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Magazine Chapter 7: Problem 11
Suppose that \(f\) is an analytic function such that \(f^{(n)}(0)=n\). Find \(f(1)\).
Short Answer
Expert verified
The value of \( f(1) \) is \( e \).
Step by step solution
01
Understand Analytic Functions
An analytic function can be represented by a power series around a point. For this problem, assume the series is centered at 0 (Maclaurin series): \( f(x) = \sum_{n=0}^{\infty} a_n x^n \). Each coefficient \( a_n \) relates to the derivative of \( f \) evaluated at 0: \( a_n = \frac{f^{(n)}(0)}{n!} \).
02
Identify Given Derivatives
We are given that \( f^{(n)}(0) = n \). Therefore, the coefficients in the power series representation of \( f \) are calculated as follows: \( a_n = \frac{f^{(n)}(0)}{n!} = \frac{n}{n!} \).
03
Write the Power Series for f(x)
The power series for \( f(x) \) based on the derivatives is \( f(x) = \sum_{n=0}^{\infty} \frac{n}{n!} x^n \).
04
Calculate f(1)
To find \( f(1) \), substitute \( x = 1 \) into the power series: \[ f(1) = \sum_{n=0}^{\infty} \frac{n}{n!} \cdot 1^n = \sum_{n=0}^{\infty} \frac{n}{n!} \]This series can be manipulated using the exponential function: - Consider \( e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \).- Differentiate to get: \( e^x = \sum_{n=0}^{\infty} \frac{n x^{n-1}}{n!} \).- Multiply by \( x \): \( x e^x = \sum_{n=0}^{\infty} \frac{n x^n}{n!} \).For \( x = 1 \), this simplifies to \( e \cdot 1 = e \).
05
Conclusion
Therefore, \( f(1) = e \). This solution comes from recognizing the pattern in the power series and linking it with the properties of the exponential function.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Power Series
A power series can be thought of as an infinite polynomial. It's a way to express functions in terms of an infinite sum of powers of a variable, typically written as: \[ f(x) = \sum_{n=0}^{\infty} a_n x^n \] where each \(a_n\) represents a coefficient. This expression allows complex functions to be analyzed and utilized more easily.
- The interpretations of power series are particularly useful for approximating functions.
- One key aspect is its radius of convergence, within which the series converges to the function.
Derivative Evaluation
When we talk about evaluating a derivative, we are looking at how a function behaves as it changes. Derivatives give us the rate of change of a function at any point. Here, multiple derivatives are evaluated at zero to find the coefficients of our power series.For a function \(f(x)\), the \(n\)-th derivative evaluated at a point \(x = 0\) is denoted \(f^{(n)}(0)\). This helps us find out the coefficients for the power series:
- The general formula is \(a_n = \frac{f^{(n)}(0)}{n!}\), representing each power series term.
- By using these derivatives, you can completely determine a function's nature near the point of expansion.
Exponential Function
The exponential function, denoted \(e^x\), is a fundamental mathematical function characterized by its constant rate of growth. It's comprehensively expressed as:\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]This series expansion is identical to the power series we derived in this problem.
- The connection between the exponential function and our power series is made by relating the terms of the series.
- By differentiating and manipulating the exponential function鈥檚 series, we confirmed our desired outcome.
Maclaurin Series
A Maclaurin series is a specific type of power series with a focus around zero. It simplifies the expression of functions and is great for approximations.The Maclaurin series is given by:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \]In this exercise, because we were given \(f^{(n)}(0)=n\), we could form an accurate representation of the function as an infinite polynomial.
- Maclaurin series can approximate a variety of functions, especially when other function representations are complex.
- This series helps build the groundwork for Taylor Series, where the expansion is around any point \(a\) instead of zero.
