Hello sir, my name is Atiqur Rahman. I am a trader. I want to build up an AI Algorithm trading software system on trading, I have been trying to solve a math calculation on trading for a long time. I request your cooperation in this regard
My first exposure to this was of the form: i^i and the question asked to give the results....a real number. I was fascinated with that ever since. An imaginary number raised to an imaginary number gives a real number result.
@@herbie_the_hillbillie_goat Or basically complex numbers operations are rotations and scaling in the complex plane so it's not really incredible that some times they fall onto the real axis
Imaginary to imaginary power even *always* yields a real result. Compare to irrational number to irrational power, which can yield an integer, but when picked at random remains irrational.
@@quantumsoul3495 Exactly! It is pretty fascinating early on when you're first encountering complex numbers. These days when I see something "inexplicable" happing on the real numbers, I look to the complex plane to see where the peculiarity is coming from. Rotations on the plane is also why it's not surprising to see pi everywhere either.
The challenge question is a trick question! The principle root of (-1)^pi is a transcendental number in its simplest form! Sure you could give decimal approximations like -0.9026853619 - 0.430301217i, but I think we can all agree that just like pi is usually best left as pi, and e is usually best left as e, (-1)^pi is best left as is. Fun fact, if you graph all possible roots of this number, it makes a circle in the complex plane with radius 1.
Thank you for posting a lot of videos and engaging with us in the comments, especially knowing the fact that you have over 1 million subscribers is amazing. I've been having a rough time pursuing my college degree because I am so intimidated of taking Calculus 1 but the more I watch your videos it made me feel that it's not that bad after all. If only you could make more videos about getting into calculus as a complete beginner that would be nice! But anyways, You're blessing in this world and thank you for keeping up the grind of all of us!
(-1)^pi = cos(pi^2) + i sin(pi^2) is the principal solution, but I think the general form would be (2n+1)pi^2 inside the cos and sin, n in integers, right?
I'm from Brazil and I'm really enjoying your content and with the videos having the option of subtitles in Portuguese it's even better. Congratulations on the quality of the classes and I hope you continue with the great work.
1:57 Ah, I'm glad we are all still adults :) Most of the things I have seen on your channel, I think I learned in high school or first year university. Most of those things I forgot until you reminded me in your videos. But some things are new to me, such as the Lambert W function. Either way, I like being educated or reminded :)
The math was cool! But I couldn't help but be amazed at the magic he pulled off with his markers! I only ever noticed two markers in his hand, yet he was able to write with 3 colours!
Hello ! I hope you see my comment I saw this nice question so that I recommend it The question is : solve the system of equations a = exp (a) . cos (b) b = exp (a) . sin (b) It can be nicely solved by using Lambert W function after letting z = a + ib Hope you the best ... your loyal fan from Syria
I've been watching your videos for a while and usually there's a lot I don't understand, but I started pre-calculus this year and I'm slowly starting to understand more and more of what you're saying lol
(-1)^π e^iπ=−1 and e^2πki=1 for integer k =(e^iπe^2πki)^π =e^iπ2^(1+2k) =cos(π^2(1+2k))+i sin(π^2(1+2k)) (i.e. infinite complex values for k) Let k=0: (principal) = cos π^2 + i sin π^2 ≈ -0.903... - (0.430...)i
I haven't been studying these things for over 10 years and it's still interesting to see and think about, try to solve even though you have no idea how or even what to do
the multiple answers connects my brain to those spiral shapes I saw in a video about graphing x^x for all values of k, when plotting the imaginary axis off into a third dimension perpendicular to x and y.
I failed my high school pre-calc courses almost 15 years ago, so it's nice to see that these concepts can make a minor amount of sense to me all of this time later.
Many of these videos feel like me playing with my graphing calculator in high school. "Hey, what's the i th root of -1?" "Let's try it!" "What does this even mean?" "I don't know!"
