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We are asked to find the coefficients \(c_n\) for the Taylor series of \( e^x \) centered at \( x = 2 \). The Taylor series expansion of a function \( f(x) \) around \( a = 2 \) is given by:

\[
f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x - a)^n
\]

Here, \( f(x) = e^x \), so all the derivatives of \( e^x \) are still \( e^x \). Therefore, at \( x = 2 \), \( f^{(n)}(2) = e^2 \) for all \( n \).

Thus, the general form for the Taylor series is:

\[
e^x = \sum_{n=0}^{\infty} \frac{e^2}{n!}(x - 2)^n
\]

In this case, the coefficients \( c_n \) correspond to \( \frac{e^2}{n!} \). Let's compute the values for \( c_0 \) through \( c_4 \).

1. \( c_0 = \frac{e^2}{0!} = e^2 \)
2. \( c_1 = \frac{e^2}{1!} = e^2 \)
3. \( c_2 = \frac{e^2}{2!} = \frac{e^2}{2} \)
4. \( c_3 = \frac{e^2}{3!} = \frac{e^2}{6} \)
5. \( c_4 = \frac{e^2}{4!} = \frac{e^2}{24} \)

So, the coefficients are:

\[
c_0 = e^2, \quad c_1 = e^2, \quad c_2 = \frac{e^2}{2}, \quad c_3 = \frac{e^2}{6}, \quad c_4 = \frac{e^2}{24}
\]

Would you like a detailed explanation on how to find the general formula for Taylor series or any specific part? Here are 5 relative questions to expand:

1. How do you compute higher-order Taylor coefficients?
2. Why does the Taylor series of \( e^x \) have all derivatives equal to \( e^x \)?
3. What does the radius of convergence mean for a Taylor series?
4. How can you approximate functions using a truncated Taylor series?
5. What is the error bound when truncating the Taylor series at a certain degree?

**Tip:** When working with Taylor series, keep in mind that the closer \( x \) is to the center \( a \), the faster the series converges to the actual value of the function.

Find the coefficients for the Taylor series of e^x centered at x = 2.

Discover how to compute the coefficients of the Taylor series for the function e^x, centered at x=2. This problem explores the derivatives of exponential functions and their Taylor expansions, perfect for advanced calculus students.

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