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Magazine Chapter 11: Problem 33
Use the binomial series to expand the function as a power series. State the radius of convergence. $$ \frac{1}{(2+x)^{3}} $$
Short Answer
Expert verified
The series is \(1 - \frac{3x}{2} + \frac{3x^2}{2} - \frac{5x^3}{4} + \cdots\), with radius of convergence 2.
Step by step solution
01
Identify the function in binomial form
To use the binomial series, identify the function in the form \((1+x)^n\) or transform it as necessary. In this problem, we have \(\frac{1}{(2+x)^3}\), which can be rewritten as \((2+x)^{-3}\). Thus, if we rewrite this as \((1 + \frac{x}{2})^{-3}\), we can use the binomial series.
02
Write the binomial series formula
The binomial series for expressing \((1 + x)^n\) is given as:\[(1 + x)^n = \sum_{k=0}^{\infty} \binom{n}{k} x^k,\]where \(\binom{n}{k}\) is the binomial coefficient, \(\binom{n}{k} = \frac{n(n-1)(n-2)...(n-k+1)}{k!}\) for integer \(k\).
03
Apply the binomial series to \((1 + \frac{x}{2})^{-3}\)
Here, \(n = -3\) and \(x = \frac{x}{2}\). Therefore, substitute into the binomial series:\[(1 + \frac{x}{2})^{-3} = \sum_{k=0}^{\infty} \binom{-3}{k} \left(\frac{x}{2}\right)^k.\]
04
Calculate the binomial coefficients
The binomial coefficient \(\binom{-3}{k}\) is calculated as:\[\binom{-3}{k} = \frac{(-3)(-3-1)(-3-2)...(-3-k+1)}{k!}.\]Using this formula, we find the first few coefficients:\ \(-3\) for \(k=1\),\ \(6\) for \(k=2\),\ \(-10\) for \(k=3\), and so on.
05
Write out the first few terms of the series
Substitute the computed coefficients into the series expansion:\[(1 + \frac{x}{2})^{-3} = 1 - \frac{3x}{2} + \frac{6x^2}{4} - \frac{10x^3}{8} + \cdots\]This simplifies to:\[1 - \frac{3x}{2} + \frac{3x^2}{2} - \frac{5x^3}{4} + \cdots\].
06
Modify back to original function's series form
Recall that \(\frac{1}{(2+x)^3} = (1 + \frac{x}{2})^{-3}\), so the power series can be written as:\[\frac{1}{(2+x)^3} = 1 - \frac{3x}{2} + \frac{3x^2}{2} - \frac{5x^3}{4} + \cdots\].
07
Determine the radius of convergence
For the binomial series expansion \((1 + u)^n\), where \(u=\frac{x}{2}\), the series converges if \(|u|<1\). Thus, \(|\frac{x}{2}|<1\) implies \(|x|<2\). Therefore, the radius of convergence is 2.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Radius of Convergence
The radius of convergence is a crucial concept when dealing with power series. It tells us the range of values for which the series converges to a specific function. For the binomial series expansion, we used the relation \( 1 + u \) where \( u = \frac{x}{2} \).
This setup implies that for the series to converge, the absolute value of \(|u| < 1\). Transladted to our formula, this becomes \(|\frac{x}{2}| < 1\).
From here, we derive that \(|x| < 2\). Hence, the radius of convergence for this specific problem is 2. This means our power series converges for any \(x\) value between -2 and 2.
This setup implies that for the series to converge, the absolute value of \(|u| < 1\). Transladted to our formula, this becomes \(|\frac{x}{2}| < 1\).
From here, we derive that \(|x| < 2\). Hence, the radius of convergence for this specific problem is 2. This means our power series converges for any \(x\) value between -2 and 2.
- Converging within \(-2 < x < 2\) ensures the function behaves predictably.
- Beyond this interval, the series does not necessarily represent our function.
Power Series
A power series is an infinite series of terms, each involving a power of a variable. In our exercise, we looked at expanding \((1+\frac{x}{2})^{-3}\) into a power series. This allows us to express complex functions as an infinite sum of simpler terms.
Power series take the form \( \sum_{k=0}^{\infty} a_k x^k \), where \( a_k \) are coefficients each multiplied by a power of \( x \). This is particularly useful for approximating functions.
Power series take the form \( \sum_{k=0}^{\infty} a_k x^k \), where \( a_k \) are coefficients each multiplied by a power of \( x \). This is particularly useful for approximating functions.
- Power series can provide good approximations for functions in complex analysis and other fields.
- It expresses calculations in a more manageable way by breaking down functions into their building block terms.
Binomial Coefficient
The binomial coefficient is a key component in the binomial series expansion. In mathematical terms, it's denoted as \( \binom{n}{k} \). It represents the number of ways to choose \( k \) elements from a set of \( n \) elements without considering the order.
In our formula \( \binom{-3}{k} = \frac{(-3)(-3-1)(-3-2)...(-3-k+1)}{k!} \), the coefficients determine the weighting of each term in the power series. The coefficient considers each time it chooses a term in the series expansion.
In our formula \( \binom{-3}{k} = \frac{(-3)(-3-1)(-3-2)...(-3-k+1)}{k!} \), the coefficients determine the weighting of each term in the power series. The coefficient considers each time it chooses a term in the series expansion.
- Each binomial coefficient corresponds to a specific term in the expanded series and changes with \( k \).
- It鈥檚 a key player that defines the contribution of each term to overall expansion.
Function Transformation
Function transformation involves rewriting functions in a way that simplifies them or makes them fit special forms, like the binomial series form in our case. It is an important concept when expanding functions into a power series as sometimes a direct fit isn't possible.
For the problem, we began with \( \frac{1}{(2+x)^3} \) and transformed it into \( (1 + \frac{x}{2})^{-3} \). Such transformations facilitate the application of series or algorithms that require a specific starting format. Modifying this form allowed us to use the binomial series directly.
For the problem, we began with \( \frac{1}{(2+x)^3} \) and transformed it into \( (1 + \frac{x}{2})^{-3} \). Such transformations facilitate the application of series or algorithms that require a specific starting format. Modifying this form allowed us to use the binomial series directly.
- Always look to transform functions to a familiar form when expanding them.
- Modification may involve factoring, changing variables, or algebraic manipulation.
