Help with Fourier coefficients for sin 2x

[KFS]

Hello.
(Refer to the picture, Exercise 75.)
I don't think I really understand these formulas. I integrated and found
$$a_m= \frac{1}{\pi} \int_{0}^{2\pi} \sin 2x cos mx \, dx= \frac{2cos({2\pi m})-2}{{\pi}(m^2-4)}$$Now, I don't know how to find m. I read that for any odd function $$a_m=0$$So in this case, for $$f(x)=sin(2x)$$ $$a_m=0$$ I believe this is because areas cancel each other
My real struggle starts with $$b_m$$ From the formula given by the book I know that $$b_m= \frac{1}{\pi} \int_{0}^{2\pi} f(x)sin(mx)dx$$Integrating this with $$f(x)=sin(2x)$$ I obtain $$b_m= \frac{2sin(2\pi m)}{\pi (m^2-4)}$$Again my problem is with m. I don't know how to find m. And once I know m what do I have to do? Plug m in both formulas for $$a_m$$ and $$b_m$$ and find both integrals?
Thank you for your time, and please feel free to write back if I typed something wrong or I didn't provide enough information.

 

[Steven G]

You should know exactly what 2cos(2πm)−2 and 2sin(2πm) equals given that m is a natural number (possibly including 0).
You have a_1, a_2, a_3, ..., a_m, ...

 

[blamocur]

Did you notice that for m=2 the denominator of your result is 0 ?

 

[KFS]

Did you notice that for m=2 the denominator of your result is 0 ?

Yes I did, that's something I don't understand either. According to the book for $$b_m$$ m=2 is a valid solution because "it is not zero there".

 

[KFS]

You should know exactly what 2cos(2πm)−2 and 2sin(2πm) equals given that m is a natural number (possibly including 0).
You have a_1, a_2, a_3, ..., a_m, ...

The book didn't give any more information thant that on the picture. I find it strange since, until now, the book has not mentioned anything about Fourier coefficients nor Fourier series. Is there any way to find "all m's values"?

 

[blamocur]

The book didn't give any more information thant that on the picture. I find it strange since, until now, the book has not mentioned anything about Fourier coefficients nor Fourier series. Is there any way to find "all m's values"?

The way I understand the task you are not expected to find $m$. Instead, you are supposed to find $a_m, b_m$ for every non-negative $m$.

Did you understand post #2 by @Steven G ?

 

[KFS]

The way I understand the task you are not expected to find $m$. Instead, you are supposed to find $a_m, b_m$ for every non-negative $m$.

Did you understand post #2 by @Steven G ?

How do I do it? I've thought using double angle trigonometric formulas. Am I right?
Thanks for helping.

 

[BigBeachBanana]

How do I do it? I've thought using double angle trigonometric formulas. Am I right?
Thanks for helping.

Think about the periodicity of $\sin(x)$ and $\cos(x)$
$$\sin(2\pi)=?\\ \sin(2\pi \times 2)=?\\ \sin(2\pi \times 3)=?\\ \vdots \\ \sin(2\pi \times m)=?\\$$

---

$$\cos(2\pi \times 1)=?\\ \cos(2\pi \times 2)=?\\ \cos(2\pi \times 3)=?\\ \vdots\\ \cos(2\pi \times m)=?\\$$

 

[KFS]

Think about the periodicity of $\sin(x)$ and $\cos(x)$
$$\sin(2\pi)=?\\ \sin(2\pi \times 2)=?\\ \sin(2\pi \times 3)=?\\ \vdots \\ \sin(2\pi \times m)=?\\$$

---

$$\cos(2\pi \times 1)=?\\ \cos(2\pi \times 2)=?\\ \cos(2\pi \times 3)=?\\ \vdots\\ \cos(2\pi \times m)=?\\$$

So for $$a_m=\frac{2cos(2\pi m)-2}{\pi (m^2-4)}$$ it is always 0 because $$cos(2\pi m)$$ is 1 for 0, $$2\pi$$, $$4\pi$$... Making the numerator 2-2=0.
Correct? But for m=2 the denominator is 0, I don't understand that.
Also for $$b_m=\frac{2sin(2\pi m)}{\pi (m^2-4)}$$ $$2sin(2\pi m)$$ is 0 for m=0, 1, 2, ... .
But again the denominator becomes 0 for m=2.
Am I correct?
Thank you.

 

[BigBeachBanana]

So for $$a_m=\frac{2cos(2\pi m)-2}{\pi (m^2-4)}$$ it is always 0 because $$cos(2\pi m)$$ is 1 for 0, $$2\pi$$, $$4\pi$$... Making the numerator 2-2=0.
Correct? But for m=2 the denominator is 0, I don't understand that.
Also for $$b_m=\frac{2sin(2\pi m)}{\pi (m^2-4)}$$ $$2sin(2\pi m)$$ is 0 for m=0, 1, 2, ... .
But again the denominator becomes 0 for m=2.
Am I correct?
Thank you.

Yes, it's 0 everywhere except for m =2 as pointed out above.

$$a_2 = \dfrac{1}{\pi} \int_{0}^{2\pi} \underbrace{\sin(2x)\cos(2x)}_{\text{odd function}} ~ dx = 0$$
Can you integrate when m=2?

$$b_2 = \dfrac{1}{\pi} \int_{0}^{2\pi} \sin(2x)\sin(2x)~ dx = ?$$

 

[KFS]

Yes, it's 0 everywhere except for m =2 as pointed out above.

$$a_2 = \dfrac{1}{\pi} \int_{0}^{2\pi} \underbrace{\sin(2x)\cos(2x)}_{\text{odd function}} ~ dx = 0$$
Can you integrate when m=2?

$$b_2 = \dfrac{1}{\pi} \int_{0}^{2\pi} \sin(2x)\sin(2x)~ dx = ?$$

Yes, sure I can integrate that. Let me write it to see if I finally understood:
First I must check if $$\int_{0}^{2\pi}f(x)cos(mx)dx$$ and $$\int_{0}^{2\pi}f(x)sin(mx)dx$$ are odd so I can know if $$a_m=0$$Then I integrate both $$a_m$$ (if it's not odd) and $$b_m$$ and then I find for what values of m the integrals do not exist/are defined. And finally integrate with the corresponding value of m.
Is that correct?

 

[Steven G]

... Making the numerator 2-2=0.

I hope that you know that there is at least one fraction whose numerator equals 0, yet the fraction is not 0.
So just checking for the numerator to verify that a fraction equals to 0 is not a good thing to do.