(-1)^π We know e^(iz)=cos(z) + isin(z) So e^(iπ)=-1 Then we elevate it to π (e^(iπ))^π=(-1)^π So (-1)^π=e^(iπ²) But we know that with complex numbers e^(iz)=e^(i(z+2π)) Then (-1)^π=e^(i(π²+k2π)), k=0,±1,±2... In polar form: radius=1 Angle=π²+2kπ rad, k=0,±1,±2... In binomial form cos(π²)+isin(π²)
Please tell me why we can't do this to the exponent: 1/i = 1/SQRT(-1) = SQRT(1)/SQRT(-1) = SQRT(1/-1) = SQRT(-1) = i. So exponent is just i since 1/i = i
2:46 You use the rule to convert from radical form to exponential form with a negative base. Forget about the i-part for a moment. This conversion has to be dealt in a very delicate manner when the base is negative as there are a plethora of examples where this move is unjustified. While I am not disputing your work, this "quick" move doesn't sit well. What justification is there to make this move when a complex power is involved along with a negative base?
Thank you both!! ;) ;) Anne Frank is one of my autistic special interests along with mathematics which is why I have a pic Miss Frank as my profile photo and why I enjoy blackpenredpen's videos so much! :) :) :) :)
I'm Japanese junior high school student. This movie is very interesting. It make me like math than before. I want my friends to watch this movie…Oh, I forget that I don't have friends who understands complex number.
Isn't there technically only 1 solution to the problem, not infinite? I get that (-1)^(1/i) would have infinite solutions due to the argument being of the form pi + 2n*pi, with n being an integer. However, since the initial statement used the radical symbol, it means the principal value only, right?
The initial statement (-1)^(1/i) didn't use a branch cut making it the principle root. The "known" statement is there to show the only information expected of us to know for this problem but isn't the initial statement. We can make a branch cut to make the solution finite but without a branch cut there are infinite solutions due to periodicity of the complex exponential.
I don't think that i = \sqrt{-1} is really Kosher though! Complex numbers as far as I know do not really have "principal roots" because i is neither positive nor negative, much like the complex number 3 - 2i. I think it would be better to say "there are two solutions to z^2 = -1 and they are +i and -i."
-1^pi could sweep along the complex unit circle, with the principle value being e^(pi^2)i. It can't be the entire complex unit circle as you'd end up with a countably infinite set against the continuum that is the complex unit circle.
Really nice showcase how everything connects in maths and has impact on each other. One thing about roots in this field of math tho, using trygonomic notation one always get infinite number of results since used agle is periodic. The more interesting thing is that in complex numbers you have actually as many results as the number before root symbol. For example if you have sqrt of 1 it can be 1 or -1, sqrt of -1 is i or -i but 3rd root of -i will give you 3 results (i and 2 others results, (-sqrt(3) + i)/2 and (sqrt(3) - i)/2). I talk about principal values ofc because mentioning infinite amount of results every single time is unnecessary
I find it so interesting that even though i and -1 both have unit length, taking the ith root can so drastically affect the magnitude (the primary root being ~23.14... and getting closer and closer to infinity as you go up, while going the opposite direction starts at 0.04321... and gets closer and closer to 0). When taking roots using real numbers, taking the 1st root keeps the same magnitude, going more positive increases it towards infinity, going more negative decreases it towards zero, and going into fractional roots shifts it into the complex plane but still the magnitude goes down. When taking imaginary (or complex) roots all that intuition goes out the window... I wonder if a similar operation to the root (or exponentiation in general) can be formulated where taking the ith "root" has the same invariant properties as taking the 1st root of a number. Almost like an abstract inverse or complex conjugate to exponentiation.
I thought i was useless. But it comes up in quantum physics. The momentum of a particle comes from an operator on the wave function. There is an i in the formula. So they are useful.
Here's another approach, and how i solved it before watching the vid (i wanted something to brag about lol) if the i-th root of -1 = x, x^i = -1. by definition, e^ln(x) = x, so e^ln(x^i) = -1. more properties of logarithms says that ln(a^b) = b ln(a), so e^(i ln x) = -1. then i realised: if e^(i ln x) = -1, then ln(x) must be pi becuase of euler's identity, e^(i*pi) = -1. so if ln(x) = pi, x = e^pi by definition (the principal value anyway). Edit: just realise you just divide the exponent of euler's formula by i to get the answer instantly xD
x^a = b (a,b,x complex numbers) in general has multiple solutions for x, sometimes infinitely many (e.g. when a=i like in this case). How is "the" answer for b^(1/a) selected?
