Fourier Series: f(x) = 2(x-1) for 1<=x<2, 0 for 2<=x<=3; find value using series

[moonIight]

Fourier Series: f(x) = 2(x-1) for 1<=x<2, 0 for 2<=x<=3; find value using series

For a periodic function where in one period,

. . . . .\(\displaystyle f(x)\, =\, \begin{cases}2\, (x\, -\, 1),&\,1\, \leq\, x\, < 2\\0,&\,2\, \leq\, x\, \leq\, 3\end{cases}\)

What is the value of the following series, using the Fourier Series derived from above?

. . . . .\(\displaystyle 1\, -\, \dfrac{1}{3}\, +\, \dfrac{1}{5}\, -\, \dfrac{1}{7}\, +\, \dfrac{1}{9}\, -\, \dfrac{1}{11}\, +\, ...\, \dfrac{(-1)^{(n-1)/2}}{n}\)

(-1)^[(n-1)/2] / n

Really struggling to solve this problem.. Would greatly appreciate it if full solution/guide would be provided

 

[HallsofIvy]

Since the problem says "using the Fourier series derived from above", have you determined what that Fourier series is?

Is it finding that Fourier series that is the difficulty? It has been a long time since I have worked with Fourier series but I looked on "Wikipedia"
https://en.wikipedia.org/wiki/Fourier_series
where it gave the formula for the coefficients of "cosine" to be
\(\displaystyle a_n= \frac{2}{P}\int_{x_0}^{x_0+ P} f(x)cos\left(\frac{2\pi nx}{P}\right) dx\)
and the formula for the coefficients of "sine" to be
\(\displaystyle b_n= \frac{2}{P}\int_{x_0}^{x_0+ P} f(x)sin\left(\frac{2\pi nx}{P}\right) dx\)

Here, since the period of f, from 1 to 3 is 2, \(\displaystyle x_0= 1\) and P= 2. Technically, the integral goes from 1 to 3 but since f is defined to be 0 from 2 to 3, we really only integrate from 1 to 2 (but still use P= 2):
\(\displaystyle a_n= \int_1^2 2(x- 1) cos\left(\frac{2\pi nx}{P}\right) dx\) and
\(\displaystyle b_n= \int_1^2 2(x- 1) sin\left(\frac{2\pi nx}{P}\right) dx\)

I would integrate those using "integration by parts, taking \(\displaystyle u= 2(x- 1)\) and \(\displaystyle dv= cos\left(\frac{2\pi nx}{P}\right) dx\) in the first and \(\displaystyle dv= sin\left(\frac{2\pi nx}{P}\right) dx\) in the second.