TAYLOR’S THEOREM FOR
FUNCTIONS OF TWO VARIABLES
AND
JACOBIANS
PRESENTED BY
PROF. ARUN LEKHA
Associate Professor in Maths
GCG-11, Chandigarh
Statement:
Taylor’s Theorem in two variables
If f (x,y) is a function of two independent variables x and
y having continuous partial derivatives of nth order in
some neighbourhood of the point (a,b) and if (a + h, b +
a) is any point of this neighbourhood, then there exists
some  in (0,1) such that
f (a+h, b + k) = f (a,b) + df (a,b) + ½! d2f (a,b)+…
1/n-1! dn-1 f(a,b) + 1/n! dn f(a+h,b+ k)
Where
 
 

 
n
d   h  k  and d  h  k 
y 
y 
 x
 x
n
Remainder after n terms denoted by Rn
Where
1 n
1  
 n
 h  k  f (a  h, b   k );
Rn  d f (a  h, b   k ) 
n!
n!  x
y 
Statement of Maclaurin’s Theorem (Two Variable)
 
 
1 
 
f ( x, y )  f (0,0)   x  y  f (0,0)   x  y 
y 
2!  x
y 
 x
1  
 
 x  y 

n  1!  x
y 
n 1
1
f (0,0).  Rn where Rn 
n!
2
f (0,0).......
 
 
 x  y 
y 
 x
n
f (x  y ), 0    1
• Using Taylor’s Theorem, find the approximate
value of log [(1.03)1/3
+ (0. 98)1/4 – 1]
Solution:
Let f (x+h, y + k ) = log [(x+h)1/3 +(y+k)1/4 – 1]
Putting h = 0 = k in (1), we get
f(x,y ) = log (x1/3 + y¼ - 1)
1 2 / 3
1 3 / 4
x
y
3
f x  1/ 3
, fy  1 / 34 1 / 4
1/ 4
x  y 1
x  y 1
f(x+h,y+k)=f(x,y)+(hfx+kfy)
(1)
log [(x+h)1/3+(y+k)1/4-1)=log (x1/3+y1/4-1)
1 2 / 3
1 3 / 4 

x
y


3
4

  h, 1/ 3
,

k
1/ 4
1/ 3
1/ 4
x  y 1 
 x  y 1




Put x = 1, h = 0.03, y = 1, k = -0.02
log [(1.03)1/3 + (0. 98)1/4 – 1] = log (1+1-1)
1
1 





3
4
 (0.03).
 (0.02).

1

1

1
1

1

1






= log 1 + 0.01 – 0.005 = 0.005
DEFINITION OF JACOBIAN
If u, v, w are the function of x, y, z having first order partial
derivatives w.r.t. x, y, z then the determinant
u
x
 (u , v, w)
(u , v, w) v
J

 ( x, y , z )
( x, y, z ) v
w
x
y
y
v
y
w
y
is called Jacobian of u, v, w w.r.t. x, y, z
z
z
v
z
w
z
CHAIN RULE FOR JACOBIAN
STATEMENT:
If u, v, w are the functions of two independent variable x and y
which are themselves functions of two independent variable r
and s, then
(u, v) (u, v) ( x, y)

x
(r , s) ( x, y ) (r , x)
FUNCTIONAL DEPENDENCE OR NOT INDEPENDENT
Theorem
If n function u1, u2, u3….un, of n independent variable are
functionally dependent i.e., there exists a relation of the
form F (u1, u2…….un) = 0
iff
(u1 , u 2 .........u n )
 0 identicall y
( x1 , x 2 ,....x n )
Example: Show that the functions
U = x2 + y2 + z2 – 2xz,
v = x + y – z,
w=x–y–z
are not independent of one another. Also, find the relation
between them
2x  2x 2 y 2z  2x 2x  2z 2 y 0
 (u, v, x)

1
1
1 
1
1 0
 ( x, y , z )
1
1
1
1
1 0
2x  2x 2 y 2z  2x 2x  2z 2 y 0
 (u, v, x)

1
1
1 
1
1 0
 ( x, y , z )
1
1
1
1
1 0
Now, v2 + w2 = (x+y-z)2 + (x-y-z)2
= (x2+y2+z2 + 2xy-2yz-2zx) + (x2+y2+z2 – 2xy + 2yz-2zx)
= 2 (x2+y2+z2 – 2zx) = 2u
v2 + w2 = 2u is the required relation between u,v,w.