@@ateium2409 Yes, but Arithmetic and Algebra and Calculus all have so-called Fundamental Theorems. This was the first time I had seen the term used for this trig identity. As so often with Papa Flammy-isms, I like it !
In the IIT Genius interview, I was asked to express sin(t) in terms of tan(t/2). Since I did not know it previously, I had to work it out on the spot. Never forgot it since then :)
For those wondering about case y^2 > 1, write n^2 + 1 - y^2 = n^2 - (y^2 - 1) = (n - sqrt(y^2-1))(n+\sqrt(y^2-1)) and use partial fractions. Jens did table integral of the form 1/(x^2+a^2), and the other guy is table integral of the form 1/(x^2-a^2).
Took me around 2 hours of being stuck before doing something random near the end of the proof for that trigonometric identity at the end and finally getting an epiphany on how to finalize it. It's super subtle and very intense as papa suggested. I've had enough math for today.
That final identity is not that bad actually, spoilers ahead: ~let α=atan((y+1)/sqrt(1-y^2))-atan(y/sqrt(1-y^2)) ~using atan addition formula we get α=atan(sqrt((1-y)/(1+y))) ~similarity between triangles on the unit circle gives cos(α)=sqrt((y+1)/2) ~finally, double angle formula gives cos(2α)=y
This was so much interesting. I solved this integral and used this to solve the basel problem without using hyperbolic functions(the way the book does). Anyways thank you for introducing the book.. :)
I would use Euler sustitution with roots to calculate this integral sqrt(1-x^2)=(1+x)u then i would consider two cases |y|=1 In first case i will get arctan In second case i will need partial fraction
All you have to do to verify the trig identity arctan((1+y)√(1-y²))-arctan(y/√(1-y²))=1/2arccos(y) is apply the identity arctan(x)±arctan(y)=arctan((x±y)/(1-±xy)) twice, then draw a reference triangle using the resulting inverse tangent function. When I attempted this, I knew the arctan identity, and applied it once, receiving arctan(√((1-y)/(1+y)))=1/2arccos(y). However, it took me a bit to realize that I could multiply both sides of the equation by 2 and then apply it again. By the way, you can derive this identity by using the sum angle formula for tangent, tan(a±b)=(tan(a)±tan(b))/(1-±tan(a)tan(b)), and substituting a=arctan(x) and b=arctan(y), then take the arctan of both sides.
34ºC and papa flammy is getting a bit too hot, I feel you boi, it's been 32ºC here in south of Sweden and no wind at all. No rain either. It has just been quite balmy and truly annoying to work in. I've been sweating profusely all day. I'm just glad that I know how to survive a hot day, Math videos.
i instantly went and ripped off a pdf somewhere on the web, then i came back to the video papa: "DON'T rip off the pdf somewhere on the web" ~Oh crap he is right, what have i done! *deletes the pdf and buys it the right way*
@@PapaFlammy69 you didnt specify if y is a constant or if y is a funxtion of x? If y is a function of x then we have to rewrite y in terms of x to solve it..
If you're willing to take a trip into the complex numbers that solution to the integral will actually work for all real y values (even y=1 in the limit).
Flammable Maths I know you’re a pure mathematician but if you’ve got time I’d love to see how you might approach Tensor Calculus as applied to General Relativity. I know it’s not your usual thing. I kinda got hooked on your channel and now it’s almost a nightly ritual. I’ll have coffee and toast and watch a Flammable Maths video!
Eyy you checked it out The whole PDF deal is from a massive google drive folder I have of advanced math and physics textbooks' pdfs that I had no hand in creating. I know they are probably pirated. If I intend to pursue one of the textbooks' topic deeply, I buy a print copy. If you're gonna pirate, pirate responsibly. Maybe it fails the categorical imperative. But if I don't use the folder to find textbooks, then no one is supported. If I do use it, then the authors often get supported. Net positive for the authors and for me.