Hi! I was wondering if you could give me some information about the whiteboard you're using. I'm looking for one rn and it would really help me out. Thanks
Check out quaternions. There are *4* axes: real plus three kinds of imaginary numbers: i, j, and k. I believe quaternions are used in computer graphics to perform arbitrary 3D rotations without gimbal locking.
Interesting! ..Alt soln - Take Euler's Identity e^ipi + 1 = 0... subtract 1 from each side and take the ith root of both sides and get e^pi = ith root of (-1)... but of course that only gives you the primary answer, not the 2pi*n...
If a root is a function, then it cannot have infinite results. What you gave is the result to the equation z^i = - 1 Rather than z=i th root of - 1 But still, for the rest, interesting!
@@Syndicalism yes, a root is a function. What you are thinking about is the inverse *expression* of a power, rather than a function. A function is a one to one relation or a many to one relation, which a root is.
Ok I kinda did something I dont really understand, So what i did was take the sequence of natural numbers as such 1, 2, 3, 4, 5, 6, 7, 8....... and then formed another sequence by taking the first member of the sequence that is 1 and adding the numbers in the sequence to it meaning that first I would add 1 and then 2 and then 3 and so on, which makes 1, 1+1, 1+1+2, 1+1+2+3....... or 1, 2, 4, 7, 11, 16, 22, 29, 37..... then I would repeat the same step to form the next sequence 1, 1+1, 1+1+2, 1+1+2+4, 1+1+2+4+7 .... or 1, 2, 4, 8, 15.... and suppose I repeat this infinitely, the sequence it seems to approach is 2^0, 2^1, 2^2 and so on. essential 2^n why is it so that it approaches this sequence? please explain
Because 2^{n+1}=(2^n + 2^{n-1} + ... 2^1 + 2^0) + 1, it's easy to prove that if the first terms of the sequence are 1, 2, 4, ..., 2^n, then at the next step the sequence will start with 1, 2, 4, ..., 2^n, 2^{n+1} Since the initial sequence starts with 1, 2, it follows that after n steps it will start with 1, 2, ... 2^{n+1}
Interestingly,, you can also get the general solution using De-Moivre's root formula: -1 = cis(pi) -> -1^(1/i) = cis((pi + 2pi*k)/i) = e^i*((pi+2pi*k)/i) = e^(pi + 2pi*k) Though, I gotta wonder about the range - the formula limits us to a integer k in [0,n), yet here we have infinitely many solutions. I guess since i is imaginary, no integer can be defined as larger or smaller than it. I wonder how to justify that it works, lol.
Learn more complex numbers from Brilliant: 👉 brilliant.org/blackpenredpen/ (20% off with this link!)
Hello sir, my name is Atiqur Rahman. I am a trader. I want to build up an AI Algorithm trading software system on trading, I have been trying to solve a math calculation on trading for a long time. I request your cooperation in this regard
If theta is pi, then the complex number is just negative r.
Formula for
√x - √y = √xy
What's your opinion on this
ruclips.net/video/TdYi3FHV7B4/видео.html
About his theories? 🤔
Is he right?!🤯
What's your opinion on this guy ruclips.net/video/LKYa5e0GVSg/видео.html and his theories is he saying the truth?🤔 Ty VM sifu.
My first exposure to this was of the form: i^i and the question asked to give the results....a real number. I was fascinated with that ever since. An imaginary number raised to an imaginary number gives a real number result.
That blew my mind too. Then I remembered the product of two negative numbers is positive and i^i kind of lost its charm.
@@herbie_the_hillbillie_goat Or basically complex numbers operations are rotations and scaling in the complex plane so it's not really incredible that some times they fall onto the real axis
Imaginary to imaginary power even *always* yields a real result.
Compare to irrational number to irrational power, which can yield an integer, but when picked at random remains irrational.
1.559605... Let's call it x. Then x^x would be 2, even tho x is trancendental. This thing DOES happen.
@@quantumsoul3495 Exactly! It is pretty fascinating early on when you're first encountering complex numbers. These days when I see something "inexplicable" happing on the real numbers, I look to the complex plane to see where the peculiarity is coming from. Rotations on the plane is also why it's not surprising to see pi everywhere either.
"This rotation which is 180 degrees, but we're adults now, so let's use Pi"
I didn't expect to laugh in a math video
Like pie
360° = 2pi?
@@chillx656 ye
The challenge question is a trick question! The principle root of (-1)^pi is a transcendental number in its simplest form!
