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Math  /  Calculus

Question3. [15] a) Find the 5th 5^{\text {th }} degree Taylor polynomial of 8x5−6x4+3x3+27x2+32x−128 x^{5}-6 x^{4}+3 x^{3}+27 x^{2}+32 x-12 about x=0x=0 and [6] about x=20x=20, then simplify each result. What do you observe? b) Find the 100th 100^{\text {th }} degree Taylor polynomial of 8x5−6x4+3x3+27x2+32x−128 x^{5}-6 x^{4}+3 x^{3}+27 x^{2}+32 x-12 about x=0x=0 [6] and about x=20x=20, then simplify each result. What do you observe? c) What can you conclude about the nth n^{\text {th }} degree Taylor polynomial about x=ax=a of a polynomial of [3] degree mm, where n≥mn \geq m ?

Studdy Solution

STEP 1

1. The Taylor polynomial of a function is an approximation of the function as a polynomial centered around a specific point.
2. The degree of the Taylor polynomial is the highest power of x x in the polynomial.
3. For a polynomial function, the Taylor polynomial of degree equal to or greater than the degree of the polynomial is the polynomial itself.

STEP 2

1. Find the 5th 5^{\text{th}} degree Taylor polynomial about x=0 x=0 .
2. Find the 5th 5^{\text{th}} degree Taylor polynomial about x=20 x=20 .
3. Find the 100th 100^{\text{th}} degree Taylor polynomial about x=0 x=0 .
4. Find the 100th 100^{\text{th}} degree Taylor polynomial about x=20 x=20 .
5. Conclude about the nth n^{\text{th}} degree Taylor polynomial for n≥m n \geq m .

STEP 3

To find the 5th 5^{\text{th}} degree Taylor polynomial of 8x5−6x4+3x3+27x2+32x−12 8x^5 - 6x^4 + 3x^3 + 27x^2 + 32x - 12 about x=0 x = 0 , we note that the polynomial is already of degree 5. Therefore, the Taylor polynomial of degree 5 about x=0 x = 0 is the polynomial itself:
P5(x)=8x5−6x4+3x3+27x2+32x−12 P_5(x) = 8x^5 - 6x^4 + 3x^3 + 27x^2 + 32x - 12

STEP 4

To find the 5th 5^{\text{th}} degree Taylor polynomial about x=20 x = 20 , we need to express the polynomial in terms of (x−20) (x - 20) . However, since the polynomial is of degree 5, the Taylor polynomial of degree 5 about any point is the polynomial itself. Therefore:
P5(x)=8x5−6x4+3x3+27x2+32x−12 P_5(x) = 8x^5 - 6x^4 + 3x^3 + 27x^2 + 32x - 12

STEP 5

For the 100th 100^{\text{th}} degree Taylor polynomial about x=0 x = 0 , since the original polynomial is of degree 5, the Taylor polynomial of degree 100 is again the polynomial itself:
P100(x)=8x5−6x4+3x3+27x2+32x−12 P_{100}(x) = 8x^5 - 6x^4 + 3x^3 + 27x^2 + 32x - 12

STEP 6

For the 100th 100^{\text{th}} degree Taylor polynomial about x=20 x = 20 , similar reasoning applies. The polynomial is of degree 5, so the Taylor polynomial of degree 100 is the polynomial itself:
P100(x)=8x5−6x4+3x3+27x2+32x−12 P_{100}(x) = 8x^5 - 6x^4 + 3x^3 + 27x^2 + 32x - 12

STEP 7

For a polynomial of degree m m , the nth n^{\text{th}} degree Taylor polynomial about any point x=a x = a where n≥m n \geq m is the polynomial itself. This is because the Taylor polynomial of degree n n captures all terms up to degree n n , and since n≥m n \geq m , it includes all terms of the original polynomial.
The observations for parts (a) and (b) show that the Taylor polynomial of a polynomial function, when the degree of the Taylor polynomial is greater than or equal to the degree of the polynomial, is the polynomial itself, regardless of the center point. This leads to the conclusion in part (c).

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