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
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
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.
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.
That love and hate of separation of variables is sooo to the point! 👍👍 Great vid again, Dr. P!!
It's always a pleasure to watch your videos Dr Peyam. It's like I always were in my Calculus class in the uni. ☺️
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.
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 =/
Almost shed a tear :))))) wonderful result
Dr.Peyam: this video was beautiful!
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.
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.
That was a good differential equation.
An integral more moist than a tres leches cake.
I love tres leches omg
"Spicy Special" I LITERALLY DIEDDDD
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.)
Amazing!!
Wow that trig substitution simplification really was miraculous
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!
Sir,can u plz make a video on differentiation of matrix
Insane video love it
Nice solution to a hard problem
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 👍🏻
Secant you shall find.
😂😂
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
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.
Merits on this one surely equals a caramel flan. Cheers.
Thank you!!!
Thanks sir
Next question is whether you can adjust it to include the surface areas of the ends.
❤
The men who overcome the machine
Wow genial
On to derivative = volume.
nice problem~
I thought this would lead us to a sphere.
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