1. ROOTS OF EQUATIONS GRAPHIC METHOD

2. GRAPHIC METHOD This method is used primarily for find an interval where function has any root. Tounderstandbetterwegoingto do anExample. fx=arctan(x)+(x−1)

3. solution To findtherood of f(x), we do arctanx+x−1=0, wherewehavearctanx=1. Thus, theproblemistofindthepoint of intersection of thegraphs of thefunctionsgx=arctanx , y, hx=1−x. So, wegraphthis.

4. Graphictofindtherood g(x)= arctan x h(x)= 1-x Wherewe can seecrearlythatanintervalwheretherootisonliinterval (0,1)

5. Someconsiderations If fa∗fb<0, is probably to find an odd number of roots for this equation. If fa∗fb>0, is probably to find an even number of roots or that there aren’t.

6. OPEN METHODS FIXED POINT

7. FIXED POINT This methodisappliedtosolveequations of theformx=gx Iftheequationisfx=0, thenyou can eithercleared x oradded x in bothsides of theequationtoput in theproperly. CLOSE METHODS

8. Graphicallywehave.. f1x=x , f2=g(x) y=f(x) Raíz Iteratively x1+1=g(xi) - CLOSE METHODS

9. Someconditionstoconvergence If f2(x)≤f1x y f2x>0: we’llhave a monotonicallyconvergentsolutionbecauseeachsolutionisobtainedclosertotheroot. If f2(x)≤f1x y f2x<0: it has a convergentoscillatorysolutionbecauseeachsolutionisobtained in a mannerclosertotherootoscillatory. Iff2(x)≥f1x y f2x>0, so it has a divergentsolution. CLOSE METHODS

10. example We needfindtherootsforthisequiation: fx=e−x−lnx So gx=e−x−lnx+x then you have to do a table like: Fixedpoint Tolerance = 0.01

11. CLOSE METHODS NEWTON-RAPHSON

12. Newton-Raphson This method, which is an iterative method is one of the most used and effective. Newton-Raphson method does not work on a range bases his formula in an iterative process.Suppose we have the approximation xi to the root xr of f(x), f(x) tangente this line intersects the axis x, at a point xi+1that will be our next approximation to the root xr. Xi Xr Xi+1

13. To calculatedthepointxi+1, first we have to find equation of the tangent line. We know that id has pending m=f′(xi) So, theequationis: y−fxi=f′xix−xi Afterwe do Y=0 −fxi=f′xix−xi And solvefor x: x=xi−f(xi)f′(xi)thisistheiterativeformto Newton – Rapson. Newton-Raphson

14. Newton-Raphson To apreciatedbetterwegonnasolveranexample. fx=e−x−ln(x) , xo=1 tol= 0.01

16. METODOS NUMERICOS PhD EDUARDO CARRILLO – UNIVERSIDAD INDUSTRIAL DE SANTANDER 2010

17. CHAPRA, Steven C. “Métodos Numéricos para Ingenieros”. Edit. McGraw Hil.