Complete two iterations of Newton’s method for the function using the initial guess.

Question

4.) f(x)=x⁵-5, x₁= 1.8

5.) f(x)= cos x, x₁= 1.6

Answer 1

Great question Caleb! Let's make this as simple as possible!

Newton's Method, as you might already know, approximates the x-value at which the function has f(x) = 0.

We know, by linear approximation, that f(x2) - f(x1) = d/dx( f(x1) ) * (x2 - x1). Because we want f(x2) = 0, and in order to get the new x value (which is x2 by the way), we rearrange the equation to get:

x1 - ( f(x1) / f ' (x1) ) = x2

Let's try #4 first.

Iteration #1: If we let x1 = 1.8, then f(1.8) = 13.89568 and f ' (1.8) = 5 * (1.8) ^ 4 = 52.488

Then, x2 = 1.8 - ( 13.89568 / 52.488 ) = 1.5352599....

Iteration #2: Use the answer we just got as x2 as x2 = 1.5352599...

f(x2) = 3.5292234.... and f ' (x2) = 27.777....

Therefore, x2 - ( f(x2) / f ' (x2) ) will be about 1.41 >>>>>>>>>>> (ANSWER for #4)

#5: (it's not that different from #4, just a different function)

Iteration #1: x1 = 1.6, then f(x1) = -0.0292.... and f ' (x1) = -0.99957...

Then, x2 = 1.6 - ( 0.0292..../-0.99957... ) = 1.57079...

Iteration #2: If x2 = last answer we just got, and f(x2) = 8.3 * 10^(-6) and f ' (x2) = essentially -1, then

x2 - ( f(x2) / f ' (x2) ) = 1.571 >>>>>>>>>>>>>>>>>>>>> (ANSWER for #5) Hope this helps!