Fourier series of T periodic function

Question

Calculate the fourier series of

$$ f(t)= \begin{cases} 2t+1, -1 \le t < 1\\ \text{continued periodically}\\ \end{cases} $$

f is neither even nor odd, so we must assume that $a_k$ and $b_k$ are not equal to $0$.

$a_0 = \int_{-1}^{1}(2t+1)dt = 2$

Since $T=2$ we have

$$a_k = \frac{2}{2} \int_{-1}^{1}2t \cos(k \pi t)dt + \int_{-1}^{1}\cos(k \pi t)dt$$

The second integral becomes = $0$ since,

$$\int_{-1}^{1}\cos(k \pi t) = \frac{\sin(\pi k t)}{\pi k} |_{-1}^{1} = 0 $$ because $\sin(\pi k) = 0$ for all k and so is $\sin(-\pi k) = 0$.

The first integral

$$ \int_{-1}^{1}2t \cos(\pi k t)dt = 2t \frac{\sin(\pi k t)}{\pi k}\bigl|_{-1}^{1}-2\int_{-1}^{1}\frac{\sin(\pi k t)}{\pi k}$$

$$2t \frac{\sin(\pi k t)}{\pi k}\bigl|_{-1}^{1} = 0$$ because $\sin(\pi k)$ and $\sin(-\pi k)=0$.

Now

$$-2\int_{-1}^{1}\frac{\sin(\pi k t)}{\pi k} = 2 ( \frac{\cos(\pi k t)}{(\pi k)^2} - \frac{\cos(-\pi k t)}{(\pi k)^2} \bigl)_{-1}^{1}$$

since $\cos(x) = \cos(-x)$ we get

$$2 ( \frac{\cos(\pi k t)}{(\pi k)^2} - \frac{\cos(-\pi k t)}{(\pi k)^2} \bigl)_{-1}^{1}=0$$

regarding $b_n$

$$b_n = \int_{-1}^{1}2t \sin(k \pi t)dt + \int_{-1}^{1}\sin(k \pi t)dt $$

Now the second integral $$\int_{-1}^{1}\sin(k \pi t)dt =0$$ because

$$\int_{-1}^{1}\sin(k \pi t)dt = -\frac{\cos(\pi k t)}{\pi k} \bigl|_{-1}^{1} = \frac{-\cos(\pi k)}{\pi k} + \frac{\cos(-\pi k)}{\pi k} = 0$$

and the last integral

$$\int_{-1}^{1} 2t \sin(k \pi t)dt = -2t\frac{\cos(\pi k t)}{\pi k} \bigl|_{-1}^{1} + \frac{2 \sin(\pi k t)}{(\pi k)^2} \bigl|_{-1}^{1}$$

again we now that $ \sin=0$

so we only have

$$-2t\frac{\cos(\pi k t)}{\pi k} \bigl|_{1}^{1} = -2\frac{\cos(\pi k )}{\pi k}- 2\frac{\cos(-\pi k )}{\pi k} =-4\frac{\cos(\pi k )}{\pi k}$$

and since cos is -1 or 1, we have

$$b_k = \frac{-4(-1)^{n}}{\pi k} = \frac{4(-1)^{n+1}}{\pi k}$$

So to sum it up the fourier coefficients are

$$a_0 =2$$ $$a_k = 0$$ $$b_k = \frac{4(-1)^{n+1}}{\pi k}$$

is everything correct?

Answer 1

Initial remark : if you subtract initially 1 to your function, it becomes odd. This is why, beyond $a_0$, all the other $a_k=0$ ; this can spare a lot of effort.

Here is a way to check that "the rest is okay" by writing a little program with one of the most powerful (public domain) scientific software, SAGE :

f = piecewise([((-1,1),2*x+1)])
s=[f.fourier_series_sine_coefficient(k) for k in range(7)];show(s)
s7 = f.fourier_series_partial_sum(7);
plot(f, (-1,1)) + plot(s7, (x,-1,1), color='red') 

which will give you, as expected :

$$0, \frac{4}{\pi}, -\frac{2}{\pi}, \frac{4}{3 \, \pi}, -\frac{1}{\pi}, \frac{4}{5 \, \pi}, -\frac{2}{3 \, \pi}$$

together with the graphical representation :