@@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.
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.
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.. :)
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.
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!
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
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).
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
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..
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
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
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 ;)
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
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.
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..
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))
@@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.
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.
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....
this man deadass said the fundamental theorem of trig
That's a good name for sin^2(x)+cos^2(x) = 1. 😁 Wish I'd thought of it.
@@bowtangey6830 isn't that just the Pythagoras theorem ?
@@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 !
Me: Bruh, It can't be that hard.
Jens: now let z = tan (t/2)
Me: oh boi, here we go again......
"Going through all those Integehral,sums & serious"
God i love the accent....😁
He was actually mimicking someone else's accent.
Very cool/10
Ah quality integrals... What I love this channel for.
Check my channel then, i do lots of integrals,sums,limits,,,
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 :)
Nice to get such an easy question on a scholarship interview.
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).
34°C
What's that in Fahrenheit Radians?
It's non-elementary
Idk ask your dog
first let's convert °C to °F, that's 93.2°F. now ° to radians, that's 1.627 rad F (rounded)
12:18 When you kill boss but boss music doesn't stop
This is a beautiful example of how so much math is just disguised in a different form. So cool!
>supporting springer
papa 😢
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.
When will you resume this awesome series?
hopefully soon! :D
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.
@@PapaFlammy69 Come to Huntington Beach, CA. 24 deg Celsius.
When he said "im sweating like filthy pig" I drooled a little...
Papa, I can't concentrate. Your guns are distracting. 😉
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.. :)
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.
Hyperbolic secant would also work :D
I love your videos my friend. Many more to come I hope. Apart from the mathematics you’re a really entertaining guy. I salute you sir.
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!
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
Flammable Maths atan formula -sure, I can see it not being common knowledge, but everything else is pretty straight forward.
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).
very nice substitution! by the way we should also mention Paul Nahin's book (Inside Interesting Integrals)
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.
Welcome & LOVE U Sir from India
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
Make more videos on this book! 😄
"nice human individual".......... that's an extremely specific thing to say now is it
Speaking about pigs, I'm here for the pie.......
Andrew's meme wars brought me here.
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*
:000
@@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..
Leif 13:46 or so
And what's about the accident when |y|>1? Therefore we've got another variant, haven't we?
Pigs don't sweat very well.
Temperature: π² ∙ 6. Answer: π² / 6.
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
*Simple stuff, it's just a matter of grasping the concepts & applying them.Nice vid. still tho.*
Pappas burning through some integrals
93.2 degrees Fahrenheit! That is a crazy way to lose weight!
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
Ayy is there no one gonna talk about the complex z & 1/z substitution then calculating the residue huh?
Sir great 👍👍👍👍👍
My mind just went into *some closed curves* in complex plane at 5:56
:D
Ayyyyy papa euleroid boi
Flammy, soo many Integrals. Das Ist genug fur heite ' boi . . .
Sounds like an interesting book
Class work:
Exam : ^^^^
1 = sec - tan ?
Trigonometry wants to know your location.
sec^2, tan^2
@@alexting827 I know, the thing I did is called a joke to point out hilarious things Jens said in the video xD.
Hey, where is the derivation of the transformation from Arctan to ArcCos? I was trying to do it and couldn't figure it out. Danke mein Herr
Yesss, a "Čau" ending
Was y treated as a constant during the whole exercise? Or as a function itself..
Do you think you will be doing anymore from this book?
Hi
Can you help me in this integral
Int (e^x cosx )^1/2 dx
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 ;)
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
And if y > 1 partial fractions can be used
I'm gonna read this book before we all die from global warming
> mfw only holds for -1
Can you also do the case when 1-y^2
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.
@@Ennar Yep, but they both fail when y^2=1. Which is another table integral as well
@@h4c_18 But you didn't ask bout that one :P
Just want to know what he did with "y"
What about the absolute value that happens when you square and then square toot, it could be negative.
Try this integration out:
int_0_to_1 ((x^a) - 1)/lnx dx
Would love to see some Papa techniques to solve this ?😉
I cna do it on my channel, i do a lot of integrals: D ruclips.net/video/vqbL5j4GLtM/видео.html
i feel dumb rn
Sec(x)= excite funczionen. Yayay
Almost impossible? The thumbnail suggests it is easier than integrating 1 :)
Tell that to sqrt(tan x), it seems too inocent...
I love you Flammy
日本ってこーゆー数学の本が体感少ない気がする。
俺が知らんだけかもだけど、
OP
Pls do them all
One Romanian boi
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..
"y is an element of (-1,1)"
It's right next to the problem in the book.
@@desertrainfrog1691 the book? What book? And so y is a constant value in that interval?
does the book only give the problems or does it explain the solutions too?
@@PapaFlammy69 cheers mate
Where's dotson daddy
May we expect polylogarithm ?
+C
:)
How many probhlems are there in the book?
What about whrn y2 is biger then one do the arctanh one
I have one imposible sum n^k/n! for any integer k
Wow these are really good but too hard for me papa
What about if y>1?
What if we want to find the original integral when y=2?
Easy man x = sin x => 1/(1+ysin x) => 1/(1+Yx) * 1/sqrt(1-x^2) pretty easy from here
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
An welcher Uni studierst du ?
Kelvin, please
You went crazy in just 34°C ? Come to India you'll see how people live here in 50°C.....
too much of an exaggeration buddy, I live in India and it’s 38-43 on average in peak summer
papa youre so hot more like daddy flammy
This is awesome sauce x very sexi too 😍
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))
I did it i got arccos(x)/2 :)
Lean 1 Let me guess, P(x)=sqrt((1-x)/2)? 😏
@@erikchinellato2405 yeees XD
Who disliked the video!😡
It's quite unfair for your other sponsors. They get their products eaten by daddy but not this one?? Please eat this book
Shpringah? I think it's called "Springer"
Springer was founded by Julius Springer, a German publisher. I'm with Jens on the pronunciation.
@@Ennar I thought everyone just speaks American.
@@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.
@@Ennar Chill, I'm just kidding, I know Springer should be pronounced the German way.
Hm it wasn't all that hard. Common Papa let's go harder
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.
Its just one form of beta function, y as constant, your hard work is useless...
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.
@@grandstrategos1144 lol,, keep it up...never be the smart one....
@@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....
@@reetanshukumar1865 Well, I’d rather take the words of someone who taught mathematics for multiple years.
he is on facebook as Cornel> What's his full name?