Chapter 8: Q. 3 (page 692)

If the series ${\sum a_{k} x^{k}}_{k - 0}^{\infty}$ converges to the function $f (x)$ on the interval (−2, 2), provide a formula for $a_{k}$ in terms of the function f .

Short Answer

Expert verified

The formula for $a_{k}$ is $\frac{f^{(k)} (0)}{k!}$

Step by step solution

Step 1. Given information

The series is ${\sum a_{k} x^{k}}_{k - 0}^{\infty}$ is convergent to f(x) on interval (-2,2).

Step 2. Explanation.

The series is converges to the function so, Maclaurin polynomial for f(x) is in the form $P_{n} (x) = \sum_{k = 0}^{n} \frac{f^{(k)} (0)}{k!} x^{k}$ .

For series ${\sum a_{k} x^{k}}_{k - 0}^{\infty}$ , $a_{k} = \frac{f^{k} (0)}{k!}$

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If the series ${\sum a_{k} x^{k}}_{k - 0}^{\infty}$ converges to the function $f (x)$ on the interval (−2, 2), provide a formula for $a_{k}$ in terms of the function f .

Short Answer

Step by step solution

Step 1. Given information

Step 2. Explanation.

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If the series ∑akxkk-0∞converges to the function f(x) on the interval (−2, 2), provide a formula for akin terms of the function f .

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If the series ${\sum a_{k} x^{k}}_{k - 0}^{\infty}$ converges to the function $f (x)$ on the interval (−2, 2), provide a formula for $a_{k}$ in terms of the function f .