Sure you could give decimal approximations like -0.9026853619 - 0.430301217i, but I think we can all agree that just like pi is usually best left as pi, and e is usually best left as e, (-1)^pi is best left as is.
Fun fact, if you graph all possible roots of this number, it makes a circle in the complex plane with radius 1.
a circle only if pi is normal
How would you graph the roots of a number?
@@dinosaric4862 He means all solutions, like how square root of any number has 2 solutions
And (-1)^e ??? 😀😉
A transcendental number powered by another transcendental number + 1 equals absolutely nothing. How crazy is that?
I love how you started with the basics of the coordinate systems. So many people jump to the equalities and the basic principles are lost. A+
Thank you for posting a lot of videos and engaging with us in the comments, especially knowing the fact that you have over 1 million subscribers is amazing. I've been having a rough time pursuing my college degree because I am so intimidated of taking Calculus 1 but the more I watch your videos it made me feel that it's not that bad after all. If only you could make more videos about getting into calculus as a complete beginner that would be nice! But anyways, You're blessing in this world and thank you for keeping up the grind of all of us!
Thank you very much!
@@blackpenredpen 4:21 shouldn`t it be 2npi/i?
@@getnie6867 parenthesis
(-1)^pi = cos(pi^2) + i sin(pi^2) is the principal solution, but I think the general form would be (2n+1)pi^2 inside the cos and sin, n in integers, right?
Yes!
(-1)^π = -e^(π+2k)
@@hervesergegbeto3352 There is no negative.
I'm from Brazil and I'm really enjoying your content and with the videos having the option of subtitles in Portuguese it's even better. Congratulations on the quality of the classes and I hope you continue with the great work.
Challenge 1:)prove leibnitz theorem of geometry by using complex numbers
Is there any libnitz theorem in geometry ?
If so then state it
@@Mathologix now I've a challenge for u
Find the statement..😅lmaooo
@@j.u.4.n620 lmfao
It's very straightforward and therefore left as an exercise for the reader.
Did you mean “A Geometrical Theorem of Leibniz”? Because that’s the only one I found on Google.
I have no idea what this guy is talking about but its addictive.
Coolest thing I have seen all day. I wondered about this but never sat down to try to figure it out myself. Thanks so much. Love your videos!
Great articulation, great explanations for conceptual techniques, and an overall easy to follow presentation!
1:57 Ah, I'm glad we are all still adults :)
Most of the things I have seen on your channel, I think I learned in high school or first year university. Most of those things I forgot until you reminded me in your videos. But some things are new to me, such as the Lambert W function. Either way, I like being educated or reminded :)
real chads use tau/2 lol
I am in 7th class and 12 years only and you taught me calculus. Thnx
OMG knowing Euler's identity and looking at the answer it all made sense! i'th root just means dividing the power by i, so e^pi(i) became e^pi
A really fun fact is that, as a french fan, when I'm enjoying math class I write my equations with your voice intonation in my head 😂
Chapters:
0:00 what’s the i-th root of -1
5:30 check out Brilliant
6:08 challenging question (-1)^pi
Wow! Somebody liked my comment already!
Your teaching not just improve my knowledge
They really motivate me
First of all love your videos, they are very informative and makes math enjoyable. Math is the biggest "what if" ever 😁
The math was cool! But I couldn't help but be amazed at the magic he pulled off with his markers! I only ever noticed two markers in his hand, yet he was able to write with 3 colours!
The tiny mic is an added bonus :)
Hello ! I hope you see my comment
I saw this nice question so that I recommend it
The question is : solve the system of equations
a = exp (a) . cos (b)
b = exp (a) . sin (b)
It can be nicely solved by using Lambert W function after letting z = a + ib
Hope you the best ... your loyal fan from Syria
Your are an incredible teacher. Even I got it! thanks!
I've been watching your videos for a while and usually there's a lot I don't understand, but I started pre-calculus this year and I'm slowly starting to understand more and more of what you're saying lol
(-1)^π
e^iπ=−1 and e^2πki=1 for integer k
=(e^iπe^2πki)^π
=e^iπ2^(1+2k)
=cos(π^2(1+2k))+i sin(π^2(1+2k))
(i.e. infinite complex values for k)
Let k=0: (principal)
= cos π^2 + i sin π^2
≈ -0.903... - (0.430...)i
Dude my vision blurred I got dizzy and started drooling 2 minutes into this and by the end I could smell something burning.....
