There are two ways that we can generalize the example of Taylor's theory for `e^x`.

(1) Use a function other than `e^x`.
(2) Investigate the function at a point `x=a` different from `0`.

Definition IX.B.1: A function `f ` is called `C^{oo}` ("C- infinity") on an open interval if for every natural number `n`, `f^{(n)}(x)` exists for each `x` in the interval, i.e.,`f ` has derivatives of order `1,2,3,4,...` on the interval.

ASSUMPTION: FOR THE REMAINDER OF THIS SECTION, ASSUME THAT `f ` IS A  `C^{oo}` FUNCTION ON AN OPEN INTERVAL I THAT CONTAINS 0.
Proposition IX.B.2: Let `P_n(x) = f(0)+f'(0)x + {f''(0)}/2 x^2 +{f'''(0)}/{3!} x^3+...+{f^{(n)}(0)}/{n!} x^n`. Then
`P_n(0)=f(0), P_n'(0)=f'(0), P_n''(0)=f''(0)` ,... , and `P_n^{(n)}(0)=f^{(n)}(0)`.

Proof:. `P_n'(x) = f '(0)+f ''(0)x + { f '''(0)}/2 x^2 + ...+{f^{(n)}(0)}/{(n-1)!} x^(n-1)`. So `P_n'(0)=f '(0)`.

The other equations are demonstrated similarly. EOP.

Definition: Let `P_n(x,f) ` denote the polynomial of the last proposition, so

`P_n(x,f) `  is called the Maclaurin's Polynomial of degree n for `f`.

Remark: With this notation we have shown in the proof above that the derivative of the Maclaurin's Polynomial of degree `n` for`f` is the Maclaurin's Polynomial of degree `n-1` for the derivative of `f`, that is,`P_n'(x,f)= P_{n-1}(x,f')`. 

Futhermore, since `P_n^{(n)}(x,f)=f^{(n)}(0), P_n^{(n+1)}(x,f)=0` for all x. Thus `P_n(x,f)` actually solves the differential equation `g^{(n+1)}(x)=0` with initial conditions `g(0) = f(0), g'(0) = f '(0), g''(0)=f''(0)`, ... , and `g^{(n)}(0)=f^{(n)}(0)`.

Example IX.B.1: Find the Maclaurin's polynomial of degree 4 for `f(x)=e^{3x}`.

Solution: This is a matter of some systematic use of the calculus of derivatives.
`f(x)=e^{3x}` so `f(0)=1`.
`f'(x) = 3e^{3x}` so `f '(0) = 3`.
`f''(x)= 9e^{3x}` so `f ''(0) = 9`.
`f '''(x)=27e^{3x}` so `f '''(0) = 27`.
`f^{(iv)}(x) = 81e^{3x}` so `f^{ (iv)}(0) = 81`.
Here is a table that can help organize the work.
 

Therefore `P_n(x,e^{3x}) = 1+3x + 9/2 x^2 +27/6 x^3+81/24 x^4 = 1 +3x+9/2x^2+9/2x^3+27/8x^4`.

Proposition IX.B.3: (Corollary) If  `f ` is a polynomial function of degree `m` on an open interval I that contains `0` with `f(x) = a_0 + a_1x+a_2x^2 + ... +a_kx^k+ ... +a_mx^m` then 
for  `0<= n<=m,   P_n(x,f) = a_0 + a_1x+a_2x^2 + ... +a_kx^k+ ... +a_nx^n` and in particular `f(x) = P_m(x,f)` for any `x` in I.

Comment: This result merely notes that the Maclaurin Polynomial of degree `n` for a poynomial `f` corresponds to the sum of the terms of `f` of degree less than or equal to `n`.

Proof Outline:  For the details of the proof here apply the result of Proposition IX.B.2.

Theorem IX.B.1: If `f` is a `C^{oo}` function on an open interval I that contains `0` then `f(b)` is approximately `P_n(b,f)` for any `b` in I, and if
`R_n(b)=f(b)-P_n(b,f)` then for some `c` between `0` and `b`, `R_n(b)= {f^{(n+1)}(c)}/{(n+1)!} b^{(n+1)}`.

Remark: You should check for yourself that when `f(x) = e^x`, this is merely a restatement of Proposition IX.A.1 .

