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This Problem is Smooth like Butter
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- Äas pĆidĂĄn 12. 08. 2024
- This Putnam integral (Serret's Integral) was given on the Putnam exam in (2005 A5). Let's figure out a way to deal with this Putnam problem!
For those that don't know, the Putnam math competition features some super interesting and challenging problems!
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/ @brithemathguy
Disclaimer: This video is for entertainment purposes only and should not be considered academic. Though all information is provided in good faith, no warranty of any kind, expressed or implied, is made with regards to the accuracy, validity, reliability, consistency, adequacy, or completeness of this information.
Putnam Exam 2005 A5
#math #brithemathguy #putnam
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Equation solving like this makes me love maths so so much
me too!!
Ah this reminds me of my innocent days when I though all of math was this elegant.
Fun side note :
There's a nice and quick geometric way to see what A and B are in the equation (sin x + cos x) = A cos (x - B). Consider a square of side 1 unit and a line outside the square. The projection of the adjacent sides of the square on the line is (sin x + cos x) , which is also equal to the projection of the diagonal of the square on the line. The maximum value of the projection is clearly equal to the length of the diagonal or sqrt2 and this occurs when the angle x is equal to pi/4( since the diagonal is parallel to the line in that case ). Therefore A=sqrt2, B = pi/4
Love this method, way easier than using the Feynman technique to do it x3
Glad you think so!
From how he said "a Calc 2 student would try..." I kept thinking the trig substitution was going to hit a wall and he'd have to use Feynman's technique
4 minutes!
A hard integral!
So impressed!
Thanks so much for watching!
@@BriTheMathGuy thank you as well!
Pffft
I can do that in l less than two
@@prajodcp7145 no
@@prajodcp7145 clownin
I loved this. No bizarre tricks, just do whatever the problem is screaming at you.
Thank you.
Really glad you enjoyed it!
Shorter method :
We know integral fx from a to b is same as integral f(a+b-x) from a to b
once we got I=ln(1+Tanx)
I=ln(1+(tan(45-x)) .. expand tan(a-b)=(tana-tanb)/1+tanAtanB
=Ln(1+(1-tanx)/1+Tanx)
=Ln(2/1+Tanx)
I=ln(2)-I
I=1/2ln(2)
Now substitute limits
I= (1/2)ln(2)*(Ï/4-0)
Done
nice usage of King's property. I like that
You're a JEE Aspirant, aren't you? đ
I got⊠so excited when I saw that the integrals would cancel out
2:42 In JEE questions, to solve sinx + cosx we usually divide and multiply by â2 to get â2 sin(x+Ï/4) (common in SHM) it's same but easier.
thats exactly what i was thinking
That trick is widely used in Suneung too
suiiii one of my peeps are here
I knew it I will find someone and see! i did xD
Johnny Sins of the math world, thanks! I love this video.
đđ
*Johnny Sinx lol
@@Surajsharma-sj9gx Johnny Sins of the math world = Johnny Sinx it is. Let's make this viral.
@@malexmartinez4007 together we can đ
â@@Surajsharma-sj9gx * ln(e) sinx
Amazing presentation. Done like a real Math Prof!
I didn't really find a way to solve this integral without using advance techniques, and you've just showed you're solution like a champ. I'm shookt. Here's a quick sub for ya!
Easier than Feynmanâs trick, and also shorter ! Great
Glad you enjoyed it!
Literally man, you made this question super tough. Just apply â«f(x)=â«f(a+b-x) for limits a to b. Apply formula of tan(A-B) for A=Ï/4 and B=0. Add both equations and apply ln(a)+ln(b)=ln(ab). Only ln2 will left. Answer would come (Ï/8)ln2. Seriously this is a question from one of the easiest books of Indian Mathematics, NCERT.
It will be much more easier
Put I =lim 0 to Ï/4 ln (1+tan theta)
= lim 0 to Ï/4 ln{1 + tan (Ï/4- theta) }
Next put the formula of tan (Ï/4-theta)
Then I = ln (2/(1+ tan theta))
2I = lim 0 to Ï/4 (ln 2)
And thus I = ln 2 by Ï/8
kings rule?
Yep I did exactly the same when I first saw this problem
Your videos are amazing ! Calculus is my favourite topic from Math. I wish I'd seen them while I was still in school. You're a big inspiration
Wow, thank you!
@@BriTheMathGuy no, thank â€you for these amazing videos !
