Monday

October 20, 2014

October 20, 2014

Posted by **Tracy** on Tuesday, September 12, 2006 at 5:34pm.

The secant method of finding a root of any function y=f(x) is illustrated in the figure below. Given two points p1 =(x1 , y1 ) and p2 =(x2 , y2 ) on the curve y=f(x), it finds a point p3 =(x3 , y3 ) which is a better approximation to the root as follows: Let x3 be the intersection of the secant line through p1 and p2 with the x-axis, and let y3 =f(x3 ).

**Basically the graph looks like a half of a cosine curve with 3 points labeled and a secant line drawn between the first 2 that intersects the x axis**

Suppose that

f(x) = x^2 - 2

x1 = 3, x2 = 2

Now I know that based on this you can find the coordinates of the first 2 points, and then a slope but after that I am just....drawing a blank.

Thanks again for any of your help.

Well slope is change in y over change in x

The secant line through p1,p2 = (f(x2)-f(x1))/(x2-x1)= (2-7)/(2-3)=-5/-1=5

I think this is what you have.

The equation of that line through either point is y=5x-8 and it intercepts the x-axis at 8/5

Now I'm not entirely sure what the question is...

One thing you should know is that Newton's formula is the limiting form for the secant method. The idea there is to project down to the x-axis along a secant line, evaluate the corresponding function value, determine another secant line, etc. until the secant line passes through the function's root. If you work a couple iterations it shouldn't be too hard to see what's happening.

Hmm, I don't really understand what your saying. I know the question is that I have to find out what the point P3 actually is, I need to find the coordinates, I just have no idea how. I don't really dont understand waht your saying about it though.

Find the equation of the line through p1 and p2. Then find the intersection of that line with the x-axis. If I did the arithmetic correctly it's 8/5 so x3=8/5 and y3=f(8/5)=(8/5)^2-2=.56 so p3 is (1.6,.56). Now repeat the steps using p2 and p3. eventually you'll get real close to the aqrt(2). That's what we're trying to find for this problem:f(x)=0 or x^2=2.

Does that help?

Okay I see that part. But why would I do it again for P2 and P3 because don't we already know those points since they give us what the first coordinate is and we know the equation?

Ohhh okay. I see now that that is exactly how I did the question and then didn't connect that that was the answer, but I understand now. Thanks alot! So that means that x3=8/5 and y3=14/25.

Yes, the secant method is a root finding algorithm and the question was just asking for the coordinates of the next iteration.

You're welcome.

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