Taylor Polynomials

Definition: The degree n Taylor polynomial of the function f at the point x = a is

We will let denote the k ^th derivative of f with respect to x.
We will also focus on the special case where a = 0.

Example

Here we have defined a function t(n, z) of two variables, and n is the degree of the polynomial.

You can use the applet above to see the values of other polynomials.

Try it yourself ...

Click the mouse on the 5 in the t(5, 0.1) above. A box with a blinking cursor should appear around the 5. Use the right arrow, if necessary, to put the cursor to the right of the 5, then backspace to delete the 5 and type 7. Now reevaluate the function by typing <ctrl> z. (If the box is around more than just the 5, then use the up, right and left arrow keys to "navigate" in the formula until only the 5 is boxed.)
If you evaluate the degree 9 polynomial you will get the same value that is reported for sin(0.1). These two values are not actually equal, but when rounded to 15 decimal places, they agree.
Question: Why did I use only odd degrees? Try evaluating t(4, 0.1) and compare with t(3, 0.1).

Now we will take a closer look at the polynomials. In the applet below I defined a vector v with 8 components. Initially each component is set equal to 0. Next I set the k^th component of v equal to the k^th derivative of f. Notice the pattern. Finally, I redefined the components of v to be the derivatives evaluated at x = 0. Now you see why I used only odd degree polynomials. The even degree terms have coefficient 0.

So far we have compared sin(x) and it's Taylor polynomials only at x = 0.1. What happens with other values of x?

Try it yourself ...

Go back and replace 0.1 with 1 and reevaluate the functions with <ctrl> z.
At x = 1, the degree 9 Taylor polynomial agrees with sin to only 5 decimal places.
Try x = 2, and you will see that the degree 9 polynomial is not even close to sin(2).

The Taylor polynomials at x = a can effectively approximate a function for values of x that are near a. In our example, a is 0, and the Taylor polynomials are very close to sin at x = 0.1, but very far off at x = 2.

Looking at the graphs will help to illustrate this point.

Here I have defined a polynomial taylor of just one variable. The n that appears in the formula for taylor is the degree of the polynomial (when n is odd), and it is initially set equal to 3.

The graph shows the the sine function f and the degree 3 taylor polynomial of f at 0.

Clearly, taylor is quite close to f near zero. However, away from 0 the values of the polynomial and f are not close.

You can use the applet to change the value of n and then graph the new polynomial.

Try it yourself ...

Click on the 3 to select that node of the formula. As before, use the arrow keys as necessary to box only the 3 and position the cursor to the right of the 3. Now backspace and replace the 3 with a 5. You must reevaluate the n formula by typing <ctrl> z. After you do so, the formula will be n:=5=5. This is the applet's way of indicating that it has registered the new value of n.
You do not need to reevaluate the formula for taylor. Click the update button to the right of the graph to show the new polynomial.
You can drag the mouse in the graph window to change the view. Either right-click in the graph window, or click the window button to get zoom options.
Question: What does the degree one Taylor polynomial look like?