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# 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

- az_lenderLv 78 years agoFavorite 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.