Yo, your boi here could use some help. I'm trying to upgrade my recording set up to a tablet with s pen capability so I can have a bigger screen to write on. That way the viewer can have better visuals of the solution development and I don't have to put so much strain on my eyes and back while writing using my phone. A decent mid range upgrade will cost me about 500 dollars so it would be awesome if you guys could subscribe to my patreon for just this month and unsubscribe later as I don't need the monetary support on a regular basis and this is just a one time expense anyway. Here's the link to my patreon and thank you so much in advance! www.patreon.com/Maths505
@@benwhitaker9888 well did you really think I'd be sane and yet solve these calculus problems😂😂😂 Jokes aside gotta work with what I have. But the strain is getting to me.
@@maths_505 Man, I have nothing but my deepest of respects for you. I've also made a couple of math videos in a similar style to this in 9th grade and even though I had a wacom drawing tablet the process was still very much straining when done over a long period. Doing this on a phone is simply incredible! I hope you get your drawing tablet as soon as possible :D
You can also express this with the Lemniscate constant (also called Gauss's constant), denoted by the letter G. It actually simplifies the answer here very nicely as ( G * Pi ) / Sqrt[2].
@@r2k314 Yeah, it's not that different from integral representations of Gauss's Constant. If you're familiar with the constant it'd be an obvious next step.
Thr original integral = 1/2 integral between -inf to +inf (just substitute x by -x then sum). After that, substituting u^2=t/(1-t) would lead straight to the beta function.
When you got to 4:05, I spotted a substitution u = pi/2 - x to convert the limits to 0 -> pi/4. Then 2*I would be the integral 0 -> pi/2. But we would have missed all the fun exploration using the reflection formula.
Out of topic but I think I manage to find the area of an elipses let kx^2+y^2=c, k ∞ (-1)^n*cos^(2n)(θ)*(k-1)^n} dθ I switch up the integration and summation, leaving us with {sum n from 0 -> ∞ (1-k)^n * [integral(0 to 2 pi) cos^(2n)(θ)dθ]} I evaluate the integral separately, using the limits from (0 to pi/2), (pi/2 to pi), (pi to 3pi/2) , (3pi/2 to 2pi) integral(0 to pi/2) cos^(2n)(θ)dθ+ integral(pi/2 to pi) cos^(2n)(θ)dθ [call this P1]+ integral(pi to 3pi/2) cos^(2n)(θ)dθ [call this P2]+ integral(3pi/2 to 2pi) cos^(2n)(θ)dθ [call this P3] for P1, substituting θ = x-pi/2 gets us integral(0 to pi/2) sin^(2n)(θ)dθ for P2, substituting θ = x-pi gets us integral(0 to pi/2) cos^(2n)(θ)dθ for P3, substituting θ = x-3pi/2 gets us integral(0 to pi/2) sin^(2n)(θ)dθ add them up all results in 2*integral(0 to pi/2) cos^(2n)(θ)dθ+2*integral(0 to pi/2) sin^(2n)(θ)dθ for 2*integral(0 to pi/2) cos^(2n)(θ)dθ, this evaluates to B(1/2,n+1/2) for 2*integral(0 to pi/2) sin^(2n)(θ)dθ, this evaluates to B(n+1/2,1/2) summing this up gets 2*sqrt(pi)*Γ(n+1/2)/Γ(n+1) put this result back in the summation gets me {sum n from 0 -> ∞ (1-k)^n*2*sqrt(pi)*Γ(n+1/2)/Γ(n+1)} I got stuck here and i use wolfram alpha and it got 2pi/sqrt(k) putting it back in the beginning and got c/2*2pi/sqrt(k) the 2 cancel and expand c as sqrt(c)*sqrt(c) pi*[sqrt(c)]*[sqrt(c)/sqrt(k)] and finally i got the area of ellipse is pi*a*b CMIIW
If you extend the original integral to all of R (since it's even), you end up with only the I_1 integral; or, if you do a x -> π/2 - x substitution, you get only I_2.
You'll learn about the gamma and beta functions in complex analysis and mathematical physics. Contour integration from complex analysis. And some of the series expansions from Taylor series in cal1. The rest can be derived using compiled knowledge.
Yo, your boi here could use some help. I'm trying to upgrade my recording set up to a tablet with s pen capability so I can have a bigger screen to write on. That way the viewer can have better visuals of the solution development and I don't have to put so much strain on my eyes and back while writing using my phone. A decent mid range upgrade will cost me about 500 dollars so it would be awesome if you guys could subscribe to my patreon for just this month and unsubscribe later as I don't need the monetary support on a regular basis and this is just a one time expense anyway. Here's the link to my patreon and thank you so much in advance!
www.patreon.com/Maths505
Wait your telling me you've been doing this on your phone the whole time?? My god your a madman.
@@benwhitaker9888 well did you really think I'd be sane and yet solve these calculus problems😂😂😂
Jokes aside gotta work with what I have. But the strain is getting to me.
@@maths_505 Man, I have nothing but my deepest of respects for you. I've also made a couple of math videos in a similar style to this in 9th grade and even though I had a wacom drawing tablet the process was still very much straining when done over a long period. Doing this on a phone is simply incredible! I hope you get your drawing tablet as soon as possible :D
Btw do you also edit on your phone? If you have a pc, I feel like a wacom tablet would be better for your use case
Your videos are on par, if not even greater in educational value than that of bprp. You have huge potential sire!
