Taylor polynomials Find the nth-order Taylor polynomial for the following functions centered at the given point a. ƒ(x) = e^(sin x), n = 2, a = 0
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Identify the function and the point of expansion: here, the function is \(f(x) = e^{\sin x}\) and the center is \(a = 0\).
Recall the formula for the nth-order Taylor polynomial centered at \(a\):
\[T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!} (x - a)^k,\]
where \(f^{(k)}(a)\) is the \(k\)th derivative of \(f\) evaluated at \(x = a\).
Calculate the derivatives of \(f(x)\) up to order 2:
- First, find \(f(0) = e^{\sin 0}\).
- Then, find \(f'(x)\) using the chain rule and evaluate \(f'(0)\).
- Next, find \(f''(x)\) by differentiating \(f'(x)\) and evaluate \(f''(0)\).
Substitute the values of \(f(0)\), \(f'(0)\), and \(f''(0)\) into the Taylor polynomial formula:
\[T_2(x) = f(0) + f'(0)(x - 0) + \frac{f''(0)}{2!}(x - 0)^2.\]
Write the final expression for the 2nd-order Taylor polynomial centered at 0, which approximates \(f(x)\) near \(x=0\).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Taylor Polynomials
A Taylor polynomial approximates a function near a point a by using the function's derivatives at that point. The nth-order Taylor polynomial includes terms up to the nth derivative, providing a polynomial that closely matches the function's behavior near a.
To find the Taylor polynomial of e^(sin x), you must compute derivatives of a composite function. This involves applying the chain rule repeatedly to differentiate e^(sin x), since it is a composition of the exponential and sine functions.
After finding the derivatives, evaluate each at the center point a=0. These values are used as coefficients in the Taylor polynomial, ensuring the polynomial matches the function and its derivatives at that point.