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Complex Analysis
Functions of Continuous Variables
The function f(z) tends to the limit L as z z 0 if, for
any positive number e , there exists a positive number
d (depending on e ), such that f (z ) - L < e for all z
such that z - z 0 < d .
The function f(z) is continuous at the point z = z0 if
lim f (z ) = f (z 0 ) .
z z 0
The function f(z) is bounded as z z 0 if there exists
positive numbers K and d such that f (z ) < K for all
z with z - z 0 < d .
To find the definitions when z 0 = ¥ , we simply
replace the z - z 0 < d statement in each of those by
z >R.
O notation:
o f (z ) = O(g(z )) as z z 0 means that f (z )/ g(z )
is bounded as z z 0 .
o f (z ) = o(g(z )) as z z0 means that
f (z )/ g(z ) 0 as z z 0 .
o f (z ) ~ g(z ) as z z 0 means that f (z )/ g(z ) 1
as z z 0 . [This means that f is asymptotically
equal to g – but this shouldn’t be written
f (z ) g(z ) !]
o Notes:
These also apply when z 0 = ¥ .
f(z) = O(1) means that f(z) is bounded.
Both the latter of these relations imply
the former.
f (z ) ~ g(z ) is a symmetric relation.
Taylor’s Theorem for functions of a real variable
states that:
Maths Revision Notes © Daniel Guetta, 2007
, Page 2 of 6
h2 h n -1 (n -1)
f (x 0 + h ) = f (x 0 ) + hf ¢(x 0 ) + f ¢¢(x 0 ) + + f (x 0 ) + Rn
2! (n - 1)!
Where
x 0 +h (x 0 + h - x )n -1 (n )
Rn = ò x0 (n - 1)!
f (x )dx
Is the remainder after n terms of the Taylor series.
(Which can apparently be proved by multiplying Rn
by parts, n times). Lagrange’s expression for the
remainder is:
h n (n )
Rn =
f (x )
n!
Where x is unknown, in the interval x 0 < x < x 0 + h .
Therefore:
Rn = O(h n )
If f(x) is infinitely differentiable in the interval, then it
is a smooth function, and we can write an infinite
Taylor Series:
¥
h n (n )
f (x 0 + h ) = å f (x 0 )
n =0 n !
Functions of Complex Variables
The derivative of a function f(z) at the point z = z0 is
f (z ) - f (z 0 )
f ¢(z 0 ) = lim
z z 0 z - z0
f (z + dz ) - f (z )
f ¢(z ) = lim
dz dz 0
Requiring a function of a complex variable to be
differentiable is a surprisingly strong constraint,
because it requires the limit to be the same when
dz 0 in any direction in the complex plane.
Maths Revision Notes © Daniel Guetta, 2007
Complex Analysis
Functions of Continuous Variables
The function f(z) tends to the limit L as z z 0 if, for
any positive number e , there exists a positive number
d (depending on e ), such that f (z ) - L < e for all z
such that z - z 0 < d .
The function f(z) is continuous at the point z = z0 if
lim f (z ) = f (z 0 ) .
z z 0
The function f(z) is bounded as z z 0 if there exists
positive numbers K and d such that f (z ) < K for all
z with z - z 0 < d .
To find the definitions when z 0 = ¥ , we simply
replace the z - z 0 < d statement in each of those by
z >R.
O notation:
o f (z ) = O(g(z )) as z z 0 means that f (z )/ g(z )
is bounded as z z 0 .
o f (z ) = o(g(z )) as z z0 means that
f (z )/ g(z ) 0 as z z 0 .
o f (z ) ~ g(z ) as z z 0 means that f (z )/ g(z ) 1
as z z 0 . [This means that f is asymptotically
equal to g – but this shouldn’t be written
f (z ) g(z ) !]
o Notes:
These also apply when z 0 = ¥ .
f(z) = O(1) means that f(z) is bounded.
Both the latter of these relations imply
the former.
f (z ) ~ g(z ) is a symmetric relation.
Taylor’s Theorem for functions of a real variable
states that:
Maths Revision Notes © Daniel Guetta, 2007
, Page 2 of 6
h2 h n -1 (n -1)
f (x 0 + h ) = f (x 0 ) + hf ¢(x 0 ) + f ¢¢(x 0 ) + + f (x 0 ) + Rn
2! (n - 1)!
Where
x 0 +h (x 0 + h - x )n -1 (n )
Rn = ò x0 (n - 1)!
f (x )dx
Is the remainder after n terms of the Taylor series.
(Which can apparently be proved by multiplying Rn
by parts, n times). Lagrange’s expression for the
remainder is:
h n (n )
Rn =
f (x )
n!
Where x is unknown, in the interval x 0 < x < x 0 + h .
Therefore:
Rn = O(h n )
If f(x) is infinitely differentiable in the interval, then it
is a smooth function, and we can write an infinite
Taylor Series:
¥
h n (n )
f (x 0 + h ) = å f (x 0 )
n =0 n !
Functions of Complex Variables
The derivative of a function f(z) at the point z = z0 is
f (z ) - f (z 0 )
f ¢(z 0 ) = lim
z z 0 z - z0
f (z + dz ) - f (z )
f ¢(z ) = lim
dz dz 0
Requiring a function of a complex variable to be
differentiable is a surprisingly strong constraint,
because it requires the limit to be the same when
dz 0 in any direction in the complex plane.
Maths Revision Notes © Daniel Guetta, 2007
