Nasty Math Problem of the Day...

Move one of the arctans to the other side of the equation, take the tangent of both sides using the formula for tan(a+b), and solve it algebraically.

Cheers

Phil Hobbs

The arctans are on the one side.

As I see it, applying TAN to both sides gives

TAN(THETA) = TAN(arctan1 + arctan2) ...Jim Thompson

There are different ways to do it, but you'll need the tan(a+b) formula. That'll make it an algebraic equation rather than a transcendental one.

Cheers

Phil Hobbs

A _messy_ Algebraic equation :-)

Oooops! I just realized the advantage of your approach... ding-dong

Not sure, but whatever it is, it's literally ahead of its time. Four years? Getting a bit presumptuous in your old age there. :)

Tim

-- Seven Transistor Labs Electrical Engineering Consultation Website:

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

Aarrrgh :-(

Knock on wood, I don't seem to be as frail as my father was when he was 74, and he made it to 90. ...Jim Thompson

-- | James E.Thompson | mens | | Analog Innovations | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | San Tan Valley, AZ 85142 Skype: skypeanalog | | | Voice:(480)460-2350 Fax: Available upon request | Brass Rat | | E-mail Icon at

Seriously , your approach is all wrong.

years? Getting a bit presumptuous in your old age there. :)

was 74, and he made it to 90.

The optimum BMI for people older than 65 seems to be about 28. The photos Jim has posted of himself suggest he's somewhat heavier than that, and once your BMI goes over about 30, life-expectancy starts dropping fast.

Jim's father was probably on the skinny side - most of his generation were.

"Jim Thompson" wrote in message news: snipped-for-privacy@4ax.com...

tan(a) + tan(b) tan(a+b) = ---------------- 1 - tan(a)*tan(b)

I think your only hope is a numerical approach. I suggest you state the problem in the form: a = fa(x, y, p), b = fb(x, y, p) and post your query on the SciPy newsgroup:

I thinks Hobbs' move satisfies the Algebra. I'll report back later... this is all about behavioral modeling/curve fitting. ...Jim Thompson

This is too messy to attempt an algebraic solution, particularly since you're starting with transcendental functions. Use numerical methods to find a solution. If you're not prepared to do it discretely, Matlab or Mathcad might be able to solve it for you.

A few small substitutions will help you see the wood for the trees.

If I have read it right then the following will turn it into a recognisable quadratic form.

Let t = tan(theta) x = Xw y = 1-Yw^2 p = P/w

with the tan formula above simplifying and then later z = x/y to get a quadratic form which subject to algebra slips I get to be :

(1-pt)z^2 + 2(p+t)z - 1 + pt = 0

Hence an expression for z = x/y as a function of p

Quick and dirty approx starting solution from sqrt(1+x) = 1 + 2x/(4+x) x