The infamous "arctan puzzle"
First, some history
Before this website was even created, I was on a vacation to europe for family, and I solved at last \(2\) puzzles. After that, I said that I solved the strand puzzle and the arctan puzzle. First, I solved the strand puzzle in praha, but I only solved fir \(x_n\) and not \(y_n\), the time that I truly solved the strand puzzle was while I was writing it down on the page. I solved the strand puzzle first by the way. I know what you might be thinking,
what even is the arctan puzzle?
\[ z = a + bi = re^{i\theta} = r(cos(\theta) + i \text{ } sin(\theta)) = rcos(\theta) + irsin(\theta) \]
\[ a = rcos(\theta) \]
\[ b = rsin(\theta) \]
Good, you can convert from polar to cartesian coordinates, but what about back from cartesian to polar? (\(r\) Is the distance from the origin at \(0\), and \(\theta\) is the angle counerclockwise from the positave \(x\) axis. I'm saying this, because either I didn't say it earlier, or as a reminder) You would probably use the pythagorian therom to prove this, but this website dosen't heve illustrations, so I will prove the pythagorian therom without geometry. First, take the complex conjegate.
\[ a - bi = rcos(\theta) - irsin(\theta) \]
\[ cos(\theta) = cos(-\theta) \]
\[ -sin(\theta) = sin(-\theta) \]
\[ a - bi = rcos(-\theta) + irsin(-\theta) \]
\[ a - bi = re^{-i\theta} \]
\[ \text{Side note!} \]
\[ \text{ccong} (e^{i\theta}) = e^{-i\theta} = \frac{1}{e^{i\theta}} \]
\[ \text{End of side note.} \]
\[ (a + bi)(a - bi) = a^2 - abi + bia + b^2 = a^2 + b^2 = re^{i\theta} re^{-i\theta} = r^2 \]
\[ \text{Side note!.. Again.} \]
\[ r = |z| \text{ The distance from } 0 \text{, get it?} \]
\[ z \text{ ccong} (z) = |z|^2 \]
\[ \text{ccong} (z) = \frac{|z|^2}{z} \]
\[ \frac{1}{z} = \frac{|z|^2}{\text{ccong} (z)} \]
\[ \text{End of side note... Again.} \]
\[ a^2 + b^2 = r^2 \]
Okay, that is the pythagonian therom done, bou I was looking for \(r\).
\[ r \text{ (Incert is greater than or equal to symbol here.) } 0 \]
\[ r = \sqrt{a^2 + b^2} \]
That's \(r\), but what about \(\theta\)?
\[ \frac{b}{a} = \frac{rsin(\theta)}{rcos(\theta)} = \frac{sin(\theta)}{cos(\theta)} = tan(\theta) \]
\[ \text{Good, } r \text{ dosen't change } \theta! \]
\[ \theta = arctan(\frac{b}{a}) = ? \]
There's the arctan. By the way, this was when I used \(h\) instead of \(dx\). If \(arctan = tan^{-1}\), than I tried to find the derivitves of the inverse for the taylor seiries. I accedentally proved the chain rule with the proof that is now on the calculus page as a bonus. I thought "different method for finding \(\theta\). First, normalise \(z\)."
\[ ln(z) = ln(e^{i\theta}) = i\theta \]
\[ i^2 = -1 \]
\[ (i) (-i) = 1 \]
\[ \frac{1}{i} = -i \]
\[ \theta = -i \text{ } ln(z) \]
\[ \text{But what about } ln(z) \text{?} \]
\[ \text{You might remember the formula } \frac{a^{dx} - 1}{dx} \text{ for } ln(a) \text{, so let's try that!} \]
\[ \theta = -i \frac{z^{dx} - 1}{dx} \]
\[ \text{But then, what about }z^{dx} \text{?} \]
\[ \theta = -i \text{ } \frac{((1 + i dx)^{\frac{\theta}{dx}})^{dx} - 1}{dx} = \frac{1}{i} \text{ } \frac{(1 + i dx)^\theta - 1}{dx} = \text{ } \frac{(1 + i\theta dx) - 1}{i dx} \]
\[ \theta = \theta \]
So, all that we have proved is that \(\theta = \theta\). Yay!
