Question

find the Fourier series to represent f(x) = x² -2 for -2≤x≤2.​

Answer 1

Answer:

Zeroes of Polynomial :

x = √2 and x = (-√2)

Step-by-step explanation:

If f(x) = x² - 2 is a quadratic polynomial

then, let us factorise it now :

We will apply identity here i.e., explained as follows :

(a - b)(a + b) = a² - b²

Now, let us solve it accordingly :

= x² - 2

= (x - √2)(x + √2)

Such That,

x - √2 = 0

x = √2

And,

x + √2 = (-√2 )

Verification :

a (coefficient of x²) = 1

b (coefficient of x) = 0

c (constant term) = (-2)

Consider,

@ = alpha and, ß = beta

Sum of Zeroes = - b/a

= - (0)/1

= 0/1

= 0

@ + ß = √2 - √2

= 0

Product of Zeroes = c/a

= (-2)/1

= -2

@ß = (-√2)(√2) = (-2)

Hence,

Sum of Zeroes = @ + ß = 0

Product of Zeroes = @ß = (-2)

Answer 2

Step-by-step explanation:

equation of Fourier series

f(x)= a°/2 +£xturns to 1to infinity an cos(n pi x/l)+ sigma 1to infinity bn sin n pi x /l)

where a° = 1/l Integral (limit) xdx

an = 1/l Integral (limit) x cos(n pi x /l) dx

bn = 1/l Integral (limit) x sin (n pi x/l) dx

here x^2- 2 is an even function