Question #204175

Find a root of the following equations correct to three decimals using Newton's raphson method: x³ -5x +3=0


1
Expert's answer
2021-06-08T04:25:11-0400
x35x+3=0x^3-5x+3=0

f(x)=x35x+3f(x)=x^3-5x+3

f(x)=3x25f'(x)=3x^2-5

xn+1=xnf(xn)f(xn)x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}

x0=2x_0=-2

stepx0f(x0)x1f(x1)12.000052.71433.425722.71433.42572.51400.318732.51400.31872.49120.003942.49120.00392.49090.000052.49090.00002.49090.0000\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} step & x_0 & f(x_0) & x_1 & f(x_1) \\ \hline 1 & -2.0000 & 5 & -2.7143 & -3.4257 \\ \hdashline 2 & -2.7143 & -3.4257 & -2.5140 & -0.3187 \\ \hdashline 3 & -2.5140 & -0.3187 & -2.4912 & -0.0039 \\ \hdashline 4 & -2.4912 & -0.0039 & -2.4909 & 0.0000 \\ \hdashline 5 & -2.4909& -0.0000 & -2.4909 & 0.0000 \\ \hdashline \end{array}



xI=2.491x_{I}=-2.491



x0=1x_0=1

stepx0f(x0)x1f(x1)11.00001.00000.50000.625020.50000.62500.64710.035630.64710.03560.65660.000240.65660.00020.65660.0002\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} step & x_0 & f(x_0) & x_1 & f(x_1) \\ \hline 1 & 1.0000 & -1.0000 & 0.5000 & 0.6250 \\ \hdashline 2 & 0.5000 & 0.6250 & 0.6471 & 0.0356 \\ \hdashline 3 & 0.6471 & 0.0356 & 0.6566 & 0.0002\\ \hdashline 4 & 0.6566 & 0.0002 & 0.6566 & 0.0002 \\ \hdashline \end{array}



xII=0.657x_{II}=0.657




x0=2x_0=2

stepx0f(x0)x1f(x1)12.00001.00001.85710.119521.85710.11951.83480.002831.83480.00281.83420.000041.83420.00001.83420.0000\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} step & x_0 & f(x_0) & x_1 & f(x_1) \\ \hline 1 & 2.0000 & 1.0000 & 1.8571 & 0.1195 \\ \hdashline 2 & 1.8571 & 0.1195 & 1.8348 & 0.0028 \\ \hdashline 3 & 1.8348 & 0.0028 & 1.8342 & 0.0000\\ \hdashline 4 & 1.8342 & 0.0000 & 1.8342 & 0.0000 \\ \hdashline \end{array}



xIII=1.834x_{III}=1.834



Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS