Whenever I'm feeling burnt out from university I always come here, your enthusiasm is contagious!! Thank you so much for all your videos!!

@drpeyam

2 жыл бұрын

Thanks so much!!!

@milan.matejka

2 жыл бұрын

9:54 - 10:15 made my day.

It's somewhat fitting that the real part of the solution involves π/4 + 2πn, since it's these are the real maxima points of the function

@snowflake5204

2 жыл бұрын

@abigmonkeyforme Makes for a good story to me!

Haven't had the chance to work this out for myself, but I'm curious to get feedback about using this approach: The LHS of the equation is merely a linear combination of a sine and a cosine with the same angular frequency. As such, it can be represented as one sinusoid with an appropriate phase shift as follows: sin(x)+cos(x) = sqrt(2)*sin(x+pi/4) Then solve sqrt(2)*sin(x+pi/4) = 2 using the complex exponential method that was so eloquently derived in this video.

In my head, I'm thinking: cos(x) + sin(x) =2 cos²(x) + 2cos(x)sin(x) + sin²(x) = 4 2cos(x)sin(x) = 3 sin(2x) = 3 2x = arcsin(3) x = 0.5arcsin(3) I hope this is consistent with the video. I also don't know what complex thing arcsin(3) is equal to, but I'm leaving it like this because I can claim it's exact form.

@drpeyam

2 жыл бұрын

Cool!!

@meroepiankhy183

2 жыл бұрын

sin(2x) = 3 ??? In the real world, sin is bounded by 1

@gamerdio2503

2 жыл бұрын

@@meroepiankhy183 Thats why the solution is complex

@allozovsky

2 жыл бұрын

@@meroepiankhy183 Objection! "Imaginary Numbers Are Real" - Welch Labs

@meroepiankhy183

2 жыл бұрын

@@allozovsky Sorry, but for the equation sin(2x) = 3 2x = arcsin(3) x = 0.5arcsin(3) You cannot solve this problem like that. This equation has Complex solutions only

14:46 oh so that's what complex exponentials sound like. Good to know!

@drpeyam

2 жыл бұрын

😂😂😂😂

We can go a little further. (√2 + 1)(√2 - 1) = (√2)^2 - 1^2 = 2 - 1 = 1 so (√2 - 1) = 1 / (√2 + 1) so ln(√2 - 1) = - ln(√2 + 1) so ln(√2 +/- 1) = +/- ln(√2 + 1) So the solution is: x = (π/4 + 2πM) +/- i ln(√2 + 1)

@nschloe

2 жыл бұрын

Indeed, this is more instructive.

@alexfekken7599

2 жыл бұрын

And more importanttly this reflects the symmetry of the original equation around x = π/4. It is clear from the basic properties of sin and cos that the solution must have this symmetry (in addition to the periodicity) as well.

It can also be done by first squaring both sides, which leads to solving sin2z = 3, which is a straightforward complex trig equation. As a fun alternative to exponentials, one may also solve it using a combination of trig and hyp functions, given sin2z = sin(2x+ 2iy) and the addition formula.

@Nameless-qe9hu

Жыл бұрын

I did that when I solved before watching. It led me to get two extra solutions that when plugged into Geogebra result in -2.

Yes, I am using this at the moment in electronics, where it could be cos(x) + sin(x) = Ψ or cos(x) - sin(x) = Ψ, based on if I am working with the sum or difference of the signals. Where Ψ is the output waveform from two input sine waves at any point in time, using 0° and 90° offset that makes up the real and imaginary parts.

For √i, multiply 2/2 inside square root, you get √(2i)/√2 = √(1+i)^2/√2 = (1+i)/√2.

cos(x)+sin(x)=2 1+sin(2x)=4 sin(2x)=3 u=2x sin(u)=3 (e^(iu)-e^(-iu))/2i=3 (e^(iu))^2-6i(e^(iu))-1=0 e^(iu)=i(3(+ or -)2sqrt(2)) iu=i×pi/2+ln(3(+ or -)2sqrt(2)) u=pi/2-i×ln(3(+ or -)2sqrt(2)) x=pi/4-i/2×ln(3(+ or -)2sqrt(2))

@siddharthamondal4346

2 жыл бұрын

I did the same way. Yayy

Same Spiel at 1:38. 😂 Many greetings from Austria.

