Solve cosx+3-2x=0, correct to three decimal places by using fixed point iteration method.
Answers
Answer:
What is a positive root of the equation cos(x)-3(x) +1=0 by using the iteration method?
Step-by-step explanation:
Two points to begin with:
(1) You need a good starting point. You need a good initial guess for x.
(2) You need to edit your question and specify how much accuracy is required; 1 decimal place? 2 decimal places? 50 decimal places? Yes, it matters. Otherwise your question is too vague. The answer will be a numeric approximation, so accuracy has to be specified up front.
Usually the iterative method is based on getting the part that is linear in x isolated first:
cos(x)+1=3xcos(x)+1=3x
Then make it just “x” (divide by 3):
i. x=cos(x)+13x=cos(x)+13
Now you need to get your initial guess. I will call this value x0x0 . I choose 0, I will let you decide if that is a good choice.
On the right side of Equation i. substitute x0x0 . Call the result on the left side x_1 , like this:
x1=cos(x0)+13=23x1=cos(x0)+13=23
If you use a calculator that is fine, you will have 0.666666667 or something. Let’s assume, for example, that you are supposed to find x to 3 decimal places accuracy. Then do not round this result to three decimal places. I personally would keep 5 or 6 for now. I choose to keep 6. x1=0.666667x1=0.666667 .
Now we “iterate” the process by substituting x1x1 on the right side and calling that answer x2x2 :
x2=cos(x1)+13x2=cos(x1)+13
x2=cos(0.666667)+13=0.595296x2=cos(0.666667)+13=0.595296
Keep iterating, define your new answer with the next subscript, and use your previous answer on the right side:
x3=cos(0.595296)+13=0.609328x3=cos(0.595296)+13=0.609328
I calculated up to x4x4 in Excel:

I am not going to finish for you. You finish the work.
If you keep going (Hint: at least 4 more steps) you should notice that the two numbers will start to agree to 4 and 5 decimal places. Because of rounding you cannot stop just because they agree to three. Make sure they agree to at least 4, and make sure they don’t change in the 4th place, if you want three decimal places accuracy. For example, if the fourth place stabilized at “8”, you must round the third place up.