Exercise 9.30

Exercise 30: Let f C ( m ) ( E ) , where E is an open subset of n . Fix ǎ E , and suppose x ˇ n is so close to 0 ˇ that the points p ˇ ( t ) = ǎ + t x ˇ lie in E whenever 0 t 1 . Define

h ( t ) = f ( p ˇ ( t ) )

for all t 1 for which p ˇ ( t ) E .

(a) For 1 k m , show (by repeated application of the chain rule) that

h ( k ) ( t ) = D i 1 i k f ( p ˇ ( t ) ) x i 1 x i k .

The sum extends over all ordered k -tuples ( i 1 , , i k ) in which each i j is one of the integers 1 , , n .

(b) By Taylor’s theorem (5.15),

h ( 1 ) = k = 0 m 1 h ( k ) ( 0 ) k ! + h ( m ) ( t ) m !

for some t ( 0 , 1 ) . Use this to prove Taylor’s theorem in n variables showing that the formula

f ( ǎ + x ˇ ) = k = 0 m 1 1 k ! D i 1 i k f ( ǎ ) x i 1 x i k + r ( x ˇ )

represents f ( ǎ + x ˇ ) as the sum of its so-called “Taylor polynomial of degree m 1 ,” plus a remainder that satisfies

lim x ˇ 0 ˇ r ( x ˇ ) | x ˇ | m 1 = 0 .

Each of the inner sums extends over all ordered k -tuples ( i 1 , , i k ) , as in part (a); as usual, the zero-order derivative of f is simply f , so that the constant term of the Taylor polynomial of f at ǎ is f ( ǎ ) .

(c) Exercise 29 shows that repetition occurs in the Taylor polynomial as written in part (b). For instance, D 113 occurs three times, as D 113 , D 131 , D 311 . The sum of the corresponding three terms can be written in the form

3 D 1 2 D 3 f ( ǎ ) x 1 2 x 3 .

Prove (by calculating how often each derivative occurs) that the Taylor polynomial in (b) can be written in the form

D 1 s 1 D n s n f ( ǎ ) s 1 ! s n ! x 1 s 1 x n s n .

Here the summation extends over all ordered n -tuples ( s 1 , , s n ) such that each s i is a nonnegative integer, and s 1 + + s n m 1 .

Answers

(a) I am going to show this by induction on k . For the case k = 1 , we have by Theorem 9.15 and Theorem 9.17

h ( t ) = f ( p ˇ ( t ) ) p ˇ ( t ) = i = 1 n D i f ( p ˇ ( t ) ) x i

which is the assertion in the case k = 1 . Now assume the assertion is true for the case k 1 . Then we have

h ( k ) ( t ) = d dt h ( k 1 ) ( t ) = d dt i 1 , , i k 1 = 1 n D i 1 i k 1 f ( p ˇ ( t ) ) x i 1 x i k 1 = i 1 , , i k 1 = 1 n d dt D i 1 i k 1 f ( p ˇ ( t ) ) x i 1 x i k 1 = i 1 , , i k 1 = 1 n ( i k = 1 n D i k D i 1 i k 1 f ( p ˇ ( t ) ) x i k ) x i 1 x i k 1 = i 1 , , i k = 1 n D i 1 i k f ( p ˇ ( t ) ) x i 1 x i k

where the last equality follows from Exercise 29.

(b) Plugging in the results of part (a), we get, for some t ( 0 , 1 ) .

f ( ǎ + x ˇ ) = h ( 1 ) = k = 0 m 1 h ( k ) ( 0 ) k ! + h ( m ) ( t ) m ! = k = 0 m 1 1 k ! i 1 , , i k = 1 n D i 1 i k f ( ǎ ) x i 1 x i k + 1 m ! i 1 , , i m = 1 n D i 1 i k f ( ǎ + t x ˇ ) x i 1 x i m

Since f C ( m ) ( E ) , there is a bound M such that | D i 1 i m f ( ǎ + t x ˇ ) | < M for all t ( 0 , 1 ) and all partial derivatives of f of order m . Hence,

| r ( x ˇ ) | 1 m ! i 1 , , i m = 1 n | D i 1 i k f ( ǎ + t x ˇ ) | | x i 1 | | x i m | 1 m ! m ! ( M | x ˇ | m ) lim x ˇ 0 ˇ | r ( x ˇ ) | | x ˇ | m 1 = lim x ˇ 0 ˇ M | x ˇ | = 0

(c) By simple combinatorics, the number of ways to arrange k distinct objects in an ordered sequence is k ! . If s of these objects are identical, this reduces the number of distinct ordered sequences by a factor of s ! , since there are s ! ways of rearranging the identical objects in a given sequence. Hence the number of times a given partial derivative D 1 s 1 D n s n of order k = s 1 + + s n occurs in the Taylor polynomial is k ! ( s 1 ! s n ! ) , so we can rewrite the Taylor polynomial as

k = 0 m 1 1 k ! i 1 , , i k = 1 n D i 1 i k f ( ǎ ) x i 1 x i k = k = 0 n 1 k ! s 1 + + s n = k k ! D 1 s 1 D n s n f ( ǎ ) s 1 ! s n ! x 1 s 1 x n s n = k = 0 n s 1 + + s n = k D 1 s 1 D n s n f ( ǎ ) s 1 ! s n ! x 1 s 1 x n s n
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2023-08-07 00:00
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