@@zinzhao8231 Said differently: 1. Whenever z is real, then Γ(z) is also real 2. For every w, there is a power series about w and a positive number R, such that the series converges to Γ(z) whenever |z-w| < R. Because these two properties are true (and the domain is connected), we can conclude that Γ(conjugate(z)) = conjugate(Γ(z)) for all z. This is a lemma in complex analysis.
I had a lecturer ask me to figure out the bandwidth of an oscillator. He didn't have a formula for it, but was curious what we could come up with. The bandwidth is (b - a), where a and b are where the function is 50% of it's peak. I know where the peak is, and I got (b^2 - a^2) = thing, at which point I got stuck. I came up with an approximation based on the idea that the peak was at the halfway point (so, (a+b)/2) and presented it to the lecturer. Later, I realised I could get past it without the approximation and came back with the exact formula. He said he preferred the approximation.
The modulus of the approximation converges fast on the true value of ~ 0.521564, but the approximation itself converges more slowly (so successive approximants lie approximately on a circular arc in the complex plane).
That's bad news for us engineers (whose axioms seem to tell us that convergence is always exponential, lol). I also fear that the speed of convergence using the Euler infinite product would be even much worse.
By the fundamental theorem of programming (I googled "i!") it's about 0.5+0.15i. So you got much closer to the absolute value than to the real and imaginary parts independently.
I am in a similar boat. I have done some personal study, but it was in number theory. I would love if the channel wanted to tackle some analytic number theory (like drawing connections between the reimann zeta function and prime numbers, and getting cool results from that), but I suspect that isn't his interest.
While stupidly looking for a closed form, this is the closest I got: We know the magnitude of i!, so it makes sense to go for the argument next, which is the imaginary part of ln(i!). The digamma function has a closed form for z = iy, which is 1/2y + coth(piy). You have to integrate the digamma function then to get the argument, with the boundary of the integration ending at z = i. The start point however must also lie on the line z = iy, so if you manage to get the arg((iy)!) for some y0, you can get it for any point on the z = iy line. int from (y0 to 1) 1/(2y) + coth(piy) dy + 1/2*ln(y0) + 1/pi*ln(sinh(piy0)) = arg(gamma(i)) And i! = i*gamma(i) And i! = i*sqrt(pi/sinhpi)*e^i*(arg(gamma(i)) It's too bad y0 cannot be 0 or infinity because the integral blows up😢. The integral of 1/2y + coth(piy) is 1/2*lny + 1/pi*ln(sinh(piy)). So theres a pole at y=0.
I have solved em. Problem is, the result is expressed in terms of Γ(1+i) so that would be cheating in this content perhaps a future video can be made related to those integrals.
Thus far. We know exact forms of it, but in series and integral forms. Mainly we focus on ArgΓ(1+i) since that's what we need to go along with |Γ(1+i)| to exactly know Γ(1+i). Do you want some forms?
But why did you say you use the second term in the approximation? You just used the k=1 term and the multiplication did not start from 0 so it is actually a first term approximatiln. Did I lost something?
Uhm, what would you define as the first term in this context? I am just curious because I would like to understand how you obtained the 50% error you mentioned
(Aleph Zero)! = (Aleph One). And .... this is zero% error, isn't it? Down With Vaccinations, Globalization, Digitization, Collectivization and Electrification!
Mathematicians only care about approximation when they are trying to get a handle on something, while engineers don’t care about exact solutions ever.
What's the difference between mathematicians and engineers? One's actually useful.
Pi = 10^0 for engineers
My PhD is in numerical analysis and I would have to disagree with you.
bro got the wrong answer and had to switch subjects to justify it 💀💀 (i'm in eng and have turned in lab reports with 200% error)
@@whenelvescry2625 the answer was in fact within the acceptable margins of error and so deemed correct by the axioms of engineering.
The “ok, cool” is very comforting at this point
Yes, suppose the engineer's bridge or cable is 8% short. No problem, it's within tolerance :)
Indeed 😂
The Closed Form Gods are crying in the club right now!
f(conjugate(z)) = conjugate(f(z)) provided that f is analytic and real on the real axis
i f analytic and real on the real axis ?
@@zinzhao8231 Said differently:
1. Whenever z is real, then Γ(z) is also real
2. For every w, there is a power series about w and a positive number R, such that the series converges to Γ(z) whenever |z-w| < R.
Because these two properties are true (and the domain is connected), we can conclude that Γ(conjugate(z)) = conjugate(Γ(z)) for all z. This is a lemma in complex analysis.
Hi,
13:00 It would be better to get an interval. For the modulus and for the argument as well.
"ok, cool" : 2:44 , 5:43 , 9:05 , 9:20 .
Amazing as usual, I love results like these so much!
Hey not sure if you've noticed me on your channel but I'm a huge fan😂
I had a lecturer ask me to figure out the bandwidth of an oscillator. He didn't have a formula for it, but was curious what we could come up with. The bandwidth is (b - a), where a and b are where the function is 50% of it's peak. I know where the peak is, and I got (b^2 - a^2) = thing, at which point I got stuck. I came up with an approximation based on the idea that the peak was at the halfway point (so, (a+b)/2) and presented it to the lecturer. Later, I realised I could get past it without the approximation and came back with the exact formula. He said he preferred the approximation.
