we can solve this using complexification technique. we should consider the function g(x)=e^(2x) cos(3x) +i e^(2x) sin(3x)=e^(2x+3ix) now we can calculate the n th derivative of g. it is easy. but still i like linear algebraic method. many students will benifit by understanding the solution of this problem. linear algebra is a powerful subject used in science and engineering. they will learn so many things if understand this solution.
I’ll say this, it’s useful, but like markedly unnecessary to do this much to arrive at an answer that is otherwise pretty straightforward. But still cool 😎
For those looking for how it turns into a real number, note that sin(x)=(e^(ix)-e^(-ix))/(2i) This means that if you write 2+3i in polar form, you can turn beta into 13^1515×sin(3030×arctan(3/2)) since the magnitude is the square root of 13, and the angle is arctan(3/2)
Interesting method. 22:14 Personally, I think *β = 13¹⁵¹⁵ sin(3030 · arctan(3/2))* is even better. The original task has sine, and this form has the advantage that it uses only real numbers. You can get my expression from (-i/2)(2+3i)³⁰³⁰ + (i/2)(2-3i)³⁰³⁰ using polar form or de Moivre.
The disadvantage of this way it's not easy to see that the answer is in fact an integer. You could argue that it being a real number is closer to it being an integer, but to show that the answer is in fact the imaginary part of (2+3i)³⁰³⁰ seems a lot easier than to show that sin(3030 · arctan(3/2)) is rational (and in fact a non-positive power of 13, since the first factor is 13¹⁵¹⁵).
@@barutjeh It's pretty straightforward to show that cos(atan(x)) = 1/sqrt(1 + x^2) and sin(atan(x)) = x/sqrt(1 + x^2) (just think of a right triangle with shorter sides having lengths of 1 and x). From this, we have sin(atan(3/2)) = 3 sqrt(13)/13 and cos(atan(3/2)) = 2 sqrt(13)/13. Using the angle sum formulae for sine and cosine, we get sin(2 atan(3/2)) = 12/13 and cos(2 atan(3/2)) = - 5/13. Since the angle sum formulae are polynomials in sin(x) and cos(x), if you have rational values for sin(x) and cos(x), you will also have rational values for sin(k x) and cos(k x), for any positive integer k. Further, it's easy to establish by induction that sin(2 k atan(3/2)) = m/13^k, for some integer m.
Thank you for β. :) My solution was Im[ (2+3i)^3030 ] and I was hoping for Michael P. to get some formulas/simplification for (2+3i)³⁰³⁰ from Jordan decomposition...
At 15:00, I think the third layer of the commutative diagram should be C² -> C², and there should be some discussion why PinvDP maps R² to R² despite taking a detour into C².
The aproximate answer is -1.5 × 10^1687 And that is the calculated exact answer: -15018255505063763257087672670709830706621447608990211783269486428013011453559304704180313050291715215424340044168893744476500362245682528403047785586557182697833303424319495343261638479908442941457950507062295565330841752403362771528128679101816061530173846075950314137424690962115852949501512302859651778278255235726892360166045065905930096733842057581718458064428985516626392077710287235350664155415391405587168171614324047854339887473862335585608322420104069884939883199181004805433050290925448965257738757988699891524693275967481854222285195323271231879995590409838311205965156178286673620939661452988891623608690677839262067219938754701843412720840233261915661138964590911610820585782720719584900311258982941391373884926843974520216534592154744662052176911839876334494052258307963557527351037665864714160533041099932114971026737491595926803316049215661121083103021154948754029788416261318828197875321936936871779964935291799071394815848503041933918322238783514127555366715175685601546635563264091592043846459254910298783135358287725838400612526839046008782349039266146804411247048515286420816714176869731412760896663688772529612955695633724241060095697903486439036905058906251017157927740734180504264514938889152001938614369429576967228579148023156450659595744505904648991760099816458992262378134883392248415663225704452218662928910498786161269399801517100791471862979570551973287054167920129507370948135710472897395347181181191698692712930025905527486041017261069074793847913444927870735475169153383155390389057721273891657640145726526485493745070698712980097514216809556189562905070539795930536026820218060858834753826417565912950898819384331586023746036852071822726519739024539868
Interesting, also because in quantum mechanics switching from calculus to linear algebra avoids solving difficult differential equations and integrals. It's not just a math trick its a powerful instrument.
