a. Use the definition of a Taylor series to find the first four nonzero terms of the Taylor series for the given function centered at a.
f(x)=sin x, a = π/2
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Recall the definition of the Taylor series of a function \(f(x)\) centered at \(a\):
\[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n,\]
where \(f^{(n)}(a)\) is the \(n\)-th derivative of \(f\) evaluated at \(x = a\).
Identify the function and center: here, \(f(x) = \sin x\) and \(a = \frac{\pi}{2}\). We will need to compute the derivatives of \(\sin x\) and evaluate them at \(x = \frac{\pi}{2}\).
Calculate the first four derivatives of \(f(x) = \sin x\):
- \(f(x) = \sin x\)
- \(f'(x) = \cos x\)
- \(f''(x) = -\sin x\)
- \(f^{(3)}(x) = -\cos x\)
- \(f^{(4)}(x) = \sin x\) (which repeats the cycle).
Write the first four nonzero terms of the Taylor series using the formula:
\[f(x) \approx f(a) + \frac{f'(a)}{1!}(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f^{(3)}(a)}{3!}(x - a)^3 + \cdots\]
Substitute the values found and simplify, including only the nonzero terms.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Taylor Series Definition
A Taylor series represents a function as an infinite sum of terms calculated from the derivatives of the function at a single point. Each term involves the nth derivative evaluated at the center point, multiplied by (x - a)^n and divided by n!. This series approximates the function near the point a.
To find the Taylor series of sin(x), you need to compute its derivatives at the center point. The derivatives of sin(x) cycle every four steps: sin(x), cos(x), -sin(x), -cos(x), then back to sin(x). Evaluating these at x = π/2 helps determine the coefficients of the series.
Derivative of the Natural Exponential Function (e^x)
Centering the Series at a = π/2
Centering the Taylor series at a = π/2 means expanding the function around this point, so terms involve powers of (x - π/2). This shifts the approximation focus to values of x near π/2, requiring evaluation of derivatives specifically at π/2 to find accurate coefficients.