i need series representation for the equation of erfc(x) and maching with it in plot

i need series representation for the equation of erfc(x) and maching with it in plot. there are series representation. taylor series is series expansion. in wolfram website (www.wolframalpha.com) you can write series erfc(x). you will find three or more series representation. i need one of these match with erfc(x) with function by k not x. like equation imges below (which one is good for matching plot figure )

Answer 1

[Chunru]

x = 0.5; 
format long
erfc(x)

ans = 
   0.479500122186953

for n=1:20
    y(n) = erfc_by_series(x, n);
end

y'

ans = 20×1

0.482826215081223 0.479300030184050 0.479509922142215 0.479499719060915 0.479500136459696 0.479500121743713 0.479500122199208 0.479500122186648 0.479500122186960 0.479500122186953

function y= erfc_by_series(x, n)
   
    t = x;      % term inside sum for k=0
    s = t;
    for k=1:n
        t = -t*x.^2/k;
        s = s + t/(2*k+1);
    end
    y = 1 - 2/sqrt(pi)*s;
end

[sura]

thanks alot chunru kindly i need plot (x,y) because the function is not plot in my program of matlab

[sura]

y(n) = erfc_by_series(x, n); is not work in my version matlab 2019b. it is notrecognised

[Sam Chak]

Hi @sura

Because it is custom function created @Chunru. You can save it as m-file.

function y= erfc_by_series(x, n)
   
    t = x;      % term inside sum for k=0
    s = t;
    for k=1:n
        t = -t*x.^2/k;
        s = s + t/(2*k+1);
    end
    y = 1 - 2/sqrt(pi)*s;
end

[sura]

ok sam it same no response as in m -file . i need the plot (x,y) for above kindly can you plot that and give the figure here

[sura]

kindly i need plot figure of (x,y). it should match with erfc(x) by using

y = 1 - 2/sqrt(pi)*s;

[Torsten]

x = 0:0.1:2;
n = 20;
y = arrayfun(@(x)erfc_by_series(x,n),x);
hold on
plot(x,y)
plot(x,erfc(x))
hold off

function y= erfc_by_series(x, n)
  t = x;      % term inside sum for k=0
  s = t;
  for k=1:n
    t = -t*x.^2/k;
    s = s + t/(2*k+1);
  end
  y = 1 - 2/sqrt(pi)*s;
end

[sura]

figure should give the figure for erfc(x) and y = 1 - 2/sqrt(pi)*s

[sura]

Torsten thank alot but this is not figure should appear

[sura]

when i apply the function above y = 1 - 2/sqrt(pi)*s; the figure is become like this so ineed to match the two figure

[sura]

x=-5:0.1:5

and k=0:2:100

[Torsten]

y returned from erfc_by_series is an approximation to erfc(x).

You should now be able to plot what you want to plot.

[sura]

series matching with erfc(x)

[sura]

not approximation

[Torsten]

Series matches as you can see. But quite a few terms are needed.

x = -5:0.1:5;
n = 80;
y = arrayfun(@(x)erfc_by_series(x,n),x);
hold on
plot(x,y)
plot(x,erfc(x))
hold off

function y= erfc_by_series(x, n)
  t = x;      % term inside sum for k=0
  s = t;
  for k=1:n
    t = -t*x.^2/k;
    s = s + t/(2*k+1);
  end
  y = 1 - 2/sqrt(pi)*s;
end

[sura]

good that is for erfc(x)

it should the y = 1 - 2/sqrt(pi)*s; make the same figure

[Torsten]

erfc(x) is 1 - 2/sqrt(pi)*s.

Or didn't you look at what you posted as the first representation of erfc(x) ?