Question

Calculate the taylor polynomials 2() and 3() centered at =0 for ()=sin().

Answer 1

To find the Taylor polynomials
`\(T_2(x)\)` and
`\(T_3(x)\)` for
`\(f(x) = \sin(x)\)` centered at
`\(x = 0\)`, we use the general formula. Calculating up to the third derivative, we construct
`\(T_3(x) = x - \frac{{x^3}}{6} + \frac{{x^5}}{120}\)`, capturing the sinusoidal behavior with increased precision.

Step-by-step explanation:

In order to calculate the Taylor polynomials
`\(T_2(x)\) and \(T_3(x)\)` centered at
`\(x = 0\)` for
`\(f(x) = \sin(x)\),` we utilize the general formula for the Taylor polynomial:

`\[ T_n(x) = f(0) + f'(0)x + \frac{{f''(0)}}{2!}x^2 + \ldots + \frac{{f^((n))}(0)}}{n!}x^n \]`

For
`\(T_2(x)\)`, the function values and derivatives up to the second order at
`\(x = 0\)` are:

`\[ f(0) = \sin(0) = 0 \]`

`\[ f'(0) = \cos(0) = 1 \]`

`\[ f''(0) = -\sin(0) = 0 \]`

Therefore,
`\(T_2(x) = 0 + 1 \cdot x + 0 \cdot \frac{{x^2}}{2!} = x\)`.

For
`\(T_3(x)\)`, we need to calculate the third derivative at
`\(x = 0\)`:

`\[ f'''(x) = -\cos(x) \]`

`\[ f'''(0) = -\cos(0) = -1 \]`

Now, we can construct
`\(T_3(x)\)`:

`\[ T_3(x) = \sin(0) + \cos(0) \cdot x + \frac{{-1}}{2!} \cdot x^2 = x - \frac{{x^3}}{6} \]`

Therefore,
`\(T_3(x) = x - \frac{{x^3}}{6}\)` is the Taylor polynomial of degree 3 for
`\(f(x) = \sin(x)\)` centered at
`\(x = 0\)`.

copmlete Question here:

Calculate the Taylor polynomials T2(x) and T3(x) centered at x = 0 for f(x) = sin(x). (Use symbolic notation and fractions where needed.) T3(x) = T: (x) = ?