Asked by Devin
                Find all root of ln(4-x^2)=x correct to 8 decimal places. List all successive approximations.
            
            
        Answers
                    Answered by
            Reiny
            
    by definition:
4 - x^2 = e^x
by making a rough sketch of
y = 4-x^2 and y = e^x, I can see that there are roots at appr. 1 and -2
setting up for Newton's Method:
let y = 4 - x^2 - e^x
y' = -2x - e^x
His method says:
newx = x - y/y' = x - (4-x^2 - e^x)/(-2x - e^x)
= x + (4-x^2 - e^x)/(2x+e^x)
so start with x = 1
newx ---- x
???? ----- 1
1.059707788 ---1.059707788
1.058007813 --- 1.058007813
1.058006401 --- 1.058006401
1.058006401 YEahhhhh! correct to 9 decimal places
start with x = -2
------ -1.9649813...
-1.964635631 --- -1.964635631
-1.964635597 --- -1.964635597
-1.964635597 Yeah again,
x = 1.058006401 or x = -1.964635597
    
4 - x^2 = e^x
by making a rough sketch of
y = 4-x^2 and y = e^x, I can see that there are roots at appr. 1 and -2
setting up for Newton's Method:
let y = 4 - x^2 - e^x
y' = -2x - e^x
His method says:
newx = x - y/y' = x - (4-x^2 - e^x)/(-2x - e^x)
= x + (4-x^2 - e^x)/(2x+e^x)
so start with x = 1
newx ---- x
???? ----- 1
1.059707788 ---1.059707788
1.058007813 --- 1.058007813
1.058006401 --- 1.058006401
1.058006401 YEahhhhh! correct to 9 decimal places
start with x = -2
------ -1.9649813...
-1.964635631 --- -1.964635631
-1.964635597 --- -1.964635597
-1.964635597 Yeah again,
x = 1.058006401 or x = -1.964635597
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