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I found the third degree taylor polynomial $$f(x,y,z) = xy^2z^3$$ at $$(1,0,-1)$$

The answer i got $$p_3(x,y,z)=(\frac{1}{2!}(-2y^2))+ (\frac{1}{3!}(-2(x-1)y^2))+(\frac{1}{3!}(6(z+1)y^2))$$

I am not very confident about my answer as some of the summations get pretty harry. Thanks guys!

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up vote 2 down vote accepted

Well, both Dominic's and Ittay's answers are good but to add to Ittay's answer, Ittay is absolutely right that for any polynomial the Taylor expansion is the polynomial itself BUT in this case writing $f(x,y,z)$ as $xy^2z^3$ is the Taylor expansion of $f(x)$ around the origin. The OP I think wants to expand $f(x)$ around the point (1,0,-1) in which case the Taylor expansion isn't $xy^2z^3$.

Just to give an example, if you want to expand $g(x)=x^2$ around $x=0$ then the Taylor expansion is $g(x)=x^2$. But if you want the Taylor expansion around $x=1$, it would instead be $g(x)=(x-1)^2+2(x-1)+1$. And this case we can use the trick


The polynomials are identical. If you simplify the series you do get $x^2$ but the FORM is different because a Taylor polynomial expanded at $x=a$ must be written as a polynomial in $(x-a)$.

So for your original question $f(x,y,z)=xy^2z^3$ is NOT the Taylor expansion around (1,0,-1). Either use the definition of the multivariate Taylor expansion carefully...which as you correctly noted is quite hairy...or we can use the same trick again. Write


and then just expand this polynomial keeping the parenthesis grouped together and you will have a polynomial in $(x-1)$, $y$ which is the same as $(y-0)$, and $(z+1)$.

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It IS the taylor exponsion just multiplied out ... – Dominic Michaelis Mar 8 '13 at 9:34
(copying my comment on my own answer) I understand what you mean (and I agree this is a subtle issue), but it is important to realize that the Taylor polynomial is xy^2z^3, it's just the you might say the question implicitly expects the polynomial to be presented in a certain way rather than another. – Ittay Weiss Mar 8 '13 at 9:35
that seems to be what I have... except i have y^2s everywhere since the only partials that exist have $$\frac{d^2}{dy^2}$$ – Neo Mar 8 '13 at 9:35

It is a multivariate polynomial of degree $3$, so the taylor polynomial of order $3$ is exact the function $f$. As your highest power of $z$ is only $1$ there must be something wrong.

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I guess I am confused about how to calculate them... As there is only one nonzero third derivative with z... – Neo Mar 8 '13 at 8:43

This is a classical example where pausing to think about what it is you are about to compute before actually doing any computation, very quickly gives you the answer. Plunging into performing routine things algorithmically is a dangerous habit.

So, what is it that the third degree Taylor polynomial supposed to be? Well, it's to be a polynomial in the variables $x,y,z$, of degree not more than $3$, that best approximates the given function $f$. But, $f$ is such a polynomial, and so the best approximation to it is, of course itself. So, without a single computation, it follows that the required Taylor polynomial is $xy^2z^3$.

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I mean isn't that the case for all functions? Oh because x is already a polynomial... guhh – Neo Mar 8 '13 at 8:47
then why would a book have multiple questions asking for taylor polynomials of taylor polynomials? – Neo Mar 8 '13 at 8:50
perhaps in order to test the reader for understanding how to do it quickly. Or, to practice with simple cases. I'm not sure. – Ittay Weiss Mar 8 '13 at 8:52
Makes sense. I suppose. I guess i can just graph the two in matlab to see how accurate it is. – Neo Mar 8 '13 at 8:57
The purpose of the exercise is to expand around different points. The required Taylor polynomial is not $xy^2z^3$. And you still have to do a bit of work, albeit it is just simple algebra. – Fixed Point Mar 8 '13 at 9:29

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