The constant polynomial P₀ is just f(0), which is 1. However, when you ask your helper application for P₀, it may respond 1 + O(x). This means that, for small values of x, the "error" [i.e., the difference between P₀ and f(x)] is roughly proportional to x.

Similarly, when you ask for P₁(x), you may see 1 + x + O(x²) instead of just 1 + x. This means that P₁(x) actually is 1 + x, and (again, for small x) the error in approximating f(x) is roughly proportional to x². To create a formula for a Taylor polynomial, you need to delete the "O" term. One way to do this is to highlight and copy all of the output formula except that term, and then paste wherever you need the formula.

1. Enter the formula for f(x) in your worksheet, and generate the first five Taylor polynomials, P₀, P₁(x), P₂(x), P₃(x), and P₄(x). Record these in your worksheet.

2. Guess the general form for the n-th degree Taylor polynomial, P_n(x), for f(x). Check your answer by calculating P₅(x). If your answer does not agree with your conjectured form, change your conjecture, and verify your new conjecture by calculating P₆(x).

3. Ask your computer algebra system to calculate the sum 1 + x + x₂ + ^... + xⁿ in the form
   

   Explain the following formula:

   

   [Hint: Consider the expression you obtained in step 2 for P_n(x). Multiply by x to obtain an expression for x P_n(x), and subtract the result from the expression for P_n(x).]

The sum in step 3, in which each term is x times the preceding term, is called a geometric sum. The fact that we can also write this sum as the quotient of two (short) polynomials is very important. We will put it to use in the next part. Since a geometric sum with x as ratio of each term to the preceding one is also a polynomial in x, it is sometimes called a geometric polynomial.