# After calculating Fourier series coefficients for $x(t)=2 cos(4t) + 4 sin(10t)$, why am I getting all zeroes for all coefficients?

I am trying to find the Cosine/Sine Fourier series coefficients for the given equation:

$$x(t)=2\cos(4t) + 4\sin(10t)$$

$\cos(4t)$ has a period of $T=\frac{\pi}{2}$, and $\sin(10t)$ has a period of $T=\frac{\pi}{5}$. Therefore, the fundamental period of $x(t)$ is:

$$T_0=\text{LCM}\left(\frac{\pi}{2},\frac{\pi}{5}\right)=\pi$$

And the fundamental angular frequency is:

$$\omega_0=\frac{2\pi}{T_0}=2$$

I know that the equations to find all the coefficients are: $$a_0=\frac{1}{T}\int_0^Tx(t)dt$$ $$a_n=\frac{2}{T}\int_0^Tx(t)\cos(\omega nt)dt$$ $$b_n=\frac{2}{T}\int_0^Tx(t)\sin(\omega nt)dt$$

However, after a long calculation showed that:

$$a_0=0$$ $$a_n=\frac{2}{\pi}\left(\frac{n\sin(2\pi n)}{n^2-4}-\frac{20\sin^2(\pi n)}{n^2-25}\right)$$ $$b_n=\frac{4}{\pi}\left(\frac{n\sin^2(\pi n)}{n^2-4}+\frac{5\sin(2\pi n)}{n^2-25}\right)$$ But since n is an integer, you can see that $a_n=0$ and $b_n=0$. But isn't that impossible because any function can be formed with a Fourier series? Did I make a mistake somewhere?

• "But since n is an integer, you can see that $a_n=0$ and $b_n=0$." No, what I can see is that the formulas are absurd if $n^2=4$ or $n^2=25$ hence these $a_n$ and $b_n$ should be computed more rigorously. – Did Nov 26 '14 at 6:40
• Oh wow, I didn't notice that. So I need to add a limit when $n^2$ approaches 4 or 25? – vxs8122 Nov 26 '14 at 6:58
• No, n2 does not "approach 4 or 25", rather you have to redo the computation in this case. – Did Nov 26 '14 at 7:00
• $a_0=0$ is correct since the function average is zero. In the formulas you gave the result is zero unless n=-2,2,-5,5. You have to either do the computation by scratch or take the limit as $n$ tends to one of the values above, that should work too. Hint: as n->0, Sin(n)/n->1 – Georgy Nov 26 '14 at 7:43

The formulae you have derived are invalid for $n=2,5$ but are correct otherwise. You would need to compute the coefficients individually in the cases where $n=2$ or $5$. However, you are making this more difficult than it needs to be.
Since your original function is of the form $$x(t)=a\cos(2nt)+b\sin(2mt),$$ you should be able to see that your original function is already a Fourier expansion with most coefficients equal to zero. By the uniqueness of Fourier expansions this going to be the Fourier expansion of $x$.