Halving The Interval (1 Viewer)

[Kutay]

Hey i was wondering if someone could help me understand the Halfving the Interval Method of Approximation.... i dont seem to be geting it !

Thanks in advance kutay

 

[maths > english]

1. put the equation into the form f(x) = 0

2. obtain 2 estimates of the root x₁ and x₂ such that f(x₁) < 0 and f(x₂) > 0 (therefore a root exists between x₁ and x₂)

3. find x₃ = ¹/₂ * (x₁ + x₂)

4. if x₃ < 0 the root is between x₃ & x₂ or if x₃ > 0 the root is between x₁ & x₃

5. repeat steps 3 & 4 until the solution is accurate enough

e.g. sin x = ¹/₃

1. sin x - ¹/₃ = 0

2. sin 15 - ¹/₃ < 0 & sin 30 - ¹/₃ > 0 therefore x₁ = 15 and x₂ = 30

3. x₃ = ¹/₂ * (15 + 30) = ⁴⁵/₂

 

[withoutaface]

Hey i was wondering if someone could help me understand the Halfving the Interval Method of Approximation.... i dont seem to be geting it !

Thanks in advance kutay

1. Take an estimate interval and verify that one bound is positive, the other negative.
2. Take a value half way between the two bounds and find its sign, if it is positive then the new interval is between that and the negative bound, if it is negative then etc
3. Repeat until you get a sufficiently small interval.

 

[clive]

*you probably should make sure that the function is continuous on the estimate interval first. (probably wouldn't be asked at HSC level though)

eg. f(x) = 1/x does not have a root on the interval [-1,1] even though it satisfies f(-1) = -1 and f(1) = 1 as it is discontinuous at x = 0.

 

[gordo]

lol y are u doing maths at 2 am on the weekend

 

[withoutaface]

Because year 12 maths is easier than non-linear equations in matlab.

 

[Kutay]

Thankyou heaps everyone it really helped in my understanding