Taylors theorem for manyvariable functions

PARTIAL DIFFERENTIATION
Thus, from (5.17), we may write
∂
∂
sin φ ∂
= cos φ
−
,
∂x
∂ρ
ρ ∂φ
∂
∂
cos φ ∂
= sin φ
+
.
∂y
∂ρ
ρ ∂φ
Now it is only a matter of writing
∂
∂ ∂f
∂2 f
∂
=
f
=
2
∂x
∂x ∂x
∂x ∂x
∂
∂
sin φ ∂
sin φ ∂
= cos φ
cos φ
g
−
−
∂ρ
ρ ∂φ
∂ρ
ρ ∂φ
∂g
∂
sin φ ∂
sin φ ∂g
cos φ
= cos φ
−
−
∂ρ
ρ ∂φ
∂ρ
ρ ∂φ
= cos2 φ
+
∂2 g
2 cos φ sin φ ∂g
2 cos φ sin φ ∂2 g
+
−
∂ρ2
ρ2
∂φ
ρ
∂φ∂ρ
sin2 φ ∂g
sin2 φ ∂2 g
+
ρ ∂ρ
ρ2 ∂φ2
and a similar expression for ∂2 f/∂y 2 ,
∂
∂
cos φ ∂
cos φ ∂
∂2 f
sin φ
g
= sin φ
+
+
2
∂y
∂ρ
ρ ∂φ
∂ρ
ρ ∂φ
2 cos φ sin φ ∂2 g
∂2 g
2 cos φ sin φ ∂g
−
+
2
2
∂ρ
ρ
∂φ
ρ
∂φ∂ρ
cos2 φ ∂g
cos2 φ ∂2 g
+
.
+
ρ ∂ρ
ρ2 ∂φ2
= sin2 φ
When these two expressions are added together the change of variables is complete and
we obtain
∂2 f
∂2 g
1 ∂g
∂2 f
1 ∂2 g
+ 2 = 2 +
.
+ 2
2
∂x
∂y
∂ρ
ρ ∂ρ ρ ∂φ2
5.7 Taylor’s theorem for many-variable functions
We have already introduced Taylor’s theorem for a function f(x) of one variable,
in section 4.6. In an analogous way, the Taylor expansion of a function f(x, y) of
two variables is given by
∂f
∂f
∆x +
∆y
∂x
∂y
2
∂2 f
∂2 f
1 ∂ f
2
2
∆x∆y
+
(∆x)
+
2
(∆y)
+
+ ··· ,
2! ∂x2
∂x∂y
∂y 2
f(x, y) = f(x0 , y0 ) +
(5.18)
where ∆x = x − x0 and ∆y = y − y0 , and all the derivatives are to be evaluated
at (x0 , y0 ).
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5.7 TAYLOR’S THEOREM FOR MANY-VARIABLE FUNCTIONS
Find the Taylor expansion, up to quadratic terms in x − 2 and y − 3, of f(x, y) = y exp xy
about the point x = 2, y = 3.
We first evaluate the required partial derivatives of the function, i.e.
∂f
= y 2 exp xy,
∂x
∂2 f
= y 3 exp xy,
∂x2
∂f
= exp xy + xy exp xy,
∂y
∂2 f
= 2x exp xy + x2 y exp xy,
∂y 2
∂2 f
= 2y exp xy + xy 2 exp xy.
∂x∂y
Using (5.18), the Taylor expansion of a two-variable function, we find
!
f(x, y) ≈ e6 3 + 9(x − 2) + 7(y − 3)
"
+ (2!)−1 27(x − 2)2 + 48(x − 2)(y − 3) + 16(y − 3)2 . It will be noticed that the terms in (5.18) containing first derivatives can be
written as
∂f
∂
∂
∂f
∆x +
∆y = ∆x
+ ∆y
f(x, y),
∂x
∂y
∂x
∂y
where both sides of this relation should be evaluated at the point (x0 , y0 ). Similarly
the terms in (5.18) containing second derivatives can be written as
2
∂2 f
∂
∂2 f
1 ∂2 f
1
∂
2
2
∆x∆y
+
+
∆y
(∆x)
+
2
(∆y)
f(x, y),
=
∆x
2! ∂x2
∂x∂y
∂y 2
2!
∂x
∂y
(5.19)
where it is understood that the partial derivatives resulting from squaring the
expression in parentheses act only on f(x, y) and its derivatives, and not on ∆x
or ∆y; again both sides of (5.19) should be evaluated at (x0 , y0 ). It can be shown
that the higher-order terms of the Taylor expansion of f(x, y) can be written in
an analogous way, and that we may write the full Taylor series as
f(x, y) =
n
∞
∂
1
∂
+ ∆y
f(x, y)
∆x
n!
∂x
∂y
x0 ,y0
n=0
where, as indicated, all the terms on the RHS are to be evaluated at (x0 , y0 ).
The most general form of Taylor’s theorem, for a function f(x1 , x2 , . . . , xn ) of n
variables, is a simple extension of the above. Although it is not necessary to do
so, we may think of the xi as coordinates in n-dimensional space and write the
function as f(x), where x is a vector from the origin to (x1 , x2 , . . . , xn ). Taylor’s
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