How to code erf(z) summation series?

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Henry B. 2021-5-25

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https://ww2.mathworks.cn/matlabcentral/answers/839705-how-to-code-erf-z-summation-series

评论： Torsten 2021-5-28

采纳的回答： Torsten

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Torsten 2021-5-26

Please insert the code you have so far.

If you did not start yet, make an attempt.

Henry B. 2021-5-27

编辑：Henry B. 2021-5-27

在 MATLAB Online 中打开

%Final Lab
clear all 
clc
z = 0;%intialize variable z
n = 0;%intialize variable n
N0 = zeros(1,21); %preallocate array for solutions of 1-erf(z)
fh = @(n,z) 1-erf(z)%create function handle for 1-erf(z)
N0(1) = n;%set n to first position in array
z = 0:0.1:2%array of z values stepping in 0.1 to 2
i = 0;%intialize index
for (i=1:20)%loop counter
  while(abs(N0(i+1)) < eps(N0(i)))%loop condition
    N0(i+1) = N0(i)+fh(n,z)(i)%sum series
    end
  end
  plot(N0, z)%plot graph
  z2 = [0.1 0.5 1 2]%values for erfc(z2)
  errgraph=erfc(z2)%errgraph function
  hold%place lines on same plot
  plot(errgraph)%plot graph on same plot to see convergence.
 
  

I first attempted to use the the scientific expansion but I couldnt figure out how to implement the sigma notation to properly have n= 0 to infinity, so I tried using 1-erf(z) but it gives me a plot in the opposite direction of erfc function.

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Torsten 2021-5-27

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编辑：Torsten 2021-5-27

在 MATLAB Online 中打开

function main
z = [0.1 0.5 1 2];
for i=1:numel(z)
   [erfc1(i),n(i)] = myerfc1(z(i));
end
erfc_matlab = erfc(z);
X = [z.',erfc1.',n.',erfc_matlab.'];
disp(X)
end
function [erfc1,n] = myerfc1(z)
erf1 = 0.0,
term = z;
n = 0;
while abs(term) >= eps
  erf1 = erf1 + term;
  n = n+1;
  term = term * (-1)/n * z^2 * (2*n-1)/(2*n+1);
end
erf1 = erf1 * 2/sqrt(pi);
erfc1 = 1-erf1;
end

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Henry B. 2021-5-27

I still dont get the what my teacher wants as result because when I expand the coded and remove the semicolons and look in the command line the n ouputs from the function dont match values for the erfc(z)

but somehow do in the output?

Torsten 2021-5-28

Include your final code to see what's going wrong.

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Sulaymon Eshkabilov 2021-5-27

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Hi,

Here is the corrected code:

clearvars; clc; close all
N0 = zeros(1,21);   %preallocate array for solutions of 1-erf(z)
fh = @(z) 1-erf(z);  %create function handle for 1-erf(z)
z = 0:0.1:2;               %array of z values stepping in 0.1 to 2
for ii=1:20%loop counter
  while(abs(N0(ii+1)) < eps(N0(ii))) %loop condition
    N0(ii+1) = N0(ii)+fh(z(ii+1)) ;     %sum series
    end
  end
  plot(N0, z)                   %plot graph
  z2 = [0.1 0.5 1 2]         %values for erfc(z2)
  errgraph=erfc(z2)      %errgraph function
  hold                              %place lines on same plot
  plot(errgraph)            %plot graph on same plot to see convergence

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Henry B. 2021-5-27

Thanks and I was wondering about the convergence and now I feel silly because that was the first function handle I used. Are the slopes supposed to go in the opposite direction of each other?

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