MIXED METHOD FOR EQUATIONS SOLVING

SERGIU CORLAT, “ORIZONT” LYCEUM

Algebraic equations

Necessary conditions

Let the equation f (x)=0 is givenTo solve the equation on [a,b] the following conditions are necessary:

1. f(x) – continious on [a,b]2. f(a) f(b) < 03. f ’(x), f ”(x) on [a,b]4. The sign of f ’(x), f ”(x) on [a,b] is constant.

Used methods

Chords Method Finding fixed extreme

point Finding solutions using

chords proprieties and the formula:

1

( )( )

( ) ( )i

i i ii

f xx x e x

f e f x

Newton’s Method

Finding initial approximation

Finding solutions using tangent’s proprieties and the formula:

1

( )

( )i

i ii

f zz z

f z

x0 x1 x2

… … z2 z1

z0

Uuupppsss….

1 11

1i i

M mx x

m

221 1

1

( )2i i i

Mx x x

m

Error approximation in Chord’s Method

Error approximation in Newton’s Method

Error approximation in mixed Method i ix z

Method description

Find first approximation x1, using chords method

Find first approximation z1, using Newton’s method

The exact solution is placed between x1 and z1

Repeat previous steps until | xi – zi | becomes less then given precision , or a given number of times.

Illustration

x0 x1 x2

z0 z1z2

…

Algorithm

1. Finding starting point z0 for Newton’s method: c a - (f(a))/(f(b) - f(a))(b - a); If f(c)f(a)>0 then z0 b; x0 a; else z0 a; x0 b;

2. i 0

3. Calculating

4. Calculating

5. i i + 1

6. If |zi – xi| > Then go to step 3, else – to step 7.

7. The final solution s will be . END.

1

( )

( )i

i ii

f zz z

f z

1 11

( )( );

( ) ( )i

i i i ii i

f xx x z x

f z f x

2i ix z

s

Example:

Let we have the equation

Find the approximate solution of equation on [2; 6] with precision = 0.00001, using mixed method.

2( ) 1 7.f x x x

Solving. 1. Math processing:

1 1 1 1 1 22 2 2 2 2 22 2 2 2 2

2

2 1( ) ( 1) ( 1) ( 1) ( 1) ( 1)

1

xf x x x x x x x x x

x

Solving

2. Program

program cn011;var a,b,c,x,z,e : real; function f(x:real):real;begin f:=x*sqrt(1+x*x) -7 ;end; function fd1(x:real):real;begin fd1:=(2*x*x+1)/ sqrt(1+x*x) ;end;

Variables

f(x) description

f ’ (x) description

Solving

begin a:=2; b:=6; e:=0.0001;c:=a-(f(a))/(f(b)-f(a))*(b-a); if f(c)*f(a)>0 then begin z:=b; x:=a; end else begin z:=a; x:=b; end;while abs(z-x)>e do begin z:=z-f(z)/fd1(z); x:= x-(f(x))/(f(z)-f(x))*(z-x); writeln (’z=’,z:9:5,’ f(z)=’,f(z):9:5, ’ x=’, x:9:5, ’ f(x)=’, f(x):9:5); end; writeln((z+x)/2:9:5);end.

{z},{ x}

initializa

tion

zi,xi calculus

and printing

Results

z= 3.54218 f(z)= 6.03747 x= 2.45514 f(x)= -0.49146z= 2.69058 f(z)= 0.72307 x= 2.55041 f(x)= -0.01326z= 2.55649 f(z)= 0.01787 x= 2.55300 f(x)= -0.00001z= 2.55301 f(z)= 0.00001 x= 2.55300 f(x)= -0.000002.55301