How to code erf(z) summation series?

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Henry B.
Henry B. on 25 May 2021
Commented: Torsten on 28 May 2021
  2 Comments
Torsten
Torsten on 26 May 2021
Please insert the code you have so far.
If you did not start yet, make an attempt.
Henry B.
Henry B. on 27 May 2021
Edited: Henry B. on 27 May 2021
%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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Accepted Answer

Torsten
Torsten on 27 May 2021
Edited: Torsten on 27 May 2021
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
  12 Comments
Henry B.
Henry B. on 27 May 2021
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
Torsten on 28 May 2021
Include your final code to see what's going wrong.

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More Answers (1)

Sulaymon Eshkabilov
Sulaymon Eshkabilov on 27 May 2021
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
  1 Comment
Henry B.
Henry B. on 27 May 2021
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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