What about rewriting cosx+sinx as sqrt(2)cos(pi/4-x)?

I never thought that equation in such a clever way, I study math by my own and this motivates me a lot, thanks and greetings from Perú!!!

@brendanlawlor2214

2 жыл бұрын

amo a todos en Peru.,...ustedes son mis favoritos ....hola desde Australia 🇦🇺 🤪👋 jaja jiji jeje

@Sir_Isaac_Newton_

2 жыл бұрын

@@brendanlawlor2214 no le sabes al español 🥵🤌🥶🤌

@Helpadoggoreachsubs

2 жыл бұрын

@@Sir_Isaac_Newton_ i don't know Spanish too but I got google translate 🤬😎

@liemnguyenchi6170

2 жыл бұрын

Perfect

the lazy rooster at the end won me over ;-)

Love the clarity.

حلقة ممتازة دكتور پايام وبارك الله فيك

I've always wondered how one would define cos and sin for a generic vector. But it's actually pretty obvious, you just need to define the trig of vector a for the vector b. I feel like it would just end up at something we already know and just call by a different name though xD.

In (-1+i)/(1+i) you could also factor out i from the numerator to get (i(1+i))/(1+i) which simplifies to just i

I have no idea how I ended up here, but your captivating energy kept me here for the entire video, and I loved every second of it

@drpeyam

2 жыл бұрын

Thanks so much!!

I loved this video, I'm excited to see more videos! Lots of thanks for making people to enjoy maths :)

Complex solutions can make life easier, even it is not clear at first glance. 🙂

More and more crazy. And this is really what I love in maths. Thank you for your enthusiasm

To solve this problem, I would first divide by sqrt(2), to give sin(x)cos(pi/4)+cos(x)sin(pi/4)=sqrt(2) since sin(pi/4)=cos(pi/4)=1/sqrt(2). Hence sin(z)=sqrt(2) where z=x+pi/4. Thus cos²(z)=1-sin²(z)=-1, and hence, exp(iz)=cos(z)+i sin(z)= i(sqrt(2)+1) or i(sqrt(2)-1). It follows that x=z-pi/4=-i[ log(sqrt(2)+1) + i pi/2 + 2 N pi] - pi/4 =-i log(sqrt(2)+1) + (2 N + 1/4) pi or x=z-pi/4=-i[ log(sqrt(2)-1) + i pi/2 + 2N pi] - pi/4 = -i log(sqrt(2)-1) + (2 N + 1/4) pi for any integer N. Now, (sqrt(2)-1)(sqrt(2)+1)=2-1=1, and hence x = (2N +1/4) pi (+/-) i log(sqrt(2)+1), for any integer N. These are complex conjugate solutions.

this just showed up and i already love ur channel, congrats

I wish my university lecturers had a half of your enthusiasm and energy 🤩

"How could it have a solution?" "Yes"

You can also notice by simple operations that ln(1+sqrt(2))i=ln(1+sqrt(1^2+1))i=arsh(1)i=arcsin(i). So the answer is pi/4+2pim-arcsin(i). Pretty nice, huh. Maybe you have a bit different sign for the arc functions, but I'm russian.

I have tried to graph the function and the result is really surprising.. sin((π/4)-(ln(sqrt(2)+-1)i)+(2πD))+cos((π/4)- (ln(sqrt(2)+-1)i)+(2πD))

Your energy is so contagious!! I love it

could you not solve this with t results or auxiliary method ?

Haha 11:47, LN the-generous. BRILLIANT!

How do you solve X raised to the X power equals a particular number?

once again i think theres likely an easier way to d this using hyperbolic functions and the identities that relate them to the regular trig functions. not exactly sure what the working out would look like off the top of my head

I think this suggests that a treatment of what the algebra means in terms of geometry would be of great interest.

Dear Sir: At time 7:10 when you completed the square..Could you add a few more steps I do not follow what you did.. Thanks Richard

In your first video in this year you talked about properties of the number 2021. Will you do so with the number 2022 beginning the next year?

