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I am trying to understand the derivation here

kiwi.atmos.colostate.edu/group/dave/pdf/TaylorSeries.pdf

I understand how first, second total differentials are derived. I do not understand how they are plugged into a form that is compatible with the single variable Taylor Series, which is

$f(x) = f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a) (x-a)^2 + ...$

The coefficients of multivariate expansion are exactly the same as the single variable version. Sure given the total differential, $(x-a)$ is substituted for $dx$, $(y-b)$ for $dy$, but it's not shown why, or how the rest of $f(x,y)$ expansion should mirror the single variable form based on the total differential. Expansion for $f(x,y)$ is below

$f(x,y) = f(a,b) + \bigg[(x-a)\frac{\partial}{\partial x} + (y-b)\frac{\partial}{\partial x}\bigg]f + \frac{1}{2!} \bigg[(x-a)\frac{\partial}{\partial x} + (y-b)\frac{\partial}{\partial x}\bigg]^2 f + ... $

If you know a better 2D Taylor Series derivation, that would be welcome as well.

Thanks,

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This is not a rigorous proof: Suppose that one can write $f(x,y)=f(a,b)+A(x-a)+B(y-b)+C(x-a)^2+D(x-a)(y-b)+E(y-b)^2+\cdots$, how can the values $A, B, C, D, E, \dots$ be determined? –  pharmine Sep 27 '11 at 11:08
    
I think I came across a similar idea here physics.miami.edu/~nearing/mathmethods/series.pdf pg 7. Thanks. –  BB_ML Sep 27 '11 at 11:26
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2 Answers 2

up vote 3 down vote accepted

So we want to express $f(a+u,b+v)$ in terms of $f(a,b)$, the increment $(u,v)$ and the partial derivatives of $f$ at $(a,b)$. To this end we look at the auxiliary function $$\phi(t):=f(a+t u, b+t v)\qquad(0\leq t\leq 1)\ .$$ By the one-variable Taylor theorem we have $$f(a+u,b+v)=\phi(1)=\sum_{r=0}^n{1\over r!}\phi^{(r)}(0)+{\rm error\ term}\ .\qquad(*)$$ Therefore we have to compute the successive derivatives of $\phi$ at $t=0$.

Obviously $$\phi^{(0)}(t)=\phi(t)=f(a+tu,b+tv)$$ and therefore $\phi^{(0)}(0)=f(a,b)$. To compute $\phi'(t)$ we need the chain rule: $$\phi'(t)={d\over dt}\phi(t)={d\over dt}f(a+t u,b+ t v)=f_1(a+t u,b+ tv)u+f_2(a+ tu, b+ tv) v\ ,$$ whence $\phi'(0)=f_1(a,b)u+f_2(a,b)v$. Proceeding in this manner we get $$\eqalign{\phi''(t)&={d\over dt}\bigl(f_1(a+t u,b+ tv)u+f_2(a+ tu, b+ tv) v\bigr) \cr &=f_{11}(a+tu,b+tv)u^2+2f_{12}(a+tu,b+tv) uv +f_{22}(a+tu,b+tv)v^2\cr}$$ and therefore $\phi''(0)=f_{11}(a,b)u^2 + 2f_{12}(a,b)uv+f_{22}(a,b) v^2\ .$ And so on for the higher derivatives of $\phi$. The values $\phi(0)$, $\phi'(0)$, $\phi''(0)$, $\ldots$ so obtained have to be plugged into the equation $(*)$

Of course all of this could be written up more elegantly using vector notation.

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Hi Christian, with $1$ in $f_1$ did you mean the first block of variables, which are $a+tu$? –  BB_ML Sep 28 '11 at 6:01
    
Similar idea math.uc.edu/~halpern/Calc.4/Handouts/Taylorseries.pdf –  BB_ML Sep 28 '11 at 6:02
    
@user6786: $f$ is a function of two real inputs, whatever their name. By $f_1$ I mean the partial derivative with respect to the first input. It seemed strange to write $f_x$ when there is nowhere an $x$ in the argument. –  Christian Blatter Sep 29 '11 at 13:26
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If you are looking for a physical intuition behind multivariate Taylor series, remember that $ T_x = \partial_x$ is the generator of translations, as for suitable class of functions: $$ (T_x f)(x) = \lim_{\epsilon \to 0} \frac{ f(x+\epsilon) - f(x)}{\epsilon} $$ Thus $\mathcal{T}_a = \mathrm{e}^{a T_x}$ is the operator of translation by $a$: $$ (\mathcal{T}_a f)(x) = f(x+a) $$ Writing defining series of exponential give univariate Taylor expansion: $$ f(x+a) = (1 + a T_x + \frac{a^2}{2!} T_x \circ T_x + \ldots ) \circ f(x) $$ Taking this to mulrivariate, the translation by vector $\vec{u}$ becomes $$ \mathcal{T}_\vec{u} = \exp( \sum_{i=1}^n u_i T_{x_i} ) = \circ_{i=1}^n \mathcal{T}_{u_i} $$ You can now develop each univariate translation operator into a series, giving the multivariate Taylor expansion formula or, alternatively, you could expand the combined exponential:

$$ f(\vec{x} + \vec{u}) = \sum_{n=0}^\infty \frac{1}{n!} \left( \langle \vec{u}, \mathcal{T}_{\vec{x}} \rangle \right)^n \circ f(\vec{x}) $$

Hope this helps.

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