BRO THIS WAS FREAKING INSANE! each move was so unexpected lol! thank you for reminding me that I skippped the infinite product chapter from my complex analysis book.
69.420? You sir are the Shelon musky of calculus. I wrote jee in 2002. Got a 4200 rank which wasn’t good enough to go to any iits. But I get a ton of joy watching your videos. Hats off to you!
@@arxalier2956 I’m so sorry, I have been out of touch with jee for 20 years, this channel brings back those good memories. Back in 2000, you had to have a sub 4000 rank to get anywhere, even in ism dhanbad as it was called then. I lived very close to iit Mumbai, and you needed a sub 1500 rank to get in the integrated ms in civil engineering.😅. So I went to a local engg college called Spce, 2 yrs with infosys, and pgp mba from Iim ahd. Then I emigrated to USA, now a citizen after 12 yrs. I wish you well on the journey. Remember, when you get there, know where you came from!
OMG I almost did this myself except I made the mistake of thinking you could telescope it instantly (mistaking the initial k as a constant) getting pi/2 -1/2 ln2, although I felt intuitivey that it was wrong lol thanks for the proper solution I feel I'm improving just by watching your vids
Great solution from start to finish! Loved the sinh result for log..reminds me of using the sin version and taking the logarithmic derivative to get a cotangent generating series for even values of the Zeta function. Your approach to the frac/floor makes me wonder if it generalizes. It seems like a good strategy to take the inverse function of whatever's under the floor operator, then run thru the integer inputs to that inverse function.
What level of calculus is this? The most I've taken is AP Calc BC in high school, I'm wondering what this compares to that, because this ALMOST seems like something I could do, but there are a few things I don't recognize.
If we subtract 1/2*ln(2pi) from the result and uses the sinhx expontential form, we will get the difference is pi-1/2*ln(e^(2pi)-1), the reason why they are so close is because if we remove -1 in the ln (which is very small difference because it is inside a logarithm), we get the difference is exactly 0 because pi-1/2*2pi=0
If tan(φ) = y, then this means that we can represent this as a right-angled triangle with height y and base 1 (remember that tan(x) = OPP/ADJ). From this we can conclude that the hypothenuse needs to have lenght L = sqrt(1²+y²) (from phytagorean theorem). Thus sin(φ) = sin(arctan(y))= OPP/HYP = y/sqrt(1+y²)
BRO THIS WAS FREAKING INSANE! each move was so unexpected lol! thank you for reminding me that I skippped the infinite product chapter from my complex analysis book.
Thank you Maths 505 for solving the integrals
Logging the product to obtain a series was a really nice touch.
Thanks. Playing around with infinite products often yields surprisingly beautiful results
Your channel is like a breath of fresh air for me)
Thanks!😄
Simply brilliant!!
69.420? You sir are the Shelon musky of calculus. I wrote jee in 2002. Got a 4200 rank which wasn’t good enough to go to any iits. But I get a ton of joy watching your videos. Hats off to you!
I figured I should use the fundamental constant of nice for a change😂
Sir I am also JEE aspirant, was 4200 rank below cutoff in 2002?
@@maths_505 I’d call it the bronze ration, if there isn’t one already
@@arxalier2956 I’m so sorry, I have been out of touch with jee for 20 years, this channel brings back those good memories. Back in 2000, you had to have a sub 4000 rank to get anywhere, even in ism dhanbad as it was called then. I lived very close to iit Mumbai, and you needed a sub 1500 rank to get in the integrated ms in civil engineering.😅. So I went to a local engg college called Spce, 2 yrs with infosys, and pgp mba from Iim ahd. Then I emigrated to USA, now a citizen after 12 yrs. I wish you well on the journey. Remember, when you get there, know where you came from!
Damn awesome bro 😎
OMG I almost did this myself except I made the mistake of thinking you could telescope it instantly (mistaking the initial k as a constant) getting pi/2 -1/2 ln2, although I felt intuitivey that it was wrong lol thanks for the proper solution I feel I'm improving just by watching your vids
Points for the effort
brilliant example of the fractional part
Love these interesting integrals
I was able to get to the ln series by myself, and thats where I got stuck. amazing integral!
This is so cool, my goodness!
Beautiful result indeed!
Great solution from start to finish! Loved the sinh result for log..reminds me of using the sin version and taking the logarithmic derivative to get a cotangent generating series for even values of the Zeta function.
Your approach to the frac/floor makes me wonder if it generalizes. It seems like a good strategy to take the inverse function of whatever's under the floor operator, then run thru the integer inputs to that inverse function.
I thought the same while penning this down.
@@maths_505 bro do there is any to evaluate this for every power of tanx
did pretty the same with the original remainder function and bunch of telescoping sums and limits, but yours looks much cleaner!
It took me a solid few seconds to realize why you chose that particular decimal 😂
😂
Nice integrals bro 🎉
Cannot lie...need an integral a day now!
Inshallah. I try to upload everyday or atleast 5 videos a week.
Really Smart solution. Thank you.
Alternatively for the sum you could have used the sums you got for cot(x), put in an imaginary x and integrated.
So cool! Now do the same integral, just with the integrand being {cotx}, please
right it is standard strategy when you have floor function integral_, worth to remember
Quite enlightening ！
Thank you
I am amazed
Amazing ❤
Thats some legit example for floor function
Can you try to make an axproximation for the catalan constant ?
Catalan = Top G
Wouldn't dare approximate it😂
@@maths_505 😂😂
Nice video. Had to speed up to 1.5x to follow.
What level of calculus is this? The most I've taken is AP Calc BC in high school, I'm wondering what this compares to that, because this ALMOST seems like something I could do, but there are a few things I don't recognize.
Beautiful problem, elegant solution!
I notice the numerical value is strikingly close to (1/2) ln(2 pi). Is this a coincidence?
If we subtract 1/2*ln(2pi) from the result and uses the sinhx expontential form, we will get the difference is pi-1/2*ln(e^(2pi)-1), the reason why they are so close is because if we remove -1 in the ln (which is very small difference because it is inside a logarithm), we get the difference is exactly 0 because pi-1/2*2pi=0
Noice
Can you solve integral of x^(1/x) from 0 to 1 ?
That's a Bernoulli integtal. I think Dr. Trefor already made a video on that.
Are you able to link the video having trouble finding it ?
@@roshanmadhav8876
ua-cam.com/video/PxyK_Tsnz10/v-deo.html
So is there anyway to solve the integral x^(1/x) 0 to 1?
ua-cam.com/video/PxyK_Tsnz10/v-deo.html
crazy
Intg => π/4• dx
can you solve e^(1/x) ln(x) Already wasted 3 hrs, got nothing.
I think you'll need a special function for that
I do not understand your explanation for sin arctan x = x/sqrt(1+x²)
If tan(φ) = y, then this means that we can represent this as a right-angled triangle with height y and base 1 (remember that tan(x) = OPP/ADJ).
From this we can conclude that the hypothenuse needs to have lenght L = sqrt(1²+y²) (from phytagorean theorem).
Thus sin(φ) = sin(arctan(y))= OPP/HYP = y/sqrt(1+y²)
My dumbass thought it was smart to cancel the tanx💀💀💀
69.420? Really? 😂
Whaaaat....
Its nice 😂