@drpeyam

2 жыл бұрын

I couldn’t find any, lol, but there will be another fun 2022 video

Dr. Peyam, this was a fun derivation. It is one of those problems that would not normally be considered even in math studies. The idea of using a complex number solution for the equation is novel and your development was interesting. Your later comments about digital signal processing was also a nice addition. This problem would be an interesting question in a PhD oral exam or a MS oral exam to assess how the candidate would solve the problem by casting the solution as possible only with a complex value for “x”. Very nice video. Thank you.

@drpeyam

2 жыл бұрын

Thank you!!

Perhaps others have written this already, but you can also use the substitution t = tan(x/2). Then cos(x) = (1 - t²) / (1 + t²) and sin(x) = 2·t / (1 + t²). The equation for t is quadratic: 3·t² - 2·t + 1 = 0, whose solutions are t = (1 +- sqrt(2))/3. Then x = 2*arctan(t). Three steps from start to solution.

@willowwedemeyer5152

2 жыл бұрын

Oops, left out the i: t = (1 +- i*sqrt(2))/3

Very nice & clear development

Sir, this was really great, literally blew my mind!! it was like lambasingi !!!

Hey I signed up of real analysis in the spring and was wondering what it was about. Is it anything like abstract algebra? Thanks guys!!

@drpeyam

2 жыл бұрын

Completely different, but sooooo much better!!!

Can it reach any complex number?

Very helpful indeed. Hope we had this 15 years ago

@14:39 - Where do I learn about how complex exponentials are used in sound processing? I must know. The brain can process sound, so does that mean the brain is processing complex numbers?

Great job Dr. Peyam!👍

In physics we use exponentials (exp(ix) for instance ) because they are easier to manipulate but whenever we want the final answer or something meaningful , we must take the real part of the expressions This idea is used throughout any wavy thing

@brandonklein1

2 жыл бұрын

Perhaps in classical electrodynamics this is the case, as the complex parts of waves always cancel in the classical wave equation, but this is simply false in quantum mechanics, for example.

Can you help me with a physics equation I've been seeking for years and don't have the math language to create? Starts off simply as follows. A circle around a line where the circle has a point rotating that dictates the tangent of the angle of the line through the center. The rotation of the circle is caused by a 1 dimensional object (just +) interacting with a 2 dimensional (dipole of +&-) object as such that this causes this energetic system to rotate. Now the line would be just a 2 dimensional object (+&-) by itself. When you combine the two, the 3-brane system (1brane+2brane) acts on the 2-brane system which adjusts the vector angle at the center point of the circle. (To what degree can be played with). The line is obviously in some sort of Sin wave and the circle by itself in a void is more like a corkscrew but ends up stablelizing when it has the line through the center. The line I call Es, the 2 I call Et, and the 1 I call Ep. The entire system together is Esys. So that Esys=Ep+Es+Et What I am looking for is the mathematical modal of the path of Es, and the overall vector modal of Esys. It's like a little dimensional piston basicly. Its been driving me crazy for years.

@tomctutor

2 жыл бұрын

Bit lost with your explanation, if this is not what your talking about then DrPeyam can delete it. in Physics (oh no Dr Peyam recoils) we teach the helical path taken by a charged particle about a magnetic field line: The circular component, is solved by central Lorentz pseudoforce f=qvBsin(θ) now here vBsin(θ) is simply the vector product of v⋆B, remember v is tangent to circular path of charged particle and is its instantaneous velocity. θ is the "angle" of the helix measured from the B-vector direction. B is the strength of the magnetic field line parallel to the helical path. The central force is mv^2/r = mrω^2 where ω= 2π/T, the T being the period of rotation, r is the radius of the particle's circular path. Solving both forces the same get r= mvsin(θ)/qB. The pitch along the direction (B) is simply d= vcos(θ)T. I have linked to image to here (from Physexams.com).. www.google.com/url?sa=i&url=https%3A%2F%2Fphysexams.com%2Fblog%2FMotion-of-a-charged-particle-in-a-uniform-magnetic-field_13&psig=AOvVaw20b4Qn8wysgp4GZ3imfDg0&ust=1640289563246000&source=images&cd=vfe&ved=0CAsQjRxqFwoTCIitu_WY-PQCFQAAAAAdAAAAABAD

