How to code erf(z) summation series?

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Henry B.
Henry B. 2021 年 5 月 25 日
コメント済み: Torsten 2021 年 5 月 28 日
  2 件のコメント
Torsten
Torsten 2021 年 5 月 26 日
Please insert the code you have so far.
If you did not start yet, make an attempt.
Henry B.
Henry B. 2021 年 5 月 27 日
編集済み: Henry B. 2021 年 5 月 27 日
%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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採用された回答

Torsten
Torsten 2021 年 5 月 27 日
編集済み: Torsten 2021 年 5 月 27 日
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 件のコメント
Henry B.
Henry B. 2021 年 5 月 27 日
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 2021 年 5 月 28 日
Include your final code to see what's going wrong.

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その他の回答 (1 件)

Sulaymon Eshkabilov
Sulaymon Eshkabilov 2021 年 5 月 27 日
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 件のコメント
Henry B.
Henry B. 2021 年 5 月 27 日
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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