply however for some reason I see your last message only now even if I check this thread very often... Yes, I think your suggestion will work, thanks again!

-Arrigo

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