256
III - Convergence: Continuous variables
between binomial coefficients. For example
(fg)'" = f"'g + 3/"g' + 3!,g" + g"'.
After these algebraic rules, here are two very different formulae: for the
composition of differentiable maps, and for the inverse map of a differentiable
function.
(D 4) (Chain rule) Let I and J be two intervals in JR., let f be a map
from I into J, let 9 be a map from J into C, and let c(x) = g[J(x)] be the
composite map of I into C. Assume that f is differentiable at a E I and that
9 is differentiable at f(a) = b E J. Then c is differentiable at a, and
(15.5)
c' (a) = g' (b)!, (a) = g'[f(a)].f' (a).
Recall the relation
f(a + h) = f(a) + !,(a)h + o(h)
and put k = f'(a)h + o(h), whence f(a + h) = f(a) + k = b + k. It is clear
that k tends to 0 with h. Now
c(a+h)
g[f(a + h)] = g(b + k) = g(b) + g'(b)k + o(k) =
c(a) + g'(b)[!,(a)h + o(h)] + o(k).
The term g' (b )o( h) is o( h). Since, on the other hand,
k = h[!,(a) + o(h)/h]
and since, in this relation, the factor between [ ] tends, by definition of o(h),
to f'(a) as h tends to 0, one has Ikl ~ Mlhl for Ihl small, for some constant
M > o. Every o(k) function is thus also 33 o(h), in particular the last term of
the expansion of c( a + h). Finally
c(a + h) = c(a) + g'(b)f'(a)h + o(h),
qed.
Example 1. Since we believe we know that the derivative of the function
log x is l/x by (14.15) or Chap. II, nO 10, Theorem 3, and since it is not very
difficult to establish, by applying the general formulae of Chap. II, nO 19,
that that of the function expx = LX[n j is Lx[n-l[ = expx, we see that
33 If Ikl < 10 100 lhi for Ih~ sufficiently small one can be sure that Icp(k)1 < lO-looolhl
once tcp(k) I is < 10- lOolkl, which is the case for Ikl < r', so for Ihl < r =
10- 10 r'o More generally: if 0 = O('Ij;) and if'lj; = o(cp), then 0 = o(cp). See the
general rules of Chap. VI, nO 1, which follow directly from the definitions.
III - Convergence: Continuous variables
between binomial coefficients. For example
(fg)'" = f"'g + 3/"g' + 3!,g" + g"'.
After these algebraic rules, here are two very different formulae: for the
composition of differentiable maps, and for the inverse map of a differentiable
function.
(D 4) (Chain rule) Let I and J be two intervals in JR., let f be a map
from I into J, let 9 be a map from J into C, and let c(x) = g[J(x)] be the
composite map of I into C. Assume that f is differentiable at a E I and that
9 is differentiable at f(a) = b E J. Then c is differentiable at a, and
(15.5)
c' (a) = g' (b)!, (a) = g'[f(a)].f' (a).
Recall the relation
f(a + h) = f(a) + !,(a)h + o(h)
and put k = f'(a)h + o(h), whence f(a + h) = f(a) + k = b + k. It is clear
that k tends to 0 with h. Now
c(a+h)
g[f(a + h)] = g(b + k) = g(b) + g'(b)k + o(k) =
c(a) + g'(b)[!,(a)h + o(h)] + o(k).
The term g' (b )o( h) is o( h). Since, on the other hand,
k = h[!,(a) + o(h)/h]
and since, in this relation, the factor between [ ] tends, by definition of o(h),
to f'(a) as h tends to 0, one has Ikl ~ Mlhl for Ihl small, for some constant
M > o. Every o(k) function is thus also 33 o(h), in particular the last term of
the expansion of c( a + h). Finally
c(a + h) = c(a) + g'(b)f'(a)h + o(h),
qed.
Example 1. Since we believe we know that the derivative of the function
log x is l/x by (14.15) or Chap. II, nO 10, Theorem 3, and since it is not very
difficult to establish, by applying the general formulae of Chap. II, nO 19,
that that of the function expx = LX[n j is Lx[n-l[ = expx, we see that
33 If Ikl < 10 100 lhi for Ih~ sufficiently small one can be sure that Icp(k)1 < lO-looolhl
once tcp(k) I is < 10- lOolkl, which is the case for Ikl < r', so for Ihl < r =
10- 10 r'o More generally: if 0 = O('Ij;) and if'lj; = o(cp), then 0 = o(cp). See the
general rules of Chap. VI, nO 1, which follow directly from the definitions.