You should delete this post as it is covered in spam.

Don't get me wrong, I am sympathetic to your irritation and do agree that the -1 notation for inverse is a pain. Life soldiers on. Happy New Year.

Well, I have never seen a shortcut for tan[tan(x)], so I think you are alone on that one. Good luck trying to change that. The notation , for example, in sin^2(x) + cos^2(x) = 1, is here to stay. f(f(x)) written as f^2(x) is just never going to happen.

Can you change the definition? No, unfortunately, you can't.

Take n = 3 and (i1,i2,i3) = (5,6,7) 5,6,7: i1*i2 + i2*i3 + i3*i1 = 30 + 42 + 35 5,7,6: i1*i2 + i2*i3 + i3*i1 = 35 + 42 + 30 6,5,7: i1*i2 + i2*i3 + i3*i1 = 30 + 35 + 42 etc ... They are all the same so divide by n! to get the average.

It means 1/(tan x)^2. The -1 exponent for inverse trig functions is here to stay whether you like it or not. It's just too ingrained in the global culture. If you want to express 1/tan(x) then (tan x)^-1 is about your only option. Also the -1 notation is used for a the general representation for "f inverse" that goes well beyond just the trig functions. Some use "arc" specifically for trig to avoid the confusion and that's about as good as you are going to get. I think you are just wasting your time as no petition is likely to have any real acceptance in the world wide mathematics community.

No there isn't. Your equation reduces to tan(x) = x and there is no solution in terms of elementary operations. i.e. you can't separate out and isolate x. Having said that it's clear, by inspection, that x = 0 is a solution. But there is also another solution between 0 and 2pi. The best you can do is set up an iterative process (Bisection or Newton's Method) to get an approximate solution to the desired accuracy. If you graph y = x and y = tan(x) on the same plot you can see x := 4.49341 is an approximate solution in [0,2pi]

No closed form since x is buried inside the inverse tangent. No way to easy way to isolate x - at least as far as I can see. You can differentiate this to lose the inverse tan but now you are stuck with a differential equation (which by the way your original is a solution). Even an infinite series for inverse tan is not going to help much. Do as far as I can see you are out of luck.

The only think I could come up with is: 1225 = 35^2 = (7.5)^2 So for a^2 + b^2 = c^2, I knew that (3,4,5) is a Pythagorean Triplet (PT) So multiplying all of them by 7 will also be a PT. This gives the solution (21,28,35) This relies on c = 5 being a solution to a^2+b^2=25 where a=3,b=4 There are no PTs having c = 7. So a^2 + b^2 <> 49 (easy to verify by exhaustion as there aren't that many even/odd pairs to check). If there was we could multiply them all by 5 and have a solution to your equation. So I am going to say that (21,28,35) is the only one: a= 21,b=28,c=35 You can 1) interchange a and b and 2) juggle negative signs using all 3 to get a total of 2(1+3+3+1) = 16 solutions in the integers. Maybe you can look at squares modulo 8 - I didn't. I think (even)^2 + (odd)^2 is congruent 1 or 5 (mod 8) but as I said I did not pursue this.

It is one equation in 2 variables - so there are an infinite number of solutions. Graphically it is a circle with radius sqrt(1225)=35. Are you looking for only integer solutions?

That's exactly what I did. Except you have missed one case. 2same 2same. You can see why 1 - P(4 different) is much easier.

What? No takers?

That is a very good question Wasi. At first I was thinking exponential but by manipulating I got: [dx/dt]^2 = -2A * d/dt(1/x) and then guessed at a power function: x = C*t^m So x^2 = C^2*t^(2m) and x'' = Cm(m-1) * t^(m-2) That gave me 2m+m-2 = 0 and m = 2/3. Similar for C by equating co-efficient to A. It reminded me of Newton's Law of Gravitation: acceleration is prop. the inverse square of position.

Convert log (base 10) to a natural log log(a) = ln(a) / ln(10) The 10 can be replaced by any positive number b. log a(base b) = ln(a) / ln(b) ======================== y = log a (baseb) ====> a = b^y ====> ln(a) = ln(b^y) ====> ln(a) = y*ln(b) ====? y = ln(a) / ln(b)

x(t) = [-9A/2]^(1/3) * t^(2/3)

That works. Well done. Try this: Use the polar form of a + bi = R*e^it where R = sqrt(a^2 + b^2) and t = arctan(b/a) So ln(a + bi) = ln(R*e^it) = ln(R) + it = 1/2*ln(a^2 + b^2) + i* arctan(b/a) Of course, in the complex numbers, any integer multiple - 2npi - added to t also works. What we both have done here is find the principal value when n= 0.

That's a somewhat difficult and long explanation so as to really understand it. In a function of 1 variable y = f(x), if you go back to basic, basic fundamentals it all comes down to what the "diffrential of y" and the "differential of x" really mean and how both of them compare to f'(x). And why is it that when you perform the operation "dy by dx" you get the same answer when you find " the differential of y divided by the differential of x. The truth is that if you look at the definition of the "diffrential of x" there is NO requirement that it be infinitely small. So this leads to a conversation that takes up way too much space here to do it justice. =================================================================== Try this. Let's use a function of 2 variables. z = f(x,y). The total change in z (dz) is achieved by combining how much the change in x contributes to dz with how much the change in y contributes to dz. pz/px ( y is held constant) gives you how much z has changed per unit of x( it's a rate). So multiplying by dx gives you the total change of z as contributed by x. pz/px times dx Similarly pz/py (x is held constant) gives you how much z has changed per unit of y. Multiplying by dy then gives you the the total change in z as contributed by y. pz/py times dy So when we combine them: dz = (pz/px)dx + (py/pz)dy

You missed: dx = du/2cos^2(x). Line 3 should have du. After that it's game over.

It does not exist. The sum is divergent. I get the feeling you're just yanking our chain.

my guess is there is no closed form