Answer to Question #287596 in Quantitative Methods for beki

Question #287596

approximate the real root to four decimal places of x3+5x-3=0 (newton raphson method)


1
Expert's answer
2022-01-17T07:02:36-0500

Solution:

Here x3+5x3=0x^{3}+5 x-3=0

Let f(x)=x3+5x3f(x)=x^{3}+5 x-3

ddx(x3+5x3)=3x2+5f(x)=3x2+5\\ \\\frac{d}{d x}\left(x^{3}+5 x-3\right)=3 x^{2}+5 \therefore f(x)=3 x^{2}+5

Here


Here f(0)=3<0 and f(1)=3>0f(0)=-3<0\ and\ f(1)=3>0

\therefore Root lies between 0 and 1

x0=0+12=0.5x0=0.5x_{0}=\frac{0+1}{2}=0.5 \\x_{0}=0.5

1st1^{s t} iteration :

f(x0)=f(0.5)=0.53+50.53=0.375f(x0)=f(0.5)=30.52+5=5.75f\left(x_{0}\right)=f(0.5)=0.5^{3}+5 \cdot 0.5-3=-0.375 \\f^{\prime}\left(x_{0}\right)=f(0.5)=3 \cdot 0.5^{2}+5=5.75

x1=x0f(x0)f(x0)x1=0.50.3755.75x1=0.5652\begin{aligned} &x_{1}=x_{0}-\frac{f\left(x_{0}\right)}{f\left(x_{0}\right)} \\ &x_{1}=0.5-\frac{-0.375}{5.75} \\ &x_{1}=0.5652 \end{aligned}

2nd2^{n d} iteration :

f(x1)=f(0.5652)=0.56523+50.56523=0.0067f(x1)=f(0.5652)=30.56522+5=5.9584\begin{aligned} &f\left(x_{1}\right)=f(0.5652)=0.5652^{3}+5 \cdot 0.5652-3=0.0067 \\ &f^{\prime}\left(x_{1}\right)=f(0.5652)=3 \cdot 0.5652^{2}+5=5.9584 \end{aligned}

x2=x1f(x1)f(x1)x2=0.56520.00675.9584x2=0.5641\begin{aligned} &x_{2}=x_{1}-\frac{f\left(x_{1}\right)}{f\left(x_{1}\right)} \\ &x_{2}=0.5652-\frac{0.0067}{5.9584} \\ &x_{2}=0.5641 \end{aligned}

3rd 3^{\text {rd }} iteration :

f(x2)=f(0.5641)=0.56413+50.56413=0f\left(x_{2}\right)=f(0.5641)=0.5641^{3}+5 \cdot 0.5641-3=0

f(x2)=f(0.5641)=30.56412+5=5.9546x3=x2f(x2)f(x2)x3=0.564105.9546x3=0.5641 Approximate root of the equation x3+5x3=0 using Newton Raphson mehtod is 0.5641 (After 3 iterations) \begin{aligned} f^{\prime}\left(x_{2}\right)=f(0.5641)=3 \cdot 0.5641^{2}+5=5.9546 \\ x_{3}=x_{2}-\frac{f\left(x_{2}\right)}{f\left(x_{2}\right)} \\ x_{3}=0.5641-\frac{0}{5.9546} \\ x_{3}=0.5641 \\ \text { Approximate root of the equation } x^{3}+5 x-3=0 \\\text { using Newton Raphson mehtod is } 0.5641 \text { (After } 3 \text { iterations) } \end{aligned}


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!

Leave a comment

LATEST TUTORIALS
New on Blog