Ok, so that's the end of that train of thought. I googled "derivitave of arctan" for the result. When I found the awnser \(\frac{1}{1 + x^2}\), I diden't find an awnser why. So, of course I asked ChatGPT, the proof went something like this:
Actually, I forgot it and this is my best guess, with \(\frac{}{f}\) as the inverse of \(f\):
derivitave of arctan
\[ f(\frac{}{f} (x)) = : x \]
\[ \frac{d}{dx} f(\frac{}{{}^f} (x)) = f\prime (\frac{}{f} (x)) \frac{}{f}\prime (x) = 1 \]
\[ \frac{}{f}\prime (x) = \frac{1}{f\prime (\frac{}{f} (x))} \]
\[ f(x) = tan(x) \]
\[ \frac{}{f} = arctan(x) \]
\[ arctan\prime (x) = \frac{1}{tan\prime (arctan(x))} \]
\[ \text{Oh! I forgot to tell you about (well, for one, it's pi day today) the other trigonometric functions other than } \text{sin} \text{ pronounced "sine", and } cos \text{ pronounced "cosine" or "cos", but I prefer cosine, theres } tan = \frac{sin}{cos} \text{ pronounced "tan" or "tangent", which I use interchangibly in real life, and it's inverse } arctan \text{ pronounced execly how it is spelled, but theres also } sec = \frac{1}{cos} \text{ pronounced "secant" or "sec", but I prefer secant, but theres also } csc = \frac{1}{sin} \text{ counterintuitivly, but it's pronounced "cosec" or "cosecant", but I prefer cosec. You can also square any of these for a } 2 \text{ superscript in front of the function instead of the parenthasies} \]
\[ tan\prime (x) = (\frac{sin(x)}{cos(x)}) \prime = \frac{sin\prime (x) cos(x) - sin(x) cos\prime (x)}{cos^2(x)} = \frac{cos(x) cos(x) + sin(x) sin(x)}{cos^2(x)} \]
\[ \text{Okay, we have } cos^2 + sin^2 \text{. Eccept, I still have to derive that.} \]
\[ z = e^{i\theta} = r(cos(\theta) + i \text{ } sin(\theta)) = rcos(\theta) + irsin(\theta) \]
\[ r = 1 \]
\[ z = cos(\theta) + isin(\theta) = a + bi \]
\[ a = cos(\theta) \]
\[ b = sin(\theta) \]
\[ r = \sqrt{a^2 + b^2} \]
\[ cos^2 (\theta) + sin^2 (\theta) = 1 \]
\[ \text{Okay, time to keep going.} \]
\[ tan\prime (x) = \frac{1}{cos^2(x)} \]
\[ tan\prime (x) = sec^2 (x) \]
\[ tan^2 (x) = \frac{sin^2 (x)}{cos^2 (x)} = \frac{sin^2 (x) + cos^2 (x) - cos^2 (x)}{cos^2 (x)} = \frac{sin^2 (x) + cos^2 (x)}{cos^2 (x)} - \frac{cos^2 (x)}{cos^2 (x)} \]
\[ tan^2 (x) = sec^2 (x) - 1 \]
\[ arctan\prime (x) = \frac{1}{sec^2 (arctan(x))} = \frac{1}{1 + sec^2 (arctan(x)) - 1} = \frac{1}{1 + tan^2 (arctan(x))} = \frac{1}{1 + tan(arctan(x))tan(arctan(x))} \]
\[ arctan\prime (x) = \frac{1}{1 + x^2} \]
Many weeks later...
I asked ChatGPT for awnsers, and it solved it immediety. This is why I called the arctan puzzle the "tricky puzzle that took \(1\) ChatGPT search to solve". Warning: This counts as a taylor seiries (404 page not found) for \(arctan(x)\) in the interval \(-1 < x < 1\) (or equal to \(1\)!)
\[ arctan(0) = 0 \]
\[ arctan(x) = \int_{0}^{x} \frac{1}{1 + t^2} dt \]
Quick tangent (no pun intended, plus this is more about arctan than tan)
\[ \sum\limits_{n=0}^{\infty}x^n = ? \]
\[ \text{By the convention in the harmonic page below, if } -1 < x < 1 \text{, than the powers of } x \text{ approach } 0 \]
.
\[ ? x = x \sum\limits_{n=0}^{\infty}x^n = \sum\limits_{n=0}^{\infty}x^{n + 1} = \sum\limits_{n=1}^{\infty + 1}x^n = \sum\limits_{n=0}^{\infty}x^n - x^0 + x^{\infty + 1} \]
\[ x^0 = 1 \]
\[ x^{\infty + 1} \to 0 \]
\[ ? x = ? - 1 \]
\[ ? (x - 1) = -1 \]
\[ ? = \frac{-1}{x - 1} \]
\[ \sum\limits_{n=0}^{\infty}x^n = \frac{1}{1 - x} \]
back on track
\[ \frac{1}{1 + x^2} = \frac{1}{1 - (-x^2)} = \sum\limits_{n=0}^{\infty} (-x^2)^n = \sum\limits_{n=0}^{\infty} (-1)^n x^{2n} \]
\[ arctan(x) = \int_{0}^{x} \sum\limits_{n=0}^{\infty} (-1)^n t^{2n} dt = \sum\limits_{n=0}^{\infty} \int_{0}^{x} (-1)^n t^{2n} dt \]
\[ \int_{0}^{x} (-1)^n t^{2n} dt = (-1)^n \frac{x^{2n + 1}}{2n + 1} \]
\[ arctan(x) = \sum\limits_{n=0}^{\infty} (-1)^n \frac{x^{2n + 1}}{2n + 1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} \dots \]