bisection and newton-raphson methods

Outlines

Bisection and Newton-Raphson Methods

Mike Renfro

September 7, 2004

Mike Renfro Bisection and Newton-Raphson Methods

OutlinesPart I: Review of Previous LecturePart II: Bisection and Newton-Raphson Methods

Review of Previous Lecture


OutlinesPart I: Review of Previous LecturePart II: Bisection and Newton-Raphson Methods


Bisection MethodProblem SetupBisection Method ProcedureBisection Method Advantages and DisadvantagesBisection Method Example

Newton-Raphson MethodProblem SetupNewton-Raphson Method ProcedureNewton-Raphson Method Advantages and Disadvantages

Homework


Part I




Sample problems solved with numerical methods

Natural frequencies of a vibrating barStatic analysis of a scaffoldingCritical loads for buckling a columnRealistic Design Properties of Materials

Solution of nonlinear equations

IntroductionExample: fluid mechanicsIncremental search method


Bisection MethodNewton-Raphson Method

Homework

Part II




Homework

Problem SetupBisection Method ProcedureBisection Method Advantages and DisadvantagesBisection Method Example

Bisection Method

One problem with the incremental search method is its lack ofefficiency in finding a root. If a root is expected on the interval0 < x < 1, it will require between 1 and 10 loops through themethod to bracket the root with 0.1 uncertainty:

Calculate f (0.0),

Calculate f (0.1),

...

Calculate f (1.0)

You may get lucky and only have to perform two functionevaluations. You may be unlucky and have to perform 11evaluations. In general, you’ll probably have to perform 6-7evaluations to find the solution. There has to be a more efficientway to find a solution.



Homework


Problem Setup

Start with a function f (x) andtwo values of x (a and b) suchthat f (a) and f (b) have oppositesigns. These values of a and bmay be the final interval of anincremental search method witha relatively large step size.



Homework


Bisection Method Procedure

Evaluate f (x) at themidpoint of the interval, atxmid = a+b

2 .

If f (xmid) 6= 0, then the signof f (xmid) will match thesign of f (a) or the sign off (b).

If f (xmid) matches thesign of f (a), then seta = xmid and repeat.If f (xmid) matches thesign of f (b), then setb = xmid and repeat.



Homework


Bisection Method Procedure

Eventually, this method will limitthe root of f (x) to a sufficientlysmall interval, or |f (xmid)| ≤ ε,where ε is the error tolerance forthe problem.



Homework


Bisection Method Advantages

Since the bisection method discards 50% of the current interval ateach step, it brackets the root much more quickly than theincremental search method does.To compare:

On average, assuming a root is somewhere on the intervalbetween 0 and 1, it takes 6–7 function evaluations to estimatethe root to within 0.1 accuracy.

Those same 6–7 function evaluations using bisection estimatesthe root to within 1

24 = 0.625 to 125 = 0.031 accuracy.



Homework


Bisection Method Disadvantages

Like incremental search, the bisection method only finds rootswhere the function crosses the x axis. It cannot find rootswhere the function is tangent to the x axis.

Like incremental search, the bisection method can be fooledby singularities in the function.

Like incremental search, the bisection method cannot findcomplex roots of polynomials.



Homework


Bisection Method Example

Find the root of f (x) = x3 − 2 on the interval where a = 1 andb = 2, and ε = 0.05:

f (1) = 13 − 2 = −1, f (2) = 23 − 2 = 6,f (1.5) = 1.53 − 2 = 1.375. Since 1.375 and 6 have the samesign, the root must be between 1 and 1.5.

f (1) = −1, f (1.5) = 1.375, f (1.25) = −0.047. Since| − 0.047| < ε, the root is assumed to be at x = 1.25.

If we used incremental search to find the same root, we wouldhave required 4 function evaluations using a step size of 0.1,followed by 6 function evaluations using a step size of 0.01.



Homework

Problem SetupNewton-Raphson Method ProcedureNewton-Raphson Method Advantages and Disadvantages

Problem Setup

Given: a function f (x), itsderivative f ′(x), a starting pointx1, and an error tolerance ε.We assume that both the heightof the function f (x) and its slopecan help us make a moreeducated guess of the root x∗.



Homework


Newton-Raphson Method Procedure

Draw a line tangent to thefunction at the point (x1, f (x1)).The point where the tangent linecrosses the x axis should be abetter estimate of the root thanx1. Call that point x2.Calculate f (x2), and draw a linetangent to the function at thepoint (x2, f (x2)). The pointwhere the new tangent linecrosses the x axis should be abetter estimate of the root thanx2. Call that point x3.Repeat until |f (x)| < ε.



Homework


Newton-Raphson Method Advantages

Unlike the incremental search and bisection methods, theNewton-Raphson method isn’t fooled by singularities.

Also, it can identify repeated roots, since it does not look forchanges in the sign of f (x) explicitly.

It can find complex roots of polynomials, assuming you startout with a complex value for x1.

For many problems, Newton-Raphson converges quicker thaneither bisection or incremental search.



Homework


Newton-Raphson Method Disadvantages

The Newton-Raphson method only works if you have a functionalrepresentation of f ′(x). Some functions may be difficult toimpossible to differentiate. You may be able to work around thisby approximating the derivative f ′(x) ≈ f (x+∆x)−f (x)

∆x .



Homework



The Newton-Raphson method isnot guaranteed to find a root.For example, if the starting pointx1 is sufficiently far away fromthe root for the functionf (x) = tan−1 x , the function’ssmall slope tends to drive the xguesses further and further awayfrom the root.



Homework



If the derivative of the functionat any tested point xi issufficiently close to zero, the nextpoint xi+1 will be very far away.You may still find the root, butyou will be delayed.



Homework



If the derivative of the functionchanges sign near a tested point,the Newton-Raphson methodmay oscillate around a pointnowhere near the nearest root.



Homework

Homework

Find the solution of f (x) = x2 − 10 = 0 using the followingmethods and starting points:

Bisection method, with a = 1, b = 3, and ε = 0.01.

Newton-Raphson method, with x0 = 0 and ε = 0.01.

Will both of these problem statements yield a solution? If not,what can you do to change the problem setup to allow you to finda solution?


bisection and newton-raphson methods

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