How to find positive x-root of this function?

Hi All,

I want to find positive root x of this function;

x^10=1;

I wrote a program using modified false postion algorythm but i don't know how to use it to solve equation written above. Here is the Code;

function ModFalsePos=eqn(xl,xu,es,xr,ea)
f=@(x)x^10-1
xl=0
xu=1.3
es=0.01
fl=f(xl)
fu=f(xu)
while (1)
    xr=xu-fu*(xl-xu)/(fl-fu)
    xrold=xr
    fr=f(xr)
    if xr<0
    elseif xr>0
        ea=abs((xr-xold)/xr)*100
    end
    test=fl*fr
    if test<0
        xu=xr
        fu=f(xu)
        iu=0
        il=il+1
        if il>=2
            fl=fl/2
        end
    elseif test>0
        xl=xr
        fl=f(xl)
        il=0
        iu=iu+1
        if iu>=2
            fu=fu/2
        end
    else
        ea=0
    end
    if ea<es
        break
    end
end
ModFalsePos=xr
end

Could anyone please help me for solving this equation using this code? What's wrong here?

Thanks for any Help!

[Jonathan Epperl]

if make_your_code_readable(you)
   help(we, you)
else
   why
end

[Otto]

I'm beginner in this forum. I don't know how to make my code "readable". Could you please help me also for this?

[Jonathan Epperl]

• Paste your code
• Highlight your code
• Hit the "{ } Code" button right above the field your a typing the text into.

[Otto]

ok now you can see the code clearly.

[Jonathan Epperl]

See, much better! Do you get an error, an infinite loop, or simply a wrong result?

[Otto]

I want to get a loop that stops when the error criterion is met. (if Ea<Es) and finding results while initial guesses are xl=0,xu=1.3 let say xr1=.. xr2=...,fu1=... fu2=..., continious results until criterion is met. but i can't get this results. if you copy this code to your MATLAB screen you will understand what i mean.

[Gang-Gyoo]

 % I moved two statements before 'while'. The code results the answer
 % (x= 1) but you still check your code whether is as same as 
 % the algorithm.

 function main
   ModFalsePos= eqn(0,1.01,0.01)
  end

function ModFalsePos=eqn(xl,xu, es)
f=@(x)x^10-1;

fl=f(xl);
fu=f(xu);
xold= xl;
iu=0;

while (1)
    xr=xu-fu*(xl-xu)/(fl-fu);
    xrold=xr;
    fr=f(xr);
    if xr<0
    elseif xr>0
        ea=abs((xr-xold)/xr)*100;
    end
    test=fl*fr;
    if test<0
        xu=xr;
        fu=f(xu);

          il=il+1;
          if il>=2
              fl=fl/2;
          end
      elseif test>0
          xl=xr;
          fl=f(xl);
          il=0;
          iu=iu+1;
          if iu>=2
              fu=fu/2;
          end
      else
          ea=0;
      end
      if ea<es
          break
      end
  end
  ModFalsePos=xr;
  end

[Jonathan Epperl]

You never change xold in the loop, so this will only work, if test<0 through every iteration.

[Otto]

this code finds the root of equation ModFalsePos=1 but i want to see the change in other results, for example fl,fu,iteration number,xr, xl,xu in tabulated form. this code gives just the root of polynomial. What i want to see is "change" in parameters until the criterion is met.

Thank you for your interest!

[Gang-Gyoo]

you shound put the fprintf statement inside the while statement as

fprintf('%f %f %f \n', fl, fu, xr);

Answer 1

[Jonathan Epperl]

Okay, in this line here:

ea=abs((xr-xold)/xr)*100

ea is always going to be zero, since a few lines above, you assigned xold=xr. Thus your algorithm always exits after the first iteration. So you need to flip the first lines after while:

    xrold = xr;
    xr = xu - fu*(xl-xu)/(fl-fu);

Then, you need to initialize iu, il and xr outside of the loop.

xu=1.3;
xr=xu;
iu = 0; il = 0;

Also, remove the function line and the last end, what you have here is a script, not a function. If you want it to be a function, at least f should be an input argument, too, otherwise it doesn't make much sense.

And now it should be running.

[Otto]

what you suggested caused an infinite loop. and all I get is zeros again. what should I do now?

Thank you for your interest!

[Jonathan Epperl]

Ah yeah, I also forgot that I had to change this line

ea=abs((xr-xold)/xr)*100

to this

ea=abs((xr-xrold)/xr)*100

assuming that you just made a typo, calling on xold instead of xrold. Here is the entire code, and it runs. If you want more outputs, then just remove the semicolon ";" from the end of the respective line.

Another thing that you didn't do: Initialize ea.

Without further ado, here's my complete code, that should really run now, with different values for xl and xu:

f=@(x)x^10-1
xl=-.5
xu=14
xr = xu;
es=0.01
fl=f(xl)
fu=f(xu)
ea = inf;

iu = 0; il = 0;
while (1)
    xrold = xr;
    xr = xu - fu*(xl-xu)/(fl-fu)

      fr = f(xr);
      if xr<0
      elseif xr>0
          ea=abs((xr-xrold)/xr)*100 % relative change in xr?
      end
      test = fl*fr
      if test<0
          xu = xr
          fu = f(xu)
          iu = 0
          il = il+1
          if il>=2
              fl=fl/2
          end
      elseif test>0
          xl=xr
          fl=f(xl)
          il=0
          iu=iu+1
          if iu>=2
              fu=fu/2
          end
      else
          ea=0
      end
      if ea<es
          break
      end
  end
  ModFalsePos=xr

Answer 2

[Matt Fig]

Why not just use:

R = roots([1 0 0 0 0 0 0 0 0 0 -1]);