Taylor's Theorem

For two variable functions, Taylor's Theorem looks like this:


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Is there a compact way of writing this without using matrices? (Particularly not that awful-looking Hessian matrix.) At first, I was thinking about the below:



But then I thought about it and expanding it seemed to imply this



:/

So is there some kind of operator that states "collect the terms in order"? The reason I ask is that I want to generalize this into 3 variables and it doesn't seem possible to do so with the matrix notation. Also, the formula just looks awful, is difficult to memorize for anything beyond the second order, and over-complicates this. I'm really just looking for simplicity.

Any help is appreciated.

______


Why does this paper state a different formula? o.o

http://user.gs.rmit.edu.au/rod/files/pub...er%207.pdf

*cricket cricket*

*waits for Rust or Corn*

Why does this paper state a different formula? o.o

http://user.gs.rmit.edu.au/rod/files/pub...er%207.pdf

it doesn't? (equations 7.5 / 7.6)
they just use a, b instead of x_0, y_0 and assume every f is evaluated at the point

looks like it has your 3-d variant too, which is about what i'd expect for it.
i can't think of a more elegant way to write it though.

It's not pretty, but it's working.



for clarity:

 Spoiler

Noah

*cricket cricket*

*waits for Rust or Corn*

I was thinking of Noah. Maybe Russt might know, but I think Corn is still studying single-variable calculus.

it doesn't? (equations 7.5 / 7.6)

For (7.5), they have f''(a)(x-a)^2+f'(a)f'(b)(x-a)(y-b)+f''(b)(y-b)^2. There should be a coefficient of 2 in the middle term. They do something similar for (7.6). I'm not sure why they left the coefficients out.

Hmm... doesn't that run into the same problem as mine? That gives me an idea though:



How would I express this for variables? In the notation I used in the first post, I can just add a (z-z0)(d/dz) term inside the brackets without losing anything, but I don't think I can do the same thing here. For example, if I did use this notation, if a=x-x0, b=y-y0, and c=z-z0, I would lose the 6abc term the expansion of (a+b+c)^3. Or am I supposed to lose that term?

Hmm... doesn't that run into the same problem as mine? That gives me an idea though:

Well, those two are essentially the same though, but your expression would be a bit more clear, perhaps? I mean,

,

so in theory, they express the same thing. Though I get what you mean, but I don't know if it really matter. It's the same for the function



If we treat the symbols as unknown variables, it will be equivalent to our expressions. I'm not entirely sure, but I think it is okay to write it in this way.

How would I express this for variables? In the notation I used in the first post, I can just add a (z-z0)(d/dz) term inside the brackets without losing anything, but I don't think I can do the same thing here. For example, if I did use this notation, if a=x-x0, b=y-y0, and c=z-z0, I would lose the 6abc term the expansion of (a+b+c)^3. Or am I supposed to lose that term?

That's true, you can't exactly just add another variable in here without breaking it into pieces. It's tricky without matrices or lambda-calculus, but I'll try without making it too messy: We need some recursive function in order to generate our expression. (Or, there is probably some kind of function out there, but I don't know of that one as of right now)

Let



be the function we're trying to make a Taylor series to. Then, let



be our recursive function, which has the possibility to take in a variable amount of values.

Then, the Taylor series will equal to



Symbolically equivalent (which is easier to remember):



Now, I don't know with you, but I'd use the latter if I'm supposed to remember how to expand a Taylor series.

Noah

The differential operator by itself doesn't actually mean anything... it's purely symbolic. So considering that, I think your



is valid enough, if you do treat it as a symbolic notation and apply it in the understood way.

Edit: According to http://en.wikipedia.org/wiki/Taylor_seri..._variables (edited to match your notation and the 2-variable case):



I believe it's equivalent, and has the operators in the right order. Also doesn't require foiling which is nice.

Edit: Yup, it ends up being a nicer looking (and generalizable) version of yours in post #5, since your choose(n, k) cancels out the n! in the denominator and leaves (n-k)!k! there.

See? I said "RUSTT" and he came and said "OB3LISK I heard a damsel in distress. You must need me. Let me do this calculus (which will not help you in real life) for you!"

So whatever he said I take credit for.

Same thing if Corn comes in.

For (7.5), they have f''(a)(x-a)^2+f'(a)f'(b)(x-a)(y-b)+f''(b)(y-b)^2. There should be a coefficient of 2 in the middle term. They do something similar for (7.6). I'm not sure why they left the coefficients out.

Guess i was out of it yesterday. I'm gonna go with wikipedia on this one and say that paper is wrong. Also, their mixed derivative notation is terrible. In fact, the "Geospatial Science" header should be a very big warning that no proper math person wrote that. While they may know what they're doing in practice, I bet they couldn't and wouldn't explain the theory to you.

Thank you both, Noah and Russt. :]

Screw you OB3LISK. You didn't help at all, and you spelled Russt incorrectly. And Taylor series for multi-variable functions is just as useful as Taylor series for single variable functions. They prevent your electronics from dying during computation.