Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}=\sum_{n=0}^{\infty} b_{n} x^{n}\) for \(|x|

Step-by-Step Solution

Answer

Coefficients are equal: for all \(n\), \(a_n = b_n\).

Step 1: Set Equation at Zero

Substitute \(x = 0\) into the given equation: \(f(x) = \sum_{n=0}^{\infty} a_{n} x^{n} = \sum_{n=0}^{\infty} b_{n} x^{n}\). At \(x = 0\), both sides simplify to \(a_0 = b_0\). This shows that the coefficients for the constant term are equal.

Step 2: Differentiate Both Series

Differentiate the function defined by the series term by term. By differentiating, we get \(f'(x) = \sum_{n=1}^{\infty} a_{n} n x^{n-1} = \sum_{n=1}^{\infty} b_{n} n x^{n-1}\). This holds for \(|x|

Step 3: Evaluate the Derivative at Zero

Let \(x = 0\) in the differentiated equation. This gives us \(f'(0) = a_1 = b_1\). At \(x = 0\), only the first derived terms \(a_1 x^0\) and \(b_1 x^0\) contribute, so the first-order coefficients are equal.

Step 4: Repeat Differentiation

Continue differentiating the series to find higher coefficients. The second derivative is \(f''(x) = \sum_{n=2}^{\infty} a_{n} n(n-1) x^{n-2} = \sum_{n=2}^{\infty} b_{n} n(n-1) x^{n-2}\). Evaluate at \(x = 0\) to get \(f''(0) = 2! a_2 = 2! b_2\), leading to \(a_2 = b_2\).

Step 5: Generalize the Process

By continuing this process, the \(k\)-th derivative evaluated at \(x = 0\) is \(f^{(k)}(0) = k! a_k = k! b_k\). This implies that \(a_k = b_k\). Since this holds for all \(k\), we deduce that \(a_n = b_n\) for all \(n\).

Key Concepts

Differentiation

The concept of differentiation is fundamental in calculus and it allows us to find the rate at which a function is changing at any given point. In the case of power series, we can differentiate the series term by term. This particular exercise involves setting the functions equal and differentiating with respect to the variable, to explore the coefficients' behaviour.

• The power series is a way to represent a function as an infinite sum of terms. Each term is of the form \(a_n x^n\).
• By differentiating term by term, we obtain: \(f'(x) = \sum_{n=1}^{\infty} a_n n x^{n-1}\). This is because the differentiation of \(x^n\) becomes \(n x^{n-1}\).

Differentiation helps discover relationships between coefficients when we evaluate the derivatives at specific points, such as \(x = 0\). This enables us to compare and equate coefficients from two different series.

Derivatives

Derivatives measure change and are central to not only calculus but also in this problem of determining equality in power series. They are calculated by repeatedly differentiating the original function.

• The first derivative of a power series gives the slope or rate of change which is represented as another power series.
• Subsequent derivatives, like the second derivative, result in terms like \(f''(x) = \sum_{n=2}^{\infty} a_n n(n-1) x^{n-2}\), giving insight into higher-order behavior.

By employing derivatives at \(x=0\), we can extract individual coefficients from a power series. Doing this systematically confirms that each corresponding coefficient in the two given series is identical.

Infinite Series

An infinite series is a series that extends indefinitely. Terms are systematically derived from a function, often giving a concise way to express complex functions in mathematical terms. This exercise employs such series to explore function equality.

• A power series is a specific type of infinite series, focused on powers of a variable \(x\).
• They are essential in function representation, especially within a radius of convergence \(|x| < R\), ensuring the series captures the function accurately.

For the exercise, we had a condition \(|x| < R\) ensuring convergence. By looking at the sum of the series, we were able to employ techniques such as differentiation to verify equality term by term, demonstrating how powerful the manipulation of infinite series can be when analyzing functions.

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Problem 30

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Problem 31

Other exercises in this chapter

Problem 30

\(\sum_{n=1}^{\infty} \frac{5^{2 n}}{n !}\)

View solution

Problem 30

In Problems 13–30, classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=1}^{\infty}(-1)^{n+1} \sin \frac{\pi}{n} $$

View solution

Problem 31

Find the radius of convergence of $$ \sum_{n=1}^{\infty} \frac{1 \cdot 2 \cdot 3 \cdots n}{1 \cdot 3 \cdot 5 \cdots(2 n-1)} x^{2 n+1} $$

View solution

Problem 31

By writing \(1 / x=1 /[1-(1-x)]\) and using the known expansion of \(1 /(1-x)\), find the Taylor series for \(1 / x\) in powers of \(x-1\).

View solution

Key Concepts

Differentiation

The concept of differentiation is fundamental in calculus and it allows us to find the rate at which a function is changing at any given point. In the case of power series, we can differentiate the series term by term. This particular exercise involves setting the functions equal and differentiating with respect to the variable, to explore the coefficients' behaviour.

• The power series is a way to represent a function as an infinite sum of terms. Each term is of the form \(a_n x^n\).
• By differentiating term by term, we obtain: \(f'(x) = \sum_{n=1}^{\infty} a_n n x^{n-1}\). This is because the differentiation of \(x^n\) becomes \(n x^{n-1}\).

Differentiation helps discover relationships between coefficients when we evaluate the derivatives at specific points, such as \(x = 0\). This enables us to compare and equate coefficients from two different series.

Derivatives

Derivatives measure change and are central to not only calculus but also in this problem of determining equality in power series. They are calculated by repeatedly differentiating the original function.

• The first derivative of a power series gives the slope or rate of change which is represented as another power series.
• Subsequent derivatives, like the second derivative, result in terms like \(f''(x) = \sum_{n=2}^{\infty} a_n n(n-1) x^{n-2}\), giving insight into higher-order behavior.

By employing derivatives at \(x=0\), we can extract individual coefficients from a power series. Doing this systematically confirms that each corresponding coefficient in the two given series is identical.

Infinite Series

An infinite series is a series that extends indefinitely. Terms are systematically derived from a function, often giving a concise way to express complex functions in mathematical terms. This exercise employs such series to explore function equality.

• A power series is a specific type of infinite series, focused on powers of a variable \(x\).
• They are essential in function representation, especially within a radius of convergence \(|x| < R\), ensuring the series captures the function accurately.

For the exercise, we had a condition \(|x| < R\) ensuring convergence. By looking at the sum of the series, we were able to employ techniques such as differentiation to verify equality term by term, demonstrating how powerful the manipulation of infinite series can be when analyzing functions.