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Magazine Chapter 9: Problem 31
Find the Taylor series generated by \(f\) at \(x=a.\) $$f(x)=e^{x}, \quad a=2$$
Short Answer
Expert verified
The Taylor series of \( e^x \) at \( x = 2 \) is \( \sum_{n=0}^{\infty} \frac{e^2}{n!} (x-2)^n \).
Step by step solution
01
Understand the Problem
We need to find the Taylor series of the function \( f(x) = e^x \) centered at \( a = 2 \). This means expressing \( f(x) \) in an infinite series around this point.
02
Recall the Taylor Series Formula
The general formula for a Taylor series centered at \( x = a \) is: \[T(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots\]
03
Compute Derivatives of \(f(x)\)
The function \( f(x) = e^x \) has the property that all its derivatives are the same as the original function, meaning \( f(x) = e^x, \) \( f'(x) = e^x, \) \( f''(x) = e^x, \ldots \) Thus, \( f^{(n)}(x) = e^x \) for any \( n \geq 0 \).
04
Evaluate the Derivatives at \(x = a = 2\)
We evaluate these derivatives at \( x = 2 \):\[ f(2) = e^2, \, f'(2) = e^2, \, f''(2) = e^2, \ldots \] For any \( n, \) we have \( f^{(n)}(2) = e^2 \).
05
Write the Taylor Series Using Evaluated Derivatives
Substitute these values into the Taylor series formula:\[ T(x) = e^2 + e^2(x-2) + \frac{e^2}{2!}(x-2)^2 + \frac{e^2}{3!}(x-2)^3 + \ldots \]This represents the series expansion of \( f(x) \) around \( x = 2 \).
06
Present the Final Taylor Series
The Taylor series for \( f(x) = e^x \) at \( x = 2 \) is: \[T(x) = \sum_{n=0}^{\infty} \frac{e^2}{n!} (x-2)^n\] This is the infinite series that represents the function \( e^x \) centered at \( x = 2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Function Derivatives
In mathematics, derivatives help describe how a function changes as its input, or variable, changes. They are fundamental in finding Taylor series as they capture the function's behavior around a specific point. Here鈥檚 what you need to know:
- The first derivative of a function gives the slope at any point, essentially showing whether the function is increasing or decreasing.
- Higher-order derivatives (second, third, etc.) provide more detailed information about curvature and changes in the slope of the function.
- For the exponential function \( f(x) = e^x \), something extraordinary happens: the derivative of \( e^x \) is still \( e^x \). This uniqueness simplifies the computation of its Taylor series, as all derivatives remain the same.
The Concept of Infinite Series
An infinite series is a sum of infinitely many terms. It's a core concept of calculus used to represent functions as sums. The beauty of infinite series is in its approximation power.
- A finite sum can approximate the function to a certain degree of accuracy, and adding more terms brings it even closer to the exact function value.
- By considering an infinite series, especially the Taylor series, you can express complex functions in terms of simpler polynomial expressions.
- This is particularly useful in cases like the Taylor series for \( f(x) = e^x \), where the series provides an exact representation that can be evaluated at various points around a center.
Exploring the Maclaurin Series
The Maclaurin series is a special case of the Taylor series where the series expansion is centered at \( x = 0 \). For any function \( f(x) \), the Maclaurin series is:\[f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots\]
- It's especially useful when analyzing functions that naturally pass through the origin, simplifying the computations involved.
- Maclaurin series allow for approximating functions around \( x = 0 \) using basic calculus and algebraic skills without the need for more complex calculus techniques.
- Our function, \( e^x \), has the interesting property where all derivatives are equal, making the Maclaurin series simply \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \), identical to the function itself, illustrating the concept's practical elegance.
