Taylor expansion of (1-x)(1-y).

I am trying to compute the taylor expansion to second order of the following function about point (x,y)=(a,a)(x,y)=(a,a), with a<1a<1, f(x,y)=(1−x)(1−y).f(x,y)=(1-x)(1-y). If I am not mistaken, this expansion can be given by (1−a)(1−a)−(x−a)(1−a)−(y−a)(1−a),(1-a) (1-a)-(x-a)(1-a)- (y-a)(1-a), where it can be seen that e.g. there are not terms in (x−a)2(x-a)^2, which I think it results from the fact that ∂2f∂x2=0\frac{\partial^2 f}{\partial x^2}=0. Is my analysis correct ? and if so, can I claim that the taylor expansion to order n≥2n \ge 2 will be the same as for n=1n=1 ? ================= ================= 2 Answers 2 ================= Recall that the Taylor expansion of order kk around a point p∈Rdp \in \mathbf R^d of a CkC^k-function f:U⊆Rd→Rd′f \colon U \subseteq \mathbf R^d \to \mathbf R^{d'} is given by Tpkf(x)=∑|α|≤k1α!∂αf(p)(x−p)α T^p_kf(x)= \sum_{|\alpha|\le k} \frac{1}{\alpha!}\partial^\alpha f(p)(x-p)^\alpha for k=2k = 2 we have T(a,a)2f(x)=f(a,a)+∂xf(a,a)(x−a)+∂yf(a,a)(y−a)+12∂2xf(a,a)(x−a)2+∂x∂yf(a,a)(x−a)(y−a)+12∂2yf(a,a)(y−a)2 T^{(a,a)}_2 f(x) = f(a,a) + \partial_x f(a,a)(x-a) + \partial_y f(a,a)(y-a) + \frac 12 \partial_x^2 f(a,a)(x-a)^2 + \partial_{x}\partial_{y}f(a,a)(x-a)(y-a) + \frac 12 \partial_y^2f(a,a)(y-a)^2 We have ∂xf(x,y)=−(1−y)∂yf(x,y)=−(1−x)∂x∂yf(x,y)=1∂2xf=∂2yf=0\begin{align*} \partial_x f(x,y) &= -(1-y)\\ \partial_y f(x,y) &= -(1-x)\\ \partial_x\partial_y f(x,y) = 1\\ \partial_x^2 f = \partial_y^2 f &= 0 \end{align*} This gives T(a,a)2f(x,y)=(1−a)(1−a)−(1−a)(x−a)−(1−a)(y−a)+(x−a)(y−a) T^{(a,a)}_2f(x,y) = (1-a)(1-a) - (1-a)(x-a) - (1-a)(y-a) + (x-a)(y-a) You are missing the xx-yy-term, note that ∂x∂yf≠0\partial_x\partial_y f\ne 0. You are correct in saying, that there is no (x−a)2(x-a)^2 term due to ∂2xf=0\partial_x^2 f = 0. As ∂αf=0\partial^\alpha f = 0 for indices α\alpha of order ≥3\ge 3, we have that all expansions of order ≥3\ge 3 are equal to the one for k=2k = 2.      So if f(x,y,z)=(1−x)(1−y)(1−z)f(x,y,z)=(1-x)(1-y)(1-z), the expansion can be given by (1−a)3−(1−a)2(x−a)−(1−a)2(y−a)−(1−a)2(z−a)+(1−a)(x−a)(y−a)+(1−a)(x−a)(z−a)+(1−a)(y−a)(z−a)(1-a)^3-(1-a)^2(x-a)- (1-a)^2(y-a)-(1-a)^2(z-a)+(1-a)(x-a)(y-a)+(1-a)(x-a)(z-a)+(1-a)(y-a)(z-a) – din 2 days ago The function f(x,y)f(x,y) is a polynomial in two variables and following identity holds: f(x,y)=(1-x)(1-y)=(1-a)(1-a)-(1-a)(x-a)-(1-a)(y-a)+(x-a)(y-a).f(x,y)=(1-x)(1-y)=(1-a)(1-a)-(1-a)(x-a)-(1-a)(y-a)+(x-a)(y-a). Therefore your expansion for n=1n=1 is correct, but for n=2n=2 something is missing. Note that, since the degree of f(x,y)f(x,y) is two, the expansions for n\geq 2n\geq 2 will be all the same.      Could you please tell me if this identity can be generalized as the following: f(x_1,\ldots,x_k)=(1-a)^{k}-(1-a)^{k-1} \sum_i (x_i-a)+\prod_i (x_i-a)f(x_1,\ldots,x_k)=(1-a)^{k}-(1-a)^{k-1} \sum_i (x_i-a)+\prod_i (x_i-a) – din 2 days ago      With f=\prod_i (1-x_i)f=\prod_i (1-x_i)? No. Take for example x_1=a\not=1x_1=a\not=1 and 00 the others. – Robert Z 2 days ago