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$$F(x,y)= (x_2-x_1^2) (x_2-2x_1^2)= 2 x_1^4+x_2^2-3x_1^2x_2$$

Where $x^*=[x_1 \ \ x_2]' = [0 \ \ 0]'$

I want to show Taylor expansion of the function for third degree.


What I did is that;

Taylor expansion for vectors

$$F(x)= F(x^*)+ \frac{1}{1!} dF(x^*)+ \frac{1}{2!} d^2F(x^*)+\frac{1}{3!} d^3F(x^*)$$ For $h:=x-x^*$

$$dF(x^*)=\sum_1^3 h_i F_{x_i}(x*)=h_1F_{x_1}(0,0)+ h_2F_{x_2}(0,0)=h_1(8x_1^3-6x_1x_2)+h_2(2x_2-3x_1^3)=h_1. 0 +h_2. 0=0$$

$$d^2F(x^*)=\sum_{I=1}^3\sum_{j=1}^3 h_i h_jF_{x_i x_j}(x*)=h_1h_1F_{x_1 x_1}(0,0)+ h_1h_2F_{x_1x_2}(0,0)+h_2h_1F_{x_2x_1}(0,0)+h_2h_2F_{x_2x_2}=h_1h_1(24x_1^2-6x_2)+h_1h_2(-6x_1)+h_2h_1(-6x_1)+h_2h_2(2)=......$$

$$d^3F(x^*)=\sum_{i=1}^3\sum_{j=1}^3\sum_{k=1}^3h_ih_jh_kF_{x_ix_jx_k}(x^*)$$

Firstly I cannot expand $d^3F(x^*)$ and secondly,I cannot calculate $h_1, h_2, h_3$

Please help me to properly complete this question. Thank you for helping.

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  • $\begingroup$ you don't need to calculate the $h$'s , they are your two variables - all you need to do now is to calculate 8 third order partials ( the upper bound on all the sums should be 2 rather than 3 ) $\endgroup$ – WW1 Oct 26 '15 at 22:47
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You are not meant to calculate $h_1$ and $h_2$: they are the independent (input) variables of your polynomial. That is, the Taylor polynomial of f at (0,0) is a polynomial function of $h = (h_1, h_2)$ that approximates $f(h_1, h_2)$, where $h$ is the vector displacement from (0,0) in the domain of f. There is therefore no such thing as $h_3$, since the domain of f is $\mathbb{R^2}$.

One more thing you may note is the the Taylor polynomial of a polynomial function is designed to be exactly the same polynomial. When we take the nth derivative of a polynomial, and evaluate it at x = 0, the only number that remains is the coefficient of the nth degree term, multiplied by n!. This is the nth derivative of the polynomial at x = 0. So to recover the coefficient of the original polynomial, we can divide this term by n!, which is where the formulation of the Taylor polynomial comes from.

Therefore, the 3rd degree Taylor polynomial of a polynomial would contain precisely all of the same coefficients of the same terms, up to the third degree monomials. So, without doing any work whatsoever, we know that the Taylor polynomial of degree 3 of f at (0,0) must be $$P_{f,(0,0)}^3(h_1,h_2) = h_2^2 -3h_1^2 h_2$$ where we have simply evaluated the polynomial $f(h_1,h_2)$ and omitted any monomial terms of higher than 3rd degree. When we have any non-polynomial terms, then you have to go into the derivative based formulation.

In case you are not entirely convinced, we can go into the derivative based formulation and compare the result. To be more concise, I prefer to use the multi-index notation, where $I = (m, n)$ is a multi-exponent that lets us write multivariable monomials with a single index: $x^I = x_1^m x_2^n$. Also, $\mathcal{I}_n^m$ is a multi-index that refers to the set of all multi-exponents of degree m over n variables. In addition, if $I = (m, n)$, we define $I! = m!n!$ and define $f_I = f_{x_m x_n}$, which neatly handles the equality of mixed partial derivatives.

