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# MTH603 Numerical Analysis Friday, November 13, 2015 to Monday, November 16, 2015 GDB

Solve the equations x3 – 3x+ 0 in the interval [0,1] using the Regula Falsi Method(Method of False Position) up to two iterations.

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### Replies to This Discussion

at this point i'm also confused

f(x)=x3−9x+1
f(2)=23−9(2)+1=8−18+1=−9
f(4)=43−9(4)+1=64−36+1=29
Sincef(2) andf(4) are of opposite signs, the root off(x) = 0 lies between 2 and 4.Takingx1 = 2,x2 = 4 and using Regula−Falsi method, the first approximation is given by
x3=x2−x2−x1f(x2)−f(x1)f(x2)=4−4−229−(−9)=4−2(29)38
=4−5838=4−1.5263=2.4736
Now
f(x3)=2.47363−9(2.4736)+1=−6.12644
Sincef(x2) andf(x3) are of opposite signs, the root lies betweenx2 andx3.
The second approximation to the root is given as
x4=x3−x3−x2f(x3)−f(x2)f(x3)=2.4736−2.4736−4−6.12644−29(−6.12644)
=2.4736−−1.5264−35.12644(−6.12644)=2.4736−(0.04345)(−6.12644)
=2.4736+0.26619=2.73989
Therefore
f(x4)=2.739893−9(2.73989)+1=20.5683−24.65901+1=−3.090707
Now, sincef(x2) andf(x4) are of opposite signs, the third approximation is obtained
from
x5=x4−x4−x2f(x4)−f(x2)f(x4)=2.73989−2.73989−4−3.090707−29(−3.090707)=2.86123
=2.73989−−1.26011−32.090707(−3.090707)=2.73989+0.039267(3.090707)=2.73989+0.121363=2.86125
Now
f(x5)=2.861253−9(2.86125)+1=23.42434−25.75125+1=−1.326868

May be two iterations are f(x3) and f(x4)

not sure

but i know that we have to solved just f(x3) and f(x4)

aoa

Hira Anmol thanks to pointing out a mistake

Thanks to All :)

make change

Attachments:

not correct

plz share your gdb es mein copy ka drr na krein plz sb smjdar hain

Attachments:

not correct

why

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