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Hello how does one apply the Taylor polynomials in a function of three variables?

If we consider the function $f: \mathbb{R}^3$ to $\mathbb{R}$ with $(x_1, x_2, x_3)$ mapped to $\sin(x^2_1) + \exp(x_2) + \cos(x_1x_3)$.

How can one calculate the Taylor polynomials of order one, two and three evaluated at the point $x = 0$?

Can someone please give me an idea as to how one tackles it.

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1 Answer 1

Computationally simplest is using the known Taylor expansions of the functions $\sin$, $\exp$, and $\cos$. This would give $$\eqalign{\sin(x_1^2)&=x_1^2-{1\over6}x_1^6 +{1\over120}x_1^{10}-\ldots\ ,\cr \exp(x_2)&=1+x_2+{1\over 2}x_2^2+{1\over 6}x_2^3 +\ldots\ ,\cr \cos(x_1x_3)&=1-{1\over2}x_1^2x_3^2 +{1\over24}x_1^4x_3^4 -\ldots\ .\cr}$$ Now collect all terms up to the desired order. E.g., the third order taylor expansion of $f$ at ${\bf 0}=(0,0,0)$ is given by $$f(x_1,x_2,x_3)=2+x_2+x_1^2+{1\over2}x_2^2+{1\over6}x_2^3+ R_3({\bf x})\ ,$$ where the remainder term is $o(|{\bf x}|^3)$ when ${\bf x}\to{\bf 0}$.

Of course there is also a general procedure for the Taylor expansion in several variables. When $f:{\mathbb R}^n\to{\mathbb R}$ is sufficiently smooth then at any point ${\bf p}$ it has differentials of order $0$, $1$, $2$, $3$, etc.. The differential of order $r$ at ${\bf p}$, denoted by $d^r f({\bf p})$, is a homogeneous polynomial in the auxiliary variable ${\bf X}=(X_1, \ldots, X_n)$ and is given by $$d^r f({\bf p}).{\bf X}=\sum_{k_1,\ldots, k_r} f_{.k_1\ldots k_r}({\bf p}) X_{k_1}\cdot\ldots\cdot X_{k_r}\ .$$ Here the summation variables $k_j$ run independently from $1$ to $n$, so formally there are $n^r$ terms in this sum. The zeroth differential is just the constant function with value $f({\bf p)}$, and $d^1f({\bf p}).{\bf X}=f_{.1}X_1+\ldots+f_{.n}X_n$ (the $f_{.k}$ evaluated at ${\bf p}$) is the usual "differential" of $f$ at ${\bf p}$.

The Taylor expansion of $f$ at ${\bf p}$ can then be written as $$f({\bf p}+{\bf X})=\sum_{r=0}^N{1\over r!}d^r f({\bf p}).{\bf X} + R_N$$ where $R_N$ denotes the remainder term. The Taylor theorem says, e.g., that $R_N=o(|{\bf X}|^N)$ when ${\bf X}\to{\bf 0}$.

When ${\bf p}={\bf 0}$ we can replace ${\bf X}=(X_1,\ldots ,X_n)$ by ${\bf x}=(x_1,\ldots,x_n)$ and simply write $f({\bf x})=\sum_{r=0}^N\ldots + R_N$.

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Considering the first method, for 1, 2 and 3, we get x2, (x1)^2, (1/6)(x2)^3 respectively. So in this question, why have we been given the third term (i.e., cos(x1.x3)? –  carla Jul 10 '12 at 10:31
    
I mean, we get the polynomials of order 1, 2 and 3 as 2 + x2,, 2 + x2 + (x1)^2, 2 + x2 + (x1)^2 +(1/6)(x2)^3. So how does the third term help? –  carla Jul 10 '12 at 10:47
    
@carla: The third term in $f$ is given for whatever reason. In the third order Taylor polynomial of $f$ at ${\bf 0}$ it only contributes $1$ to the constant term but nothing further. –  Christian Blatter Jul 10 '12 at 11:30

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