Use Newton’s method to find all real-valued solutions of y4 +2y = 1.?

4Y + 2Y = 6y

6y = 1

y = 1/6

good luck

Newtons method x(i+1) = x(i) - f(x(i)/f '(x(i)), where f(x) is ideally 0. We do this until out f(x(i))=0.

First get one side to equal zero.

y^4 + 2y -1 = 0. Set this to f(y)

f(x) = y^4 + 2y -1. Find f ' (y)

f ' (y) = 4*y^3 + 2

We pick a starting guess. We have to be careful about where we start, since there are possibly 4 solutions to this. We generally pick a starting guess near a known solution, but I will choose an arbitrary one. Lets let y(1) = 2.9234.

Make a table with i, y(i), f(y(i)), f ' (y(i)) to keep you organized. When you get a number close to zero for f(y(i)) you have found a solution. The first ten numbers for y(i) that I got are:

2.92340000000000

2.15934211529859

1.56654223059180

1.09723084156316

0.734255400014375

0.522399429075830

0.475993930037884

0.474627660850473

0.474626617563212

0.474626617562606

And the first ten f(y(i)) that I got are:

77.8854892071084

25.0599998560155

8.15546824575709

2.64387426164646

0.759173027049163

0.119273903190634

0.00332192531592535

2.53276780770584e-06

1.47126755223326e-12

0

Note that the last one is 0, meaning we have found our solution and it is .474626617562606. Note that if we take .4746^4+2*.4746 we get 1.

MATLAB code to do this is:

y(1)=2.9234

for i =1:9

y(i+1)=y(i) - (y(i)^4 + 2*y(i) -1)/(4*y(i)^3 + 2)

end