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.
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)
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.
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.
@@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
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.
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.
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))
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 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))
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.
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.
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.
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)
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'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.
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.
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.
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
@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?
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.
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
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.
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.
What we can do is taking note that sin(x)+cos(x)= sqrt(2)*sin(x+pi/4) If that = 2 Then sin(x+pi/4)= 2/sqrt(2) = sqrt(2) now we just have one formula we have to work with
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
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
To avoid log of a complex number cos(x-pi/4) = cos(x)cos(pi/4)+sin(x)sin(pi/4)) = (sqrt(2)/2)(sin(x)+cos(x)) = sqrt(2) Set y=x-pi/4, u=exp(iy) and you get the quadratic u^2 - 2 sqrt(2) u + 1 = 0 u has positive real solutions u = sqrt(2)+/-1 y = ln(sqrt(2)+/-1)/i x = pi/4 - i ln(sqrt(2)+/-1) x ~ 0.7854 -/+ 0.8814 i + 2 pi n
@@drpeyamHi Dr Peyam. In my notation y=x-pi/4 and cos(y) = sqrt(2) cos(y) = (exp(iy)+exp(-iy))/2 Put exp(iy)=u so sqrt(2) = (u+1/u)/2 Multiply through by 2u and rearrange u^2 - 2 sqrt(2)u + 1 = 0 u = (2 sqrt(2)+/-sqrt(8-4))/2 u = sqrt(2) +/- 1
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 👋 🤪
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….
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
"there is nothing special about this 2, you could replace it with any positive value" Im guessing its undefined for negative values? Or infinite solutions? Im curious
Whenever I'm feeling burnt out from university I always come here, your enthusiasm is contagious!! Thank you so much for all your videos!!
Thanks so much!!!
9:54 - 10:15 made my day.
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.
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)
Indeed, this is more instructive.
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.
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.
Cool!!
sin(2x) = 3 ??? In the real world, sin is bounded by 1
@@meroepiankhy183 Thats why the solution is complex
@@meroepiankhy183 Objection! "Imaginary Numbers Are Real" - Welch Labs
@@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
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.
I did that when I solved before watching. It led me to get two extra solutions that when plugged into Geogebra result in -2.
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
@abigmonkeyforme Makes for a good story to me!
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.
What about rewriting cosx+sinx as sqrt(2)cos(pi/4-x)?
14:46 oh so that's what complex exponentials sound like. Good to know!
😂😂😂😂
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))
I did the same way. Yayy
Thanks!
Thanks so much for the super like, you're always the best :)
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!!!
amo a todos en Peru.,...ustedes son mis favoritos ....hola desde Australia 🇦🇺 🤪👋 jaja jiji jeje
@@brendanlawlor2214 no le sabes al español 🥵🤌🥶🤌
@@Sir_Isaac_Newton_ i don't know Spanish too but I got google translate 🤬😎
Perfect
Complex solutions can make life easier, even it is not clear at first glance. 🙂
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.
We have different approaches
@@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))
Love the clarity.
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.
حلقة ممتازة دكتور پايام وبارك الله فيك
5:54 if im wrong someone correct me but i believe
Minus i square equals plus one !!
He said it equals minus one
Your energy is so contagious!! I love it
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
Thanks so much!!
Haha 11:47, LN the-generous. BRILLIANT!
Same Spiel at 1:38. 😂 Many greetings from Austria.
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.
Thank you!!
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
could you not solve this with t results or auxiliary method ?
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))
Very nice & clear development
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.
Oops, left out the i: t = (1 +- i*sqrt(2))/3
the lazy rooster at the end won me over ;-)
How do you solve X raised to the X power equals a particular number?
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
Can do better!
Great job Dr. Peyam!👍
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)
Sir, this was really great, literally blew my mind!! it was like lambasingi !!!
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.
this just showed up and i already love ur channel, congrats
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
"How could it have a solution?"
"Yes"
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.
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.
The solution is correct
More and more crazy. And this is really what I love in maths. Thank you for your enthusiasm
What is "M" in 2piM?
Very helpful indeed. Hope we had this 15 years ago
Can it reach any complex number?
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.
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
U missed ln(root2+-1)i-pi/4+2pim
Cos(x) +sin(x) =sin(x+pi/4)sqrt(2) =2
So
Sin(x+pi/4)=sqrt(2)
This has solutions in complex numbers
What's that sound @11:42
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
Why can't we square LHS and RHS?
