I started watching your channel because I needed help with the differential equations homework but I really come to love watching you break down math problems Your enthusiasm is infectious You are an excellent teacher
I like both methods. If I recall, the power series regarding sinx and cosx is a good way to "prove" Euler's identity right? Edit: never mind, I found an old video you made haha
First take on this - that looks like a Taylor Series of some trig function. For cos(x), the signs would alternate, and x would have only even powers. So this series is cos(i√x) = cosh(√x) And since cosh y ≥ 1 for all real y, the solutions (y) are all complex, non-real. But x can be real and negative, while y is non-real. In fact, we get cos(i√x) = 0; i√x = (k+½)π, all integers, k x = -(k+½)²π², all integers, k There are infinitely many solutions; the largest is when k=0, x = -¼π² Fred
Hey Steve Sir, Here is a small challenge for you (actually pretty big), Can you make a video on the number of permutations of a Rubik's cube.? This was a great video by the way...although I understood only the real part as I don't know hyperbolic functions 😅
Oh man! So simple equation for you took me to Riemann hypothesis. Very good job! I like the way you enjoy maths. Go ahead you´re a really encouraging example for new generations.
Hi blackpenredpen I am a big fan of yours and obviously maths . I am a student of class 11 and I made a problem.. If y is a solution to the equation y+y'+y''+.... +y''( n dashes {nth derivative}) =e^x.then prove that y equals the (n+1) th derivative of itself. Love from India...
Done. Take the derivative of the original equation, and subtract the new one from the original, everything else cancels and you get y - (n+1)th derivative of y = 0.
Doli Paik Yes. Let [D^n]y(t) denote the nth derivative of y(t). To solve [D^(n + 1)]y(t) = Dy(t), integrate over the interval (0, x), resulting in [D^n]y(x) = y(x) + y(0) - [D^n]y(0). Letting x = 0 implies that y(0) = [D^n]y(0), simplifying the equation to simply [D^n]y(t) = y(t). Therefore, y(t) is the sum from m = 0 to m = n -1 of the terms of the sequence a(n, t) = C(m)·e^[ω(n)^m·t], where ω(n) = e^(2πi/n).
Instantly thought of the complex way because I know the expansion of cosh, but it's very smart the way you solved it on the left side. Would have never thought of it
i sqrt(x) = pi/2 + z pi = {..., -3pi/2, -pi/2, pi/2, 3pi/2, ...} When you square both sides, -x = {pi^2/4, 9pi^2/4, 25 pi^2/4, 49 pi^2/4, ...} The given solution is the largest (i.e. least negative), when you multiply by -1 to get it.
Clarification of biggest soln. We have Ln (-1)= i (1+2n)Pi, also cos(sqrt(-x)) = 0 has solutions sqrt(-x)=Pi (1+2n)÷2 for integer n=0,1,2,... . So general soln is x = - [Pi(1+2n)]^2/4. However, since x is negative, the biggest soln corresponds to n=0, therefore x = - (Pi/2)^2.
Most universities in Japan offer online classes, so one has more time than usual. I am watching your videos in this situation. I am always grateful for your help. Let's overcome this together!
Serious question: What is the domain of cosh(sqrt(x))? On the one hand, it should be x>=0 because of the ‘x’ under the root, but on the other hand, the function has a real number output for all x, so it should be all real x.
Hey bprp! I have a question Why, in the complex way, replacing ln(-1) with ln(i²) wouldn't work? This would lose us the Pi we need to get to the final answer. Maybe someone can explain it to me :)
sir, i have a question. we can write ln1=0, can we write ln1=2iπ?? ln1=ln i^4= 4 ln i = 4.iπ/2 = 2iπ another one, e^iπ=-1 iπ= ln(-1) 2iπ=2ln(-1) 2iπ=ln1=0 that means 2iπ=0 is it right or wrong????
