[Plaumpholavup]

From most basic trigonometric equation:
cos**2 + sin**2 = 1
you can deduct
-> cos**2 = 1 - sin**2
-> cos**2 = (1-sin)(1+sin)
-> 1+sin = cos**2 / (1-sin)

by definition:
sec = 1/cos
tan = sin/cos

So your left expression:

-> 1/(1/cos + sin/cos) -> 1/((1+sin)/cos) -> cos/(1+sin)
replace 1+sin with the expression above
-> cos/(cos**2 / (1-sin) )
-> (1-sin)/cos
-> 1/cos - sin/cos
-> sec - tan

[investor]

thank you

[qQVXpYM6]

Originally posted by Ramo


Terrible way of doing proofs (since operations to algebraic equations aren't necessarily if and only if, i.e. multiplications by 0). But not a bad way to come up with proofs, as long as you check that you can legitimately reverse the argument.

[XinordiX]

Also, N subscript 0 usually refers to the positive integers + 0, right? No negatives?

[BalaGire]

The surrounding symbols would be nonsensical if it were aleph-null, since the symbol is denoting a set. Just checking, thanks.

[SOgLak]

That's a funny way of saying it.
Obviously the solution doesn't have to be polynomial.

Yes it does, that's guaranteed by how I got q and r and p.

PLUS, the q[x+1] in the notation is unneeded as far as I can tell, since you might as well assume it's q[x].

It's needed when you know the origin of those three polynomials.

(For context, I'm performing Gosper's algorithm to find the closed form of a hypergeometric sum, if it exists. This is just one step.)

[selayeffethy]

Oh, that's helpful, I didn't realize there was a maximum order for f. I had considered an algorithm going in the other direction (trying progressively higher orders of f) but didn't see that it would necessarily ever stop or reach a contradiction.

xpost