arctangent

The inverse function to the tangent.

If y is the tangent of theta, then theta is the arctangent of y.

supposing a function inverse to tan(x). Those of us that have gone through high school trig have learned to refer to this function as tan^-1(x), but for the sake of brevity we shall call it atan(x). Thus we have f(x)=tan(x), f^-1(x)=atan(x). It is important to note that such a function is the inverse of a periodic oscillator, and thus in order to be a proper function, must have a restricted range. (A function may not map one domain value to two range values, and because a function y=tan(x) would map many values in the domain(x) to a single range value(y), then its inverse y=atan(x) would do the opposite. Thus in order to be a proper function it must have a restricted domain)

Well what is the derivative of this function. We know that the first derivative of tan(x) is sec²(x). This can be derived with an analysis of limits and the sum formula for tan: tan(A+B)=(sinAcosB+cosAsinB)/(sinAsinB-cosAcosB). But that can be saved for another node. So for now we assume that d(tan(x))/dx = sec²(x) or 1/cos²(x). Well then now we imagine the inverse function y=atan(x), and then we rewrite tan(y) = x then d(x)/dy = sec²(y)then dx/dy=1/sec²(y) (we can do this if we look at limits, which calculus is all about). Now using an identity tan²(x) + 1 = sec²(x) we see that dy/dx = 1/(1+tan²(x)), however our original equation was tan(y) = x which is the same as tan²(y) = x². Thus our final result is dy/dx=1/(1+x²)

Now the interesting part comes when we apply a MacLaurin series. But for that we must no all the derivative to infinity. Well one thing we know for sure, judging by the gradient of our first derivative of atan(x), we notice a stationary point, wher the gradient is zero. Well if this exists then we know that the following derivative will intersect the x axis (where y equals zero) So then observing further the nature of such a graph we can see that the gradient gently increases toill a certain value then starts decreasing. We notice that this gradient is negative as well. Further analysis leads us to a brief description of the derivative of f(x). The nth derivative of x at zero can be characterized by -1^(n-1)/2c where c is an arbitrary constant dpendent on n and is zero for all even n. Now immediately differentiating (1+x²)^-1, using the chain rule we get -2x(1+x²)^-2, and we notice that at x = 0 this is equal to zero. The next derivative (achieved by deriving the factor, -2(1+x²)^-2 -2x(2x)(-2)(1+x²)^-3. And when x = 0 this is equal to -2. Hmmm... Now as we start to apply some logic to this we observe that the chain rule need only be use when one of the products has xⁿ where n!=0, so from here we can begin to observe that the number of units in the sum will not accumlate exponentially. Similarly when we think about the power if (1+x²), we notice that it is decreasing. and thus the modulus is increasing lineaarly. So as a sum accumulates we can begin to observe a pattern. When I made my nifty calculator program find the derivatives I discovered that the value of c always was 0 or an even factorial: -(2!), (4!), -(6!), etc. Then I applied this to the Taylor series, to get f(x)=x-x³/3+x⁵/5-x⁷/7, and so on. thus we have a nifty sum: -1^n-1x^2n-1/(2n-1).

Even cooler though is that this can be used to find the value of pi. Asuming tan(pi/4)=1then atan(1)=pi/4, 4(atan(1))=pi now one to any power is one, thus our some becomes 1-1/3+1/5-1/7+1/9...and pi=4(1-1/3+1/5-1/7+1/9...)

arctan(z) is equal to i ⁄ 2 × ln((1 + z × i) ⁄ (1 − z × i)) since tan(φ) = − i × tanh(i × φ). (Of course, you can get this directly, but as its hyperbolic counterpart is better known and more useful, this is the simplest way.)