Consider the function f(x) = 1 – (x – 1)² defined on the interval 0 < x < 2. (a) Derive a general expression for the coefficients in the Fourier Cosine series for f(x). Then write out the Fourier series through the first four nonzero terms. (b) Graph the extension of f(x) on the interval (-6, 6) that represents the pointwise convergence of the Cosine series in (a). At jump discontinuities, identify the value to which the series converges.

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.4: Values Of The Trigonometric Functions
Problem 22E
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t) ~+2 (2k – 1)n sin(2k – 1)t) = ;
1. Consider the function f(x) =1- (x – 1)? defined on the interval 0 < x < 2.
(a) Derive a general expression for the coefficients in the Fourier Cosine series for f(x). Then write
out the Fourier series through the first four nonzero terms.
(b) Graph the extension of f(x) on the interval (-6, 6) that represents the pointwise convergence of
the Cosine series in (a). At jump discontinuities, identify the value to which the series converges.
2. Consider the function f(x) defined on 0 < x < 2 (see graph).
(a) Derive a general expression for the coefficients in the
Fourier Sine series, f(x) ~ ) b, sin(nxx/2).
f(x)
n=1
(b) Graph the extension of f(x) on the interval (-6, 6) that
represents the pointwise convergence of the Sine series.
At jump discontinuities, identify the value to which the
series converges.
See page 2 for Problem 3.
3. Linear Oscillator with Periodic Forcing.
Consider the second-order differen-
1
tial equation, L[y] = y" + 2y = f(t),
where f(t) is the periodic square
wave shown in the figure.
-37
The Fourier trigonometric series for f(t) is given by,
).
2
2
sin(t) +
sin(3t)
sin(5t)
f(t) -
+..
+
k=1
(a) Derive a Fourier series representation for a particular solution to L[y] = f(t), as follows:
• For k = 0, 1, 2, .., find a particular solution y(t) satisfying y" + 2yk = f:(t), where fa(t)
is the kth term in the Fourier series for f(t).
• Sum over all yk(t) to get the solution yp(t).
(b) Write out the first four terms in the Fourier series for yp(t).
1
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Transcribed Image Text:t) ~+2 (2k – 1)n sin(2k – 1)t) = ; 1. Consider the function f(x) =1- (x – 1)? defined on the interval 0 < x < 2. (a) Derive a general expression for the coefficients in the Fourier Cosine series for f(x). Then write out the Fourier series through the first four nonzero terms. (b) Graph the extension of f(x) on the interval (-6, 6) that represents the pointwise convergence of the Cosine series in (a). At jump discontinuities, identify the value to which the series converges. 2. Consider the function f(x) defined on 0 < x < 2 (see graph). (a) Derive a general expression for the coefficients in the Fourier Sine series, f(x) ~ ) b, sin(nxx/2). f(x) n=1 (b) Graph the extension of f(x) on the interval (-6, 6) that represents the pointwise convergence of the Sine series. At jump discontinuities, identify the value to which the series converges. See page 2 for Problem 3. 3. Linear Oscillator with Periodic Forcing. Consider the second-order differen- 1 tial equation, L[y] = y" + 2y = f(t), where f(t) is the periodic square wave shown in the figure. -37 The Fourier trigonometric series for f(t) is given by, ). 2 2 sin(t) + sin(3t) sin(5t) f(t) - +.. + k=1 (a) Derive a Fourier series representation for a particular solution to L[y] = f(t), as follows: • For k = 0, 1, 2, .., find a particular solution y(t) satisfying y" + 2yk = f:(t), where fa(t) is the kth term in the Fourier series for f(t). • Sum over all yk(t) to get the solution yp(t). (b) Write out the first four terms in the Fourier series for yp(t). 1 Dashboard Calendar To Do Notifications Inbox
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