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 $x$.