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

User Nanna
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Final Answer:

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) = ?

User Hyddan
by
8.3k points

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