Use Newton’s Method to approximate the positive zero of 2 sin x = x − 2 correctly to six decimal places

Question

Use Newton’s Method to approximate the positive zero of 2sinx = x − 2 correctly to six decimal places

Answer 1

Out of curiosity, I wrote a short Basic program based on

f(x) = 2sin(x)-x+2

f'(x) = 2cos(x)-1

x_i+1 = x_i - f(x)/f'(x)

I decided to start x₀ at π and let it loop 10 times

run

1 3.14159265358979 2.76106176905986

2 2.76106176905986 2.75467908751146

3 2.75467908751146 2.75467375425005

4 2.75467375425005 2.75467375424629

5 2.75467375424629 2.75467375424629

6 2.75467375424629 2.75467375424629

7 2.75467375424629 2.75467375424629

8 2.75467375424629 2.75467375424629

9 2.75467375424629 2.75467375424629

10 2.75467375424629 2.75467375424629

OK

list

10 fnf(x)=2*sin(x)-x+2

20 fnd(x)=2*cos(x)-1

30 fnn(x) = x - fnf(x)/fnd(x)

40 x= pi

50 for i = 1 to 10

60 y = fnn(x)

70 print i,x," ",y

80 x = y

90 next i

OK

So it pretty well settles down after 4 iterations to 2.75467375424629

Answer 2

for 2sinx=x-2

I can look at when does f(x)=sinx-x/2+1 equal zero.

Newton's method says x₂=x₁-(f(x₁)/f'(x₁) (theory is mean value theorem with f(x₂)=0 and f'(x_c) set approximately to f'(x₁)

careful in that f'(x)=cosx-.5, and each iteration f(x₁) must be calculated sinx₁-x₁/2+1

Looking at graph, first approximation is 2.8 which is not accurate.

my x₂ calculates to 2.7549 (my calculator isn't set up for 6 digits)

I set x₁=2.7549

my x₂ calculates to2.7547 and I cannot make the answer more accurate.