Use Newton's Method to find the solution to the equation f(x)=0 for each of the following.?

Find the first three iterations of the sequence. a) f(x) = ((x^4)/4) + x - 1 and b) f(x) = (2^-x) - x

2 Answers

Relevance
  • 8 years ago
    Favorite Answer

    I did three iterations of (a), but I stopped after the first iteration of (b) because

    I am sure you can figure out how to continue.

    (a)

    Newton's method is useful only if your starting guess is somewhat close.

    Seeing as how f(0) = -1 and f(1) = 1/4,

    I'm going to guess

    x0 = 0.9

    y0 = f(x0) = 0.064025

    dy/dx = x^3 + 1, at x=0.9 this is 1.729.

    We need dy to be -0.064025, so dx should be -0.064025/1.729 = -0.037

    x1 = 0.863

    y1 = 0.00167

    dy/dx = 0.863^3 + 1 = 1.643

    We need dy to be -0.00167, so dx = -0.00167/1.643 = -0.001016

    x2 = 0.86198

    y2 = -0,0000043

    dy/dx = 0.86198^3 + 1 = 1.640

    We need dy to be +0.0000043, so dx = 0.0000043/1.640 = 0.00000262

    x3 = 0.8619826

    y3 < 10^(-7)

    (b) y = f(x) = 2^(-x) - x

    f(0) = 1, f(1) = -1/2

    x0 = 0.7 would be a good start.

    y0 = 1/(2^0.7) - 0.7 = -0.08443

    dy/dx = -ln(2) 2^(-x) - 1

    At x = 0.7 this is -1.4267

    dy needs to be +0.08443, so dx = -0.08443/1.4267 = -0.0592

    x1 = 0.6408

    ET CETERA ... I'm sure you have the idea by now.

  • 4 years ago

    Newton's technique does no longer artwork for this equation. What it truly is, is: you could placed random numbers in decision to x and spot, which of it truly works. as quickly as you detect the 1st answer, you could factorise the equation applying that answer to discover the different strategies.

Still have questions? Get your answers by asking now.