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In mathematics and numerical analysis, in order to accelerate convergence of an alternating series, Euler's transform can be computed as follows.

Compute a row of partial sums :

$s_{0,k}=\sum _{n=0}^k(-1{)}^n{a}_n$

and form rows of averages between neighbors,

$s_{j+1,k}=\frac{s_{j,k}+s_{j,k+1}}{2}$

The first column $s_{j,0}$ then contains the partial sums of the Euler transform.

Adriaan van Wijngaarden's contribution was to point out that it is better not to carry this procedure through to the very end, but to stop two-thirds of the way. If $a_0,a_1,\dots ,a_{12}$ are available, then $s_{8,4}$ is almost always a better approximation to the sum than $s{}_{12,0}.$

Leibniz formula for pi, $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots =\frac{\pi }{4}=0.7853981\dots$, gives the partial sum $s_{0,12}=0.8046006...(+2.4\%)$, the Euler transform partial sum $s_{12,0}=0.7854002...(+2.6\times 10^{-6})$ and the van Wijngaarden result $s_{8,4}=0.7853982...(+4.7\times 10^{-8})$ (relative errors are in round brackets).

1.00000000 0.66666667 0.86666667 0.72380952 0.83492063 0.74401154 0.82093462 0.75426795 0.81309148 0.76045990 0.80807895 0.76460069 0.80460069
0.83333333 0.76666667 0.79523810 0.77936508 0.78946609 0.78247308 0.78760129 0.78367972 0.78677569 0.78426943 0.78633982 0.78460069           
0.80000000 0.78095238 0.78730159 0.78441558 0.78596959 0.78503719 0.78564050 0.78522771 0.78552256 0.78530463 0.78547026                      
0.79047619 0.78412698 0.78585859 0.78519259 0.78550339 0.78533884 0.78543410 0.78537513 0.78541359 0.78538744                                 
0.78730159 0.78499278 0.78552559 0.78534799 0.78542111 0.78538647 0.78540462 0.78539436 0.78540052                                            
0.78614719 0.78525919 0.78543679 0.78538455 0.78540379 0.78539555 0.78539949 0.78539744                                                       
0.78570319 0.78534799 0.78541067 0.78539417 0.78539967 0.78539752 0.78539847                                                                  
0.78552559 0.78537933 0.78540242 0.78539692 0.78539860 0.78539799                                                                             
0.78545246 0.78539087 0.78539967 0.78539776 0.78539829                                                                                        
0.78542166 0.78539527 0.78539871 0.78539803                                                                                                   
0.78540847 0.78539699 0.78539837                                                                                                              
0.78540273 0.78539768                                                                                                                         
0.78540021                      

This table results from the J formula 'b11.8'8!:2-:&(}:+}.)^:n+/\(_1^n)*%1+2*n=.i.13 In many cases the diagonal terms do not converge in one cycle so process of averaging is to be repeated with diagonal terms by bringing them in a row. This will be needed in an geometric series with ratio -4. This process of successive averaging of the average of partial sum can be replaced by using formula to calculate the diagonal term.

References

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See also

Euler summation