I haven't been studying these things for over 10 years and it's still interesting to see and think about, try to solve even though you have no idea how or even what to do
i'm in the same boat XD I wish school was more like this
Love your channel bro 😊
Following this channel when it have only 20k subscribers.... amazing journey.thanku
the multiple answers connects my brain to those spiral shapes I saw in a video about graphing x^x for all values of k, when plotting the imaginary axis off into a third dimension perpendicular to x and y.
I failed my high school pre-calc courses almost 15 years ago, so it's nice to see that these concepts can make a minor amount of sense to me all of this time later.
(-1)^pi = e^(pi ln -1)
= e^(pi * i pi)
= cos (pi^2) + i sin (pi^2)
This is the primary solution.
Congrats on hitting 1 million subs!!
love your videos 🙂
Thank you!
Many of these videos feel like me playing with my graphing calculator in high school.
"Hey, what's the i th root of -1?"
"Let's try it!"
"What does this even mean?"
"I don't know!"
I have never seen polar written like that! Nor rectangular be called standard form. I just know polar as r
Can you do more double integrals in a playlist with changing the variables and polar coordinates 🌚and thx💛
I understood everything without knowing English
amazing!
I was blown away by the sudden introduction of the blue pen.
Brooo ur such a magician , how do you figure out these kinda questions
Pl read my other comment @blackpenredpen
I like that shirt with Euclid's formula with the Pi emphasized in red.
Your marker technique is just incredible lol
(-1)^π
We know e^(iz)=cos(z) + isin(z)
So
e^(iπ)=-1
Then we elevate it to π
(e^(iπ))^π=(-1)^π
So
(-1)^π=e^(iπ²)
But we know that with complex numbers e^(iz)=e^(i(z+2π))
Then
(-1)^π=e^(i(π²+k2π)), k=0,±1,±2...
In polar form: radius=1 Angle=π²+2kπ rad, k=0,±1,±2...
In binomial form
cos(π²)+isin(π²)
Please tell me why we can't do this to the exponent: 1/i = 1/SQRT(-1) = SQRT(1)/SQRT(-1) = SQRT(1/-1) = SQRT(-1) = i. So exponent is just i since 1/i = i
"It is a real no , but it is also a complex number " i dont know why , but it almost killed me
I swear mathematicians are just making things up out of boredom at this point
2:46 You use the rule to convert from radical form to exponential form with a negative base. Forget about the i-part for a moment. This conversion has to be dealt in a very delicate manner when the base is negative as there are a plethora of examples where this move is unjustified. While I am not disputing your work, this "quick" move doesn't sit well. What justification is there to make this move when a complex power is involved along with a negative base?
Steve just casually proving e^(i*pi)+1=0 in the first two minutes of the video in a way that a junior high schooler could understand
Awesome as a possum with a blossom stuff!!! I love this!!! Thanks lots for producing and posting!!! :) :) :)
Love your comment. 😁
the joyful way you commented this is what I imagine Anne Frank (your profile picture) was like before going into hiding!
Thank you both!! ;) ;) Anne Frank is one of my autistic special interests along with mathematics which is why I have a pic Miss Frank as my profile photo and why I enjoy blackpenredpen's videos so much! :) :) :) :)
I'm Japanese junior high school student. This movie is very interesting. It make me like math than before.
I want my friends to watch this movie…Oh, I forget that I don't have friends who understands complex number.
Me before: (·-·)???
Me after: (·o·)!💡
Me when I see the question at the end: (·A·)??!??
Beautiful... I didn't even "imagine" to do this... infinitely brilliant! Thank you!
I will try (-1)^Pi later!
My fav channel after learning basic calculu😊💪🏽💪🏽🔥
"Wait, is this all Eulers Identity?"
"Always has been 🔫"
I like your Film,very good
Isn't there technically only 1 solution to the problem, not infinite? I get that (-1)^(1/i) would have infinite solutions due to the argument being of the form pi + 2n*pi, with n being an integer.
However, since the initial statement used the radical symbol, it means the principal value only, right?
