MIT18_024s11_ChCnotes

MIT18_024s11_ChCnotes - Derivatives - vector functions. of...

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Derivatives - of vector functions. Recall that if x - is a point of R" and if f (5) is a scalar function of - x, then the derivative of f (if it exists) is the vector For some purposes, it will be convenient to denote the derivative of f by a - row matrix rather than by a vector. When we do this, we usually denote the derivative by Df rather than ?f. Thus If we use this notation,.the definition of the derivative takes the following form: where ( ) - 0 as - h -> - 0. Here the dot denotes matrix multiplication, so we must write h - a column matrix in order for the formula to work;
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è his is the formula that will generalize to vector functions i Definition. Let S be a sybset of R~. If - k f : S -> R , then - f(5) is called a vector function -- of a vector variable. In scalar form, we can write - f(5) out in the form Said differently, - f consists of "k real-valued functions of n variables." Suppose now that - f is defined in an open ball about the point - a. We say that - f is differentiable at - a if each of the functions fl(x), . ..,fk(~) is differentiable at - a ! (in the sense already defined) . Furthermore, we define the derivative of - f at - a to be the matrix That is, Df(=) is the matrix whose 1- Oth row is the derivative th Dfi (=) of the i- coordinate function of - f. - the' derivative 'Df (5) - of'. . - f at - - a is. the k by n matrix whose entry in row i and column j is
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it is often called the Jacobian matrix of - f (5). Another notation for this matrix the notation With this notation, many of the formulas we proved for a scalar function f (5) hold without change for a vector function - f(5). We consider some of them here: Theorem 1. - The function - f - differentiable - at - a -- if and only - if where - g(&) -> 0 as h 0. _I - - \ (Here - f, 5, and - E are written as column matrices.) Proof: Both sides of this equation represent column th matrices. If we consider the I- entries of these matrices, we have the following equation: NOW - f differentiable at - a if and only if each function is. And fi - a if and only if Ei (&) 0 as - h - 0. But Ei (h) - 0 as h - - 0, for each i, - E (h) - - 0 as h - - 0.
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Theorem 2. - If 1 -- f(x) - is differentiable - at - a, - then - f - continuous - at a. Proof. - f differentiable at - then so each function fi. Then in particular, fi continuous at - a, whence - f - a. - The general chain - rule. Before considering the general chain rule, let us take the chain rule we have already proved and reformulate it in terms of matrices. Assume that f (5) = f(xl,. . . ,x ) a scalar function n defined in an open ball about - and that - x (t) = (xl , ... , xn(t)) a parametrized curve passing through - Let - x(tO) = - f (x) - - and if - x(t) ) to, and we have shown that the composite f(~(t)) and its derivative given by the equation when t=tO. We can rewrite this formula in scalar form as follows: or we can rewrite in the following matrix form:
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Recalling the definition of the Jacobian matrix Df, - we see that the latter formula can be written in the form (Note that the matrix Df is a row matrix, while the matrix DX d by its definition a column matrix.) This the form of the chain rule that we find especially useful, for it the formula that generalizes to higher dimen- sions. Let us now consider a composite of vector functions of vector variables.
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This note was uploaded on 01/19/2012 for the course MATH 18.024 taught by Professor Christinebreiner during the Spring '11 term at MIT.

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MIT18_024s11_ChCnotes - Derivatives - vector functions. of...

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