are you sure ck=j*[(-1)k] / [k*pi]?!?!

I think you did something wrong!!  it should be in terms of sin(k*pi).

sin(k*pi) is equal to 0. Because, sin of integer multiples of pi is equal to zero, isn't it?

yes you are right, cn is equal to j*[(-1)^n] / [n*pi]=j*cos(n*pi)/(n*pi).

in fact ck=(j/(2*pi))*((exp(j*pi*n)+exp(-j*pi*n))/n). now if we want to find c(n=0)

we have .... (j/(2*pi))*((exp(j*pi*0)+exp(-j*pi*0))/0)= 0/0.... to find the real value

we have.... lim((exp(j*pi*n)+exp(-j*pi*n))/n) as n tends to 0.

we can use Hoptial rule :

lim(exp(j*pi*n)+exp(-j*pi*n))/n=lim(j*pi*exp(j*pi*n)-j*pi*exp(-j*pi*n)/1)

therefore the value is 0/1=0; then c0=0;

Dear Hossein,

You state that (j/(2*pi))*((exp(j*pi*0)+exp(-j*pi*0))/0)=0/0. But, that is not true. Because,

exp(j*pi*0)=1 and  exp(-j*pi*0)=1. Thus, exp(j*pi*0)+exp(-j*pi*0)=2.

Hence, (j/(2*pi))*((exp(j*pi*0)+exp(-j*pi*0))/0)=(j)/(pi*0) which goes to (j*infinity).

So, we cannot apply L'Hospital's rule (because,L'Hospital's rule can only be applied for indeterminate cases of (infinity)/(infinity) or 0/0.

See the attached file.

Dear Hossein,

Your suggestion seems to work. I have to add that

in calculation of cn, you forgot the (1/2) term in front of the integral.

So, the term in the denominator should be 2*x^2 instead of  x^2.

But, that does not change the result of the L'hospital's calculation.

Thank you very much for your effort.

I'd be a bit careful with first trying to compute cn and then setting n->0 and arguing with a limit. In this case it worked, but I'm not sure it works in general.

You computed Integral(t*exp(x t)) using integration by parts where the first term is t/x exp(x t). I would claim that this rule is only valid for x \neq 0.

It is safer to distinguish n=0 and n \neq 0 during the computation of cn. The case n=0 is anyways much simpler: c0 always contains the DC (i.e., the linear average). If your function is an odd function f(x)=-f(-x) you already know that c0=0.

However, a sawtooth could have c0>0 easily if you move it up or down. Imagine a sawtooth going from 0 to 2 instead of from -1 to 1. All the ck for k>0 stay the same as above but this one has c0=1.

So just treat c0 separate and remember it's the linear average. No need to compute limits or anything like that. ;-)

Dear Roemer,

we know that Cn=-(exp(x)+exp(-x))/x  +  (exp(x)-exp(-x))/x^2 ,   while x=i*n*pi

for n > 0, second term is always zero  (exp(x)-exp(-x) -> sin(n*p)), then

Cn=-(exp(x)+exp(-x))/x=i*cos(n*pi)/(n*pi)=i*[(-1)^n] / [n*pi] , n > 0

for n=0, second term is 0/0, therefore we can not remove it and only way to find it, is to use limit() .result is zero as I mentioned above.

here, questioner(Mr Akay) knows that C0 is zero(function is odd) but he wants to know why it is not zero by i*[(-1)^n] / [n*pi]. The answer is in second term of Cn formula which has been already removed!!

Dear Hossein Soleimani,

sure, I understand that this was the original question and also how you used the limit to show that the formula for Cn gives 0 at n=0 consistently. Nothing wrong with that.

I just wanted to warn that in general we should be careful with the approach. When you say "we know that Cn=..." then we should carefully look how this equation was derived. If you obtain it simply by solving the Fourier integral then you have to assume that n \neq 0 and solve the case n=0 separately because the integration by parts does not work this way for n=0.

For the special case of an odd symmetric sawtooth, solving n=0 gives a result that is consistent with solving for n \neq 0 (as you demonstrated via the limit) but that is more of a coincidence than a rule.

 I agree Co=0. Co is the average of f(t). Your attached waveform has zero average.

Dear Herr Roemer,

I really appreciate your example of shifting the sawtooth up or down. This really does not change cn for c \neq 0, but obviously changes c0 (DC average) by giving it a nonzero value (such as 1 for shifting up, or -1 for shifting down). As a result, we cannot really expect to get c0 from cn, which stays the same, however, c0 could assume any value depending on which direction and how much we shift the sawtooth vertically.

This nicely explains that we should not try to get c0 value from cn, but treat it separatly by calculating it through its own integral formula. 

I thank all the followers of this question for their insightful comments which greatly helped me in clarifying this little dilemma in my mind :-).