Suppose you know the Maclaurin series for f and that it converges to f(x) for |x|\u003c1. How do you find the Maclaurin series for f(x²) and where does it converge?

Ch. 11 - Power Series

Briggs - Calculus: Early Transcendentals 3rd Edition

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Briggs 3rd Edition

Ch. 11 - Power Series

Problem 11.3.5

Chapter 11, Problem 11.3.5

Suppose you know the Maclaurin series for f and that it converges to f(x) for |x|<1. How do you find the Maclaurin series for f(x²) and where does it converge?

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Recall that the Maclaurin series for a function \(f(x)\) is given by \(f(x) = \sum_{n=0}^{\infty} a_n x^n\), where \(a_n = \frac{f^{(n)}(0)}{n!}\) and the series converges for \(|x| < 1\).

To find the Maclaurin series for \(f(x^2)\), substitute \(x^2\) in place of \(x\) in the original series. This gives \(f(x^2) = \sum_{n=0}^{\infty} a_n (x^2)^n\).

Simplify the powers: \((x^2)^n = x^{2n}\), so the series becomes \(f(x^2) = \sum_{n=0}^{\infty} a_n x^{2n}\).

Analyze the radius of convergence for the new series. Since the original series converges for \(|x| < 1\), the substitution \(x^2\) means the new series converges when \(|x^2| < 1\), which simplifies to \(|x| < 1\).

Therefore, the Maclaurin series for \(f(x^2)\) is \(\sum_{n=0}^{\infty} a_n x^{2n}\), and it converges for \(|x| < 1\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Maclaurin Series

A Maclaurin series is a Taylor series expansion of a function about zero, expressed as an infinite sum of terms involving derivatives of the function at zero. It represents the function as f(x) = Σ (f⁽ⁿ⁾(0)/n!) xⁿ, valid within its radius of convergence.

Substitution in Power Series

To find the Maclaurin series for f(x²), substitute x² in place of x in the original series. This changes each term xⁿ to (x²)ⁿ = x^(2n), effectively altering the powers and coefficients but preserving the series structure.

Radius of Convergence

The radius of convergence defines the interval where the power series converges to the function. For f(x²), the convergence depends on |x²| < 1, which implies |x| < 1, so the radius of convergence remains the same as the original series.

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Radius of Convergence

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{Use of Tech} A different kind of approximation When approximating a function f using a Taylor polynomial, we use information about f and its derivative at one point. An alternative approach (called interpolation) uses information about f at several different points. Suppose we wish to approximate f(x)=sin x on the interval [0, π].

a. Write the (quadratic) Taylor polynomial p₂ for f centered at π/2.

b. Now consider a quadratic interpolating polynomial q(x) = ax² + bx + c. The coefficients a, b, and c are chosen such that the following conditions are satisfied:

q(0) = f(0), q(π/2) = f(π/2), and q(π) = f(π)

Show that q(x) = −(4/π²)x² + (4/π)x.

c. Graph f, p₂, and q on [0, π].

d. Find the error in approximating f(x) = sin x at the points π/4, π/2, 3π/4, and π using p₂ and q.

e. Which function, p₂ or q, is a better approximation to f on [0, π]? Explain.

Textbook Question

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Textbook Question

Radius and interval of convergence Determine the radius and interval of convergence of the following power series.

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Textbook Question

Any method

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients.

b. Determine the radius of convergence of the series.

f(x) = (1 + x²)⁻²/³