A periodic function F(t) is defined by f(t)=t^2, over -pi

Question

A periodic function F(t) is defined by f(t)=t^2, over -pi<t<pi

 A1. State the period of f(t).
 2. Sketch f(t) over the domain -3pi<t<3pi 

B. state with reasons, whether f(t) is odd , even or neither odd or even.

 C. determine the first six non zero terms of the Fourier series representing f (t).

anonymous · Answer

Do you still need help?

anonymous · Answer

i got 2 for the period, the sketch is 3times  y= pulus or minus y=x^2 the function is even .is that correct so far.

anonymous · Answer

How did you get the period?

anonymous · Answer

2pi=PI pi/2pi=2 ?

anonymous · Answer

1/2 i think?

anonymous · Answer

But it's not a trig function. I don't see how period applies to this. Perhaps the domain?

anonymous · Answer

well the first quesion asks for period , so i think it is  3, since it is 3xpi?

anonymous · Answer

Is this calculus or Differential Equations?

anonymous · Answer

Fourier series?

anonymous · Answer

Question is posted above please read.

anonymous · Answer

Is this question from a calculus class, or Differential equations?

anonymous · Answer

any ideas?

anonymous · Answer

Well the period would be 2pi, since it states that it goes from -pi to pi.

anonymous · Answer

how  did you work that out ?

anonymous · Answer

Can you answer my previous question? Is this for a calculus class, or differential equations class?

anonymous · Answer

This is for calculus class?

anonymous · Answer

On fourier series odd and even functions!

anonymous · Answer

I know the funtion is even , but the main thing im suck on is the series?

anonymous · Answer

Can you  confim that the period is 2pi ?

anonymous · Answer

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