This is the problem with using complex logarithms: since the exponential function is periodic (because it includes cos + i sin), the logarithm of a number is not uniquely defined, or in other words it’s a multi-valued relation. In order to keep the logarithm useful as a function, it’s necessary to restrict its range by taking the *principal value* of the logarithm, which means the imaginary part must be in [0, 2π), or alternatively (-π, π]. If you use a Taylor series centered at z = 1, you’ll end up with the second one, I think. Either way, you should never get 3πi as a value. Sometimes, it’s useful to have all possible values of the logarithm. In that case, you could get away with saying that 2πi is equal to 0 *up to a constant multiple* of 2πi. So, for example, you could have 11πi = πi + 2πki for some integer k, in this case k = 5. That means that ln 1 = 2πki, and ln(-1) = πi + 2πki. It’s probably better, though, to say that e^2πki = 1, rather than ln 1 = 2πki.
Great vid, as always. Aren’t your two methods really just the equivalency of cos(ix) and cosh(x)? cos(sqrt(-x)) = cos(isqrt(x)), which therefore equals cosh(sqrt(x)).
Let A be an element of the interval ]0, ♾[, and let x be an element of the real numbers except zero. x^A = A^x is equivalent to A·log(x) = x·log(A) if x > 0, and A·log(-x) = x·log(A) if x < 0. These are equivalent to log(x)/x = log(A)/A if x > 0 and log(-x)/(-x) = -log(A)/A if x < 0. It should be noted that 1/x = e^[-log(x)], while 1/(-x) = e^[-log(-x)]. Therefore, [-log(x)]·e^[-log(x)] = -log(A)/A if x > 0, and [-log(-x)]·e^[-log(-x)] = log(A)/A if x < 0. These equations translate to -log(x) = W[n, -log(A)/A] if x > 0, and -log(-x) = W[n, log(A)/A]. The W[n, x] map has a branch point at x = -1, such that W[-1, -1] = W[0, -1] = W[+1, -1], and since these are the only real valued branches, it must be the case that log(A)/A = -1 if x < 0, or log(A)/A = 1 if x > 0. log(A)/A = -1 implies log(A)·e^[-log(A)] = -1, which implies [-log(A)]·e^[-log(A)] = 1. Therefore, W(1) = Ω = -log(A) = log(1/A). This implies A = Ω. This is consistent with the domain of x |-> W(0, x), since -log(A) > 0. log(A)/A implies W(-1) = -1/e = -log(A), which implies 1/e = log(A), or simply A = e^(1/e). Therefore, A = e^(1/e), or A = Ω.
Doesn't cos(-π/2)=0? Why are we choosing the smallest positive value when negative values also work? It would yield no solution that way, but is that not also correct? Because x=(2n-1)π/2 where n is an integer would be the solution family, and it has no upper bound
anyway, could you solve the problem? Show that: lim_(n -> +inf) [ ln(2) - ( -1/2 + 1/3 - 1/4 + ... - (-1)^n / n ) ]^n = sqrt(e). source: made by D.M.Batinetu and Neculai Stanciu. p/s: I am sorry if my typing makes you read hard.
I tried to find cos 3º using cosh and I found cos 3º = (-1)^(1/60). It looks like I cant assume -1 = (-1)^15 and (-1)^(1/60)=[(-1)^15]^(1/60)=(-1)^(15/60)=(-1)^(1/4)=sqrt(i). Why not?
When you're finding 3°, you look for the *principal value* of (-1)^(1/60). But if you replace -1 with any of its odd number powers, you're not finding the principal value of it anymore. sqrt(i) is a solution of x^60+1=0, but it's just not the value you normally want
And to find cos(3°), what you want more is the angle sum (difference) formula for cosh, using 18°-15°=3° or anything you found useful to produce the 3°
LuhDooce In theory, yes, but such a thing is impossible to evaluate and calculate analytically. You can never express such an operation in any closed-form.
Technically aren't there infinte complex solutions for x? Now complex numbers are not ordered but even the magnitude of the solutions maybe taken to infinity
The solutions are infact all real, and there are infinitely many. The general form is x=-π²(n+0.5)² ∀n∈Z. In the video he has taken the special case of n=0.
