# MandysNotes

Tuesday, 15 April 2014 00:00

## A Useful Formula

By  Gideon
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Let $$f (x)$$ be a smooth (i.e., $$C^{\infty}$$) function in a convex neighborhood $$V$$ of $$\mathbb{R}^{n},$$ then:

$f(x) = f\left( x_{1}, \dots , x_{n} \right) = f\left( a_{1}, \dots , a_{n} \right) + \sum_{i = 1}^{n} \left( x_{i} - a_{i} \right) g_{i} \left( x_{1}, \dots, x_{n} \right)$

for some point $$a = (a_{1}, \dots , a_{n}) \in V,$$ some smooth functions $$g_{i}$$ defined in $$V,$$ and with:

$g_{i}(a_{i}) = \frac{ \partial f}{\partial x_{i}} (a_{i}).$

In the special case that

$a_{i} = 0,$

for all $$i,$$ and that

$f( 0 , \dots , 0 ) = 0,$

the above reduces to

$f\left( x_{1}, \dots , x_{n} \right) = \sum_{i = 1}^{n} x_{i} g_{i} \left( x_{1}, \dots, x_{n} \right).$

This result is used frequently, for example in Helgason's book on differential geometry, in Wald's book on general relativity, and in Milnor's book on Morse theory.

To prove this result we first define a smooth curve, $$x(t),$$ i.e., a $$C^{\infty}$$ mapping from $$U \in \mathbb{R}$$ to $$V \in \mathbb{R}^{n}$$ by:

$x_{i}(t) := x_{i}(0) + t \left(x_{i}(1) - x_{i}(0) \right),$

with

$x_{i}(0) = a_{i}.$

Note that $$x_{i}(0) = a_{i},$$ and $$x_{i}(1)$$ are both constant with respect to the variable $$t.$$

In terms of the curve $$x(t)$$ we write $$f\left(x(t)\right)$$ as:

$f\left(x(t)\right) = f \left( x_{1}(t), \dots, x_{n}(t) \right) = f \left( x_{1}(0) + t \left(x_{1}(1) - x_{1}(o) \right), \dots , x_{n}(0) + t \left(x_{n}(1) - x_{n}(o) \right) \right)$

$= f \left( a_{1} + t \left(x_{1}(1) - a_{1} \right), \dots , a_{n} + t \left(x_{n}(1) - a_{n} \right) \right).$

For each component of $$x(t)$$ the total derivative with respect to $$t$$ is:

$\frac{dx_{i}(t)}{dt} = \left(x_{i}(1) - x_{i}(0)\right),$

so that

$\frac{ d f \left(x(t)\right)} {dt} = \sum_{i=1}^{n} \frac{dx_{i}}{dt} \frac{ \partial f \left(x(t)\right) }{ \partial x_{i}}$

$= \sum_{i=1}^{n} \left( x_{i}(1) - a_{i}\right) \frac{ \partial f \left(x(t)\right) }{\partial x_{i}}$

and

$d f \left(x(t)\right) = \left[ \sum_{i=1}^{n} \left( x_{i}(1) - a_{i}\right) \frac{ \partial f \left(x(t)\right) }{\partial x_{i}} \right] dt.$

We integrate this total differential from zero to one to obtain:

$\int_{o}^{1} \frac{d f \left(x(t)\right) }{dt} dt = \int_{0}^{1} d f \left(x(t)\right) = f \left(x(t)\right)|_{0}^{1}$

$= f \left( a_{1} + 1(x_{1}(1) - a_{1}), \dots , a_{n} + 1(x_{n}(1) - a_{n}) \right) - f \left( a_{1} , \dots , a_{n} \right)$

$= f \left ( x_{1} , \dots , x_{n} \right) - f \left( a_{1} , \dots , a_{n} \right),$

or

$f \left ( x_{1} , \dots , x_{n} \right) = f \left( a_{1} , \dots , a_{n} \right) + \int_{o}^{1} \frac{d f(x_{i}) }{dt} dt .$

Substituting in our previous result for $$\frac{df}{dt}$$ gives:

$f \left ( x_{1} , \dots , x_{n} \right) = f \left( a_{1} , \dots , a_{n} \right) + \sum_{i=1}^{n} \left( x_{i} - a_{i}\right) \int_{0}^{1} \frac{ \partial f }{\partial x_{i}} dt$

$= f \left( a_{1} , \dots , a_{n} \right) + \sum_{i=1}^{n} \left( x_{i} - a_{i}\right) \int_{0}^{1} \frac{ \partial f \left( a_{i} + t \left(x_{i}(1) - a_{i} \right) \right) }{\partial x_{i}} dt.$

If we define

$g_{i}( x_{1}, \dots , x_{n}) : = \int_{0}^{1} \frac{ \partial f }{\partial x_{i}} dt,$

then the above result becomes:

$f \left ( x_{1} , \dots , x_{n} \right) = f \left( a_{1} , \dots , a_{n} \right) + \sum_{i=1}^{n} \left( x_{i} - a_{i}\right) g_{i}( x_{1}, \dots , x_{n}).$

At the point

$x = x_{i}(0) = a_{i},$

$g_{i}( a_{1}, \dots , a_{n}) = \int_{0}^{1} \frac{ \partial f ( a_{1}, \dots , a_{n}) }{\partial x_{i}} dt = \frac{ \partial f ( a_{1}, \dots , a_{n}) }{\partial x_{i}}\int_{0}^{1}dt = \frac{ \partial f (a_{i}) }{\partial x_{i}}.$

And the formula is proved.