-1

Apparently what I thought was a absolute value was a average down sign, I. E. if the value is 2.9 the sign will make it 2. Just got back from the professor.

In my last exam of wave physics that I didn't pass, this was one of the questions that I can't seem to get a grip on how he solved it by looking in the answer sheets, could you please help me to figure it out:

Give the fourier coefficients $a_0$,$b_1$,$a_1$,$b_2$ and $a_2$ from the signal $f(t) = (t-|t|)^2$.

Nothing about interval or anything else is mentioned and this is the solution that is given:

https://i.stack.imgur.com/pxXEa.jpg

And I don't know how he got this result which is why I'm asking here, to get help, guidance and tips of how he solved it.

ricksson
  • 9
  • @john Sorry I tried editing it but nothing happened (I'm not so used to stack exchange) and I'm very stressed for my exam on Wednesday. – ricksson Jun 07 '15 at 08:54
  • If you can, try to delete that question. –  Jun 07 '15 at 08:55
  • Well, from the solution one would assume that the interval is $[0,1]$. But I do not see the function $(t-|t|)^2$ (only $t^2$). Are you sure this is the only solution? –  Jun 07 '15 at 08:57
  • @john, this is the only solution posted by the professor. :/ – ricksson Jun 07 '15 at 09:03
  • Then either your solution is incorrect or is incomplete. –  Jun 07 '15 at 09:04
  • That doesn't make sense as $t-|t|=0$ for $t\geqslant 0$ and $t-|t|=2t$ for $t<0$. – Math1000 Jun 07 '15 at 09:05
  • @John That was exactly my thought, I will post my attempt on a solution here in 30 minutes – ricksson Jun 07 '15 at 09:13
  • Does this look correct assuming not knowing the interval I assumed that the function would be 0 for positive numbers and I did forget to add a negative sign: http://imgur.com/y0ZNnRn – ricksson Jun 07 '15 at 09:44
  • I missed to move out the value from knife the cosine while doing the integral as well – ricksson Jun 07 '15 at 09:53
  • Apparently what I thought was a absolute value was a average down sign, I. E. if the value is 2.9 the sign will make it 2. Just got back from the professor. – ricksson Jun 08 '15 at 12:38

1 Answers1

0

Apparently what I thought was a absolute value was a average down sign(truncation).

I. E. if the value is 0.9 the sign will make it 0 which in this case gives an interval of 0-1. In my opinion a horribel and not pedagogical way of writing a exam assignment but I hope I have it under control now.

An: https://i.stack.imgur.com/PdWeI.jpg

Bn: https://i.stack.imgur.com/JKoad.jpg

ricksson
  • 9