Answer to Question #139991 in Quantitative Methods for elle

Question #139991
find the roots using newton raphson. Stop until at 15th iteration or when approximate percent relative error is below 0.05%.

3x^4-8x^3-37x^2+2x+40
1
Expert's answer
2020-10-26T18:59:07-0400
"f(x)=3x^4-8x^3-37x^2+2x+40"

"f'(x)=12x^3-24x^2-74x+2"

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

Initial solution "x_0 =-0.5"

"\\begin{matrix}\n n & x_n & f(x_n) & x_{n+1} \\\\\n 0 & -0.5 & 30.9375 & -1.482143\\\\\n 1 & -1.482143 & -3.719695 & -1.295091 \\\\\n 2 & -1.295091 & 1.168422 & -1.332165 \\\\\n 3 & -1.332165 & 0.034568 & -1.333332 \\\\\n 4 & -1.333332 & 0.000037 & -1.333333 \\\\\n 5 & -1.333333 & 0.000000 & -1.333333 \\\\\n\\end{matrix}"


"\\varepsilon =\\big|\\dfrac{x_{n+1}-x_n}{x_n}\\big|\\cdot 100\\%"

"\\varepsilon =\\big|\\dfrac{-1.482143+0.5}{-0.5}\\big|\\cdot 100\\%=196.4286\\%"

"\\varepsilon =\\big|\\dfrac{ -1.295091+1.482143}{-1.482143}\\big|\\cdot 100\\%=12.6204\\%"

"\\varepsilon =\\big|\\dfrac{-1.332165 +1.295091}{ -1.295091}\\big|\\cdot 100\\%=2.8627\\%"

"\\varepsilon =\\big|\\dfrac{-1.333332+1.332165}{-1.332165}\\big|\\cdot 100\\%=0.0876\\%"

"\\varepsilon =\\big|\\dfrac{-1.333333+1.333332}{-1.333332}\\big|\\cdot 100\\%=0.000075\\%"

"x=-1.333333"


The root is "-1.333333"


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
APPROVED BY CLIENTS