November 30, 2006

Scribe Post - Day 53

So for today's class I chose to volounteer myself to do the scribe in place for Danny, because who would want to do it on their birthday? Not me! So a big happy birthday to Danny!

Anyway back on topic... We started off class with a discussion about our Pedro the Pi-Rate Panda Project, and agreed on certain criteria we are going to follow. Then we were told about the wonderful site Mr.K and Mr.H have put together called, which has many tools to make a students life much easier online, so click on it and check it out. After we talked about a couple of math contests that offer pretty decent prizes. And lastly we started with the real math.

We began to learn about Newton's Method. Newton's Method is used to approximate the root of any given function.

On the function f(x) = x3 -2x2 -5 for example,


the function graphically will look sort of like this.

Then you'd pick a value that you suspect is close to the root. In this example we see that the root is somewhere between x=2 to x=3, and closer to x=3. We chose x=2 for our chosen value, and would find the root of its tangent line at that particular point to try to zero in on our zero of the parent function.


To do this we must find the derivative of our given function which turns out to
f'(x) = 3x2 - 4x.

Just plug in the value 2 in both our derivative and parent function for
f'(2) = 4 and f(2) = -5.

Then we would use our point slope formula to determine any equation for any tangent line at a point which looks like f(x) - f(a) = f'(a)(x-a).
And a is the point we're looking at. And since we're looking for the root of the tangent line, f(x)=0 because roots exists only when the y-value is equal to 0.
Now we would just plug in our values to get 0 - (-5) = 4(x-2) and solve for x.
Which then equals to x = 13/4 which is the root of our tangent line at f'(2).

After we use this new number and repeat the process, finding yet another root of a tangent line at that point, and in this case, the root of the tangent line of where f'(13/4). So we would plug it in again in the same formula which generally looks like - f(a)/f'(a) + a = x after you balance the equation. We would repeat the process until the estimate generally stays the same.

On a calculator it is very simple to do, but you must must must know how to do the Newton's Method by hand, as it will probably show up on the exam. On the calculator though you can do this by first:
- plug in both the parent function and derivative function into Y1 and Y2 respectively
- 2nd quit to the homescreen
- compute in -Y1(2)/Y(2) + 2 and store the value in alpha a
- then 2nd enter for the same equation and press enter again, each time for a better estimate
- noticeably the value should tend to remain the same

And that is Newton's Method, which could pretty much be used for any kind of function, you name it.

However it is very important as to which number you choose to pick first to zero in on the root. It isn't the best method if you do not pick the right number as it could just put you in a circle of incorrect roots like as seen in some of these applications here.

And that comes to a closure of my scribe post. I will definitely spice things up and include graphs for a better picture. But right now I had just came back from my volounteer service, and I'm still in my outside clothes, starving to death without dinner. It's getting late, and I still have other homework to do. Sorry for any inconviences, I will fix it up tomorrow night after work. I have covered pretty much everything though, and if not you can tell me =).

Okay next scribe will be Danny because he said he'll do it =].

1 comment:

danny said...

thank you so much manny, awesome stuff man!