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Confirm Fourier co-efficient is zero despite not being odd/even function

  • Thread starter thomas49th
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  • #1
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Hi, my function is defined as

f = 1, -1<t<0
f = cos(pi*t),0<t<1
f(t+2) = f(t)

To what value does this series converge when t=1


So I need to find the Fourier series expansion and set it equal to one. Sooooo

I found the DC Value to be 1/2. Then I found a_{n} = 0 as well. Is this correct. Can a = 0 despite the function not being odd?

Thanks
Thomas
 

Answers and Replies

  • #2
LCKurtz
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Hi, my function is defined as

f = 1, -1<t<0
f = cos(pi*t),0<t<1
f(t+2) = f(t)

To what value does this series converge when t=1


So I need to find the Fourier series expansion and set it equal to one. Sooooo
You should be able to answer that question without calculating the FS at all.

I found the DC Value to be 1/2. Then I found a_{n} = 0 as well. Is this correct. Can a = 0 despite the function not being odd?

Thanks
Thomas
Yes to both. 1 + a sum of sine terms is not an odd function and neither is your original function.
 
  • #3
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Okay at t=1, the series is attempting to converge to -1 as cos(pi*1) = -1

Are you saying that DC value of a 1/2 is correct and does infact a_{n} = 0?

Is that it? Thanks
 
  • #4
LCKurtz
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Okay at t=1, the series is attempting to converge to -1 as cos(pi*1) = -1
No. The value of cos(pi*t) at t = -1 has nothing to do with this problem.

Are you saying that DC value of a 1/2 is correct and does infact a_{n} = 0?

Is that it? Thanks
Yes, a0 = 1/2 and the other an = 0 as you have calculated. But you are missing the point about what the series converges to when t = 1. What you need to do is draw a couple periods of your given function. Graph it from t = -1 to t = 3. Look at that graph at t = 1 and think about your theorem about convergence of FS. See if you can figure out whether a0 = 1/2 makes any sense.
 
  • #5
uart
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Okay at t=1, the series is attempting to converge to -1 as cos(pi*1) = -1

Are you saying that DC value of a 1/2 is correct and does infact a_{n} = 0?

Is that it? Thanks
1. The DC value of [itex]a_0=1/2[/itex] is correct.

2. The assertion that all other [itex]a_n=0[/itex] is NOT correct.
 
  • #6
LCKurtz
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1. The DC value of [itex]a_0=1/2[/itex] is correct.

2. The assertion that all other [itex]a_n=0[/itex] is NOT correct.
You are correct.
 
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  • #7
vela
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I found the DC Value to be 1/2. Then I found a_{n} = 0 as well. Is this correct. Can a = 0 despite the function not being odd?
Yes, it's possible. It's not true in this case, though. You seem to have made an error in calculating a1.

If all the ai's other than a0 were 0, you'd have

[tex]f(t) - a_0 = \sum_{n=1}^\infty b_n \sin(n\pi t)[/tex]

The RHS is an odd function, so the LHS is also odd. So to have ai=0 for i>0, you need f(t)-a0 to be odd, not just f(t) by itself.

In the particular problem, it turns out that f(t) - a0 - a1cos πt is odd.
 

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