You are not logged in.
Pages: 1
Does anyone know if there is ANY identity for sqrt ( a + b) ? It would be super usefull for integration and many other things.
A logarithm is just a misspelled algorithm.
Offline
Or try to prove that no identity exists.
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
Offline
I think not.
IPBLE: Increasing Performance By Lowering Expectations.
Offline
The only thing I've come up with is if a and b are both positive, sqrt (a + b) is the hypotenuse of a right triangle whose legs are sqrt(a), sqrt (b). The only way I know to solve this without using the pythegorean theorem is to find θ.
θ = arctan (sqrt b / sqrt a),
c = sqrt b/ sin θ OR sqrt a / cos θ = sqrt ( a + b)
but really all we get is sqrt (a + b) = sqrt b / sin [ arctan (sqrt b / sqrt a) ]
But as a substitution identity thats pretty worthless as it only makes it more complicated.
Last edited by mikau (2006-05-16 04:07:50)
A logarithm is just a misspelled algorithm.
Offline
I put a square in the center of a square and tried solving for stuff, but after quadratic formula, it came back to sqrt(a + b) again!
Last edited by John E. Franklin (2006-05-16 09:28:32)
igloo myrtilles fourmis
Offline
Yeah I wasted a lot of time going in circles with this one.
A logarithm is just a misspelled algorithm.
Offline
This may be helpful (it requires only
and :Last edited by krassi_holmz (2006-05-19 01:02:31)
IPBLE: Increasing Performance By Lowering Expectations.
Offline
Yeah but they are still a sum inside a radical. Ideally an identity similar to something like sin ( a + b) would be nice. For instance: sqrt ( a + b) = sqrt a sqrt b - 1/sqrt b sqrt a or something like that, that seperates them into individual squareroots.
A logarithm is just a misspelled algorithm.
Offline
Yeah, I know.
But I don't think we'll go further than
IPBLE: Increasing Performance By Lowering Expectations.
Offline
Pages: 1