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I have one question about Taylor theorem.

Originally, Taylor theorem is represented as

$$f(x) = f(a) + f'(a)(x-a) + \ldots$$

But my book says

Suppose that $$f : \mathbb{R}^n \to \mathbb{R}$$ is continuously differentiable and that $p\in\mathbb{R}^n$ Then we have that $$f(x + p) = f(x) + ∇f(x+tp)^Tp$$

But I don't understand how to derive this equation from Taylor theorem.

mangken
  • 63

2 Answers2

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This equality follows from the mean value theorem applied to the function $ g (s) = f (x + sp) , 0 \leq s \leq 1$.

By the mean value theorem, there exists $ t \in (0,1)$ such that $ g (1) -g (0) = g'(t) = \nabla f (x + tp)^T p$.

littleO
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5

The formula you have is incorrect (what is $t$ anyway?); it should be $\nabla f(x)^T$ at which point it is simply Taylor's theorem for functions of several variables.

A derivation of the multivariate Taylor's theorem should be covered in any introductory textbook on multivariable calculus; I'm very surprised that yours does not. There are plenty of course notes available online that also explain it, for instance here.

EDIT: Looking at Nocedal and Wright it seems that what you are actually after is the multivariate mean value theorem: for $f$ differentiable in a sufficiently large neighborhood of $x$, there exists a $t \in [0,1]$ with $$f(x+p) = f(x) + \nabla f(x+tp)^Tp.$$ You can see this directly from the single variable mean value theorem, see e.g. the Wikipedia entry.

user7530
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