### Tangent Line Approximations and Newton's Method

I haven't exactly been helpful at all. I started something and I slowed down. I guess that's what happens sometimes. I don't know wether to say sorry or not. But here's one of the missing scribes.

Wednesday, November 30, 2005

We learned half of the last section of Chapter 4.

First area it covers was on

If we were given a function.

For example f(x)= square root(x).

And we were to find an approximation value of f(26).

What do we do then?

In this particular example, we find the closest perfect square.

Which happens to be 25.

What if we were given other kind of functions?

You try to find a coordinate that you know, and is close to the value you are finding.

Now we have a set of coordinates (25, 5).

Since 26 is close to 25, we can use a tangent line to find f(26).

Differentiate f(x).

f(x)= x^(1/2)f'(x)= 1/2(x)^(-1/2)

We then find the slope at our given coordinate.

f'(25)=1/2(25)^(-1/2)f'(25)=0.1 or 1/10

Now we have all the "ingredients" to put into our slope-point formula.

y-y1=m(x-x1)

y-5=0.1(x-25)

So basically, that equation up there is all we need to find f(26).

All we have to do now is solve for y when x=26.

In the end we get, y= 5.1 or f(26)=5.1

Let me sum up the steps of

1) Identify the function.

2) Identify what value you have to find.

3) Think of a set of KNOWN coordinate (x, y), close to the value you have to find.

4) Differentiate the function.

5) Find the slope of the tangent line at the KNOWN coordinate.

6) You have the slope and set of coordinate points, plug it into the slope-point formula.

7) Let x=the value you have to find, and solve for y.

8) *optional, but recommended. Do a check. You never know you might have made a mistake somewhere along the way.

That day we were also introcuded to Newton's Method.

Newton's Method is a method of finding zeroes of a function. You are given a function, an interval or point where the xero is close to. Zeroes of Tangent lines are then used to determine if its the zero of the function as well.

At the start we are given f(x) and (Xo, f(Xo)). From the given you can find f'(x) and the derivative at the given point.

y- f(Xo) = f'(Xo)(X-Xo) <----- slope-point formula

Since we are finding a root of the tangent line, we let y=0. From this equation we are trying the find the x value of the root. So we re-arrange the equation.

0- f(Xo) = f'(Xo)(X-Xo)

-f(Xo) / f'(Xo) = x - Xo

Xo - [f(Xo) / f'(Xo)] = x

x would then be the first x value.

When using Newton's Method it may take multiple times till you finally get close to a good approximation of where the zero of the function is. That's why doing this method by hand is very very very very painful. Thank You for our stupid calculators. In our calculators we have a program stored called "NEWT". It is used for this kind of method. We can also do this method manually, as showed during next class.

Hope this was some info. Hopefully is useful for someday reference. Sorry it was very very very very very very late.

Wednesday, November 30, 2005

We learned half of the last section of Chapter 4.

First area it covers was on

**Linear Approximations**.If we were given a function.

For example f(x)= square root(x).

And we were to find an approximation value of f(26).

What do we do then?

In this particular example, we find the closest perfect square.

Which happens to be 25.

What if we were given other kind of functions?

You try to find a coordinate that you know, and is close to the value you are finding.

Now we have a set of coordinates (25, 5).

Since 26 is close to 25, we can use a tangent line to find f(26).

Differentiate f(x).

f(x)= x^(1/2)f'(x)= 1/2(x)^(-1/2)

We then find the slope at our given coordinate.

f'(25)=1/2(25)^(-1/2)f'(25)=0.1 or 1/10

Now we have all the "ingredients" to put into our slope-point formula.

y-y1=m(x-x1)

y-5=0.1(x-25)

So basically, that equation up there is all we need to find f(26).

All we have to do now is solve for y when x=26.

In the end we get, y= 5.1 or f(26)=5.1

Let me sum up the steps of

**Linear Aprroximation**.1) Identify the function.

2) Identify what value you have to find.

3) Think of a set of KNOWN coordinate (x, y), close to the value you have to find.

4) Differentiate the function.

5) Find the slope of the tangent line at the KNOWN coordinate.

6) You have the slope and set of coordinate points, plug it into the slope-point formula.

7) Let x=the value you have to find, and solve for y.

8) *optional, but recommended. Do a check. You never know you might have made a mistake somewhere along the way.

That day we were also introcuded to Newton's Method.

Newton's Method is a method of finding zeroes of a function. You are given a function, an interval or point where the xero is close to. Zeroes of Tangent lines are then used to determine if its the zero of the function as well.

At the start we are given f(x) and (Xo, f(Xo)). From the given you can find f'(x) and the derivative at the given point.

y- f(Xo) = f'(Xo)(X-Xo) <----- slope-point formula

Since we are finding a root of the tangent line, we let y=0. From this equation we are trying the find the x value of the root. So we re-arrange the equation.

0- f(Xo) = f'(Xo)(X-Xo)

-f(Xo) / f'(Xo) = x - Xo

Xo - [f(Xo) / f'(Xo)] = x

x would then be the first x value.

When using Newton's Method it may take multiple times till you finally get close to a good approximation of where the zero of the function is. That's why doing this method by hand is very very very very painful. Thank You for our stupid calculators. In our calculators we have a program stored called "NEWT". It is used for this kind of method. We can also do this method manually, as showed during next class.

Hope this was some info. Hopefully is useful for someday reference. Sorry it was very very very very very very late.

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