Odd and even in complex fourier series

[Aows]

Homework Statement

In Complex Fourier series, how to determine the function is odd or even or neither, as in the given equation
$$ I(t)= \pi + \sum_{n=-\infty}^\infty \frac j n e^{jnt} $$

Homework Equations

$Co=\pi$
$\frac {ao} 2 = \pi$
$Cn=\frac j n$
$C_{-n}= \frac {-j} n$
$an=0$
$bn=-2/n$

The Attempt at a Solution

 

[Charles Link]

To get Latex on the Physics Forums system, you need to type ## on both sides of the expression.

 

[Aows]

To get Latex on the Physics Forums system, you need to type ## on both sides of the expression.

thanks indeed, @Charles Link
how can i write sum from minus infinity to infinity ??

 

[Charles Link]

thanks indeed,
how can i write sum from minus infinity to infinity ??

Google ="Sums in Latex". Many of these I don't memorize=I need to google them as well.

 

[Aows]

Google ="Sums in Latex". Many of these I don't memorize=I need to google them as well.

I see,
thanks anyway

 

[Charles Link]

I see,
thanks anyway

In the exponent, if you put it inside of { } , it will do it properly. Also your infinity in the lower part of your sum needs a \.

 

[Aows]

In the exponent, if you put it inside of { } , it will do it properly. Also your infinity in the lower part of your sum needs a \.

yes,
I corrected now,
many thanks

 

[Aows]

In the exponent, if you put it inside of { } , it will do it properly. Also your infinity in the lower part of your sum needs a \.

where is my mistake in the exponent? @Charles Link
I(t)= \pi + \sum_{n=-\infty}^\infty \frac j n eˆ{jnt}\

 

[Charles Link]

In this one, you need $e^{jnt}=\cos(nt)+j \sin(nt)$. The $\cos(nt)$ function is even, but $\sin(nt)$ is odd.

 

[Charles Link]

where is my mistake in the exponent? @Charles Link
I(t)= \pi + \sum_{n=-\infty}^\infty \frac j n eˆ{jnt}\

You used \frac incorrectly. You need to write \frac{j}{n}

 

[Aows]

In this one, you need $e^{jnt}=\cos(nt)+j \sin(nt)$. The $\cos(nt)$ function is even, but $\sin(nt)$ is odd.

in the solution it said that the function is neither odd nor even, why is that ? @Charles Link

 

[Charles Link]

And your final "\" isn't needed in the expression.

 

[Charles Link]

in the solution it said that the function is neither odd nor even, why is that ? @Charles Link

The $\pi$ in front of other terms keeps $I(t)$ from being either even or odd regardless of anything else. We can not write $I(t)=I(-t)$ which would make it even, and we also can't write $I(t)=-I(-t)$ which would make $I(t)$ odd.

 

[Aows]

$I(t)$ because of the $\pi$ in front of other terms is likely to keep $I(t)$ from being either even or odd. In this case regardless, we can not write $I(t)=I(-t)$ which would make it even, and we also can't write I(t)=-I(-t) $which would make$ I(t) ## odd.

thanks indeed for this explanation, @Charles Link

 

[Aows]

can you solve PDE problems of FFCT (finite Fourier cosine transform) ? @Charles Link

 

[Charles Link]

The $\pi$ in front of other terms keeps $I(t)$ from being either even or odd regardless of anything else. We can not write $I(t)=I(-t)$ which would make it even, and we also can't write $I(t)=-I(-t)$ which would make $I(t)$ odd.

A minor clarification: If it had been even, the $\pi$ would be ok. The terms following the $\pi$ are a mixture of even and odd terms.

 

[Aows]

A minor clarification: If it had been even, the $\pi$ would be ok. The terms following the $\pi$ are a mixture of even and odd terms.

i didn't understand this, kindly, can you clarify more? @Charles Link

 

[Charles Link]

can you solve PDE problems of FFCT (finite Fourier cosine transform) ? @Charles Link

It's not my area of expertise. There are other mathematicians on the Physics Forums who would most likely know the subject quite well.

 

[Aows]

can we determine the value of the fundamental period for this ? @Charles Link

 

[Charles Link]

i didn't understand this, kindly, can you clarify more? @Charles Link

$I(t)=\pi+A(t)$. If $A(t)=A(-t)$, then $I(t)=I(-t)$ and the function is even. Instead though, $A(t)$ in this case has cosine terms that are even, and sine terms that are odd, and $I(t)$ is neither even or odd.

 

[Aows]

can we determine the fundamental period for this function? @Charles Link

 

[Charles Link]

can we determine the value of the fundamental period for this ? @Charles Link

$\omega_n t= n \omega_o t=n( \frac{2 \pi}{T}) t=nt$, where $T$ is the fundamental period. That part is simple.

 

[Aows]

$\omega_n t= n \omega_o t=n( \frac{2 \pi}{T}) t=nt$, where $T$ is the fundamental period. That part is simple.

the solution said that the fundamental period $T_0 = 2 \pi$

 

[Charles Link]

the solution said that the fundamental period $T_0 = 2 *pi$

Presumably, you mean $T=2 \pi$. Do the algebra on the equation I wrote, and that's what you get. (Note: You need to use \ with Latex and not /. The / is used for a fraction ).

 

[Aows]

$\omega_n t= n \omega_o t=n( \frac{2 \pi}{T}) t=nt$, where $T$ is the fundamental period. That part is simple.

actually, i can't understand this... @Charles Link

 

[Charles Link]

actually, i can't understand this... @Charles Link

In Fourier Series, the nth frequency component is $A_n e^{j \omega_n t}$, for some complex constant $A_n$. $\omega_n=n \omega_o$ where $\omega_o$ is the fundamental frequency. The $\omega_o=2 \pi f_o$ where $f_o=\frac{1}{T}$. $f$ is the actual frequency, but oftentimes $\omega$ is also called the frequency. In any case $T$ is the period over which the function is periodic. Any function that is periodic in $T$ can be expressed as a Fourier series with the $\omega_n$ , etc.

 

[Aows]

In Fourier Series, the nth frequency component is $A_n e^{j \omega_n t}$, for some complex constant $A_n$. $\omega_n=n \omega_o$ where $\omega_o$ is the fundamental frequency. The $\omega_o=2 \pi f_o$ where $f_o=\frac{1}{T}$. $f$ is the actual frequency, but oftentimes $\omega$ is also called the frequency. In any case $T$ is the period over which the function is periodic.

thanks indeed Dear Mr. @Charles Link
appreciate your help