280
III - Convergence: Continuous variables
since, u and v clearly being O(h), with an upper case 0, every function which
is o( u) or o( v), with a lower case 0, is o( h). In consequence, p is differentiable
and
(21.1)
d
p'(t) = dtg[J(t)] = D1g[f(t)].fHt) + D2g[f(t)].f~(t).
Condensed proof: put f(t + h) = f(t) + k, whence
p(t + h) = g[f(t) + k] = p(t) + Dg[f(t)]k + o(k);
but k = f'(t)h + o(h) = O(h), whence o(k) = o(h) and
p(t + h) = p(t) + Dg[f(t)lf'(t)h + o(h),
which gives (1) again, in the form
(21.1 ')
p'(t) = Dg[f(t)]!,(t),
the image of the vector f'(t) E lR? under the linear map Dg[f(t)] : ]R2 ----> ]R2
tangent to 9 at the point z = f (t).
In differential notation: the differential dp = p'(t)dt of p is obtained starting from the differential dg = D1g(z)dx + D2g(z)dy of g, substituting f(t)
for z = (x, y) in the derivatives, and replacing dx and dy by the differentials
of the functions It (t) and h (t) that have been substituted for x and y. The
analogy with Leibniz' system is complete.
Example 1. If 9 is holomorphic and if one considers f as a function with
values in C rather than in ]R2, so that f = It + ih, one finds
(21.2)
!g[f(t)] = g'[f(t)]!,(t)
since the relation D2g = iDlg allows one to exhibit
(21.3)
ff(t) + if~(t) = !,(t)
as a factor of the right hand side of (1); one could also use the fact that, for
a holomorphic function, the map Df(z) consists of multiplying each vector
hE ]R2 = C by the complex number f'(z).
In terms of differentials: in dg = g'(z)dz, replace z by f(t) and dz by
df = f'(t)dt. One should pay attention to the fact that f' is a derivative in
the real sense of nO 14, while g' is a derivative in the complex sense. The
reader can also prove (2) directly, starting from the usual definition of g' as
the limit of a quotient.
Example 2. Suppose that It(t) = x + tu, h(t) = y + tv, i.e. f(t) = z + th,
so that one is examining the behaviour of 9 along a line passing through
(x, y) = z. One finds
III - Convergence: Continuous variables
since, u and v clearly being O(h), with an upper case 0, every function which
is o( u) or o( v), with a lower case 0, is o( h). In consequence, p is differentiable
and
(21.1)
d
p'(t) = dtg[J(t)] = D1g[f(t)].fHt) + D2g[f(t)].f~(t).
Condensed proof: put f(t + h) = f(t) + k, whence
p(t + h) = g[f(t) + k] = p(t) + Dg[f(t)]k + o(k);
but k = f'(t)h + o(h) = O(h), whence o(k) = o(h) and
p(t + h) = p(t) + Dg[f(t)lf'(t)h + o(h),
which gives (1) again, in the form
(21.1 ')
p'(t) = Dg[f(t)]!,(t),
the image of the vector f'(t) E lR? under the linear map Dg[f(t)] : ]R2 ----> ]R2
tangent to 9 at the point z = f (t).
In differential notation: the differential dp = p'(t)dt of p is obtained starting from the differential dg = D1g(z)dx + D2g(z)dy of g, substituting f(t)
for z = (x, y) in the derivatives, and replacing dx and dy by the differentials
of the functions It (t) and h (t) that have been substituted for x and y. The
analogy with Leibniz' system is complete.
Example 1. If 9 is holomorphic and if one considers f as a function with
values in C rather than in ]R2, so that f = It + ih, one finds
(21.2)
!g[f(t)] = g'[f(t)]!,(t)
since the relation D2g = iDlg allows one to exhibit
(21.3)
ff(t) + if~(t) = !,(t)
as a factor of the right hand side of (1); one could also use the fact that, for
a holomorphic function, the map Df(z) consists of multiplying each vector
hE ]R2 = C by the complex number f'(z).
In terms of differentials: in dg = g'(z)dz, replace z by f(t) and dz by
df = f'(t)dt. One should pay attention to the fact that f' is a derivative in
the real sense of nO 14, while g' is a derivative in the complex sense. The
reader can also prove (2) directly, starting from the usual definition of g' as
the limit of a quotient.
Example 2. Suppose that It(t) = x + tu, h(t) = y + tv, i.e. f(t) = z + th,
so that one is examining the behaviour of 9 along a line passing through
(x, y) = z. One finds