Difference between a<t<b and t>a && t<b

Hi I am trying to verify existence of even and odd harmonics in e^(-t) periodic with period pi. For that I try to verify f(t)=f(t+pi). and for f(t+pi) i try to add shift using if-else condition according to this link answer should have even harmonics, http://www.intmath.com/fourier-series/5-harmonic-analysis.php I am trying to solve Example 2(b) in Matlab.

Now if i use the condition 0<=t(i)<pi it shows me even harmonic but if i use 0>=t(i) && t(i)<pi it gives neither even nor odd.

So, my question is why isn't Matlab showing same result for both formats. and which format is correct.

Answer 1

[Guillaume]

The first form is not going to give you the result you expect, it is equivalent to:

(0 <= t(i)) < pi

Thus, it will compare t(i) to 0, and the result is either 1 (true) or 0 (false). It then compares that 0 or 1 to pi, which is always smaller. Hence the result will always be true.

The second form is the correct one. You can't link comparisons. You have to perform them one at a time and link the results with logical operators.

[Ahmet Cecen]

You are kidding me, how have I not noticed this years an years of writing conditionals in MATLAB... I wonder if I just got lucky or there is a bug somewhere in some of my old codes...

[Guillaume]

I'm not kidding. I don't think there are many languages where chaining comparisons would work since the result of the first comparison is normally a boolean.

[John D'Errico]

If you have been doing this for years, then there are definitely bugs hidden in your code.

[Ahmet Cecen]

Thankfully I found only one in a rotation code for visualization. Apparently I like to separate out the conditionals by default. Still many thanks for pointing this out, could have been a disaster. Would give more bumps if I could!

[Maheen]

Thank you So much Guillaume, resolved my problem.

Answer 2

[Ahmet Cecen]

might be because you are supposed to write:

*t(i)>=0 && t(i)<pi*

NOT:

 *0>=t(i) && t(i)<pi*