Then, the Taylor polynomial of degree 3 of f at (0,0) is given by the more concise notation: $$P_{f,(0,0)}^3(h_1, h_2) = \sum_{m=0}^3\left(\sum_{I\in\mathcal{I}_2^m} \frac{1}{I!}f_I(0,0)h^I\right)$$

Expanding the outer sum, we get: $$P_{f,(0,0)}^3(h_1, h_2) = \left(\sum_{I\in\mathcal{I}_2^0} \frac{1}{I!}f_I(0,0)h^I\right) + \left(\sum_{I\in\mathcal{I}_2^1} \frac{1}{I!}f_I(0,0)h^I\right) + \left(\sum_{I\in\mathcal{I}_2^2} \frac{1}{I!}f_I(0,0)h^I\right) + \left(\sum_{I\in\mathcal{I}_2^3} \frac{1}{I!}f_I(0,0)h^I\right)$$ Let's do these one at a time: $$\sum_{I\in\mathcal{I}_2^0} \frac{1}{I!} f_I(0,0)h^I = \frac{1}{0!0!}f(0,0)h_1^0h_2^0 = f(0,0) = 0$$ There is only one degree 0 monomial in 2 variables. Here we use the traditional shorthand that the 0th derivative of f is f, and $0! = 1$. Next, there are 2 degree 1 monomials in 2 variables: $$\sum_{I\in\mathcal{I}_2^1} \frac{1}{I!} f_I(0,0)h^I = \frac{1}{1!0!}f_{x_1}(0,0)h_1 + \frac{1}{0!1!}f_{x_2}(0,0)h_2 = 0$$ No surprises so far, as there are no terms of degree 0 or 1 in the original polynomial function. Next, there are 3 monomials of degree 2 in 2 variables: $$\sum_{I\in\mathcal{I}_2^2} \frac{1}{I!} f_I(0,0)h^I = \frac{1}{2!0!}f_{x_1^2}(0,0)h_1^2 + \frac{1}{1!1!}f_{x_1 x_2}(0,0)h_1 h_2 + \frac{1}{0!2!}f_{x_2^2}(0,0)h_2^2 = \frac{1}{2}2h_2^2 = h_2^2$$ Above, $f_{x_1^2}(x_1, x_2) = 24x_1^2 - 6x_2$, and $f_{x_1 x_2}(x_1,x_2) = -6x_1$, so the only non-zero derivative at $(0,0)$ is $f_{x_2^2}(x_1, x_2) = 2$. Note how this term is also the only monomial of degree 2 in the original polynomial function.

Finally, there are 4 monomials of degree 3 on 2 variables: $$\sum_{I\in\mathcal{I}_2^3} \frac{1}{I!} f_I(0,0)h^I = \frac{1}{3!0!}f_{x_1^3}(0,0)h_1^3 + \frac{1}{2!1!}f_{x_1^2 x_2}(0,0)h_1^2 h_2 + \frac{1}{1!2!}f_{x_1 x_2^2}(0,0)h_1 h_2^2 + \frac{1}{0!3!}f_{x_2^3}(0,0)h_2^3$$ Now, $f_{x_1^3}(x_1, x_2) = 48x_1$, $f_{x_1^2 x_2}(x_1, x_2) = -6$, $f_{x_1 x_2^2}(x_1, x_2) = 0$, and $f_{x_2^3}(x_1, x_2) = 0$, so the only non-zero term is $\frac{1}{2}(-6)h_1^2h_2 = -3h_1^2h_2$.

Thus, $$P_{f,(0,0)}^3(h_1, h_2) = h_2^2 - 3h_1^2h_2$$ is the Taylor polynomial of degree 3 for f at $(0,0)$, as we surmised earlier from the motivation for the Taylor polynomial. Hopefully, going through the work above allows you to find Taylor polynomials for non-polynomial functions more easily.

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  • $\begingroup$ good answer! thank you!! $\endgroup$ – user315 Oct 27 '15 at 5:07
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You do not have to calculate anything, the Taylor-expansion of a polynomial is the polynomial itself. More precisely, the expansion of $p(x)$ in $x^*$ in direction $h$ is $p(x^*+h)$.

Here with $x^*=(0,0)$ you get that the Taylor-expansion is $p(h)$, the polynomial as you have it written down. You just have to throw away all terms of total degree 4 or higher to get the Taylor-polynomial of degree 3.

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