It's always fun and satisfying when things cancel out. "It blows up; it's quite nice." In my calc 1 class, things never cancelled out, and the only thing that blew up was the scribbling on my homework paper.
I like seeing \sqrt(2/\pi). Seen these normalization constants for the infinite square well problem in quantum mechanics before. At this point though I broke up physics and married math instead tbh!
Since the video "Derivative = Arch Length" i've been trying to find a curve where the Curvature = Arch lenght (k(t) = s(t)), but my calculus skills seem to be lacking. I've found a article that shows a curve where its curvature = torsion = archlenght. The derivation uses exponentials of matrixes. I couldn't understand =/
8:35 Dr. Peyam sort of debunking u-sub and using the variable u here as well- "I play on both sides and thus am the happiest youtuber ever." Alle gut sire.
This reminds me of my favorite integral: ∫ tan(x)√(sec⁴(x)+1)dx. You solve it using a double substitution: tan(u)=sec²(x), and another trig-based substitution after you simplify, some partial fractions and then back substitute, and it's totally solved. WolframAlpha takes a crap trying solve it and gives up, but using these (somewhat) standard techniques, you can solve it with a bit effort.
I wonder if Mathematica would do better. I used that in school 20 years ago... but then later it was all Wolfram. I heard that Mathematica had a huge list of integral tables they used, which sadly, were proprietary 😧..its truly sad times to see so much bs false knowledge in science, e.g. the medical industry, while real knowledge is hoarded in secret as proprietary.
@@onradioactivewaves It's possible that Mathematica would do better; it does sometimes, but mainly because WolframAlpha's parser isn't as good as using the Wolfram language with Mathematica. (If you weren't aware, Mathematica is made by Wolfram.)
If I remember : High School in Austria at the French High School, then a Ph D in US University. Dr Peyam speaks french, german, english and farsi. A genius man
Problem with differential equations is trying to find a grand theory for solutions, but there is none. Solutions are very specific and possible under certain conditions (initial and boundary) as in this example. Off the hat, to find a function whose derivative is equal to the surface area of the function is the volume of the sphere. f = V = 4/3 pi r^3. f’(V) = A = 4 pi r^2.
How do you know you need to set f(0)= the value you set? Do you think it would work with something else? And what's the logic behind such choice. It's a nice video btw 👍🏻
We are looking to check whether f(x) = sqrt(2/pi) / cos(sqrt(2pi) x) satisfies f'(x) = int_0^x 2 pi f(t) sqrt(f'(t)²+1) dt For confort, u = sqrt(2pi) x f(x) = sqrt(2/pi) / cos(u) = C1 / cos(u) f'(x) = 2 tan(u) / cos(u) (I note f'(x) as "the derivative of f with respect to x, expressed as a function of u"). There is no simplification of f(t) sqrt(f'(t)²+1) nor sqrt(f'(t)²+1) Although the expression is not a rational fraction of sinus and cosinus, I use Bioche rules to have a hint to what change of variable to perform. w(x) = f(t) sqrt(f'(t)²+1) w(-x) = w(x) Hence, we pick v = cos(u) dv = - sin(u) du = - sqrt(1-v²) du f(x) = C1 / v f'(x) = 2 sqrt(1-v²) / v² f'²(x) = 4/v^4 - 4/v² sqrt(f'²(x) + 1) = 1/v^2 sqrt(4 - v² + v^4) sqrt(f'²(x) + 1) f(x) du = -sqrt(4 - v² + v^4) /(v^3*sqrt(1-v²)) dv This method does not seem successful. We try v = tan(u/2) f(x) = C1 / cos(u) = C1 (1+v²)/(1-v²) f'(x) = 2 tan(u) / cos(u) = 4 v*(1+v²) / (1-v²)² f'(x)² = 16 v² (1+v²)² / (1-v²)^4 sqrt(f'² + 1)f(t) = C1 (1+v²)/(1-v²) sqrt(1+ 16 v² (1+v²)² / (1-v²)^4) By the magic of Wolfram Alpha, this equates: - C1(v² + 1)(v^4+6v²+1)/(v²-1)^3 By the magic of Wolfram Alpha, this has an antiderivative that conveniently simplifies for reals noted A: A = 4 C1 arctanh(x) - C1 x(x^4 - 8 x² + 