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Hello I have this function and I'm asked

1.Find the period for $f(t)$

2.Find the coefficients $a_n$ and $b_n$

$$f(t)=2(cos(2t+\frac{\pi}{4})-sin(6t-\frac{\pi}{2}))$$

I know that the period for $sin(6t-\frac{\pi}{2})$ is $\frac{\pi}{3}$

I also know that the period for $cos(2t+\frac{\pi}{4})$ is $\pi$

1.common period is :

$$\pi , \frac{\pi}{3} => \pi$$

2.I don't think $f(t)$ I need to find $a_n$ and $b_n$, can't I read them directly from $f(t)$? $a_n=2$ and $b_n=-2$ ?

Could you please tell me if I'm right/wrong for both questions ?

Oleg
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2 Answers2

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For $(2)$ use the identities

$$ \cos(A+B) = \cos(A)cos(B) - \sin(A)\sin(B) $$

$$ \sin(A-B) = \sin(A)cos(B) - \cos(A)\sin(B). $$

Added: Based on what you put in the comment down you can write $f(t)$ as

$$ f(t) = - \frac{\sqrt2}{2}\sin(2t)+\frac{\sqrt2}{2}\cos(2t) -\cos(6t) $$

and Fourier series is

$$ f(t) = a_0 + a_1\cos(t) +a_2\cos(2t)+\dots + b_1\sin(t) +b_2\sin(2t)+\dots\,. $$

Can you see the coefficients now? As I said in my comment just compare with the Fourier series and you will see only three coefficients are not $0$.

Community
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Mhenni Benghorbal
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  • well I wrote

    $$cos(2t+\frac{\pi}{4} )= \frac{\sqrt2}{2}(cos(2t) - sin(2t))$$

    and

    $$sin(6t - \frac{\pi}{2})=-cos(6t)$$

    if I substitute them in orginal function , it doesn't help me

     – Oleg Jan 24 '15 at 10:15
  • @Oleg: When you simplify it the way I said you get the Fourier series and you can readily know the coefficients which are the coefficients of the functions $\cos(2t), \sin(2t)$ and $\cos(6t)$. Just compare with Fourier series and you can notice that the rest of the coefficients are $0$. – Mhenni Benghorbal Jan 24 '15 at 10:19
  • Sir, could you please help me out with that ? – Oleg Jan 24 '15 at 10:23
  • Sorry, I think you forgot the 2 and a - :

    $$ f(t) = 2*(- \frac{\sqrt2}{2}\sin(2t)+\frac{\sqrt2}{2}\cos(2t) +\cos(6t)). $$

    and now my $a_2=\sqrt2$ , $b_2=-\sqrt2$ and $a_6=2$.

     – Oleg Jan 24 '15 at 10:32
  • @Oleg: Make sure of your calculations! – Mhenni Benghorbal Jan 24 '15 at 10:33
  • Let us continue this discussion in chat. – Oleg Jan 24 '15 at 10:40
  • I calculated once again , first time I forgot that my period is $\pi$ not $2\pi$ . So I got these coefficients for $a_1,b_1,b_3$ but not $a_2,b_2,b_6$, what do you think ? – Oleg Jan 24 '15 at 12:00
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$$f(t)=a_0+ \sum_{n=1}^{\infty}a_n \cos(n x)+\sum_{n=1}^{\infty}b_n \sin(n x)$$

Now you can compare what you have in the comment with the formula to read off coefficients $a_2,b_2,a_6,b_6$. All the other coefficients seemed to be zero.

mike
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  • L = half of Period or $\frac{\pi}{2}$ $$f(t)=a_0+ \sum_{n=1}^{\infty}a_n \cos(\frac{n\pi{x}}{L})+\sum_{n=1}^{\infty}b_n \sin(\frac{n\pi{x}}{L})$$

    and the coefficients that I will get will be $a_1,b_1,b_3$

    am I right?

    If the period were $2{\pi}$ then , I would've gotten $a_2,b_2,b_6$

     – Oleg Jan 24 '15 at 11:53
  • Yes. You are right! – mike Jan 24 '15 at 15:31