I understand that if the infinitely many solutions are clustered in about the same place, then you could have a biggest solution. But how can you be _sure_ of that? How can you be _sure_ there is even a biggest solution? Also, how can you be sure you found the biggest solution, and not just a solution in the middle?
We need Cos of something to be zero, then that something (which is √-x here) must be an odd multiple of π/2, like -π/2, 3π/2, -3π/2, 5π/2, 7π/2, etc. Squaring them all, you have the possible values for -x being π^2/4, 9π^2/4, 25π^2/4, etc. Multiply both sides by -1, you get the possible values of x: -π^2/4, -9π^2/4, -25π^2/4, etc. It's easy to see that the greatest of these is -π^2/4.
I always think that the guys who found those relations were in a quarentene or something like that. How could someone think about cosh(x) ???? What the f**k...
Oh my cosh! Also, cute bunny 🐰
Hey, nice to see you here! Love your videos.
@@green0563 yes!!!!!!!
I started watching your channel because I needed help with the differential equations homework but I really come to love watching you break down math problems
Your enthusiasm is infectious
You are an excellent teacher
Non-real way if I have to choose, but I like both.
Watch this clip
👇
th-cam.com/video/TN6uYlmc25I/w-d-xo.html
I like both methods. If I recall, the power series regarding sinx and cosx is a good way to "prove" Euler's identity right?
Edit: never mind, I found an old video you made haha
Day 100 after pandemic outbreak: talking to a toy bunny
Right what I was thinking
First take on this - that looks like a Taylor Series of some trig function.
For cos(x), the signs would alternate, and x would have only even powers.
So this series is cos(i√x) = cosh(√x)
And since cosh y ≥ 1 for all real y, the solutions (y) are all complex, non-real.
But x can be real and negative, while y is non-real. In fact, we get
cos(i√x) = 0; i√x = (k+½)π, all integers, k
x = -(k+½)²π², all integers, k
There are infinitely many solutions; the largest is when k=0,
x = -¼π²
Fred
That bunny is so freaking smart...instant solve with only a glimpse at the question. I mean wth...
Who'd expect Miffy to be such a mathematical genious?
They're best at multiplication, though.
Ramanujan 2
just noticed 3blue1brown is a patreon thats pretty cool
Ikr!
even Tibees is a patreon
Any complex solutions?
Technically real solutions are complex so yes ;)
idk, seems a pretty complex question to answer...
Boss, i have a problem for you to solve ...how can i send it to you
@Diego Marra nice
Diego Marra that’s amazing!!!
Hey Steve Sir,
Here is a small challenge for you (actually pretty big),
Can you make a video on the number of permutations of a Rubik's cube.?
This was a great video by the way...although I understood only the real part as I don't know hyperbolic functions 😅
Oh man! So simple equation for you took me to Riemann hypothesis. Very good job! I like the way you enjoy maths. Go ahead you´re a really encouraging example for new generations.
Hi blackpenredpen I am a big fan of yours and obviously maths . I am a student of class 11 and I made a problem.. If y is a solution to the equation y+y'+y''+.... +y''( n dashes {nth derivative}) =e^x.then prove that y equals the (n+1) th derivative of itself. Love from India...
Done. Take the derivative of the original equation, and subtract the new one from the original, everything else cancels and you get y - (n+1)th derivative of y = 0.
Can you find n solutions for y.
Doli Paik Yes. Let [D^n]y(t) denote the nth derivative of y(t). To solve [D^(n + 1)]y(t) = Dy(t), integrate over the interval (0, x), resulting in [D^n]y(x) = y(x) + y(0) - [D^n]y(0). Letting x = 0 implies that y(0) = [D^n]y(0), simplifying the equation to simply [D^n]y(t) = y(t). Therefore, y(t) is the sum from m = 0 to m = n -1 of the terms of the sequence a(n, t) = C(m)·e^[ω(n)^m·t], where ω(n) = e^(2πi/n).