The initial statement (-1)^(1/i) didn't use a branch cut making it the principle root. The "known" statement is there to show the only information expected of us to know for this problem but isn't the initial statement. We can make a branch cut to make the solution finite but without a branch cut there are infinite solutions due to periodicity of the complex exponential.
I don't think that i = \sqrt{-1} is really Kosher though! Complex numbers as far as I know do not really have "principal roots" because i is neither positive nor negative, much like the complex number 3 - 2i. I think it would be better to say "there are two solutions to z^2 = -1 and they are +i and -i."
I will discover USA but the radical symbol is only intended to be used with a natural power(it is defined so). Otherwise, you should use ^(1/x)
I’m more used to see the : pi + 2npi as pi(2n+1) but it’s the same ( just personal preferences)
Anyway, thank you for these explanation 😆
(-1)^π answer is e^(iπ²)
Since -1 is equal to e^iπ
From Euler's identity...
As a joke, I will say that trying to calculate the i-th root of -1 results in: YOU DESTROYED THE FABRIC OF SPACE-TIME
You are my best teacher 🐱
(-1)^(1/i) = (e^[i⋅π])^(1/i) = e^(π).
Actually fairly easy, as long as you remember the complex unit circle.
I love how he says “n” as a variable.
“2嗯派i”
I didn't know we could do this, and it's now giving me noghtmares
-1^pi could sweep along the complex unit circle, with the principle value being e^(pi^2)i. It can't be the entire complex unit circle as you'd end up with a countably infinite set against the continuum that is the complex unit circle.
Really nice showcase how everything connects in maths and has impact on each other. One thing about roots in this field of math tho, using trygonomic notation one always get infinite number of results since used agle is periodic. The more interesting thing is that in complex numbers you have actually as many results as the number before root symbol. For example if you have sqrt of 1 it can be 1 or -1, sqrt of -1 is i or -i but 3rd root of -i will give you 3 results (i and 2 others results, (-sqrt(3) + i)/2 and (sqrt(3) - i)/2). I talk about principal values ofc because mentioning infinite amount of results every single time is unnecessary
i loved this
Cool! That's why mathematics is so pretty beautiful.
you surely have a lot of black and red pens in stock!!! :)
very informative, i love it
Awesome!!
I find it so interesting that even though i and -1 both have unit length, taking the ith root can so drastically affect the magnitude (the primary root being ~23.14... and getting closer and closer to infinity as you go up, while going the opposite direction starts at 0.04321... and gets closer and closer to 0). When taking roots using real numbers, taking the 1st root keeps the same magnitude, going more positive increases it towards infinity, going more negative decreases it towards zero, and going into fractional roots shifts it into the complex plane but still the magnitude goes down. When taking imaginary (or complex) roots all that intuition goes out the window... I wonder if a similar operation to the root (or exponentiation in general) can be formulated where taking the ith "root" has the same invariant properties as taking the 1st root of a number. Almost like an abstract inverse or complex conjugate to exponentiation.
this was very neat, thank you
I thought i was useless. But it comes up in quantum physics. The momentum of a particle comes from an operator on the wave function. There is an i in the formula. So they are useful.
Nice to see. Very useful in daily life lol. But e^pi is really hard to approximate.
out of the gut I'd say (-1)^{pi} = exp(i*pi^{2})
I am in grade 11 and I solved it using euler's identity ( e to the power i π +1 = 0 )
I'm grade 5 and I know that e^i*pi=-1
@@findystonerush9339 how? What's your age
"we're all adults now, let's use pi". I'm dying xDDD
6:08 -0.902-0.4303i
Here's another approach, and how i solved it before watching the vid (i wanted something to brag about lol)
if the i-th root of -1 = x, x^i = -1.
by definition, e^ln(x) = x, so e^ln(x^i) = -1.
more properties of logarithms says that ln(a^b) = b ln(a), so e^(i ln x) = -1.
then i realised: if e^(i ln x) = -1, then ln(x) must be pi becuase of euler's identity, e^(i*pi) = -1.
so if ln(x) = pi, x = e^pi by definition (the principal value anyway).
Edit: just realise you just divide the exponent of euler's formula by i to get the answer instantly xD
ah yes, the most important, generally applicable information.
Fantastic! Thank you!
x^a = b (a,b,x complex numbers) in general has multiple solutions for x, sometimes infinitely many (e.g. when a=i like in this case). How is "the" answer for b^(1/a) selected?