Thanks sir. You always motivate me

Here is also a nice way to avoid having to deal with logarithms and exponentials. cos(z) + sin(z) = 2 where z = x+iy, x and y element of real numbers. cos(x+iy) + sin(x+iy) = cos(x)cosh(y) - isin(x)sinh(y) + sin(x)cosh(y) + icos(x)sinh(y) = 2 by trig sum identity. cos(iy) = cosh(y) by setting x = iy for cos(x) = 1/2(exp(ix) + exp(-ix)) and same for sin(iy). Factoring, we get cosh(y)(cos(x) + sin(x)) + isinh(y)(cos(x) - sin(x)) = 2 As the right hand side has nothing in terms of i then we know sinh(y)(cos(x) - sin(x)) = 0 (eq1) and cosh(y)(cos(x) + sin(x)) = 2. (eq2) If we set y = 0 then eq1 would be true but eq2 cannot as there is no x that can satisfy that equation as cosh(0) = 1. Therefore we need to find a value of x so eq1 holds which would be x = nπ/4 where n is an integer. Plugging in x = nπ/4 into eq2 we get sqrt(2)cosh(y) = 2 cosh(y) = 2/sqrt(2) y = arccosh(2/sqrt(2)). Therefore, z = nπ/4 + iarccosh(2/sqrt(2)) where n is an integer is the solution.

@drpeyam

2 жыл бұрын

We have different approaches

@RexxSchneider

2 жыл бұрын

@@drpeyam No, you don't, because the hyperbolic trig functions are just shorthand for the even and odd functions of the exponents of real variables and you used the even and odd functions of the exponents of complex variables. the solutions only look different because the inverse hyperbolic functions have log forms: arcosh(x) = ln(x + √(x² - 1)) and arsinh(x) = ln(x + √(x² + 1))

"You could use the quadratic formula everywhere, but it's more stylish not to." 🤣🤣🤣

What an enjoyable problem to see get worked out.

Cos(x) +sin(x) =sin(x+pi/4)sqrt(2) =2 So Sin(x+pi/4)=sqrt(2) This has solutions in complex numbers

But why Peyam...because I say so...hilarious, that made me laugh out loud, thank you 🤣

5:54 if im wrong someone correct me but i believe Minus i square equals plus one !! He said it equals minus one

i really liked the "but yyyyy peyam...bcs i say so "😬😂😂

Well done Dr!

awesome.....the real challenge is to imagine the problem in the first place......the solution is easy once the complex form is used. Another beauty by Doc the Guru 👋 🤪

When I tried to solve it i didnt use periods because with periods exp and ln are no longer inverses

Square both sides to get 1+sin2x = 4, so sin2x = 3 and work from there. Much simpler.

Why can't we square LHS and RHS? The procedure can be much simplified in terms of no. of steps...

The field behinds every clickbait title: complex numbers

What I did was sinx+cosx=2 square both sides (sinx+cosx)^2=2^2 (sinx)^2+(cosx)^2+2sinxcosx=4 using trigonometric identities, 1+sin(2x)=4 sin(2x)=3 x=(arcsin3)/2 I find using the identities simpler to change forms x=-i*ln(3i+2i(sqrt2))/2 x= -i*(ln(i)+ln(3+2sqrt2))/2 x= -i* (i*pi/2+ln(3+2sqrt2))/2 x= (pi/2-i*ln(3+2sqrt2))/2 x= pi/4- i*ln(3+2sqrt2)/2

What's that sound @11:42

Incorrect solution as x is meaning less. Can you provide x as some range in degrees? (0 - 360)? Logs can't be negative and hence it's very close to 45 degree.

@drpeyam

2 жыл бұрын

The solution is correct

U missed ln(root2+-1)i-pi/4+2pim

7:21 I think you meant y^2-2(1+i)y+i=0 -> (y-(1+i))^2-(1+i)^2/2+i , NOT (y-(1+i))^2-(1+i)^2+i . Correct me if I'm wrong please

Yo diria senx^2 + cosx^2 = 1 jajaja, no me imaginaba resolver esto, pero escribiendo al seno y coseno en su forma exponencial salen muchas posiblidades

just saw this on yt home, first reaction: he is a younger version of electroBOOM bruh btw cool video

Great job, and good "hook" too - "How can (cos(x)+sin(x)==2?????")

