You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Engineering Mathematics

Fourier Series 1 (Convergence Problems)

Let g: [0, ∞ ...

Question

Let g: $[0, \infty) \to [0, \infty)$ be a function defined by g(x) = x - [x], where [x] represents the integer part of x. (That is, it is the largest ineger which is less than or equal to x). The value of the constant term in the Fourier series expansion of g(x) is
0.5

Open in App

Solution

The correct option is A 0.5
Here we will use Case I of main definition of Fourier series because g(x) = x -[x] is periodic function with period 1 so we have

$(0.2 l) \approx (0, 1) \Rightarrow l = \frac{1}{2}$
So, $a_{0} = \frac{1}{l} \int_{0}^{2 l} f (x) d x$
$= \frac{1}{1 / 2} \int_{0}^{1} (x - [x]) d x$
$= 2 \int_{0}^{1} x d x = 1$
Hence, constant term in Fourier series $= \frac{a_{0}}{2} = \frac{1}{2}$

Suggest Corrections

Similar questions

Let f: R → R be the Signum Function defined as

and g: R → R be the Greatest Integer Function given by g(x) = [x], where [x] is greatest integer less than or equal to x. Then does fog and gof coincide in (0, 1]?

Q. Let

f : R \to R

be the Signum function defined as

⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 1, & x > 0 0, & x = 0 - 1, & x < 0 \end{matrix}

and

g : R \to R

be the Greatest Integer Function given by

g (x) = [x]

, where

[x]

is greatest integer less than or equal to

x

. Then, does

f o g

and

g o f

coincide in

(0, 1]

Let $f : R \to R$ be the signum function defined as

$f (x) = ⎧ ⎨ ⎩ \begin{matrix} 1, & x > 0 0, & x = 0 - 1, & x < 0 \end{matrix}$

and $g : R \to R$ be the greatest
integer function given by $g (x) = [x]$ , where $[x]$ is greatest
integer less than or equal to $x$ . Then, does $f o g$ and $g o f$ coincide in
$(0, 1]$ ?

Q. Let f : R → R be the Signum Function defined as and g : R → R be the Greatest Integer Function given by g ( x ) = [ x ], where [ x ] is greatest integer less than or equal to x . Then does f o g and g o f coincide in (0, 1]?

Q. Let

g (x) = 1 + x - [x] and f (x) = \{\begin{cases} - 1, & x < 0 \\ 0, & x = 0, \\ 1, & x > 0 \end{cases}

, where [x] denotes the greatest integer less than or equal to x. Then for all

x, f (g (x))

is equal to
(a) x
(b) 1
(c) f(x)
(d) g(x)

Join BYJU'S Learning Program

Let g: [0,∞)→[0,∞) be a function defined by g(x) = x - [x], where [x] represents the integer part of x. (That is, it is the largest ineger which is less than or equal to x). The value of the constant term in the Fourier series expansion of g(x) is0.5

The correct option is A 0.5Here we will use Case I of main definition of Fourier series because g(x) = x -[x] is periodic function with period 1 so we have (0.2l)≈(0,1)⇒l=12 So, a0=1l∫2l0f(x)dx =11/2∫10(x−[x])dx =2∫10xdx=1 Hence, constant term in Fourier series =a02=12

Let g: $[0, \infty) \to [0, \infty)$ be a function defined by g(x) = x - [x], where [x] represents the integer part of x. (That is, it is the largest ineger which is less than or equal to x). The value of the constant term in the Fourier series expansion of g(x) is
0.5