Newton-Raphson Formula question

  • Thread starter Thread starter synkk
  • Start date Start date
  • Tags Tags
    Formula
AI Thread Summary
The discussion centers on finding the smallest positive root of the equation x^4 + x^2 - 80= 0 using the Newton-Raphson method. A suggested starting point for the iteration is x = 3, which is considered appropriate for the problem. Participants note that the equation can be solved analytically, making the use of Newton-Raphson unnecessary. However, the original poster confirms their understanding of both the equation's solvability and the application of the method. The conversation emphasizes the balance between analytical solutions and numerical methods in solving equations.
synkk
Messages
216
Reaction score
0
Question:
Find the smallest positive root of the equation x^4 + x^2 - 80= 0 correct to two decimal places. Use the Newton-Raphson process.

How would I go about finding the smallest positive root? What I was thinking is using x = 3 as x = 3 is the largest number that would be close to 0 (for the equation), then finding an approximation for that, is that correct?

thanks.
 
Physics news on Phys.org
1) Independent of the question asked, can you see that the equation is trivially solvable exactly analytically (no need for Newton-Raphson)?

2) For a numeric solution, a start value of 3 looks fine. I assume (hope) you know how to apply Newton-Raphson from there.
 
Last edited:
PAllen said:
1) Independent of the question asked, can you see that the equation is trivially solvable exactly analytically (no need for Newton-Raphson)?

2) For a numeric solution, a start value of 3 looks fine. I assume (hope) you know how to apply Newton-Raphson from there.

I'm perfectly aware it is solvable from that form, and I do know how to apply Newton-Raphson. Thanks.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top