So cool Really Really good stuff 🙌

kuch samjha nahi lekin sunke acha laga :)

Dr. Peyam tks for your classes. I would like to share with you an thought about numbers. Root(2) has a representation in a geometric plane but is transcendental mathematicaly speaking. This implies, to me, that the nature of all numbers is exponencial (there is no endding) and our discretization of reality is just special cases, an kind of eigenvalues that not changes after a certain scale. PS: root(2) is irracional, I know, but the thought is the same for transcendental.

Very Very Interesting!

"multiples of myself" 2 PI M i. That's a good line

Brilliant!!!

What is "M" in 2piM?

This is a nice problem. But the solution is very lengthy and complicated. We may try using half angle formula for solving the equation in few steps as follows: Substitute t = tan(x/2); so that we will end up with the simple Quadratic Eqn 3t^2 - 2t + 1 = 0 and the two solutions will be x = 2arctan ((1 ± isqr2)/3)

Helo Dr Peyam,im 67 years old,from Buenos Aires;Argentina,it is rather difficult for me not only because of my poor english but my brain is going hard rock,but i try ;but i try; i can get no satisfaction,as old rockers Rollings,says. Really,is very nice the maths you teach us. I would like to learn some tutorial of you telling us some of problems resolved for Dr Ramanujuan. Thanks for all ;Dr Peyam ;and what work gave you the DOCTOR state.

Is it possible for cos(x)+sin(x)=i? If possible could you show how to do it? Thanks! I love watching your videos.

@drpeyam

2 жыл бұрын

Interesting! Yeah it’s possible, same approach as this video

this result is always true?

y/y-=1?

The Complex world is actually complex hahaha nice video Dr Peyam, I liked it a lot!

wow that great Dr peyam

We're solving sqrt(2) sin(z + pi/4) = 2 or sin(z + pi/4) = sqrt(2). Using arcsin w = - i log(iw +- sqrt(1 - w^2)), this gives z = - pi/4 - i log (i sqrt(2) +- i) = - pi/4 - i [log(i) + log (sqrt(2) +-1)] = - pi/4 + pi/2 + log(sqrt(2) +- 1) = pi/4 + ln(sqrt(2) +- 1) + 2pi m.

@drpeyam

2 жыл бұрын

But that arcsin formula is cheating

@michaelz2270

2 жыл бұрын

@@drpeyam Not really, you can derive it by solving (e^{iz} - e^{-iz}) / 2i = w for z using the quadratic formula pretty easily.

@drpeyam

2 жыл бұрын

@@michaelz2270 So basically the same thing as the video

@michaelz2270

2 жыл бұрын

@@drpeyam It's more direct and doesn't involve taking square roots of i, completing the square, etc, even if you want to derive the arcsin z formula.

Very Wholesome Video...

why didn't u just square both sides and then used sin double angle

double angle formula Rsin(alpha+b)=2

Would have been easier to use the fundamental equality of trigonometry that sq of cosine plus sq sin of cosine is zero to transform everything into a simple quadratic equation where x is either sin or cosine….

@drpeyam

2 жыл бұрын

Zero?

14:39 This is why i love maths

You could use complex plane to calculate root i but I guess we could do it this way lol

But if cos(x) + sin(x) = (2/sqrt(2)) * (cos(x)sin(π/4) + sin(x) cos(π/4)) = 2/√2 * sin(x + π/4), doesn't it means that cos(x) + sin(x)

Amazing

Cos(x) +sin(x) =2 1/2 (cos(x)+sin(x))=1 rad2 /2 (sin(x) + cos(x)= rad 2 Sin(x +pi/4) =rad 2 X +pi/4 = indefinit

@drpeyam

2 жыл бұрын

Can do better!

please help me🙏🙏 I have a request to you. How to find maxima and maxima of the type: y=k/f(x) Where f(x)=ax²+bx+c or,|x-a|+|x-b|

@drpeyam

2 жыл бұрын

I have a similar video, interesting minimization problem

very clear

It is really great mathematician

cos x + sin x = √2 cos (x-π/4) = 2, cos (x-π/4) = √2 cos x = cosh ix x = ±i(arccosh √2 + π/4)

@drpeyam

2 жыл бұрын

Keep simplifying

But the signals turned out to be only showing us the real part when being played out...

In Vietnam,I was study cos,sin,tan,co tan ad thk for teach me

You teach so nice