Given information:
The equation is given as:
x2(4−x2)=4x2+1 .
Formula used:
Newton’s Method:
We seek a solution of f(x)=0 , starting from an initial estimate x=x1 .
For x=xn , compute the next approximation xn+1 by
xn+1=xn−f(xn)f'(xn) and so on.
Calculation:
Consider the equation,
x2(4−x2)=4x2+1
The initial approximation x1 can be guessed by sketching both the graphs
y=x2(4−x2) and y=4x2+1 .
Both the graph intersect at four points whose x-coordinate can be seen in graph.
Let x1=0.843 where
f(x)=−x6+3x4+4x2−4f'(x)=−6x5+12x3+8x
Now, at n=1
x2=x1−f(x1)f'(x1)
x2=x1−{−x16+3x14+4x12−4−6x15+12x13+8x1}
x2=(0.843)−(−(0.843)6+3(0.843)4+4(0.843)2−4−6(0.843)5+12(0.843)3+8(0.843))x2=(0.843)−(−0.35889338+1.515066004+2.842596−4−2.554401282+7.188925284+6.744)x2=(0.843)−(−0.00123137611.378524)
x2=0.843−(−0.0001082193086)x2=0.843+0.0001082193086x2=0.843108219
The second approximation is x2=0.843108219 .
Let n=2 ,
x3=x2−f(x2)f'(x2)
x3=x2−{−x26+3x24+4x22−4−6x25+12x23+8x2}
x3=(0.843108219)−(−(0.843108219)6+3(0.843108219)4+4(0.843108219)2−4−6(0.843108219)5+12(0.843108219)3+8(0.843108219))x3=(0.843108219)−(−0.359169903+1.515844132+2.843325876−4−2.556041292+7.191694246+6.744865752)x3=(0.843108219)−(0.00000010511.38051871)
x3=0.843108219−(0.00000000922629299)x3=0.843108219−0.00000000922629299x3=0.843108209
The third approximation with six decimal places is x3=0.843108209 .
Therefore, the roots with eight decimal places are :-
x2=0.84310821
x3=0.84310820