Hi! I was wondering if you could give me some information about the whiteboard you're using. I'm looking for one rn and it would really help me out.
Thanks
You know what would be really cool? A third plane of numbers, basically just making the number line/plane into 3D space.
Check out quaternions. There are *4* axes: real plus three kinds of imaginary numbers: i, j, and k. I believe quaternions are used in computer graphics to perform arbitrary 3D rotations without gimbal locking.
@@eryqeryq Neat! I’ll check that out. I guess I just haven’t have too much exposure to the technical side of math in digital form so yeah, but thanks!
Interesting! ..Alt soln - Take Euler's Identity e^ipi + 1 = 0... subtract 1 from each side and take the ith root of both sides and get e^pi = ith root of (-1)... but of course that only gives you the primary answer, not the 2pi*n...
Taking the ith root does give the 2pi*n solution. If you only want a finite solution you must make a branch cut.
If a root is a function, then it cannot have infinite results. What you gave is the result to the equation
z^i = - 1
Rather than
z=i th root of - 1
But still, for the rest, interesting!
Then the root must yield "i" solutions to the problem. But how can you get "i" solutions? Tell me..
@@Bhuvan_MS the rule is applied for integer powers of z. Not complex.
A root isn't a function. However, a root with a specified branch cut can be made into a function.
@@Syndicalism yes, a root is a function. What you are thinking about is the inverse *expression* of a power, rather than a function. A function is a one to one relation or a many to one relation, which a root is.
AAh!! after thinking about it, this is how you would represent a series or a sequence of real numbers.
Can you make a video on the integral of sin(ln(x))dx ???
Please solve n^i. n is any natural number
WOW! What a great explanation.
Ok I kinda did something I dont really understand,
So what i did was take the sequence of natural numbers as such
1, 2, 3, 4, 5, 6, 7, 8.......
and then formed another sequence by taking the first member of the sequence that is 1 and adding the numbers in the sequence to it meaning that first I would add 1 and then 2 and then 3 and so on, which makes
1, 1+1, 1+1+2, 1+1+2+3.......
or
1, 2, 4, 7, 11, 16, 22, 29, 37.....
then I would repeat the same step to form the next sequence
1, 1+1, 1+1+2, 1+1+2+4, 1+1+2+4+7 ....
or
1, 2, 4, 8, 15....
and suppose I repeat this infinitely,
the sequence it seems to approach is 2^0, 2^1, 2^2 and so on. essential 2^n
why is it so that it approaches this sequence?
please explain
Because 2^{n+1}=(2^n + 2^{n-1} + ... 2^1 + 2^0) + 1, it's easy to prove that if the first terms of the sequence are 1, 2, 4, ..., 2^n, then at the next step the sequence will start with 1, 2, 4, ..., 2^n, 2^{n+1}
Since the initial sequence starts with 1, 2, it follows that after n steps it will start with 1, 2, ... 2^{n+1}
A square lawn is surrounded by a path 4m wide around it.if the area of the path is 320 square meter. Find the area of the lawn
That was neat! :D
Interestingly,, you can also get the general solution using De-Moivre's root formula:
-1 = cis(pi) -> -1^(1/i) = cis((pi + 2pi*k)/i) = e^i*((pi+2pi*k)/i) = e^(pi + 2pi*k)
Though, I gotta wonder about the range - the formula limits us to a integer k in [0,n), yet here we have infinitely many solutions. I guess since i is imaginary, no integer can be defined as larger or smaller than it. I wonder how to justify that it works, lol.
answer for your question is 'e' raised to the power of 'i' times pi squared
(-1)ᵖᶦ=cos(π² )+isin(π² )
so e raised to i pi sqr
?
at halfway i was about to comment that there were some solutions missing but it was foolish of me to assume you would not mention them 😂
Does it mean e^pi = e^3pi?
If we ln both side, does that mean pi = 3pi?
Does that mean....1 = 3?
BlackpenRedPenBluePen . . . YEAH!
aren't you supposed to also multiply the 2npi by 1/i as well?
e^pi is known as Gelfond's Constant and is provably transcendental.
1:57 BASED
(-1)^π=(e^(kπi))^π=e^(kπ^2*i)=cis(kπ^2) for an odd integer k