The procedure can be much simplified in terms of no. of steps...
y/y-=1?
@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?
I wish my university lecturers had a half of your enthusiasm and energy 🤩
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?
I couldn’t find any, lol, but there will be another fun 2022 video
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.
But that arcsin formula is cheating
@@drpeyam Not really, you can derive it by solving (e^{iz} - e^{-iz}) / 2i = w for z using the quadratic formula pretty easily.
@@michaelz2270 So basically the same thing as the video
@@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.
4:09 haha he said it
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!!
Completely different, but sooooo much better!!!
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.
Great job, and good "hook" too - "How can (cos(x)+sin(x)==2?????")
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
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.
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)
Square both sides to get 1+sin2x = 4, so sin2x = 3 and work from there. Much simpler.
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)
Keep simplifying
"multiples of myself" 2 PI M i. That's a good line
Is it possible for cos(x)+sin(x)=i? If possible could you show how to do it? Thanks! I love watching your videos.
Interesting! Yeah it’s possible, same approach as this video
Thanks sir. You always motivate me
The field behinds every clickbait title: complex numbers
At angle 45°
Sin45°+ cos45°=2
1/√2 +1√2=2
2/√2×√2×√2
2√2/2
No?
😂 wth
When I tried to solve it i didnt use periods because with periods exp and ln are no longer inverses
What an enjoyable problem to see get worked out.
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.
this result is always true?
What we can do is taking note that
sin(x)+cos(x)= sqrt(2)*sin(x+pi/4)
If that = 2
Then sin(x+pi/4)= 2/sqrt(2) = sqrt(2)
now we just have one formula we have to work with
double angle formula Rsin(alpha+b)=2
14:39 This is why i love maths
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
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
kuch samjha nahi lekin sunke acha laga :)
To avoid log of a complex number
cos(x-pi/4) = cos(x)cos(pi/4)+sin(x)sin(pi/4))
= (sqrt(2)/2)(sin(x)+cos(x))
= sqrt(2)
Set y=x-pi/4, u=exp(iy) and you get the quadratic
u^2 - 2 sqrt(2) u + 1 = 0
u has positive real solutions
u = sqrt(2)+/-1
y = ln(sqrt(2)+/-1)/i
x = pi/4 - i ln(sqrt(2)+/-1)
x ~ 0.7854 -/+ 0.8814 i + 2 pi n
I don’t get how you get your quadratic equation for u
@@drpeyamHi Dr Peyam. In my notation y=x-pi/4 and
cos(y) = sqrt(2)
cos(y) = (exp(iy)+exp(-iy))/2
Put exp(iy)=u so
sqrt(2) = (u+1/u)/2
Multiply through by 2u and rearrange
u^2 - 2 sqrt(2)u + 1 = 0
u = (2 sqrt(2)+/-sqrt(8-4))/2
u = sqrt(2) +/- 1
I think this suggests that a treatment of what the algebra means in terms of geometry would be of great interest.
sin(x)+cos(x)=i
Does x have a solution? Or its indeterminate?
It has a solution and you can use the same method to solve this
Very Wholesome Video...
see this simple solution! cos(x) +sin(x)=2 implies cos(x). 1/√2+sin(x).1/√2=√2 implies cos(x-π/4)= √2 implies x-π/4 =cos''(√2) implies x=cos''(√2) +π/4
But cos inverse root 2 does not make any sense
It becomes just another question
Well try though 👍
Very Very Interesting!
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 👋 🤪
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….
Zero?
cos(z) + sin(z) = 2 then😉
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
Arccosh?
Well done Dr!
why didn't u just square both sides and then used sin double angle
It is really great mathematician
"there is nothing special about this 2, you could replace it with any positive value"
Im guessing its undefined for negative values? Or infinite solutions? Im curious
I think it’s also infinite for negative values
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|
I have a similar video, interesting minimization problem
I would have written
cos(x)+sin(x)=sqrt(2)cos(x-pi/4)
So solve cos(x-pi/4)=sqrt(2) and apply the complex definition of the inverse cosine.
Ah asian spotted
@@lawliet2263 nah I m french
@@Ben-wv7ht haha wi wi I like french people
But the signals turned out to be only showing us the real part when being played out...
Then ((⁶√2)i)⁶+i⁶=((⁶√3)i)⁶
......Last theorem....
love how the exponents are a little satanic lol
@@XBGamerX20 🤣
Amazing
"You could use the quadratic formula everywhere, but it's more stylish not to." 🤣🤣🤣