Problem 47
Question
Find power series representations centered at 0 for the following functions using known power series. Give the interval of convergence for the resulting series. $$f(x)=\frac{1}{1+x^{2}}$$
Step-by-Step Solution
Verified Answer
The power series representation of the function is:
$$f(x) = \sum_{n=0}^{\infty}(-x^2)^n = 1 - x^2 + x^4 - x^6 + x^8 - \dots$$
The interval of convergence for this series is (-1, 1).
1Step 1: Comparing with the geometric series
Let's compare our function $$f(x) = \frac{1}{1+x^2}$$ with the nasic geometric series:
$$\frac{1}{1-r} = \sum_{n=0}^\infty r^n$$
To make our function similar to this form, we need to rewrite it as:
$$f(x) = \frac{1}{1-(-x^2)}$$
Now, our function is in the form of the geometric series with \(r=-x^2\).
2Step 2: Find the power series representation
Using the geometric series formula, we can now express our function as a power series:
$$f(x) =\sum_{n=0}^{\infty}(-x^2)^n$$
By expanding the summation, we have:
$$f(x) = 1 - x^2 + x^4 - x^6 + x^8 - \dots$$
3Step 3: Determine the interval of convergence
The geometric series converges when \(|-x^2| < 1\). Let's find the interval of x for which this inequality is true:
$$|-x^2| < 1$$
$$-1 < x^2 < 1$$
$$-1 < x < 1$$
So, the interval of convergence for the given power series is \((-1, 1)\).
4Step 4: Answer
The power series representation of the function $$f(x) = \frac{1}{1+x^2}$$ centered at 0 is:
$$f(x) = \sum_{n=0}^{\infty}(-x^2)^n = 1 - x^2 + x^4 - x^6 + x^8 - \dots$$
The interval of convergence for this series is \((-1, 1)\).
Key Concepts
Understanding the Interval of ConvergenceDecoding the Geometric SeriesExploring Series Representation of Functions
Understanding the Interval of Convergence
When we talk about the interval of convergence of a power series like the one we have here, we are referring to the range of values for which the series converges to a finite value.
For our specific function, we've written the function as a power series: \[ f(x) = \sum_{n=0}^{\infty}(-x^2)^n = 1 - x^2 + x^4 - x^6 + \cdots \]This is a geometric series, where each term is a power of \(-x^2\).
To find the interval of convergence, we need to determine where the series satisfies the convergence condition: \[ |r| < 1 \]where \(r\) is the common ratio.
In this case, since \(r = -x^2\), the inequality is \[ |-x^2| < 1 \].
Therefore, solving the inequality gives:
For our specific function, we've written the function as a power series: \[ f(x) = \sum_{n=0}^{\infty}(-x^2)^n = 1 - x^2 + x^4 - x^6 + \cdots \]This is a geometric series, where each term is a power of \(-x^2\).
To find the interval of convergence, we need to determine where the series satisfies the convergence condition: \[ |r| < 1 \]where \(r\) is the common ratio.
In this case, since \(r = -x^2\), the inequality is \[ |-x^2| < 1 \].
Therefore, solving the inequality gives:
- \(-1 < x^2 < 1\)
- This simplifies to \(-1 < x < 1\)
Decoding the Geometric Series
The geometric series is one of the simplest and most commonly used types of series in mathematics.
It takes the form:\[\frac{1}{1-r} = \sum_{n=0}^{\infty} r^n = 1 + r + r^2 + r^3 + \cdots\]This is only true when \(|r| < 1\).
In our exercise, to make the function \(f(x) = \frac{1}{1 + x^2}\) fit this form, we rewrite it as:\[f(x) = \frac{1}{1 - (-x^2)}\]Here, r is replaced with \(-x^2\).
This shows that each term in the function's series is \((-x^2)^n\).
It takes the form:\[\frac{1}{1-r} = \sum_{n=0}^{\infty} r^n = 1 + r + r^2 + r^3 + \cdots\]This is only true when \(|r| < 1\).
In our exercise, to make the function \(f(x) = \frac{1}{1 + x^2}\) fit this form, we rewrite it as:\[f(x) = \frac{1}{1 - (-x^2)}\]Here, r is replaced with \(-x^2\).
This shows that each term in the function's series is \((-x^2)^n\).
- By substituting different integer values for \(n\), you get the series terms: \(1, -x^2, x^4, -x^6, \ldots\)
- The signs alternate due to \((-x^2)^n\).
Exploring Series Representation of Functions
The series representation of a function is a way of expressing a function as an infinite sum.
This is especially useful in both theoretical and applied mathematics, providing a powerful way to approximate and interpret functions.
In the case of our series:\[f(x) = \sum_{n=0}^{\infty} (-x^2)^n = 1 - x^2 + x^4 - x^6 + \cdots\]the power series representation gives us an infinite polynomial form.
This polynomial-like form simplifies many operations, like differentiation and integration.
Such approximations help in calculations where exact forms are complex, essentially breaking down a complex function into manageable parts.
This is especially useful in both theoretical and applied mathematics, providing a powerful way to approximate and interpret functions.
In the case of our series:\[f(x) = \sum_{n=0}^{\infty} (-x^2)^n = 1 - x^2 + x^4 - x^6 + \cdots\]the power series representation gives us an infinite polynomial form.
This polynomial-like form simplifies many operations, like differentiation and integration.
- It shows the function as a sum of its simpler components: powers of \(x\).
- This is especially handy when dealing with calculus problems related to limits, integrals, and derivatives.
Such approximations help in calculations where exact forms are complex, essentially breaking down a complex function into manageable parts.
Other exercises in this chapter
Problem 47
Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. $$\cos 2$$
View solution Problem 47
Use properties of power series, substitution, and factoring to find the first four nonzero terms of the Taylor series centered at 0 for the following functions.
View solution Problem 47
a. Approximate the given quantities using Taylor polynomials with \(n=3\) b. Compute the absolute error in the approximation assuming the exact value is given b
View solution Problem 48
Use an appropriate Taylor series to find the first four nonzero terms of an infinite series that is equal to the following numbers. $$\sin 1$$
View solution