I smell jee here
Instantly thought of the complex way because I know the expansion of cosh, but it's very smart the way you solved it on the left side. Would have never thought of it
Hey man. That question was what I exactly covered when I taught series. And I like your rabbit mic haha
It's not a mic.
BrownWater oh not? Haha thought it was lol
@@drpkmath12345 😂 yeah, he's wearing the mic on his shirt.
BrownWater haha why did I think he was holding a rabbit mic lol
That was bunny...i mean, funny
cos(sqrt(x))=0
sqrt(x)=pi*n-(1/2)
square and foil and
x=(pi*n)^2-pi*n+(1/4)
general solution
i sqrt(x) = pi/2 + z pi = {..., -3pi/2, -pi/2, pi/2, 3pi/2, ...}
When you square both sides, -x = {pi^2/4, 9pi^2/4, 25 pi^2/4, 49 pi^2/4, ...}
The given solution is the largest (i.e. least negative), when you multiply by -1 to get it.
Clarification of biggest soln. We have Ln (-1)= i (1+2n)Pi, also cos(sqrt(-x)) = 0 has solutions sqrt(-x)=Pi (1+2n)÷2 for integer n=0,1,2,... . So general soln is x = - [Pi(1+2n)]^2/4. However, since x is negative, the biggest soln corresponds to n=0, therefore x = - (Pi/2)^2.
日本からこんにちは。
日本でもほとんどの大学でオンライン授業となり、従って時間がいつもより余ります。そんな中であなたの動画を見させてもらってます。 いつもありがとうございます。
共に乗り越えていきましょう！
TRANSLATION: Hi I am from Japan.
Most universities in Japan offer online classes, so one has more time than usual. I am watching your videos in this situation. I am always grateful for your help.
Let's overcome this together!
Serious question: What is the domain of cosh(sqrt(x))? On the one hand, it should be x>=0 because of the ‘x’ under the root, but on the other hand, the function has a real number output for all x, so it should be all real x.
Excelent. Both ways are so clever. Congratulation.
Very nice video explaining the trivial with a twist.
That equation looks like the series for e^x
You explain in an amazig way,,, i love it
Hey bprp! I have a question
Why, in the complex way, replacing ln(-1) with ln(i²) wouldn't work? This would lose us the Pi we need to get to the final answer. Maybe someone can explain it to me :)
itspatrick It does work, though. log(i^2) = 2·log(i) = 2·πi/2 = πi = log(-1).
@@angelmendez-rivera351 thank you! I was dumb for a moment , thanks for helping me out!
I love complex numbers. Things just work! Complex > Reals all day
Really very good professor . I like both the methods .
Brother, tell me any trick to remember cosx, sinx, logx,.... All series? I need them for jee main exam.
Cosh is beautiful......thank you ❤
sir, i have a question. we can write ln1=0, can we write ln1=2iπ??
ln1=ln i^4= 4 ln i = 4.iπ/2 = 2iπ
another one,
e^iπ=-1
iπ= ln(-1)
2iπ=2ln(-1)
2iπ=ln1=0
that means 2iπ=0
is it right or wrong????
This is the problem with using complex logarithms: since the exponential function is periodic (because it includes cos + i sin), the logarithm of a number is not uniquely defined, or in other words it’s a multi-valued relation.
In order to keep the logarithm useful as a function, it’s necessary to restrict its range by taking the *principal value* of the logarithm, which means the imaginary part must be in [0, 2π), or alternatively (-π, π].
If you use a Taylor series centered at z = 1, you’ll end up with the second one, I think.
Either way, you should never get 3πi as a value.
Sometimes, it’s useful to have all possible values of the logarithm. In that case, you could get away with saying that 2πi is equal to 0 *up to a constant multiple* of 2πi.
So, for example, you could have 11πi = πi + 2πki for some integer k, in this case k = 5.
