By Taylor's formula, the expression in the brackets can be written as
where is a value between and .
We now divide (2.11) by h and introduce the quantities
, and thus (2.11) can now be written in the form
where and is a constant
Define the function
we can look at (2.12) as the result of applying Euler's method to the solution of a new differential equation for a function
making at each step an additional error not exceeding . The initial value is zero, because .
is a value between 
and 
. 
, and thus (2.11) can now be written in the form 
and 
is a constant 
we can look at (2.12) as the result of applying Euler's method to the solution of a new differential equation for a function 
. The initial value 
is zero, because 
. 