A very elegant and analytic approach! However, I prefer Feynman's trick in this case.
Anyway: I love your condensed and concise way to present the solution.
You could have just used the identity that integral from A to B of f(x)dx = integral from A to B of f(A+B-x)dx .
Also works!
Wait thatâs a thing?
@@ridlr9299 just use u substitution and you'll see it works.
I see no reason to apply that here, as you will make the function even worse,
Or can you elaborate on that please..
@@rishu_Kumar07 I meant on the log(â2cos(pi/4-x)
wonderful explanation bruh!
I have seen lot of your math videos ,It actually helped a lot. :)
I am in my second semester currently , you are far better than my math prof. THANKS A LOT KEEP IT UP
at 1:05 you can use the king property with ln(1+tan(pi/4-theta)) and use the tan(a-b) formula
Awesome video! Thanks for sharing, Bri!
You bet! Thanks for watching!
Thanks for making and sharing another phenomenal video! I was curious: did you know that you needed to end up with the two equal and opposite integrals near the end of the problem, or was that something you stumbled onto or found in a solution writeup somewhere? Either way I think that's amazing; I was just curious about the thought process behind it and how this could help an overall approach to integrals like this. Thanks!
Wow! Thanks a lot for the awesome trick with sine and cosine addition!
I started off with by parts and understood that its gonna take me nowhere
đ
Another method: when we are at ln(1+tan(a)) [let a = theta], we can use king's rule to get ln(1+tan(pi/4 - a)). Next, we can use the tan(a-b) identity, and it will turn out to be ln(2/1+tan(a)). Next, we use the quotient property of ln to separate the integrals and turns out we get the original integral. therefore, assuming the original integral to be I, we get 2I = pi/4*ln2, which evaluates to pi/8*ln2.
this was really good. keep it up!
Thank you watching!
iâm loving your videos! super well put together! great content, great editing, and you explain things really well, youâve got yourself a new subscriber
Thanks so much!
like a criminal undercover
there is a much simpler way to solve it. When you have the ln(tan(theta)+1)) just use king property to get ln(tan(pi/4-theta)+1) and it just simplify like butter.
You can also try x = (1-t)/(1+t) it work good with log.
Thanks YT for recommending this channel! Itâs pure quality
In 1:07 instead of writing tan as sin/cos we can substitute theta for pi/4 - theta and we get the answer in 2 more steps
You are doing a good job. Thank you
So nice of you!
That was pretty amazing. Great video composition, as always
Thank you very much!
Thanks for explaining the sin x + cos x identity! This is my new favorite formula now... why didn't they teach it in schools?
(no I am not a JEE aspirant)
They did for me (in high school)
Definitely had to make use of that formula in differential equations.
They did even for me. (Jee)
Didn't learn that specific identity - but we did learn the method (harmonic form)
In india they do
If u r not fm india
Then howdo u know abt jee
This way of solving is so genius
Smooth like butter like criminal undercover
Oh why couldnât this have realised earlier when I had to do this question for my Australian hsc maths ext2 assignment. Just kidding great video, alternate format which I used however was to use the law that you used at the end of the video when you have the integral of ln(tan(theta)-Pi/4), where you can use the tan compound angle formula and do some simplifying in order to get 2x your original integral as equal to pixln(2)/4, giving the same answer as you showed here
I'm in highschool and I have no idea what have I just watched.
Maths is designed that way
The hardest butter you have ever sliced in your life
Some time ago i saw integral
Int(x/sqrt(e^{x}+(x+2)^2),x)
Try to calculate it, and it is not nonelementary as you may think (I can do it , my way involves tricks like adding zero , multiplying by one and then doing obvious substitution)
CAS programs like Mathematica or Maple are unable to calculate integral i gave
I prefer u=(1-x)/(1+x) substitution for your integral
After substituting x=tan(theta), Substitute theta=pi/4-T. To rewrite the integral then use the tan(A-B) identity then use the log properties to see the magic happen you would right away get your answere
That's really cool! Thanks BriTheMathGuy :)
This problem was a regular problem in our textbook. Another easier way would be just to use proven property of a definite integral. But got a new way to solve. Thanks! This was interesting âșïž
Great tip!
Video title is fantastic . đđ
There is also a easy way to solve this by taking
TanΊ as tan(Ï/4 -Ί)
Crazy that how i got this same integration problem for my finals in grade 12
Fantastic explanation!
The integrals have gotten cooler.
