How can you show that Sin(A)(Pi)(x) and Cos(K)(Pi)(x) are orthogonal for all A&K in R? Please keep the context within linear algebra and where the definition of orthogonality is two functions are orthogonal if the integral of their multiplication is zero within a certain bound?
Do you mean sin(Aπx) and cos(Kπx)?
Typically you’ll see orthogonality of functions defined on [0, 1], so let’s use that.
∫sin(Aπx)cos(Kπx)dx = −cos[(Aπ − Kπ)x] / [2(Aπ − Kπ)] − cos[(Aπ + Kπ)x] / [2(Aπ + Kπ)] + c
Evaluating this at 1 and 0 and subtracting the latter from the former gives:
−cos[(A − K)π] / [2_π_(A − K)] − cos[(A + K)π] / [2_π_(A + Kπ)] – 0
= −cos[(A − K)π] / [2_π_(A − K)] − cos[(A + K)π] / [2_π_(A + Kπ)]
This will equal zero for some values of A and K, but certainly not for all values.
There’s something missing here.
Yes, yes I am thinking of orthogonality of functions over the aforementioned functions.I am contemplating how would this behavior change if we bounded A & K to belong to integers and not real numbers. Maybe I am wrong. I need to go back to my notes. Also sorry for not using LaTeX here, too lazy.
Also, S2K, please post your proof of the Hodge conjecture for me (just to verify my work). I think I have finally proven it, but want to verify with your work to make sure.
Thanks!
No can do: swore off talking about cohomology in years with exactly two primes in their factorization.
Sorry.
My boy’s wicked smaht.
I love your math discussion. No cohomology.
i’m surprised there is not yet an extreme math tv show. just a couple of guys with chalkboards. steve harvey can host.
brah, you don’t need to do any integrals here - the orthogonality is obvious as long as the interval you are looking at is symmetrical around zero: (-L, L). So they are orthogonal on (-1, 1) for example but not on (0,1). Why? Because sine is an odd function, i.e. sin(-x*A*Pi)=-sin(A*Pi*x) and cosine is an even function, i.e. cos(-x*K*Pi)=cos(x*K*Pi). The product of an odd function and an even function is odd. The integral of any odd function is zero for integration bounds symmetrical around zero. Therefore the two trig functions you picked are orthogonal on any interval symmetrical around 0, for all A&K in R.
Good catch, Mobius!
I agree with your observation.But the integral is bound from 0 to 1 and NOT from -1 to 1 thus by your point it would not be orthogonal? The problem is very strict about the bound.
I know black scones derivation when I see it.
This was not stipulated; I offered that as a common interval for inner product spaces of real valued functions of a real variable. One can define the inner product using whatever interval one wants, so long as the properties of an inner product are preserved.
scones … mmm delish! 
Yeah, SCB, everyone know that.
Truth is, the first time I took level 2 I was so scared of BSM that I actually contemplated getting it tatoo’d on my chest. I was 23 then, and still no smarter today, and still have NO idea what any of the above posters are talking about.