Question

Consider the function f(x)=sin(2x)cos(x), find the third degree Taylor polynomial T

3
​
(x) of f about the point x=0. [Hint: you could find this manually using differentiation via the product rule, but you can also use the expansions of sin(x) and cos(x) directly.] T
3
​
(x)=

Answer 1

The third degree Taylor polynomial T3(x) of f(x) = sin(2x)cos(x) is 2x - 2(x^3)/6 + (2x^5)/120 - x^2 + (x^4)/24.

Step-by-step explanation:

To find the third degree Taylor polynomial T3(x) of the function f(x) = sin(2x)cos(x), we can use the expansions of sin(x) and cos(x) directly.

The expansions of sin(x) and cos(x) are:

sin(x) = x - (x^3)/3! + (x^5)/5! - ...

cos(x) = 1 - (x^2)/2! + (x^4)/4! - ...

Using the product rule, we can multiply these expansions and keep terms up to the third degree:

f(x) = (2x - (2x^3)/3! + (2x^5)/5! - ...) * (1 - (x^2)/2! + (x^4)/4! - ...)

By simplifying the terms and keeping up to the third degree, we obtain:

f(x) = 2x - 2(x^3)/6 + (2x^5)/120 - x^2 + (x^4)/24

Therefore, the third degree Taylor polynomial T3(x) of f(x) = sin(2x)cos(x) is T3(x) = 2x - 2(x^3)/6 + (2x^5)/120 - x^2 + (x^4)/24.

Answer 2

The third degree Taylor polynomial for the function f(x)=sin(2x)cos(x) at x=0 is T3(x) = 2x - x^3. This is derived by using the expansions of sin(x) and cos(x) and ignoring higher-order terms greater than x^3.

Step-by-step explanation:

The student has asked for the third degree Taylor polynomial T3(x) for the function f(x) = sin(2x)cos(x) centered at x=0. To find the Taylor polynomial, we typically calculate the derivatives of the function at the point of interest and use them to construct the polynomial. However, as hinted, we can also utilize known Taylor expansions of the sin(x) and cos(x) functions.

The Taylor expansions for sin(x) and cos(x) around 0 are given by:

sin(x) = x - x3/3! + O(x5)

cos(x) = 1 - x2/2! + O(x4)

For the third degree Taylor polynomial, higher-order terms (those involving x to a power higher than three) are ignored since they do not affect T3(x).

Since f(x) = sin(2x)cos(x), apply the angle-doubling identities for sine and cosine, and find the product to get:

f(x) ≈ (2x - (2x)3/3!) * (1 - x2/2!)

f(x) = 2x - 2x3/3 - 2x3/2 + higher-order terms

f(x) ≈ 2x - 2x3(2/3 + 1/2)

f(x) ≈ 2x - 2x3(4/6 + 3/6)

f(x) ≈ 2x - x3

Therefore, the third degree Taylor polynomial for f(x) about the point x=0 is T3(x) = 2x - x3.