Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

find the power series representation for \(f(x)\) and specify the radius of convergence. Each is somehow related to a geometric series. $$ f(x)=\frac{1}{(1+x)^{2}} $$

Step-by-Step Solution

Answer

The power series representation for \(f(x)\) is \(\sum_{n=0}^{\infty} (-1)^n(n+1)x^n\), with radius of convergence \( R = 1 \).

Step 1: Recall the geometric series

The formula for the sum of a geometric series is given by \( \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}\) for \(|r| < 1\). This formula gives us a way to rewrite functions in terms of a power series.

Step 2: Rewrite the function

Notice that \(f(x) = \frac{1}{(1+x)^2}\) can be rewritten by considering \((1+x)^{-2}\) as the function to be expanded. The task at hand is to express this function in a form similar to a geometric series.

Step 3: Use a known series expansion

The function \((1+x)^{-2}\) can be expanded using the Binomial series: \((1+x)^k = \sum_{n=0}^{\infty} \binom{k}{n} x^n\), which means: \( (1+x)^{-2} = \sum_{n=0}^{\infty} (-1)^n\binom{-2}{n} x^n \).

Step 4: Simplify the binomial coefficients

For negative binomial series, the coefficient \( \binom{-2}{n} = \frac{(-2)(-3)...(-2-n+1)}{n!} \). For small values of \(n\), these coefficients simplify to provide terms for our series expansion. Calculate these as needed.

Step 5: Calculate the Radius of Convergence

For the binomial expansion, the series converges when \(|x| < 1\), which directly gives us the radius of convergence, \( R = 1 \).

Key Concepts

Geometric Series

The concept of a geometric series is fundamental when exploring power series and series expansions. A geometric series is essentially a sum of terms where each term is a constant multiple, or ratio, of the previous one. Given by the formula \\[\sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}\]\provided \(|r| < 1\), it allows for mathematical wonder, like representing repetitive patterns or growth, to take a compact form. Here, \('a'\) is the first term and \('r'\) is the common ratio.\

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• If the series converges, it can simplify complex quantities.
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• Its convergence depends on whether the magnitude of \(r\) is less than one.
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• This criterion highlights the role of terms shrinking or growing with each addition.
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\For the function \(f(x) = \frac{1}{(1+x)^2}\), expressing it in such a manner represents slicing down complexities. Initially, it doesn't appear geometric due to the squared denominator term, which makes direct basic geometric application tricky, calling for adaptations.

Radius of Convergence

In the context of power series, especially when using a geometric or binomial approach, understanding the radius of convergence is vital. It essentially denotes the "sphere" within which the series behaves nicely and converges.\
For a series expanded in powers of \(x\), it will converge when \(|x| < R\), where \(R\) is the radius of convergence. Being crucial for ensuring valid and meaningful expansions, it plays a big role in functional representations.\
In this specific solution exploring \(f(x) = \frac{1}{(1+x)^2}\), we observe that the radius of convergence is derived from the form of binomial expansions, leading to\\[ R = 1 \] \

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• It indicates that for any \(x\) such that \(-1 < x < 1\), the series expansion converges.
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• This forms a safe boundary for legitimate function approximation.
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• Beyond this interval, the approximation could potentially "break down," losing convergence and thereby reliability.
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Binomial Series

The binomial series is a powerful tool when dealing with expressions of the form \( (1+x)^k \). This series, an extension of the geometric series, uses binomial coefficients to break down expressions, especially when \( k \) is not a positive integer.\
The generalized binomial theorem tells us that \((1+x)^k\) equals\\[ \sum_{n=0}^{\infty} \binom{k}{n} x^n \] \For the instance \( k = -2\), it helps articulate \(f(x) = \frac{1}{(1+x)^2}\) more clearly using a power series.\

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• Binomial series allow us to work conveniently with non-integer powers.
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• They enable simplifications and expansions that geometric or simple series cannot.
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• This form is especially useful in analytic contexts, lending itself to application in both pure and applied mathematics.
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Negative Binomial Coefficients

Negative binomial coefficients, such as those encountered in the expansion of \( (1+x)^{-2} \), provide a way to extend the binomial theorem to negative indices. These coefficients are computed via: \[ \binom{-2}{n} = \frac{(-2)(-3)...(-2-n+1)}{n!} \] \These are crucial when handling series with negative exponents, offering a systematic approach to deriving coefficients in such cases.\

For small values of \(n\), these coefficients give specific terms of the series:

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• When \( n = 0 \), \ \( \binom{-2}{0} = 1 \).
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• For \( n = 1 \), \ \( \binom{-2}{1} = -2 \).
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• And similarly, this pattern continues, showing precise adjustments needed due to the negative exponent.
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These coefficients provide clarity and structure, which is essential in handling expansions like the series for \( f(x) = \frac{1}{(1+x)^{2}} \) to derive interpretative and computational precision.