Newton's Method (1 Viewer)

[dawso]

Gday, i got this question from a past paper so dont hav any solutions, dont rele get what some of it is asking for so may as well post it up, how would u answer this in an exam?:

Given the two curves y = 2 cos (x/2) and x - y = 0

a) Sketch both curves on the same axis for domain 0<=x<=pi (u can sketch it for your own reference but this bit i understand....)

b) choosing x=1 as a suitable first approx. use Newton's Method to establish an x value for the intersection which is correct to 1 decimal place

c) Explain how you know that your solution is correct to 1 decimal place

d) Explain why the solution you have found represents the only intersection for the two graphs

e) State the point of intersection of the graphs correct to 1 decimal place

thanks in advance

-dawso

 

[shafqat]

Gday, i got this question from a past paper so dont hav any solutions, dont rele get what some of it is asking for so may as well post it up, how would u answer this in an exam?:

Given the two curves y = 2 cos (x/2) and x - y = 0

a) Sketch both curves on the same axis for domain 0<=x<=pi (u can sketch it for your own reference but this bit i understand....)

b) choosing x=1 as a suitable first approx. use Newton's Method to establish an x value for the intersection which is correct to 1 decimal place

c) Explain how you know that your solution is correct to 1 decimal place

d) Explain why the solution you have found represents the only intersection for the two graphs

e) State the point of intersection of the graphs correct to 1 decimal place

thanks in advance

-dawso

To find the pt of intersection you have to solve the equation

2 cos (x/2) = x
Therefore let f(x) = 2 cos (x/2) - x, and bash away

 

[shafqat]

for b, c, keep using newtons until successive apps agree to one decimal place, although this isn't strictly an error-proof method
show d graphically

 

[dawso]

so if 2 successive results have the same value for the first dec place, that is enough 2 say it is correct to 1 dec place???

also, another question shows that newton's gives an approximate root of 2.3456, but says "the actual root is exactly 2.3000, find the percentage error of the approximation"

with this, u obviously have to work out how far it is out (0.0456) but is this then divided by 2.3456, or by 2.3....does anyone get me? i think i get myself :S

anyway, ive seen this in a few situations and it is interpreted differently each time, and the syllabus is no help here....

any suggestions?