Check that the derivative of the degree 𝑛 Taylor polynomial for 𝑒𝑥 is the degree 𝑛 − 1 Taylor polynomial for 𝑒𝑥.

To check that the derivative of the degree \( n \) Taylor polynomial for \( e^x \) is the degree \( n - 1 \) Taylor polynomial for \( e^x \), let's go through the process systematically.

### Step 1: Taylor Polynomial for \( e^x \)
The Taylor series for \( e^x \) centered at \( x = 0 \) (Maclaurin series) is given by:

\[
e^x = \sum_{k=0}^{\infty} \frac{x^k}{k!}
\]

The degree \( n \) Taylor polynomial for \( e^x \) is the partial sum up to degree \( n \):

\[
T_n(x) = \sum_{k=0}^{n} \frac{x^k}{k!}
\]

Explicitly:

\[
T_n(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots + \frac{x^n}{n!}
\]

### Step 2: Take the Derivative of the Degree \( n \) Taylor Polynomial
We now compute the derivative of \( T_n(x) \):

\[
\frac{d}{dx} T_n(x) = \frac{d}{dx} \left( 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} \right)
\]

This gives:

\[
\frac{d}{dx} T_n(x) = 0 + \frac{1}{1!} + \frac{2x}{2!} + \frac{3x^2}{3!} + \dots + \frac{n x^{n-1}}{n!}
\]

Simplifying:

\[
\frac{d}{dx} T_n(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots + \frac{x^{n-1}}{(n-1)!}
\]

### Step 3: Compare with the Degree \( n - 1 \) Taylor Polynomial
The degree \( n - 1 \) Taylor polynomial for \( e^x \) is:

\[
T_{n-1}(x) = \sum_{k=0}^{n-1} \frac{x^k}{k!}
\]

Which is explicitly:

\[
T_{n-1}(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots + \frac{x^{n-1}}{(n-1)!}
\]

### Step 4: Conclusion
We observe that the derivative of the degree \( n \) Taylor polynomial, \( \frac{d}{dx} T_n(x) \), is exactly the same as the degree \( n - 1 \) Taylor polynomial for \( e^x \). Therefore, we have shown that:

\[
\frac{d}{dx} T_n(x) = T_{n-1}(x)
\]

This confirms the claim.

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Would you like more details or have any questions?

Here are some related questions to deepen your understanding:
1. What is the general formula for the Taylor series expansion of any function?
2. How do we find the remainder term in a Taylor polynomial approximation?
3. Can the same result be applied to other functions like \( \sin(x) \) or \( \cos(x) \)?
4. How does the convergence of the Taylor series for \( e^x \) compare to other functions?
5. What is the geometric interpretation of taking the derivative of a Taylor polynomial?

**Tip:** The exponential function \( e^x \) has a Taylor series that converges everywhere, meaning the series equals the function for all real numbers \( x \).

Check that the derivative of the degree n Taylor polynomial for e^x is the degree n - 1 Taylor polynomial for e^x.

Explore how the derivative of the degree n Taylor polynomial for e^x results in the degree n-1 Taylor polynomial. This calculus problem covers Taylor series, exponential functions, and the derivative process. Suitable for college-level students.

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