Question

Given that \(y=\sum_{n=0}^{\infty} a_{n} x^{n},\) compute \(y^{\prime}\) and \(y^{\prime \prime}\) and write out the first four terms of each series as well as the coefficient of \(x^{n}\) in the general term. Show that if \(y^{\prime \prime}=y,\) then the coefficients \(a_{0}\) and \(a_{1}\) are arbitrary, and determine \(a_{2}\) and \(a_{3}\) in terms of \(a_{0}\) and \(a_{1}\). Show that \(a_{n+2}=a_{n} /(n+2)(n+1), n=0,1,2,3, \ldots\)

Short answer

#Question#: Find the first and second derivatives of the power series representation of a function y, expressed as $y = \sum_{n=0}^{\infty} a_nx^n$, where the second derivative is equal to the original function. Show that the coefficients $a_0$ and $a_1$ are arbitrary, find the coefficients $a_2$ and $a_3$ in terms of $a_0$ and $a_1$, and prove the relation $a_{n+2} = \frac{a_n}{(n+2)(n+1)}$.

Step-by-step solution

Find the first derivative of y

To find the first derivative, we differentiate y with respect to x. By term-wise differentiation, we get: $$y'=\sum_{n=0}^{\infty} \frac{d(a_{n}x^{n})}{dx} = \sum_{n=1}^{\infty} a_{n}nx^{n-1}$$

Write the first four terms of the first derivative, and the general term of x^n

By substituting n=1, 2, 3, and 4, we find the first four terms of the first derivative: $$y'=a_{1}+2a_2x+3a_3x^2+4a_4x^3+\cdots$$ The coefficient of \(x^n\) in the general term is \((n+1)a_{n+1}\).

Find the second derivative of y

Next, we differentiate y' with respect to x to find the second derivative. Again, by term-wise differentiation, we get: $$y''=\sum_{n=1}^{\infty} \frac{d((n+1)a_{n+1}x^n)}{dx} = \sum_{n=2}^{\infty} (n+1)na_{n+1}x^{n-1}$$

Write the first four terms of the second derivative

By substituting n=1, 2, 3, and 4, we find the first four terms of the second derivative: $$y''=2a_{2}+3(2)a_3x+4(3)a_4x^2+5(4)a_5x^3+\cdots$$

Apply condition \(y''=y\) and find \(a_{2}\) and \(a_{3}\)

Now, we are given the condition that \(y''=y\). Equating the terms of y'' and y, we get: \begin{align*} a_0&=2a_2 \\ a_1&=6a_3 \\ a_{2}&=12a_4 \\ a_{3}&=20a_5 \\ &\cdots \end{align*} Since \(a_0\) and \(a_1\) are arbitrary, we express \(a_2\) and \(a_3\) in terms of \(a_0\) and \(a_1\): \begin{align*} a_2 &=\frac{1}{2}a_0 \\ a_3 &=\frac{1}{6}a_1 \end{align*}

Prove the given relation for \(a_{n+2}\)

Now we need to show that the following relation holds true: $$a_{n+2} = \frac{a_n}{(n+2)(n+1)}, \quad n = 0, 1, 2, 3, \ldots$$ From step 5, we have the general form $$a_{n+2} = (n+1)n a_{n+1}$$ We notice that, if we divide both sides of the above general form by another \((n+1)\), we get the required relation: $$a_{n+2} = \frac{a_n}{(n+2)(n+1)}$$ This proves the desired relation for the coefficient \(a_{n+2}\).

Key concepts

Term-wise Differentiation

Understanding how to differentiate a power series term by term is critical when dealing with differential equations. A power series is simply a series of the form \( y = \sum_{n=0}^{\infty} a_{n}x^{n} \), with each term being a multiple of a power of \( x \). Term-wise differentiation means taking the derivative of each individual term of the series, as differentiation is a linear operator and can be distributed across the summation.

Applying Term-wise Differentiation

When you differentiate each term, \( a_{n}x^{n} \) becomes \( na_{n}x^{n-1} \) for \( n \geq 1 \), since the derivative of \( x^{n} \) with respect to \( x \) is \( nx^{n-1} \). The term \( a_{0} \) disappears upon differentiation because it represents a constant term, which becomes zero. For the series given in the exercise, the first derivative \( y' \) is found this way. The patterning and behaviour of this term-wise differentiation lay the groundwork for solving differential equations using power series methods.

Power Series Representation

A power series representation of a function expresses that function as an infinite sum of terms, with each term being a coefficient times a power of \( x \). This series representation is especially powerful in solving differential equations because it can represent solutions that are otherwise difficult to express with elementary functions.

Each function has a unique set of coefficients \( \{a_n\} \) which often follow a specific pattern or rule. The reason for the effectiveness of the power series lies in its capacity to approximate functions to any desired degree of accuracy by increasing the number of terms.

Function Approximations

For example, \( e^x \) can be written as \( \sum_{n=0}^{\infty} \frac{x^n}{n!} \) which is its Maclaurin series expansion. In solving differential equations, finding such a power series solution usually involves determining a recurrence relation for the coefficients that will yield the terms of the series. Identifying this power series requires understanding the operation of differentiation and integration on these series.

Recurrence Relation

Recurrence relations are formulae that express each term of a sequence as a function of its preceding terms. They are vital in determining the coefficients \( a_n \) in the power series representation of a solution to a differential equation.

Understanding Coefficient Relationships

A recurrence relation like \( a_{n+2} = \frac{a_n}{(n+2)(n+1)} \) illustrates how each term in the power series is related to previous terms. It defines a pattern or a rule that all coefficients follow, simplifying the process of identifying all terms in the series.

In the exercise provided, after showing that \( a_{2} \) and \( a_{3} \) can be expressed in terms of \( a_{0} \) and \( a_{1} \) respectively, the recurrence relation is then proven, defining how the rest of the coefficients are determined. These relations are paramount because they allow us to describe the infinite set of series coefficients, and thereby, the function itself, by just using the first few terms and the relation itself.