False Position

7/17/2019 False Position

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False Position Method



Metode Bisection memiliki kelemahanpokok, yaitu: kecepatannya dalam

mencapai konvergensi;metode inimemiliki kelebihan yaitu: kepastian atau

jaminannya dalam menuju konvergensi.

akan dibahas suatu metode solusibaru yang memodifikasi metode bisection,yang kinerjanya lebih cepat dalammencapai konvergensi, namun masih

tetap memiliki kepastian atau jaminanmenuju konvergensi



The false-position method is a modificationon the bisection method: if it is known that

the root lies on [a, b], then it is reasonablethat we can approximate the function onthe interval by interpolating the points (a,f(a)) and (b, f(b)).



Dengan memakai segitiga Pcb dan PQR,perpotongan garis lurus dengan sumbu x

dapat ditaksir sebagai :



Given the interval [a, b], Then

• if f(c ) = 0 (unlikely in practice), then halt, as we have

found a root,

• if f(c ) and f(a) have opposite signs, then a root must

lie on [a, c ], so assign step = b - c and assign b = c ,• else f(c ) and f(b) must have opposite signs, and thus

a root must lie on [c , b], so assign step = c - a and

assign a = c .



yang apabila menggunakan notasi sepertipada bisection dapat diselesaikan :

persamaan ini digunakan untukmenggantikan persamaan pada langkah 2algoritma metode bisection

)()(

))((

a f b f

abb f b x



a b f(a) f(b) c f(c) UpdateStepSize

1.0 2.0 -2.00 1.00 1.6667 -0.2221 a = c 0.66671.6667 2.0 -0.2221 1.0 1.7273 -0.0164 a = c 0.0606

1.7273 2.0 -0.0164 1.0 1.7317 0.0012 b = c 0.0044

Consider finding the root of f( x ) = x 2 - 3. Let εstep = 0.01, εabs =0.01 and start with the interval [1, 2].

Thus, with the third iteration, we note that the last step 1.7273 →

1.7317 is less than 0.01 and |f(1.7317)| < 0.01, and therefore wechose b = 1.7317 to be our approximation of the root.



Bisection

Consider finding the root of f( x ) = x 2 - 3.

Let εstep = 0.01, εabs = 0.01 and start withthe interval [1, 2].

a b f(a) f(b) x= (a + b)/2 f(x) Galat1.0 2.0 -2.0 1.0 1.5 -0.75 0.5

1.5 2.0 -0.75 1.0 1.75 0.062 0.25

1.5 1.75 -0.75 0.0625 1.625 -0.359 0.125

1.625 1.75 -0.3594 0.0625 1.6875 -0.1523 0.0625

1.6875 1.75 -0.1523 0.0625 1.7188 -0.0457 0.03131.7188 1.75 -0.0457 0.0625 1.7344 0.0081 0.0156

1.71988 1.7344 -0.0457 0.0081 1.7266 -0.0189 0.0078



a b f(a) f(b) c f(c) Update Step Size

3.0 4.0 0.047127 -0.038372 3.5513 -0.023411 b = c 0.4487

3.0 3.5513 0.047127 -0.023411 3.3683 -0.0079940 b = c 0.1830

3.0 3.3683 0.047127 -0.0079940 3.3149 -0.0021548 b = c 0.0534

3.0 3.3149 0.047127 -0.0021548 3.3010 -0.00052616 b = c 0.0139

3.0 3.3010 0.047127 -0.00052616 3.2978 -0.00014453 b = c 0.0032

3.0 3.2978 0.047127 -0.00014453 3.2969-

0.000036998B = c 0.0009

Consider finding the root of f( x ) = e- x (3.2 sin( x ) - 0.5 cos( x )) on theinterval [3, 4], this time with εstep = 0.001, εabs = 0.001.

Thus, after the sixth iteration, we note that the final step, 3.2978 →

3.2969 has a size less than 0.001 and |f(3.2969)| < 0.001 andtherefore we chose b = 3.2969 to be our approximation of the root.In this case, the solution we found was not as good as the solution wefound using the bisection method (f(3.2963) = 0.000034799)

however, we only used six instead of eleven iterations.



a b f(a) f(b) x = (a + b)/2 F(x) Galat

3.0 4.0 0.047127 -0.038372 3.5 -0.019757 0.5

3.0 3.5 0.047127 -0.019757 3.25 0.0058479 0.25

3.25 3.5 0.0058479 -0.019757 3.375 -0.0086808 0.125

3.25 3.375 0.0058479 -0.0086808 3.3125 -0.0018773 0.0625

3.25 3.3125 0.0058479 -0.0018773 3.2812 0.0018739 0.0313

3.2812 3.3125 0.0018739 -0.0018773 3.2968 -0.000024791 0.0156

3.2812 3.2968 0.0018739 -0.000024791 3.289 0.00091736 0.0078

3.289 3.2968 0.00091736 -0.000024791 3.2929 0.00044352 0.0039

3.2929 3.2968 0.00044352 -0.000024791 3.2948 0.00021466 0.002

3.2948 3.2968 0.00021466 -0.000024791 3.2958 0.000094077 0.001

3.2958 3.2968 0.000094077 -0.000024791 3.2963 0.000034799 0.0005

Consider finding the root of f( x ) = e- x (3.2 sin( x ) - 0.5 cos( x )) on the interval [3, 4], this

time with εstep = 0.001, εabs = 0.001.

Thus, after the 11th iteration, we note that the final interval, [3.2958, 3.2968] has a

width less than 0.001 and |f(3.2968)| < 0.001 and therefore we chose b = 3.2968

to be our approximation of the root.



ALGORITMA Metode_Akolade;

DEKLARASI

x1,x2,xr,xn,akar: real;

AA: real;ES,EA:real;

i,IM,iterasi : integer;

function FngsPers(x:real): real;function MtdAkolade(xl,xu:real): real;

DESKRIPSI

read(x1); read(x2); {nilai awal yg mengurung akar}

read(ES); {estimasi kesalahan}read(IM); {iterasi maks}

if FngsPers(x1)*FngsPers(x2)>0 then

write('Tebakan X1 dan X2 tidak mengurung akar')else

xr:= MtdAkolade(x1,x2);for i:=2 to IM doAA:=FngsPers(x1)*FngsPers(xr);

if AA=0 then

akar:=xr; EA:=0;

iterasi:=i; i:=IM;if AA<0 then x2:=xr;

if AA>0 then x1:=xr;

xn:= MtdAkolade(x1,x2);if xn=0 then xr:=xn

elseEA:=abs((xn-xr)/xn)*100;

if EA<ES theniterasi:=i; i:=IM;

xr:=xn;

akar:=xr;endfor

write(akar:5:6); {cetak hasil akar}

if iterasi<>IM thenwrite(iterasi); {cetak jumlah iterasi}

write(EA:5:6); {cetak error aproksimasi}

False Position

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