© 2000 M. Flashman

Suppose we have already found `P_n(x,f (x))` and `P_n(x,g(x))` for two `C^{oo}` functions `f` and `g`. Can we use these polynomials to find other polynomials for some related functions? The answer is frequently "yes," and the next proposition and examples illustrate some of these situations.

PROPOSITION IX.C.1: Suppose `f` and `g` are `C^{oo}` functions, `n` is a natural number, and `k` is a constant.

(a) `P_n(x, f (x)+g(x)) = P_n(x, f (x)) + P_n(x,g(x))`

(b) `P_n(x,k f(x)) = k P_n(x,f(x))`

(c) `P_n'(x, f(x)) = P_{n-1}(x, f '(x))`

(d) `P_n(x,x f (x) ) = xP_{n-1}(x, f(x))`

Proof: (a) Recall that `D^n (f (x) + g(x)) = D^n f (x) + D^n g(x)`.

The result follows immediately from observing the coefficients of the polynomials.

`P_n(x,f(x))=f(0)+f'(0)x+{f''(0)}/2x^2+...+{f^n(0)}/{n!}x^n`,

`P_n(x,g(x))=g(0)+g'(0)x+{g''(0)}/2x^2+...+{g^n(0)}/{n!}x^n`

and

`P_n(x,f(x) +
g(x))=f(0)+g(0)+(f'(0)+g'(0))x+{f''(0)+g''(0)}/2x^2+...+{f^n(0)+g^n(0)}/{n!}x^n`

(b) This is left as an exercise .

(c) See the earlier discussion on the notation for MacLaurin's polynomials.

(d) This is left as an exercise.

(a) `f (x)= e^x + cos(x)`

(b) `f (x) = -5 cos(x)`.

Solutions: We use the polynomials already developed in IX.B.

`P_4(x,e^x) = 1+ x+{x^2}/2 + {x^3}/6 +{x^4}/24`

`P_4(x,cos(x)) = 1-{x^2}/2 +{x^4}/24`

(a) Using part (a) of the proposition we have `P_4(x,e^x +cos(x)) = 2+x+{x^3}/6 + {x^4}/12`.

(b) Using part (b) of the proposition we have

`P_4(x,-5cos(x)) = -5 P_4(x,cos(x)) =-5( 1-{x^2}/2 + {x^4}/24) = -5 + 5{x^2}/2 - 5{x^4}/24`.

EXAMPLE IX.C.2 : Find `P_n(x,f (x))` for the following functions with `n` as indicated:

(a) `f(x) = x cos(x)`, `n = 5`. (b) `f(x) = sin(x)`, `n = 5` (c) `f(x) = cos(x)` , `n = 6`.

Solutions: (a) Using part (d) of the proposition we have

`P_5(x,x cos(x)) = x P_4(x,cos(x)) = x (1 - {x^2}/2+ {x^4}/24) = x - {x^3}/2 + {x^5}/24`.

(b) Using part (c) of the proposition we have `P_5'(x,sin(x))
= P_4(x,cos(x)) = 1 - {x^2}/2 + {x^4}/24` .

Therefore `P_5(x,sin(x))` is an antiderivative for `1 - {x^2}/2
+ {x^4}/24`.

So `P_5(x,sin(x)) = x - {x^3}/6 + {x^5}/120
+ C`.

But `P_5(0,sin(x)) = sin(0) = 0` so `C = 0` and `P_5(x,sin(x))
= x - {x^3}/6 + {x^5}/120`.

(c) Again using part (c) of the proposition we have

`P_6 ' (x,cos(x)) = P_5(x,-sin(x)) = -x + {x^3}/6 - {x^5}/120`.Therefore `P_6 (x,cos(x))` is an antiderivative for `-x + {x^3}/6
- {x^5}/120`.

So `P_6(x,cos(x))= -{x^2}/2 +{x^4}/24-{x^6}/720 + C`.

But `P_6(0,cos(x)) = cos(0) = 1` so `C = 1` and `P_6 (x,cos(x))=
1 - {x^2}/2 + {x^4}/24 -{x^6}/720`.

**Remark:** This last example illustrates how to get the
MacLaurin's
polynomials for the sine and the cosine using simple integration with
the
facts that `sin(0)= 0` and `cos(0)=1`.

The next example illustrates how ** substitution ideas**
can
be used as well to find MacLaurin polynomials.

**Example IX.C.3.** Find MacLaurin polynomials for the following
functions:

(a) `f (x) = cos (5x)`

(b) `f (x) = cos (x^3)`

**Solution:** We start with `P_4 (x,cos(x)) = 1 - {x^2}/2+
{x^4}/24` and make the "obvious" substitutions.

(a) Substitute `5x` for `x` to obtain

`P_4(x,cos(5x)) = 1 - {(5x)^2}/2 + {(5x)^4}/24 = 1 - 25 {x^2}/2 + 5^4 {x^4}/24` .This is the desired polynomial as can be easily verified directly.

(b) Substitute `x^3` for `x` to obtain

`P_12 (x,cos(x^3)) = 1 - {(x^3)^2}/2 + {(x^3)^4}/24 = 1 - {x^6}/2 + {x^12}/24`.This too is the desired polynomial though the verification here is a little more work. You should verify this one as an exercise to see how this type of substitution will work more generally.

We can summarize some of the preceeding examples for substitutions involving polynomoials though the full justifications are not sufficiently simple at this point to include here.

Proposition IX.C.2: Suppose `f` is a `C^{oo}` function and `n` and `k` are natural numbers and `g(x)=int_0^x f(t)dt.

(a) `P_n(x, x^k f (x)) = x^k P_{n-k}(x, f (x))`.

(b) `P_{nk}(x, f (x^k)) = P_n(x^k, f (x)).

(c)`P_n(x,g(x)) = P_n(x, int_0^x f(t) dt) = int_0^x
P_{n-1}(t,f(t))dt`.

Proofs: Omitted.

Example IX.C.4: Find the 9th degree MacLaurin polynomial for `g(x) =
int_0^x t^2 cos(t^3) dt`.

Solution: Begin by finding the appropriate degree for the MacLaurin
polynomial of the cosine function, which we will call `n`.

Then `3n + 2 = 9-1 = 8`, or `n = 2`. Thus

`t^2 P_6(t, cos(t^3)) = t^2 (1 - {t^6}/2) = t^2 - {t^8}/2`.

Thus `P_9(t, g(x)) = {x^3}/3 - {x^9}/18`.

On to Chapter IX.D

Exercises IX.C: (solutions for some of the following)

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

- `f (x) = x sin(x)`; `n = 6`
- `f (x) = x^2 cos(x)`; `n = 6`
- `f (x) = x sin(x) + cos(x)`; `n = 6`
- `f (x) = x e^x`; `n = 6`
- `f (x) = sin (x^3)`; `n = 9`
- `f (x) = 1/{2 - x}`; `n = 4`. [Hint: `1/{2-x} = 1/2
1/{1-x/2}`. Example IX.B.3.]

- `f (x) = 1/{1 + x}`; `n = 4`
- `f (x) = 1/{1 + x^2}`; `n = 8`
- `f (x) = `arctan`(x)`; `n = 8`
- `f (x) = x/{2-x}`; `n = 4`
- `f (x) = e^{-2x}` ; `n = 4`