Can the Sum of Two Sums be Substituted in a Function?

[Niles]

Homework Statement

Hi all.

Lets assume that we know the following:

<br /> \sum\limits_{n = - \infty }^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t) = a_0 + \sum\limits_{n = 1}^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t),<br />

where a₀ is the contribution for n=0. Now I have an expression for a function f given by the following:

<br /> f = \sum\limits_{n = - \infty }^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t)\frac{1}{{Z(\omega _n )}}.<br />

Am I allowed to write f as this?:

<br /> f = a_0 \frac{1}{{Z(\omega _0 )}} + \sum\limits_{n = 1}^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t)\frac{1}{{Z(\omega _n )}},<br />

i.e. substitute the sum? Personally, I think yes, but I am a little unsure, which is why I thought it would be best to check here. Thanks in advance.Niles.

 

[HallsofIvy]

Homework Statement

Hi all.

Lets assume that we know the following:

<br /> \sum\limits_{n = - \infty }^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t) = a_0 + \sum\limits_{n = 1}^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t),<br />

This appears to be saying that
\sum\limites_{n=-\infty}^{-1}\varepsilon_n(t)}\exp(-i\omega_n t)= 0[/itex]<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> where a₀ is the contribution for n=0. Now I have an expression for a function f given by the following:<br /> <br /> &lt;br /&gt; f = \sum\limits_{n = - \infty }^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t)\frac{1}{{Z(\omega _n )}}.&lt;br /&gt;<br /> <br /> Am I allowed to write f as this?: <br /> <br /> &lt;br /&gt; f = a_0 \frac{1}{{Z(\omega _0 )}} + \sum\limits_{n = 1}^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t)\frac{1}{{Z(\omega _n )}},&lt;br /&gt;<br /> <br /> i.e. substitute the sum? Personally, I think yes, but I am a little unsure, which is why I thought it would be best to check here. Thanks in advance.<br /> <br /> <br /> Niles. </div> </div> </blockquote> But that does NOT mean that<br /> <br /> \sum\limits_{n = -\infty}^{-1} {\varepsilon _n (t)} \exp ( - i\omega _n t)\frac{1}{{Z(\omega _n )}}= 0<br /> since the Z(\omega_n) term may change the contribution of each indvidual term in the sum.

 

[Niles]

Aw man, I made an error in my first post. Very stupid of me, because now I made you look at something which was not my intention; I am very sorry for that.

What I meant to write is the following (I'm still really sorry, I should have taken a closer look before posting):

Lets assume that we know the following:
<br /> \sum\limits_{n = - \infty }^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t) = a_0 + \sum\limits_{n = 1}^\infty {n^2} \exp ( - i\omega _n t).<br />

Now we look at an expression for a function f given by the following:
<br /> f = \sum\limits_{n = - \infty }^\infty {\varepsilon _n (t)} \exp ( - i\omega _n t)\frac{1}{{Z(\omega _n )}}.<br />

Am I allowed to write f as:
<br /> f = a_0 \frac{1}{{Z(\omega _0 )}} + \sum\limits_{n = 1}^\infty {n^2} \exp ( - i\omega _n t)\frac{1}{{Z(\omega _n )}}.<br />

I have double-checked, and the formulas are correct now. Sorry, again.Niles.

 

[HallsofIvy]

And the answer is still "No". Because Z(\omega_n) may be different for different n, it can completely change the sum.

 

[Niles]

I had actually hoped that you would answer "Yes", because the "No" means that my calculations are wrong. But the quantity Z(\omega_n) does depend on n, so it is different for different n.

I will have to think about this. I guess I made a wrong assumption along the way.. I do that sometimes (as we can also see from this thread ).

Thanks for helping, I really do appreciate it.