3)/(x²-1)² The lower bound x = 0 becomes v = tan(u/2) = tan(sqrt(pi/2)) = C2 The upper bound tan(sqrt(pi/2)x). B = 2 pi [A]_C2^tan(sqrt(pi/2)x) B = 8 pi C1 arctanh(tan(sqrt(pi/2)x) - 8 pi C1 arctanh(C2) + right junk For the right junk, I'll use the fact that x(x^4 - 8 x² + 3)/(x²-1)² with x = tan(u) gives: 1/2 (2 cos(2u) + 3 cos(4u) + 1) tan(u) sec²(2u) I'll set C3 = sqrt(pi/2) right junk = pi C1 (2 cos(2C3 x) + 3 cos(4 C3 x) + 1) tan(C3 x) / cos²(C3 x) - pi C1 (2 cos(2C3 C2) + 3 cos(4 C3 C2) + 1) tan(C3 C2) / cos²(C3 C2) And then I give up because it doesn't seem to lead anywhere
It's always fun and satisfying when things cancel out. "It blows up; it's quite nice." In my calc 1 class, things never cancelled out, and the only thing that blew up was the scribbling on my homework paper.
There's a Stanford professor for signal processing who literally gave out an evil laugh before saying this all integrates to 0.
Almost shed a tear :))))) wonderful result
It's always a pleasure to watch your videos Dr Peyam. It's like I always were in my Calculus class in the uni. ☺️
That love and hate of separation of variables is sooo to the point! 👍👍 Great vid again, Dr. P!!
I like seeing \sqrt(2/\pi). Seen these normalization constants for the infinite square well problem in quantum mechanics before. At this point though I broke up physics and married math instead tbh!
Cheat, hope you have to pay back physics the alimony for your betrayal.
Amazing!!
Since the video "Derivative = Arch Length" i've been trying to find a curve where the Curvature = Arch lenght (k(t) = s(t)), but my calculus skills seem to be lacking. I've found a article that shows a curve where its curvature = torsion = archlenght. The derivation uses exponentials of matrixes. I couldn't understand =/
8:35 Dr. Peyam sort of debunking u-sub and using the variable u here as well-
"I play on both sides and thus am the happiest youtuber ever."
Alle gut sire.
An integral more moist than a tres leches cake.
I love tres leches omg
I'm only in high school so I don't understand any of this, but you do which makes me happy
Persevere and you'll get there pretty soon.
The integral represents a surface of revolution. That is why there is a 2pi.
Dr.Peyam: this video was beautiful!
Insane video love it
That was a good differential equation.
Wow that trig substitution simplification really was miraculous
This reminds me of my favorite integral: ∫ tan(x)√(sec⁴(x)+1)dx. You solve it using a double substitution: tan(u)=sec²(x), and another trig-based substitution after you simplify, some partial fractions and then back substitute, and it's totally solved. WolframAlpha takes a crap trying solve it and gives up, but using these (somewhat) standard techniques, you can solve it with a bit effort.
I wonder if Mathematica would do better. I used that in school 20 years ago... but then later it was all Wolfram. I heard that Mathematica had a huge list of integral tables they used, which sadly, were proprietary 😧..its truly sad times to see so much bs false knowledge in science, e.g. the medical industry, while real knowledge is hoarded in secret as proprietary.
@@onradioactivewaves It's possible that Mathematica would do better; it does sometimes, but mainly because WolframAlpha's parser isn't as good as using the Wolfram language with Mathematica. (If you weren't aware, Mathematica is made by Wolfram.)
Sir,can u plz make a video on differentiation of matrix
That’s wonderful 👏💙,Could you make a video about how you became a mathematician? That’s would be very helpful for me.
Check out the interview on my channel
If I remember : High School in Austria at the French High School, then a Ph D in US University. Dr Peyam speaks french, german, english and farsi. A genius man
@@meroepiankhy183 what!? So many languages
@@meroepiankhy183 i can only speak in English and Hindi. Dr Petam is very smart!
