Need Help With Math Problem

[BKWM]

I need a proof on this equation, it has stumped me and my dad, any help would be appreciated.

____1_____ = sec(x) - tan(x)
sec(x) + tan(x)

[Dry]

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

[Adam Smith]

Re: Need Help With Math Problem

Multiply both sides by sec(x) + tan(x) resulting in
1 = sec^2(x) - tan^2(x)
Convert fucntions to sin and cos
Apply sin^2(x) + cos^2(x) = 1
1 = 1
QED

[BKWM]

thank you

[Ramo]

Multiply both sides by sec(x) + tan(x) resulting in
1 = sec^2(x) - tan^2(x)
Convert fucntions to sin and cos
Apply sin^2(x) + cos^2(x) = 1
1 = 1
QED

Terrible way of doing proofs (since operations to algebraic equations aren't necessarily if and only if, i.e. multiplications by 0).

[civman2000]

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.

[Kuciwalker]

Re: Need Help With Math Problem

Originally posted by BKWM
I need a proof on this equation, it has stumped me and my dad, any help would be appreciated.

____1_____ = sec(x) - tan(x)
sec(x) + tan(x)

This is a lot less interesting than my current math problem ( ), which boils down to "is there a general algorithmic way to find the polynomial solution f[x] to p[x] = f[x]q[x+1] - f[x-1]r[x] where p,q,r are polynomials and f[x] is guaranteed to be polynomial if it exists". In particular, I'm too sleepy to figure out if the solution even exists for:

1 = f[k] (2k+1)^2 - f[k-1] (2k+2)^2

/me falls asleep at his desk minutes before class

[Kuciwalker]

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

[Lul Thyme]

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

If it's an N in blackboard bold, then probably (http://en.wikipedia.org/wiki/Natural_number) although personally I've never seen this notation.

You have to be careful though because an aleph can look like an N, and aleph subscript zero is the cardinality of the natural numbers (http://en.wikipedia.org/wiki/Aleph_number).

[Kuciwalker]

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

[Lul Thyme]

Re: Re: Need Help With Math Problem

Originally posted by Kuciwalker


This is a lot less interesting than my current math problem ( ), which boils down to "is there a general algorithmic way to find the polynomial solution f[x] to p[x] = f[x]q[x+1] - f[x-1]r[x] where p,q,r are polynomials and f[x] is guaranteed to be polynomial if it exists". In particular, I'm too sleepy to figure out if the solution even exists for:

1 = f[k] (2k+1)^2 - f[k-1] (2k+2)^2

* Kuciwalker falls asleep at his desk minutes before class

That's a funny way of saying it.
Obviously the solution doesn't have to be polynomial.
So I'm assuming you want an algorithm to find a solution if there exists a polynomial one?

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].

[Kuciwalker]

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.)

[Lul Thyme]

I havn't checked all the details, but I think I have shown how to approach the problem.


Assume there is a polynomial solution f(x) = sum_{i=0}^n a_i x^i.
(so f is polynomial of degree n)
Im going to assume p(x)=1 to make things easier but I think the solution works in general.

also ill write r(x)=sum_{i=0}^n r_i x^i and q(x)=sum_{i=0}^n q_i x^i.

so we need to solve
1=r(x)f(x-1)-f(x)q(x)
Note that

f(x-1)= (a_0-a_1+a_2+... ) + x(a_1-2a_2+3a_3+...) + x^2(a_2-3a_3+...)+...+x^(n-1)(a_{n-1}-n*a_n)+x^n(a_n).
where all the ... terminate at a_n and the coefficients are binomial.

In the original equation, start solving with the highest power of x. For example, suppose q(x) has degree m, r(x) has degree k and m>k.
Then we get 0=x^n(a_n)*q_m x^m (unless n+m<1)
and q_m*a_n=0, contradiction (since none of them are 0).
You get the same contradiction if k>m I think.

So there can only be a solution if k=m.
In this case, 0=x^(n+m) [(r_m)*(a_n)- (q_m)*(a_n) ]
so (a_n)(r_m-q_m)=0.
If r_m \neq q_m we get (a_n=0), a contradiction and there is no solution.
Otherwise if r_m=q_m the we can choose any nonzero value for a_n.

Now the next power yields:
0=x^(n+m-1) [(r_{m-1})*(a_n)+(r_m)*(a_{n-1}-n*a_n)- (q_{m-1})*(a_n)-(q_m)*(a_{n-1}) ]

and
[(q_m)-(r_m)] *a_{n-1}=[[(r_{m-1})-(q_{m-1})-n*r_m]*(a_n)]
so (r_{m-1})-(q_{m-1})-n*r_m=0
and (r_{m-1})-(q_{m-1})/(r_m)=n
Now if n is not a positive integer, there are no solutions to the original equation. Otherwise, you know what n is, and you can keep checking the lower degrees.
Since n is known, this will terminate, and all the equations are linear in the coefficients of f, so linear algebra will get you the solution if any.


I havn't really checked the sensitivity to p(x) but I think this procedure or something close should work in general.

Check it out some more and tell me what you think.

EDIT:
In the example you gave, the second step yields n=4, so if there is a solution, it is a fourth degree polynomial.
Any possible solution could be found pretty quickly.

[Lul Thyme]

Originally posted by Kuciwalker


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

That's not very clear.
You didn't tell me how you got q and r and p.
There are some choices for them which are allowed in the description for your problem that yield non-polynomial solutions for f[x].

Are you saying that we can assume that p,q and r are such that the only solutions would be polynomial?
Or that we are dealing with the general case but are only interested in polynomial solutions.

Originally posted by Kuciwalker
It's needed when you know the origin of those three polynomials.

The way you stated the problem, it's an unneeded complication.

[Kuciwalker]

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