Math, asked by rxjivmishra, 2 months ago

find the fourier sine and cosine series of function f(x)=x in interval 0

Answers

Answered by monikachouksey72
3

Answer:

Step-by-step explanation:

10.4 Fourier Cosine and Sine Series

To solve a partial differential equation, typically we represent a function by a trigonometric series consisting

of only sine functions or only cosine functions.

Recall that the Fourier series for an odd function defined on [−L, L] consists entirely of sine terms. Thus

we might achieve

f(x) = X∞

n=1

an sin

nπx

L

(1)

by artificially extending the function f(x), 0 < x < L to the interval (−L, L) in such a way that the extended

function is odd. That is,

fo(x) =

f(x), 0 < x < L,

−f(−x), −L < x < 0,

and extending fo(x) to all x using 2L-periodicity. fo(x) is an extension of f(x) because fo(x) = f(x) on

(0, L). This extension is called the odd 2L-periodic extension of f(x). The resulting Fourier series expansion

is called a half-range expansion for f(x) because it represents the function f(x) on (0, L).

Similarly, the even 2L-periodic extension of f(x) as the function

fe(x) =

f(x), 0 < x < L,

f(−x), −L < x < 0,

with fe(x + 2L) = fe(x).

To illustrate the various extensions, let’s consider the function f(x) = x, 0 < x < π. If we extend f(x)

to the interval (−π, π) using π-periodicity, then the extension f is given by

fe(x) =

x, 0 < x < π

x + π, −π < x < 0,

with fe(x + 2π) = fe(x). In the previous quiz we saw that the Fourier series for fe(x) is

fe(x) ∼

π

2

X∞

k=1

1

k

sin 2kx,

which consists of both odd functions (the sine terms) and even functions (the constant term), because the

π-periodic extension fe(x) is neither an even nor an odd function. The odd 2π-periodic extension of f(x) is

just fo(x) = x, −π < x < π, which has the Fourier series expansion

fo(x) ∼ 2

X∞

n=1

(−1)n+1

n

sin nx. (2)

Because fo(x) = f(x) on the interval (0, π), the expansion in (2) is a half-range expansion for f(x). The

even 2π-periodic extension of f(x) is the function fe(x) = |x|, −π < x < π, which has the Fourier series

expansion

fe(x) = π

2

4

π

X∞

k=1

1

(2k − 1)2

cos(2k − 1)x (3)

(see Example 2 in §10.3 lecture notes).

The preceding three extensions, the π-periodic function fe(x), the odd 2π-periodic function fo(x), and

the even 2π-periodic function fe(x), are natural extensions of f(x). The Fourier series expansions for fo(x)

and fe(x), given in (2) and (3) equal f(x) on the interval (0, π). This motivates the following definitions.

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