Page 306 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 11] INFINITE SERIES 297

To obtain the Lagrange form of the remainder R n ,consider the form
1 K
0 00 2 n
2! n!
f ðxÞ¼ f ðcÞþ f ðcÞðx cÞþ f ðcÞðx cÞ þ þ ðx cÞ
K
n
This is the Taylor polynomial P n 1 ðxÞ plus ðx cÞ : Also, it could be looked upon as P n except that
n!
in the last term, f ðnÞ ðcÞ is replaced by a number K such that for fixed c and x the representation of f ðxÞ is
exact. Now define a new function
n 1 j n
X
f ð jÞ ðx tÞ Kðx tÞ
j! n!
ðtÞ¼ f ðtÞ f ðxÞþ ðtÞ þ
j¼1
The function satisfies the hypothesis of Rolle’s Theorem in that ðcÞ¼ ðxÞ¼ 0, the function is
continuous on the interval bound by c and x, and exists at each point of the interval. Therefore, there
0
exists in the interval such that ð Þ¼ 0. We proceed to compute and set it equal to zero.
0
0
n 1 j n 1 j 1 n 1
X X
f f
0 0 ð jþ1Þ ðx tÞ ð jÞ ðx tÞ Kðx tÞ
j! ð j 1Þ! ðn 1Þ!
ðtÞ¼ f ðtÞþ ðtÞ ðtÞ
j¼1 j¼1
This reduces to
f ðnÞ K
0 ðtÞ n 1 n 1
ðn 1Þ! ðn 1Þ!
ðtÞ¼ ðx tÞ ðx tÞ
According to hypothesis: for each n there is n such that
ð n Þ¼ 0
Thus
K ¼ f ðnÞ ð n Þ
and the Lagrange remainder is
f ðnÞ ð n Þ n
n!
R n 1 ¼ ðx cÞ
or equivalently
1 nþ1
f ðnþ1Þ
R n ¼ ð nþ1 Þðx cÞ
ðn þ 1Þ!
The Cauchy form of the remainder follows immediately by applying the mean value theorem for
integrals. (See Page 274.)

11.53. Extend Taylor’s theorem to functions of two variables x and y.
Define FðtÞ¼ f ðx 0 þ ht; y 0 þ ktÞ,then applying Taylor’s theorem for one variable (about t ¼ 0Þ
1 1 1
0 00 2 ðnÞ n F ðnþ1Þ ð Þt nþ1 ; 0 < < t
2! n! ðn þ 1Þ!
FðtÞ¼ Fð0Þþ F ð0Þþ F ð0Þt þ þ F ð0Þt þ
Now let t ¼ 1
1 1 1
0 00 ðnÞ F ðnþ1Þ
2! n! ðn þ 1Þ!
Fð1Þ¼ f ðx 0 þ h; y 0 þ kÞ¼ Fð0Þþ F ð0Þþ F ð0Þþ þ F ð0Þþ ð Þ
When the derivatives F ðtÞ; ... ; F ðtÞ; F ðnþ1Þ ð Þ are computed and substituted into the previous expres-
ðnÞ
0
sion, the two variable version of Taylor’s formula results. (See Page 277, where this form and notational
details can be found.)
2
11.54. Expand x þ 3y 2in powers of x 1 and y þ 2. Use Taylor’s formula with h ¼ x x 0 ,
k ¼ y y 0 , where x 0 ¼ 1 and y 0 ¼ 2.