Every time I see one of your thumb nails I think, "Well there's something else I couldn't solve."
One day you're gonna solve em the minute you see the thumbnail 🔥
But even in that case please watch the video I'm trying to make a living here😂
@@maths_505😂😂😂
You can also express this with the Lemniscate constant (also called Gauss's constant), denoted by the letter G. It actually simplifies the answer here very nicely as ( G * Pi ) / Sqrt[2].
is there anything about the integral that suggests a relationship?
@@r2k314 Yeah, it's not that different from integral representations of Gauss's Constant. If you're familiar with the constant it'd be an obvious next step.
Thr original integral = 1/2 integral between -inf to +inf (just substitute x by -x then sum). After that, substituting u^2=t/(1-t) would lead straight to the beta function.
This is a very elegant solution for an appetizing integral. Thanks for sharing.
Thank you for your suitable substitutions. Great job.
When you got to 4:05, I spotted a substitution u = pi/2 - x to convert the limits to 0 -> pi/4. Then 2*I would be the integral 0 -> pi/2. But we would have missed all the fun exploration using the reflection formula.
Lost you at the beta function, but now I’ve got some special integrals to familar
I was thinking if this could be done using the binomial expansion, and then use hyperbolic sub, in any case great vid as always my dude.
You could simply say that I is (1/2)*I1 because the roles of cos and sin are simply inverted between the intervals [0,pi/4] and [pi/4,pi/2]
I am surprised that you didn't invoke the Lemniscate constant in the end. I= (√2/2)ω. 🤣
Thanks, it was beautiful!
Out of topic but I think I manage to find the area of an elipses
let kx^2+y^2=c, k ∞ (-1)^n*cos^(2n)(θ)*(k-1)^n} dθ
I switch up the integration and summation, leaving us with
{sum n from 0 -> ∞ (1-k)^n * [integral(0 to 2 pi) cos^(2n)(θ)dθ]}
I evaluate the integral separately, using the limits from (0 to pi/2), (pi/2 to pi), (pi to 3pi/2) , (3pi/2 to 2pi)
integral(0 to pi/2) cos^(2n)(θ)dθ+
integral(pi/2 to pi) cos^(2n)(θ)dθ [call this P1]+
integral(pi to 3pi/2) cos^(2n)(θ)dθ [call this P2]+
integral(3pi/2 to 2pi) cos^(2n)(θ)dθ [call this P3]
for P1, substituting θ = x-pi/2 gets us
integral(0 to pi/2) sin^(2n)(θ)dθ
for P2, substituting θ = x-pi gets us
integral(0 to pi/2) cos^(2n)(θ)dθ
for P3, substituting θ = x-3pi/2 gets us
integral(0 to pi/2) sin^(2n)(θ)dθ
add them up all results in
2*integral(0 to pi/2) cos^(2n)(θ)dθ+2*integral(0 to pi/2) sin^(2n)(θ)dθ
for 2*integral(0 to pi/2) cos^(2n)(θ)dθ, this evaluates to B(1/2,n+1/2)
for 2*integral(0 to pi/2) sin^(2n)(θ)dθ, this evaluates to B(n+1/2,1/2)
summing this up gets
2*sqrt(pi)*Γ(n+1/2)/Γ(n+1)
put this result back in the summation gets me
{sum n from 0 -> ∞ (1-k)^n*2*sqrt(pi)*Γ(n+1/2)/Γ(n+1)}
I got stuck here and i use wolfram alpha and it got 2pi/sqrt(k)
putting it back in the beginning and got
c/2*2pi/sqrt(k)
the 2 cancel and expand c as sqrt(c)*sqrt(c)
pi*[sqrt(c)]*[sqrt(c)/sqrt(k)]
and finally i got the area of ellipse is
pi*a*b
CMIIW
For I_2, all we need is to make a substitution x-> π/2-x, no bother doing the all the beta function things again.
Noise
If you extend the original integral to all of R (since it's even), you end up with only the I_1 integral; or, if you do a x -> π/2 - x substitution, you get only I_2.
Couldn't you bring the square root up from the denominator as a negative exponent, (e^x + e^-x)^-1/2 and let u equal e^-x
what writing app do you use?
Can I use the fact that cos x = (e^x + e^-x) / 2, although needs a bit of editing...
That's cosh not cos
@@maths_505 whoops.... Thanks for correcting...
hello again. im taking calculus 2 and 3 next school year and enjoy watching your videos. at what level do you learn the topics in your videos? cheers
You'll learn about the gamma and beta functions in complex analysis and mathematical physics. Contour integration from complex analysis. And some of the series expansions from Taylor series in cal1. The rest can be derived using compiled knowledge.
now that's what I call a Cool integral 😎
Yeah it's been a while since I summoned ol' beta boi
12:25 (sqrt2/2)w Where w is the leminsacte constant
I wonder if there is a contour integral that works for this problem, just looks like the type that works with a rectangular contour?
I have got solution in terms of elliptic integral
Signed up to Patreon. Good luck!
Thanks mate. Means alot
You can solve I2 much more easily with u = pi/2 - x
cooool!
nice one
Io l'ho risolto con la binomiale...=rad2/5*Somma(-1)^k(-1/2,k)
Rare case of maths505 not using Σ in his video
hc loot integral.