Merits on this one surely equals a caramel flan. Cheers.
Thank you!!!
Nice solution to a hard problem
"Spicy Special" I LITERALLY DIEDDDD
Secant you shall find.
😂😂
Problem with differential equations is trying to find a grand theory for solutions, but there is none. Solutions are very specific and possible under certain conditions (initial and boundary) as in this example. Off the hat, to find a function whose derivative is equal to the surface area of the function is the volume of the sphere. f = V = 4/3 pi r^3. f’(V) = A = 4 pi r^2.
Thanks sir
Next question is whether you can adjust it to include the surface areas of the ends.
How do you know you need to set f(0)= the value you set? Do you think it would work with something else? And what's the logic behind such choice. It's a nice video btw 👍🏻
Wow genial
The men who overcome the machine
I thought this would lead us to a sphere.
❤
We are looking to check whether f(x) = sqrt(2/pi) / cos(sqrt(2pi) x) satisfies f'(x) = int_0^x 2 pi f(t) sqrt(f'(t)²+1) dt
For confort, u = sqrt(2pi) x
f(x) = sqrt(2/pi) / cos(u) = C1 / cos(u)
f'(x) = 2 tan(u) / cos(u) (I note f'(x) as "the derivative of f with respect to x, expressed as a function of u").
There is no simplification of f(t) sqrt(f'(t)²+1) nor sqrt(f'(t)²+1)
Although the expression is not a rational fraction of sinus and cosinus, I use Bioche rules to have a hint to what change of variable to perform.
w(x) = f(t) sqrt(f'(t)²+1)
w(-x) = w(x)
Hence, we pick v = cos(u)
dv = - sin(u) du = - sqrt(1-v²) du
f(x) = C1 / v
f'(x) = 2 sqrt(1-v²) / v²
f'²(x) = 4/v^4 - 4/v²
sqrt(f'²(x) + 1) = 1/v^2 sqrt(4 - v² + v^4)
sqrt(f'²(x) + 1) f(x) du = -sqrt(4 - v² + v^4) /(v^3*sqrt(1-v²)) dv
This method does not seem successful.
We try v = tan(u/2)
f(x) = C1 / cos(u) = C1 (1+v²)/(1-v²)
f'(x) = 2 tan(u) / cos(u) = 4 v*(1+v²) / (1-v²)²
f'(x)² = 16 v² (1+v²)² / (1-v²)^4
sqrt(f'² + 1)f(t) = C1 (1+v²)/(1-v²) sqrt(1+ 16 v² (1+v²)² / (1-v²)^4)
By the magic of Wolfram Alpha, this equates:
- C1(v² + 1)(v^4+6v²+1)/(v²-1)^3
By the magic of Wolfram Alpha, this has an antiderivative that conveniently simplifies for reals noted A:
A = 4 C1 arctanh(x) - C1 x(x^4 - 8 x² + 3)/(x²-1)²
The lower bound x = 0 becomes v = tan(u/2) = tan(sqrt(pi/2)) = C2
The upper bound tan(sqrt(pi/2)x).
B = 2 pi [A]_C2^tan(sqrt(pi/2)x)
B = 8 pi C1 arctanh(tan(sqrt(pi/2)x) - 8 pi C1 arctanh(C2) + right junk
For the right junk, I'll use the fact that x(x^4 - 8 x² + 3)/(x²-1)² with x = tan(u) gives:
1/2 (2 cos(2u) + 3 cos(4u) + 1) tan(u) sec²(2u)
I'll set C3 = sqrt(pi/2)
right junk = pi C1 (2 cos(2C3 x) + 3 cos(4 C3 x) + 1) tan(C3 x) / cos²(C3 x) - pi C1 (2 cos(2C3 C2) + 3 cos(4 C3 C2) + 1) tan(C3 C2) / cos²(C3 C2)
And then I give up because it doesn't seem to lead anywhere
On to derivative = volume.
nice problem~
おはようございます。
WolframAlpha should have asked why do you want to do this
Sir, please find the general solution of y"-5y'+6y=2e^x+6x-5.Regards.
looks like a job for wolfram alpha
n+eⁿ