Obtain the Fourier series of the function defined by f(x) = x2, -1 <= x <= 1. Obtain the Fourier series of the function defined by f(t) = 1 + t/4, -4 <= t <= 0; 1 - t/4, 0 <= t <= 4 . Sketch f(t) in the interval -8 <= t <= 8. Determine the Fourier series expansion of f(t).

(a)

Step 1: The forward difference is defined as
$\Delta f(x) = f(x+1) - f(x).$

Step 2: For $f(x) = x^3$, first find $f(x+1)$:
$f(x+1) = (x+1)^3.$

Step 3: Expand $(x+1)^3$ using the binomial theorem:
$(x+1)^3 = x^3 + 3x^2(1) + 3x(1)^2 + (1)^3 = x^3 + 3x^2 + 3x + 1.$

Step 4: Substitute into the difference formula:
$\Delta f(x) = (x^3 + 3x^2 + 3x + 1) - x^3.$

Step 5: Simplify:
$\Delta f(x) = 3x^2 + 3x + 1.$

$\boxed{\Delta f(x) = 3x^{2} + 3x + 1}$

(b)

Step 1: The forward difference is defined as
$\Delta f(x) = f(x+1) - f(x).$

Step 2: For $f(x) = a^{x}$ where $a > 0$, $a \neq 1$, first find $f(x+1)$:
$f(x+1) = a^{x+1}.$

Step 3: Rewrite $a^{x+1}$:
$a^{x+1} = a \cdot a^{x}.$

Step 4: Substitute into the difference formula:
$\Delta f(x) = a \cdot a^{x} - a^{x}.$

Step 5: Factor out the common term $a^{x}$:
$\Delta f(x) = a^{x}(a - 1).$

Step 6: Note that this can also be written as
$\Delta f(x) = (a-1) f(x),$
since $f(x) = a^{x}$.

$\boxed{\Delta f(x) = (a-1)a^{x}}$