Proof: Let  `g(t) = f(t)+f'(t)(b-t) + {f''(t)}/2 (b-t)^2+...+{f^{(n)}(t)}/{n!}(b-t)^n + {R_n(b)}/{b^{(n+1)}}(b-t)^{(n+1)}` where we assume `b` is not `0` and `t` is in the interval I.

Then `g(0)=P_n(b,f) + R_n(b)= f(b)` and `g(b)= f(b)` as well. Thus, [using the Mean Value Theorem or Rolle's Theorem] for some `c` between `0` and `b` we have `g'(c) = 0`. But, after extensive use of the product rule, we find that

so `g'(t)={f^{(n+1)}(t)}/{n!}(b-t)^{n}-{R_n(b)}/{b^{n+1}}(n+1)(b-t)^{n}`. 

Now using `t = c` and `g'(c) = 0` we can conclude that

Example IX.B.2: Find `P_6(x,cos(x))`. Use this polynomial to estimate `cos(1)`.

Solution: Let `f(x) = cos(x)`. Then it is easy to verify that
`f(0) = 1`,  `f '(0) = 0`, `f ''(0)= -1`,   `f '''(0)=0`,  `f ^{iv}(0) =1`,  `f ^v(0) = 0`, and `f^{vi}(0) =-1`.
Thus `P_6(x,cos(x))=1 - {x^2}/2 + {x^4}/24 - {x^6}/720`.

The following graph visualizes both the cosine function and the polynomial `P_6(x,cos(x))`. Notice how close these graphs are when `x` is close to `0`.

We estimate `cos(1)` by evaluating `P_6(x,cos(x))` when `x = 1`, so we find that

Applying the remainder part of the theorem to this estimate allows us to say that
`R_6(1) = {f^{vii}(c)}/{7!}1^7={sin(c))/5040` where `0 < c < 1`. But since `0 < c < 1`, we have that `0 < sin(c) < 1`, so that `0<R_6(1)<1/5040`.

In conclusion we have shown that `389/720 < cos(1)<389/720 + 1/5040`. Here is a spreadsheet showing this computation and one using `n=14` as well.

Example IX.B.4: Find `P_4(x,ln(1-x))`. Use this polynomial to estimate `ln(0.9)`.
Solution: Let `f(x) = ln(1-x)`. This is a matter of some systematic use of the calculus of derivatives.
`f(x)=ln(1-x)` so `f(0)=0`.
`f'(x) = -1/{1-x}=-(1-x)^{-1}` so `f '(0) = -1`.
`f''(x)= -(1-x)^{-2}` so `f ''(0) =- 1`.
`f '''(x)=-2(1-x)^{-3}` so `f '''(0) = -2`.
`f^{(iv)}(x) = -6(1-x)^{-4}` so `f^{ (iv)}(0) = -6`.
Here is a table that can help organize the work.

Exercises IX.B: (Solutions 1-7 odd)

In exercises 1-10 , find `P_n(x,f(x))` for the specified function and `n`.

1. `f(x) = sin(x)` ; `n = 7`
2. `f(x) = cos(x)` ; `n = 6` and `n = 8`
3. `f(x) = sin(2x)` ; `n = 7`
4. `f(x) = cos(2x)` ; `n = 6` and `n = 8`
5. `f(x) = e^{2x}` ; `n = 6` and `n =7`
6. `f(x)=1/{1+x}` ; `n = 3`
7. `f(x) = ln(1+x)` ; `n = 4`
8. `f(x)=` arctan`(x)`; `n = 3`
9. `f(x) = 3 + 5x^2-3x^4+x^5` ; `n = 2,3,4,5,` and `6`.
10. `f(x) = e^{-x}` ; `n = 6`
11. Sketch the graph of `P_n(x,f(x))` together with the graph of `f(x)` on the same axes when `f(x) = sin(x)` and `n = 1,3,5,` and `7`.
12. Sketch the graph of `P_n(x,f(x))` together with the graph of `f(x)` on the same axes when `f(x) = cos(x)` and `n = 2,4,` and `6`.

13 - 22. Using the polynomials found in problems 1-10, estimate the value of `f(1)` and discuss the related error term `R_n`.
23. Estimate `int_0^1 sin(t^2) dt` using `P_5(x,sin(x))`. Discuss the error in this approximation.
[Hint: Review the work done in the last section for `int_o^1 e^{-t^2}dt`.]