Open main menu

A well-behaved function can be expanded into a power series. This means that for all non-negative integers there are real numbers such that

Let us calculate the first four derivatives using :

Setting equal to zero, we obtain

Let us write for the -th derivative of  We also write — think of as the "zeroth derivative" of  We thus arrive at the general result where the factorial  is defined as equal to 1 for and and as the product of all natural numbers for Expressing the coefficients in terms of the derivatives of at we obtain

This is the Taylor series for 

A remarkable result: if you know the value of a well-behaved function and the values of all of its derivatives at the single point then you know at all points  Besides, there is nothing special about so is also determined by its value and the values of its derivatives at any other point :

ExamplesEdit

cos Edit

 

 

 

 

 


 

 

 

 

 


     


Some basic checking:

 

   

arctan Edit

 

 

 

 . See  .

Second derivative  Edit

 

 

       

Third derivative  Edit

 

 

                   

(continued)Edit

If you continue to calculate derivatives, you will produce the following sequence:

 

 

 

 

 

 

 

 

 

 

 

 

 

 


 

 

 


Some basic checking:

 

 

arcsin Edit

 

 

 

 

Simple differential equations eliminate the square root and make calculations so much easier.

Let  

Then   where   and  


Differentiating both sides:

 

 

 

Let  

Then  


Differentiating both sides:

 

Let  

Then  


When   Calculation of more derivatives yields:

 

 

 

 

 

 

 

 

 

 

and so on.


 

 


 

 


As programming algorithm:Edit

         


As implemented in Python:Edit

from decimal import * # Default precision is 28.

π = ("3.14159265358979323846264338327950288419716939937510582097494459230781")
π = Decimal(π)

x = Decimal(2).sqrt()/2 # Expecting result of π/4

xSQ = x*x
X = x*xSQ

top = Decimal(1)
bottom = Decimal(2)

bottom1 = bottom*3
sum = x + X*top / bottom1

status = 1
for n in range(5,200,2) :
    X = X*xSQ
    top = top*(n-2)
    bottom = bottom*(n-1)
    bottom1 = bottom*n
    added = X*top/bottom1
    if (added < 1e-29) :
        status = 0
        break
    sum += added

if status :
    print ('error. count expired.')
else :
    print (x, sum==π/4, n)
0.707106781186547524400844362 True 171

In practiceEdit

If   is close to   the calculation of   will take forever.


If you limit   to   then   and each term is guaranteed to be less than half the preceding term.


If   let  

Then  

External linksEdit