That means that ln 1 = 2πki, and ln(-1) = πi + 2πki.
It’s probably better, though, to say that e^2πki = 1, rather than ln 1 = 2πki.
@@Aruthicon thanks
the expression is equal cos(i√x)
Oreo the bunny
😭
Same question as asked in kvpy exam (India)
which year?
Great vid, as always. Aren’t your two methods really just the equivalency of cos(ix) and cosh(x)? cos(sqrt(-x)) = cos(isqrt(x)), which therefore equals cosh(sqrt(x)).
Yes. And I said that in the end too : )
blackpenredpen Ah, so you did. 🤦♂️
Both methods are awesome, nice video!
Not a particularly hard problem, but a fun one to work through - find the value A such that A^x = x^A at exactly one point without crossing
Austin I am not sure that is possible, unless you are restricting yourself to a proper subset of the real numbers.
Let A be an element of the interval ]0, ♾[, and let x be an element of the real numbers except zero. x^A = A^x is equivalent to A·log(x) = x·log(A) if x > 0, and A·log(-x) = x·log(A) if x < 0. These are equivalent to log(x)/x = log(A)/A if x > 0 and log(-x)/(-x) = -log(A)/A if x < 0. It should be noted that 1/x = e^[-log(x)], while 1/(-x) = e^[-log(-x)]. Therefore, [-log(x)]·e^[-log(x)] = -log(A)/A if x > 0, and [-log(-x)]·e^[-log(-x)] = log(A)/A if x < 0. These equations translate to -log(x) = W[n, -log(A)/A] if x > 0, and -log(-x) = W[n, log(A)/A]. The W[n, x] map has a branch point at x = -1, such that W[-1, -1] = W[0, -1] = W[+1, -1], and since these are the only real valued branches, it must be the case that log(A)/A = -1 if x < 0, or log(A)/A = 1 if x > 0.
log(A)/A = -1 implies log(A)·e^[-log(A)] = -1, which implies [-log(A)]·e^[-log(A)] = 1. Therefore, W(1) = Ω = -log(A) = log(1/A). This implies A = Ω. This is consistent with the domain of x |-> W(0, x), since -log(A) > 0.
log(A)/A implies W(-1) = -1/e = -log(A), which implies 1/e = log(A), or simply A = e^(1/e).
Therefore, A = e^(1/e), or A = Ω.
Go! Go! Power Series!
Mighty Mathy Power Series!
Hey, I didn't know Miffy does Math!
Or Nijntje, as she is known in her home country, The Netherlands
Nice bro 👍
I like both, they are really interested!!.
Doesn't cos(-π/2)=0? Why are we choosing the smallest positive value when negative values also work? It would yield no solution that way, but is that not also correct? Because x=(2n-1)π/2 where n is an integer would be the solution family, and it has no upper bound
plz find general term for sum √1+√2+√3+...√n
I was too much focussed on what you were doing that I only saw the rabbit mic at the end...
That bunny senpai should write my exams
Please, integral of x/sinx definite from zero to pi over 2
Why are the factorial parts not in square roots?
If x=1, then cos(i) = 1+1/2!+1/4!+...
Does this make any sense?
Super interesting!!!
Hey bro Is Riemann hypothesis solved? Or you are still trying!
Cosh (sqrt x)
anyway, could you solve the problem? Show that: lim_(n -> +inf) [ ln(2) - ( -1/2 + 1/3 - 1/4 + ... - (-1)^n / n ) ]^n = sqrt(e).
source: made by D.M.Batinetu and Neculai Stanciu.
p/s: I am sorry if my typing makes you read hard.
The result is just unreal
the second method is like magic
I like it!
I tried to find cos 3º using cosh and I found cos 3º = (-1)^(1/60).
It looks like I cant assume -1 = (-1)^15 and (-1)^(1/60)=[(-1)^15]^(1/60)=(-1)^(15/60)=(-1)^(1/4)=sqrt(i). Why not?
