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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 11 - Power SeriesProblema 11.4.33
Capitolo 11, Problema 11.4.33

Differential equations


a. Find a power series for the solution of the following differential equations, subject to the given initial condition
b. Identify the function represented by the power series.


y′(t) − y = 0, y(0) = 2

Guida verificata passo dopo passo
1
Rewrite the differential equation in a form suitable for power series substitution: \(y'(t) - y(t) = 0\) with initial condition \(y(0) = 2\).
Assume a power series solution of the form \(y(t) = \sum_{n=0}^{\infty} a_n t^n\), where \(a_n\) are coefficients to be determined.
Differentiate the power series term-by-term to find \(y'(t) = \sum_{n=1}^{\infty} n a_n t^{n-1}\).
Substitute \(y(t)\) and \(y'(t)\) back into the differential equation to get \(\sum_{n=1}^{\infty} n a_n t^{n-1} - \sum_{n=0}^{\infty} a_n t^n = 0\).
Align powers of \(t\) by shifting indices as needed, then equate coefficients of like powers of \(t\) to form a recurrence relation for \(a_n\). Use the initial condition \(y(0) = a_0 = 2\) to find the coefficients.

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Power Series Solutions to Differential Equations

A power series solution expresses the solution of a differential equation as an infinite sum of powers of the independent variable. This method involves assuming a solution in the form of a series and determining the coefficients by substituting into the differential equation. It is especially useful when closed-form solutions are difficult to find.
Video consigliato:
05:58
Intro to Power Series

Initial Conditions and Their Role

Initial conditions specify the value of the solution and possibly its derivatives at a particular point, allowing us to find unique coefficients in the power series. For example, y(0) = 2 sets the constant term in the series, ensuring the solution matches the given starting value.
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Percorso guidato
05:03
Initial Value Problems

Identifying Functions from Power Series

Once a power series solution is found, it can often be recognized as a known function by comparing it to standard series expansions (e.g., exponential, sine, cosine). This identification helps express the solution in a closed form, making it easier to interpret and use.
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07:32
Representing Functions as Power Series
Pratica correlata
Domanda del libro di testo

L'Hôpital's Rule by Taylor series Suppose f and g have Taylor series about the point a.

a. If f(a) = g(a) = 0 and g′(a) ≠ 0, evaluate lim ₓ→ₐ f(x)/g(x) by expanding f and g in their Taylor series. Show that the result is consistent withl’Hôpital’s Rule.

b. If f(a) = g(a) =f′(a) = g′(a) = 0 and g′′(a) ≠ 0, evaluate lim ₓ→ₐ f(x)/g(x) by expanding f and g in their Taylor series. Show that the result is consistent with two applications of 1'Hôpital's Rule.

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Domanda del libro di testo

{Use of Tech} Approximating powers Compute the coefficients for the Taylor series for the following functions about the given point a, and then use the first four terms of the series to approximate the given number.

f(x) =∛x with a=64; approximate ∛60.

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Domanda del libro di testo

Manipulating Taylor series Use the Taylor series in Table 11.5 to find the first four nonzero terms of the Taylor series for the following functions centered at 0.


sinh x²

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Domanda del libro di testo

Remainders Find the remainder Rₙ for the nth−order Taylor polynomial centered at a for the given functions. Express the result for a general value of n.


f(x) = 1/(1 - x), a=0

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Domanda del libro di testo

Manipulating Taylor series Use the Taylor series in Table 11.5 to find the first four nonzero terms of the Taylor series for the following functions centered at 0.


1/(1 − 2x)

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Domanda del libro di testo

Working with binomial series Use properties of power series, substitution, and factoring to find the first four nonzero terms of the Maclaurin series for the following functions. Use the Maclaurin series


(1 + x)⁻² = 1 − 2x + 3x² − 4x³ + ⋯, for −1 < x < 1.


(1 + 4x)⁻²

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