@@chaosredefined3834 that is one hell of a plot twist 😂
@@maths_505 That is what an engineering environment looks like.
as i always hate approximations, I'll try to find a better closed form for it
The modulus of the approximation converges fast on the true value of ~ 0.521564, but the approximation itself converges more slowly (so successive approximants lie approximately on a circular arc in the complex plane).
That's bad news for us engineers (whose axioms seem to tell us that convergence is always exponential, lol). I also fear that the speed of convergence using the Euler infinite product would be even much worse.
cool (but quite simple) integral: Integral from 1 to e of: e^(-x)ln(x)dx = ln(2)
Wow!!! Your videos always surprises and shows that math is magic. Thanks for sharing
This video is gold. From the procedure to the engineers dissing.
By the fundamental theorem of programming (I googled "i!") it's about 0.5+0.15i. So you got much closer to the absolute value than to the real and imaginary parts independently.
@@farfa2937 the approximation converges quite quickly (where quickly is defined within acceptable margins of error 😂)
Love this vids even though the highest math class I've taken at the moment is calc 2 lmao
Thank you
Same
I am in a similar boat. I have done some personal study, but it was in number theory. I would love if the channel wanted to tackle some analytic number theory (like drawing connections between the reimann zeta function and prime numbers, and getting cool results from that), but I suspect that isn't his interest.
Sir its gud to be back, had some serious issues in my country( BD) . LOVE ur videos . U have just sum wonderful equations and its solutions.
Yes I have been following the news. I'm glad you're safe alhamdullilah.
@4:20 Also worth noting that e^(conj z) = conj e^z, since that's probably the least trivial fact you're relying on here (still fairly trivial though).
Blackpenredpen gets i!=.498-.1549i? Who is right?
That was Wolfram alpha not bprp💀
While stupidly looking for a closed form, this is the closest I got:
We know the magnitude of i!, so it makes sense to go for the argument next, which is the imaginary part of ln(i!). The digamma function has a closed form for z = iy, which is 1/2y + coth(piy). You have to integrate the digamma function then to get the argument, with the boundary of the integration ending at z = i. The start point however must also lie on the line z = iy, so if you manage to get the arg((iy)!) for some y0, you can get it for any point on the z = iy line.
int from (y0 to 1) 1/(2y) + coth(piy) dy + 1/2*ln(y0) + 1/pi*ln(sinh(piy0)) = arg(gamma(i))
And i! = i*gamma(i)
And i! = i*sqrt(pi/sinhpi)*e^i*(arg(gamma(i))
It's too bad y0 cannot be 0 or infinity because the integral blows up😢. The integral of 1/2y + coth(piy) is 1/2*lny + 1/pi*ln(sinh(piy)). So theres a pole at y=0.
Edit: only the imaginary part of the digamma function has the closed form, and digamma(z) = d/dz ln(gamma(z))
This needs more work.The answer is probably an irrational fraction involving pi and e
Beautiful I just got to know about your TH-cam channel ❤❤❤❤❤it's outstanding ❤❤❤❤
We have to find a closed approximation that have a zero error. This is a hope. Anyway Thank you for your fascinating video.
Why is the answer different from what I get in WolframAlpha? 🙏🙏
Wolfram Alpha is cheating but I guess the algorithm uses a few more steps in the series representation. 0.49802-0.15495i. Cool subject. 👍
Sir this might be unrelated but please may you post a vid on how to solve: Sum from 1 to inf of ((-1)^n)/((2n-1)^2)
That's just Catalans constant bro
@@maths_505 yes sorry mb
0.49802 - 0.15495 i according to Wolfram
Good job. Though I must admit I was hoping you'd try your luck with those gamma function integrals lol. Thought they'd be a breeze for you.
I have solved em. Problem is, the result is expressed in terms of Γ(1+i) so that would be cheating in this content perhaps a future video can be made related to those integrals.
Super cool, but you used one or two factors ? I saw only one factor of the product
2 terms as I used the e to the gamma tern outside as well.
So there is a closed form for the absolute value of i!, but not for its argument, is that it?
Thus far. We know exact forms of it, but in series and integral forms. Mainly we focus on ArgΓ(1+i) since that's what we need to go along with |Γ(1+i)| to exactly know Γ(1+i). Do you want some forms?
@@xinpingdonohoe3978 I do!
Interesting
Axioms of Engineering? Is that like the Power of Grayskull?
Yes but much more possible
But why did you say you use the second term in the approximation? You just used the k=1 term and the multiplication did not start from 0 so it is actually a first term approximatiln.
Did I lost something?
Well it was the term outside the product symbol and the k=1 term so technically it's 2 terms.
Uhm, what would you define as the first term in this context?
I am just curious because I would like to understand how you obtained the 50% error you mentioned
By only taking the exponential term💀 that was quite an error 😂
You mean only the term exp(-gamma z) / z ?😅😅
@@giorgioripani8469 yes 😭
😅👍
Hello can i get a reply from you
hi
@@AyushRajput-xw2ru yo
You lost me at hello
Guess who's thought about this? i!
(Aleph Zero)! = (Aleph One). And .... this is zero% error, isn't it?
Down With Vaccinations, Globalization, Digitization, Collectivization and Electrification!
calling it i! for the clickbait :\
Troppo lungo!!!...(Too long)
Not impressed.