Problem 49 - Calculus: One and Several Variables

Question

Suppose that the function \(f\) has the power series representation \(f(x)=\sum_{k=0}^{\infty} n_{k} x^{k}\) (a) Show that if \(f\) is an even function, then \(a_{2 k+1}=0\) for all k. (b) Show that if \(f\) is an odd function, then \(a_{2 k}=0\) for all \(k\)

Step-by-Step Solution

Answer

In summary, for an even function, we showed that \(n_{2k+1} = 0\) for all k using the properties of even functions, as their power series representation satisfies f(x) = f(-x). Similarly, for an odd function, we showed that \(n_{2k} = 0\) for all k since their power series representation satisfies f(-x) = -f(x). These results indicate that the coefficients of odd or even terms in the power series representation of the function will be zero depending on the parity of the function.

Step 1: Part (a) - Even function

For a function f to be an even function, it must satisfy the property: f(x) = f(-x) The power series representation of f(x) is given by: f(x) = \(\sum_{k=0}^{\infty} n_{k} x^{k}\) Now we need to find the power series representation of f(-x): f(-x) = \(\sum_{k=0}^{\infty} n_{k} (-x)^{k}\) Since f(x) is an even function, the two power series representations should be equal: \(\sum_{k=0}^{\infty} n_{k} x^{k}\) = \(\sum_{k=0}^{\infty} n_{k} (-x)^{k}\) Now, we need to analyze the coefficients of each term in the series to prove that \(a_{2 k+1}=0\) for all k. We can rewrite the power series representation of f(-x) as: \(\sum_{k=0}^{\infty} (-1)^{k}n_{k} x^{k}\) For a term in the series to be of odd degree (2k+1), k must be even. Therefore, the coefficients for odd degree terms will be: \((-1)^{2k}n_{2k+1}\) = \(n_{2k+1}\) Since we want to prove that \(a_{2 k+1}=0\) for all k, we focus on the odd degree terms in the equality: \(n_{2k+1}\) = \(-n_{2k+1}\) This equality is true only when \(n_{2 k+1}=0\) for all k.

Step 2: Part (b) - Odd function

For a function f to be an odd function, it must satisfy the property: f(-x) = -f(x) The power series representation of f(x) is given by: f(x) = \(\sum_{k=0}^{\infty} n_{k} x^{k}\) Now we need to find the power series representation of f(-x): f(-x) = \(\sum_{k=0}^{\infty} n_{k} (-x)^{k}\) Since f(x) is an odd function, the power series representation of f(-x) should be -f(x): \(-\sum_{k=0}^{\infty} n_{k} x^{k}\) = \(\sum_{k=0}^{\infty} n_{k} (-x)^{k}\) Now, we need to analyze the coefficients of each term in the series to prove that \(a_{2 k}=0\) for all k. We can rewrite the power series representation of f(-x) as: \(\sum_{k=0}^{\infty} (-1)^{k}n_{k} x^{k}\) For a term in the series to be of even degree (2k), k must be even. Therefore, the coefficients for even degree terms will be: \((-1)^{2k}n_{2k}\) = \(n_{2k}\) Since we want to prove that \(a_{2 k}=0\) for all k, we focus on the even degree terms in the equality: \(-n_{2k}\) = \(n_{2k}\) This equality is true only when \(a_{2 k}=0\) for all k.

Key Concepts

Even Function

An even function is a key concept in mathematics, especially when dealing with power series and symmetries. An even function, by definition, satisfies the condition \(f(x) = f(-x)\) for all values of \(x\). This means that the graph of an even function is symmetric about the \(y\)-axis.
When represented as a power series, \(f(x)=\sum_{k=0}^{\infty} n_{k} x^{k}\), this symmetry imposes certain conditions on the coefficients. Specifically, for the power series to maintain this symmetry, all terms with odd powers must have zero coefficients. Hence, for any even function, \(a_{2k+1} = 0\) for all \(k\).
Understanding even functions is significant because they often appear in problems involving integrals and solving differential equations, where their symmetric properties simplify calculations.

Odd Function

An odd function, contrary to even functions, adheres to the property \(f(-x) = -f(x)\). This indicates that an odd function is symmetric about the origin. In other words, if you rotate its graph 180 degrees around the origin, it looks exactly the same.
In the power series context, this definition requires all the coefficients of the even powers to be zero, resulting in only odd power terms being present. Thus, for an odd function, \(a_{2k} = 0\) for all \(k\).
This property shows that the behavior of odd functions, in conjunction with even functions, can greatly simplify mathematical analysis, such as Fourier analysis, where functions are often decomposed into sums of even and odd components.

Coefficients

Coefficients are the numerical factors that multiply each term in a power series. In a power series like \(f(x) = \sum_{k=0}^{\infty} n_{k}x^{k}\), each \(n_k\) is a coefficient. These coefficients determine the weighting and influence of each corresponding term in the series on the function.
For even and odd functions, understanding the role of these coefficients is crucial:

• In even functions, terms with odd powers must have zero coefficients (\(a_{2k+1} = 0\)).
• In odd functions, terms with even powers must have zero coefficients (\(a_{2k} = 0\)).

By analyzing coefficients, we can deduce important properties about the behavior of functions, including symmetry and periodicity.

Function Representation

Function representation is the expression of a function in a certain form, such as a power series. This form allows for a detailed understanding of the function's behavior and properties.
Power series are invaluable for representing functions because they allow us to express complex functions as infinite sums of powers of \(x\) with specific coefficients. In equations like \(f(x) = \sum_{k=0}^{\infty} n_{k}x^{k}\), the series provides insights into how each power of \(x\) contributes to the function's overall shape.
Power series representation helps analyze functions' local behavior around a point and can be used to approximate functions within a certain range. They are foundational in calculus and many areas of mathematical modeling and analysis, making them an essential tool in much of mathematics and engineering.