Question

Given that the power series \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) satisfies $$ f^{\prime}(x)=-2 x f(x), \quad f(0)=1 $$ find \(\left\\{a_{n}\right\\}\). Do you recognize \(f ?\)

Short answer

Answer: The power series function f(x) corresponds to the cosh(x).

Step-by-step solution

Take the derivative of the power series

We have f(x) = sum(a_n x^n). To find the derivative of this function, we need to differentiate each term with respect to x and then sum them. f'(x) = sum(n*a_n x^(n-1))

Substitute the given condition

We have been given that f'(x) = -2x*f(x). Substituting the given condition, we have: sum(n*a_n x^(n-1)) = -2x*sum(a_n x^n)

Expand the product on the right side

Expanding the product on the right side: sum(n*a_n x^(n-1)) = -2*sum(a_n x^(n+1))

Compare the coefficients to establish relationship

Now we compare the coefficients of x^(n-1) and x^(n+1). n*a_n = -2*a_(n-2)

Solve for the coefficients using the relationship

Here, we have a recurrence relation to find the coefficients a_n: a_n = -\frac{2}{n} a_(n-2) We are also given that f(0) = 1 which means a_0 = 1. Using this, we can find the other coefficients. a_1 = -\frac{2}{1}*a_{-1} => a_1 = 0 (since a_{-1} does not exist) a_2 = -\frac{2}{2}*a_0 => a_2 = -1 a_3 = -\frac{2}{3}*a_1 => a_3 = 0 Lets find up to a_6 using the recurrence relation: a_4 = -\frac{2}{4}*a_2 = 1/2 a_5 = -\frac{2}{5}*a_3 = 0 a_6 = -\frac{2}{6}*a_4 = -1/6 We observe an interesting pattern here: a_0 = 1, a_2 = -1, a_4 = 1/2, a_6 = -1/6,... Now, we recognize this as the Maclaurin series of the function f(x) = cosh(x): f(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ... So, the power series function f(x) corresponds to the cosh(x).

Key concepts

Derivative in Power Series

In mathematics, the derivative is a fundamental concept that represents the rate at which a function is changing at any given point. When working with power series such as \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\), finding the derivative involves differentiating each term of the series. This results in a new series:\[f'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}\]Here, each term \(a_n x^n\) becomes \(n a_n x^{n-1}\) after differentiation. Notice that the exponent drops by one and is multiplied by the original power as a coefficient. When we differentiate a power series, it often results in another power series, allowing us to analyze how the series behaves as a whole, especially under specific conditions or transformations.For instance, in this problem, we are given the condition that \(f'(x) = -2x f(x)\). Differentiating and equating the power series helps in revealing relationships between the coefficients, enabling us to find each coefficient \(a_n\) successively.

Maclaurin Series Basics

A Maclaurin series is a special kind of Taylor series that represents functions as power series. It is centered at 0, which means all the terms involve \(x\) raised to some power, starting from \(x^0\), that is:\[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n\]This formula allows us to express a broad range of functions, including trigonometric, exponential, and hyperbolic functions, as infinite sums. Each term in the Maclaurin series is determined by the function's derivatives at zero.In the current exercise, we recognize the power series from the pattern of coefficients, leading us to identify \(f(x)\) as \(\cosh(x)\). This identification relies on matching the coefficients in the series with those of a known function's series expansion.

• \(a_0 = 1\): Matches the \(0!\) term in \(1/0!\)
• \(a_2 = -1\): Matches the \(-1/2!\) term in \(-x^2/2!\)
• \(a_4 = \frac{1}{2}\): Matches \(x^4/4!\)
• \(a_6 = -\frac{1}{6}\): Matches \(-x^6/6!\)

This pattern clues us into the identity \(\cosh(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + ...\), which helps in understanding the behavior of \(f(x)\) on a deeper level.

Understanding Recurrence Relations

In the context of power series, a recurrence relation provides a powerful method for determining the coefficients \(a_n\) of the series. These relations connect each term in a sequence or function with previous terms in a prescribed way. Here, we derived a simple yet crucial recurrence relationship:\[a_n = -\frac{2}{n} a_{n-2}\]This formula implies that each coefficient \(a_n\) can be calculated using two steps back in the sequence. Given \(a_0 = 1\) and the understanding that terms at odd indices are zero, the formula can be applied successively:

• \(a_2 = -\frac{2}{2} a_0 = -1\)
• \(a_4 = -\frac{2}{4} a_2 = \frac{1}{2}\)
• \(a_6 = -\frac{2}{6} a_4 = -\frac{1}{6}\)

The pattern follows the alternating sign and decreasing magnitude observed in \(\cosh(x)\)'s series. Recognizing the recurrence relation helps solve the sequence systematically, thus defining the function's Maclaurin series completely. This approach simplifies finding coefficients for a complex function, streamlining the analysis of its behavior near the center, which in this case is \(x = 0\).