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

[thomas49th]

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

 

[LCKurtz]

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=1So 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.

 

[thomas49th]

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

 

[LCKurtz]

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 a_n = 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.

 

[uart]

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 $a_0=1/2$ is correct.

2. The assertion that all other $a_n=0$ is NOT correct.

 

[LCKurtz]

1. The DC value of $a_0=1/2$ is correct.

2. The assertion that all other $a_n=0$ is NOT correct.

You are correct.

 

[vela]

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 a_i's other than a₀ were 0, you'd have

$$f(t) - a_0 = \sum_{n=1}^\infty b_n \sin(n\pi t)$$

The RHS is an odd function, so the LHS is also odd. So to have a_i=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.