When you're finding 3°, you look for the *principal value* of (-1)^(1/60). But if you replace -1 with any of its odd number powers, you're not finding the principal value of it anymore.
sqrt(i) is a solution of x^60+1=0, but it's just not the value you normally want
And to find cos(3°), what you want more is the angle sum (difference) formula for cosh, using 18°-15°=3° or anything you found useful to produce the 3°
When he says, "Here's the DEAL"...:P
Cool,ur genius !!!!!
You ask which I like more?? I like both equally! However, I am not a yarn rabbit ;)
Odd terms of cosine of square root negative x power series: We were bad but now we're good
That's cool!!! 👍
Why always holding something in your left hand?
Usually it's a mike in his hand.
Is it possible to do something like the i-th tetration of i? Or instead, some irrational number tetration of a number
LuhDooce In theory, yes, but such a thing is impossible to evaluate and calculate analytically. You can never express such an operation in any closed-form.
Unreal!
Why can't cos(-sqrt(x)) be written as cos(i*sqrt(x)) ? Then i*sqrt(x) = pi/2 => sqrt(x) = -i*pi/2 => x = - pi^2/4
Because that's an arbitrary step that takes longer to reach the same solution
He solved it the "real way". That would be turned into a "complex way".
Imaginary way is always cooler
incredible !!!!!!!!!
Very good
Can you please solve this integral Cos^(-1) { (x+1)/√(x^2 +2x +5) } .dx
Technically aren't there infinte complex solutions for x? Now complex numbers are not ordered but even the magnitude of the solutions maybe taken to infinity
The solutions are infact all real, and there are infinitely many. The general form is x=-π²(n+0.5)² ∀n∈Z. In the video he has taken the special case of n=0.
Give this man a giant black(white)-board..
You are the best, I love you
very cool bro
I understand that if the infinitely many solutions are clustered in about the same place, then you could have a biggest solution. But how can you be _sure_ of that? How can you be _sure_ there is even a biggest solution? Also, how can you be sure you found the biggest solution, and not just a solution in the middle?
We need Cos of something to be zero, then that something (which is √-x here) must be an odd multiple of π/2, like -π/2, 3π/2, -3π/2, 5π/2, 7π/2, etc. Squaring them all, you have the possible values for -x being π^2/4, 9π^2/4, 25π^2/4, etc. Multiply both sides by -1, you get the possible values of x: -π^2/4, -9π^2/4, -25π^2/4, etc. It's easy to see that the greatest of these is -π^2/4.
Green05 Exactly.
Good .i like it
The first was more good
Yeah
the bunny mic is just great.
Tienes excelentes videos, pero dónde están los subtítulos :(
Alguien que me recomiende una página para aprender este idioma, por favor.
Me encantan estos videos
How sqrt(x^6) = x^3 ??? 1:50
6=2*3, sqrt(x^6) = sqrt(x^(2*3)) = sqrt((x^3)^2) = x^3
Nice video
无
The outro song is hype.
Nice
Hilarious with that rabbit, you remind me Mr Harrison
Is that a force of habit why are you holding that wabbit hahaha
Non real method is more Elegant
First is easy 👍
you cannot imply the "biggest" property to complex numbers
Omg the bunny is a mic, isn't it?
For wat its worth besides ur videos I love ur chenglish as well uwu
chenglish?
@@ssdd9911 I think he meant Chinese + English
x can't be positive . so we can use x = - ( t^2)
What is your real name ?
Professor Chow
Nice.....
I just remembered the hyperbolic function..............
Pozitif everybody happy😂😂
I don't get why you chose π/2 for the real solution,
ふぇああ
基本の冪級数展開の形は暗記なんですね😂ﾒﾄﾞｲ
No-real is my preference.
Ah yes, the bunny is back :)
I always think that the guys who found those relations were in a quarentene or something like that. How could someone think about cosh(x) ???? What the f**k...
The rabbit knows more math than me
You changed the thermonuclear detonator for a bunny!!! Mmmmm. Anyway, excellent video, thanks