The Fourier series of the periodic function $f(x) = |x|$, $-1 < x < 1$, $f(x + 2) = f(x)$, $x \in \mathbb{R}$ is given by $$\frac{1}{2} - \sum_{n=1}^{\infty} \frac{4 \cos((2n-1)\pi x)}{(2n-1)^2 \pi^2}$$ Using the above, the sum of the infinite series $1 + \frac{1}{3^2} + \frac{1}{5^2} + \dots$ is
The Fourier series for the periodic function $f(x) = |x|$, with period 2, is given as:
$f(x) = \frac{1}{2} - \sum_{n=1}^{\infty} \frac{4 \cos((2n-1)\pi x)}{(2n-1)^2 \pi^2}$We need to find the sum $S = 1 + \frac{1}{3^2} + \frac{1}{5^2} + \dots$. This is the sum of reciprocals of squares of odd numbers.
Consider the function value at $x=0$. Since $f(x)=|x|$, we have $f(0)=|0|=0$.
Substitute $x=0$ into the Fourier series:
$f(0) = \frac{1}{2} - \sum_{n=1}^{\infty} \frac{4 \cos((2n-1)\pi \cdot 0)}{(2n-1)^2 \pi^2}$Using $\cos(0) = 1$:
$0 = \frac{1}{2} - \sum_{n=1}^{\infty} \frac{4}{(2n-1)^2 \pi^2}$Isolate the summation term:
$\sum_{n=1}^{\infty} \frac{4}{(2n-1)^2 \pi^2} = \frac{1}{2}$Factor out constants:
$\frac{4}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{1}{2}$Solve for the summation:
$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{1}{2} \times \frac{\pi^2}{4} = \frac{\pi^2}{8}$This summation represents the series $1 + \frac{1}{3^2} + \frac{1}{5^2} + \dots$.
Thus, the sum of the series is $\frac{\pi^2}{8}$.
If we use the Fourier transform ϕ(x, y) = \(\int {{{\rm{e}}^{{\rm{ikx}}}}} {ϕ _{\rm{k}}}\left( {\rm{y}} \right){\rm{dk}}\) to solve the partial differential equation \({\rm{ - }}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {y^2}}}\, - \,\frac{1}{{{y^2}}}\frac{{{\partial ^2}ϕ \left( {x,y} \right)}}{{\partial {x^2}}} + \frac{{{m^2}}}{{{y^2}}}ϕ \left( {x,y} \right) = 0\) in the half-plane {(x, y) : -∞ < x < ∞, 0 < y < ∞} the Fourier modes ϕ k(y) depend on y as y α and y β . The values of α and β are
When a time-domain signal is converted into its Fourier representation, which of the following is/are conserved?
I. Energy
II. Power
The trigonometric Fourier series of a periodic time function can have
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The Fourier series to represent x-x2 for –π ≤ x ≤ π is given by \(x - {x^2} = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}cosnx + \mathop \sum \limits_{n = 1}^\infty {b_n}sinnx\)
The value of a0 (round off to two decimal places), is