It is clear that the integral diverges for y 1, but from the formula it seems that y > 1 is a nono (both arccos and the square root are no longer real-valued). However, for y > 1 the arccos(y)/sqrt(1-y^2) can be rewritten as arccosh(y)/(y^2 - 1), so everything is groovy again. The function is continuous in y and the value at y = 1 is the limit of arccos(y)/sqrt(1-y^2) for y -> 1 from below or arccosh(y)/sqrt(y^2-1) for y -> 1 from above and is equal to...1 :D :D
You could have directly changed the sin(t) into 2tan(t/2)/(1+tan^2 (t/2)) The half angle formula for sinx (I m not stating the half angle formula from solutions of triangles) Edit: I solved it and it comes in the form of arctan and it's a better or faster method maybe
Uff, da hab ich mir von Dir gewünscht, Dich mal in Bademoden zu sehen und schon hast Du es mit Deinem nächsten Video-Abspann verwirklicht. Insgesamt ein sehr schönes und informatives Video. Aber sag mal, gehst Du gar nicht an die Sonne ? Man, Du bist ja blasser als ich. Raus mit Dir an die Sonne und lass Dich mal ordentlich brutzeln. Immer nur schwere mathematische Aufgaben lösen wird doch auf Dauer langweilig. :)
@@PapaFlammy69 Ohje, da gehts Dir ebenso wie mir. Dann waren die letzten Tage für Dich wohl ne echte Qual. Aber zum Glück kühlt es ja jetzt wieder runter und btw. egal ob nun gebräunt oder mit adeliger Blässe, Du machst halt immer ne gute Figur *zwinker*. Und zu gut darfst Du ja am Ende dann auch nicht aussehen, sonst kann sich ja keine Frau und kein Mann mehr auf Deine Lerninhalte konzentrieren. Möchte ja wegen dem Lernstoff ins Schwitzen kommen und nicht wegen dem Lehrerpapa ;)
Do we assume y is a constant then since we are integrating with respect to x?..i assume so but if y is a function of x that is an erroneus assumption..so not enough info from the start to truly solve this..
It's easy, then you have n^2+1-y^2 = n^2 - (y^2 -1), so you can use difference of squares to factor it and partial fractions after. Actually, both of these are table integrals, one of the form 1/(x^2+a^2) which Jens did, and the other is 1/(x^2-a^2), where a is positive real.
i have just downloaded it and this books looks like it was made for this channel dont be angry cause i download a pdf, the book is expensive for me ( i am from serbia) and i am not that interested in it. also i belive that downloading books isnt immoral if you use them for studying
Sorry for the late input, but I tried the solution on my own using x = cos(t) just for fun, and i got real close to the final result. I had -2\cdot \frac{1}{1-y}\cdot \frac{1}{\sqrt{1-x^{2}}}\cdot arctan(\frac{1}{\sqrt{1-y^{2}}}) sorry for the mess, but you can use a LaTeX viewer to make it nice to look at. Anyways, I'm struggling with reducing the ARCTAN expression to match the solution. Could Papa or any of you big bois help out? Online LaTeX viewer: www.codecogs.com/latex/eqneditor.php
I tried to solve that arccos(x)/2 .It didnt worked .I only got arcsin(P(x)) .P(x) is some function (i wont say what function it is😜😜😜) but results are same . arccos(x)/2=arcsin(P(x))
the book even the kindle version is too costly. insanely and averaciously outrageously cunningly overpriced. out of reach for folks in third world. we will get this in dark web for free any way.
@@michaelz2270 Well, since all the conferences are in English, lots of names get mispronounced all the time and, unless it's misleading, who cares, it just happens. But I wouldn't correct German saying German names even if everybody else uses Americanized version.
No one actually cares how fast you can do an integral; math is exploratory; it’s not just plugging numbers in. And doing math in different ways does not make the work useless; sometimes you can learn very interesting things from dong things the long way.
@@reetanshukumar1865 Smart people don’t need to flaunt their intelligence. Math is an exploratory topic, not a game where you try and crunch numbers as fast as possible. Professor Jo Boaler is a teacher at Stanford University. She wrote a research paper on the teaching of mathematics. From it, she says: “There is a common and damaging misconception in mathematics - the idea that strong math students are fast math students.” She also writes: "Math fluency" is often misinterpreted, with an over-emphasis on speed and memorization, she said. "I work with a lot of mathematicians, and one thing I notice about them is that they are not particularly fast with numbers; in fact some of them are rather slow. This is not a bad thing; they are slow because they think deeply and carefully about mathematics." For example, the famous French mathematician Laurent Schwartz, the man who made the Theory of Distributions, Schwartz kernel theorem, and various other theorems in mathematical analysis, wrote in his autobiography that he often felt stupid in school, because he was one of the slowest math thinkers in class. Yet he was still gifted at math, and went on to become a great mathematician. So no, just because you use a faster method, it does not make you any smarter. Of course, being fast is good for competitions. But this isn’t a competition that Flaming Maths is participating in; he is just integrating a function for fun. And where are you getting the notion that I am bad at math? Is it because I reject the fact that math is just about speed, because I am pretty sure the majority of mathematicians would argue against that.
@@grandstrategos1144 I have not seen any mathematician,who loves hard works unnecessarily.btw good luck with your hard work....its final reply from my side....
![A Powerful Elementary Integral [ (Almost) Impossible Integrals #1 ]](/img/logo.svg){kind=logo}