I actually just came across this in another problem I was solving!
Very cool!
Amazing!
Glad you thought so! Thanks for watching!
You can also use king rule for solving ln cos terms
Waiting for u to get Silver Play Button đ€© You deserve it manâŠ.
Thanks so much!!
@@BriTheMathGuy đđ„°
Smooth like butter
Like a criminal under cover
There also you can do another method by not opening tan into sin and cos and only solve in tan by identity tan a + tanb. Then seperating them and put the limits
beautiful content as always
I appreciate that!!
I ran into a question like this yesterday, except it also involved partial fraction decomp, and when the terms started canceling out like that I though I did something wrong. And went about it a little different. Now Iâm gunna have to go back and look to see if I would have gotten the same answer.
The change of variable u=(1-x)/(1+x) leads to solve easily this integral.
Him: âyou probably want to do a trig subâ
Me :IBP! IBP! IBP! IBP!
The presentation just amazing
Can you make videos about what really single,double and triple integration shows?
You should have used kings rule after you get integral of ln(1+tanx) it's faster
2:55 animation gives me chills for whatever reason. Great video!! âïž
Thank you!! đ
just a recommendation: you'd love the channel 3blue1brown then, probably you already know about it
@@nikhilnagaria2672 I think Bri uses his library for visualizations
Brilliant job. You showed it so clearly.
Thank you! Cheers!
1:06, just use king's property after that and add the integrals, get the answer in 2 lines
This kind of problem solving makes me wonder how I survived 1st yr uni mathsâŠ
very well done indeed.
Thanks! Have a great day!
This was so helpful!
Really glad you thought so!
new viewer here. takes a while for the compi but gets it done:
*Integrate[Log[x + 1]/(x^2 + 1), {x, 0, 1}]*
i love how math class is just learning greek
Well, it literally looks really similar to one of the questions in HKDSE 2020 Maths Exam (The HKDSE examination is the university entrance examination for all high school students in Hong Kong)
I love your editing style brithemathguy
Thanks so much!
This is also in the "Ă©cole polytechnique de paris" oral exams (oraux X-ens analyse tome 1 pour les fr)
Wew lad solving that integral was a thrill ride
For the cos(a+b) you could Just segue that sin=cos because of the Same coefficent. Grest video
The solution for x^2 = 2 is +- root 2 and not just the positive. If the answer was carried out you'd end up with two answers, being the positive and negative version of the one you just got.
Eh, this integral had pretty much everything I hated about Calc II. And I loved i and iii.
Why wouldn't you use e to remove the natural logs and then put the natual logs back in after
What about the integral of cos(x)/(1+x^2) from 0 to 1?
I did it by setting u=ln(1+x)
Thank you so much.đ
Compute the Integral x^2024/(x^2+1)^2023 dx
Thank you that was great :)
Smooth like butter, like a criminal undercover!
like a criminal, undercover
As a jee advanced aspirant, this problem seems so easy!
I wish I knew calculus so I could understand your integral videos, but I'm still years away
You will get there!
These videos need more comments!
2 min integral for all jee aspirants
this was one of the problems on the putnam
what was the little coefficient trick at 2:30? I don't think I was ever taught that in school
That was sick
Good one.
Thank you! Cheers!
It's like you're the Bob Ross of mathematics
You should also show a plot of that function on that interval đ
Good idea!
@@BriTheMathGuy just looked for this in Wolfram Alpha and it doesn't match www.wolframalpha.com/input/?i=int+from+0+to+1+of+ln%28x%2B1%29%2F%28x%5E2%2B1%29
@@joeeeee8738 What do you mean, Wolfram gives the exact same answer
@@flyingpenandpaper6119 yeah, my bad!
A nice formula for solving such type of integrals : integral from 0 to a of ln(1+ax)/(x^2 + 1) = 1/2 * arctan*a*ln(1+a^2)
Our teacher asked us this in highschool đ
I did this with Feynman's technique, but for some reason I only see people solving it with trig subs. I don't have the harmonic addition formula memorized, I'm sorry!
It would be nice solution
Substitution x = (1-u)/(1+u) will work
nice!
is there any other way to solve it cause this suddenly became complicated
Why didn't we take the rule i.eâ« 0 to a f(x) =f(a-x) in the step of.â« 0 to a ln(1+tanÎ) please clarify
we can make another substitution t=1-x/1+x
But can we say A=negative sqrt( 2) because negative sqrt 2 square is 2.