Other expression: The value f^(3030)(0) can be written as a term of a generalized Lucas sequence. Note that f(x)=e^(2x)sin(3x) is a solution of an ODE: f''(x)-4f'(x)+13f(x) = 0. -- (☆) For particular values, f(0) = 0 and f'(0) = 3 since f'(x) = 2e^(2x)sin(3x)+3e^(2x)cos(3x). Next, differentiate (☆) n times and substitute x=0, then f^(n+2)(0)-4f^(n+1)(0)+13f^(n)(0) = 0. Define a sequence a_{n} = f^(n)(0), and then we get "a_{n+2} = 4a_{n+1}-13a_{n}, a_{0}=0, a_{1}=3". By definition of Lucas sequence of first kind, a_{n} = 3U_{n}(4, 13). ※ U_{n}(4, 13) = [(2+3i)^{n}-(2-3i)^{n}]/(6i) Therefore, f^(3030)(0) = 3U_{3030}(4, 13).
how fun! your channels have given me so much insight into the beauty of math that i totally missed in school. I always saw math just as a convenient tool for physics, but I am starting to get a glimpse of the deep and fascinating structure of mathematics. Thank you!
my first instinct on seeing this was to use cauchys integral formula for derivatives and then the residue theorem, which is a pretty efficient solution imo. love this linear algebra approach too, using the derivative as a linear mapping is great edit: a note connecting these two solutions- we find D to be (2 -3; 3 2) which is the matrix representation of the number 2+3i. thus taking the derivative is the same as multiplying by 2+3i. in the complex solution we have something looking like exp((2+3i)z) and indeed as we know of the exponential function taking the derivative each time will have the effect of multiplying an additional 2+3i. isn’t math so cool?
defining (2+3i)^n=a_n+i*b_n with two integer sequences a_n and b_n, we can infer that a_(n+2)=4a_(n+1)-13a_n and the same relation for b_n. the initial values are a_0=1, a_1=2, b_0=0, b_1=3. note that trying to find a solution using the characteristic polynomial will only result in the corresponding formulas for the real and imaginary part of (2+3i)^n. instead we can try to work out the elements of the sequence using generating functions. after some manipulations you can find A(x)=(1-2x)/(1-4x+13x²) and B(x)=3x/(1-4x+13x²) for a_n and b_n respectively. you can work out a formula for the two sequences using partial fractions and the geometric series. unlike the method using the polar form for (2+3i)^n, we get a algebraic solution. the value we want to find is b_3030 for the solution of the problem. the actual correct value turns out to be -15018255505063763257087672670709830706621447608990211783269486428013011453559304704180313050291715215424340044168893744476500362245682528403047785586557182697833303424319495343261638479908442941457950507062295565330841752403362771528128679101816061530173846075950314137424690962115852949501512302859651778278255235726892360166045065905930096733842057581718458064428985516626392077710287235350664155415391405587168171614324047854339887473862335585608322420104069884939883199181004805433050290925448965257738757988699891524693275967481854222285195323271231879995590409838311205965156178286673620939661452988891623608690677839262067219938754701843412720840233261915661138964590911610820585782720719584900311258982941391373884926843974520216534592154744662052176911839876334494052258307963557527351037665864714160533041099932114971026737491595926803316049215661121083103021154948754029788416261318828197875321936936871779964935291799071394815848503041933918322238783514127555366715175685601546635563264091592043846459254910298783135358287725838400612526839046008782349039266146804411247048515286420816714176869731412760896663688772529612955695633724241060095697903486439036905058906251017157927740734180504264514938889152001938614369429576967228579148023156450659595744505904648991760099816458992262378134883392248415663225704452218662928910498786161269399801517100791471862979570551973287054167920129507370948135710472897395347181181191698692712930025905527486041017261069074793847913444927870735475169153383155390389057721273891657640145726526485493745070698712980097514216809556189562905070539795930536026820218060858834753826417565912950898819384331586023746036852071822726519739024539868 which is approximately -1.5*10^1687
having matrix [2 -3 3 2] we already have a representation of complex number. Say, 2+3i. Thus, power of matrix actually consists from real and imag parts of (2+3*i)^3030. Latter can be evaulated using Muavre formula: norm = sqrt(2^2+3^2)=sqrt(13) arg = atan(3/2) (2+3*i)^3030 = norm^3030*exp(i*3030*arg) Thus power of matrix to be norm^3030* [cos(arg*3030) -sin(arg*3030) sin(arg*3030) cos(arg*3030)]
We can also get rid of linear algebra here and jump right into complex analysis using the following decomposition of sin(x): sin(x) = (e^(ix) - e^(-ix)) / 2i, thus e^(2x) * sin(3x) = (e^((2 + 3i)x) - e^((2 - 3i)x)) / 2i Differentiating the rhs 3030 times we get: ((2 + 3i)^3030 * e^((2 + 3i)x) - (2 - 3i)^3030 * e^((2 - 3i)x)) / 2i The only issue with this approach is that you need to combine the result to get back to real-valued functions
From the moment I saw the 3030th derivative in combination with Linear Algebra I just KNEW this is going to be about Diagonalization. It is absolutely a marvelous method. I have introductory Linear Algebra students and Diagonalization is towards the end. Unfortunately this application is just too far fetched for my students, but I am going to try this out on a similar problem: An epower with a cosine term...
I'd prefer to use the 3030th derivative of 1/2i(e^(2+3i)z-e^(2-3i)z) because mathmajor is further along in Complex Analysis than Linear Algebra right now.
Maybe explore -> take derivative once to see the two set of functions before jumping to linear algebra would help tremendously why we are doing it this way.
I found this really helpful to understand the usefulness of linear algebra. Thank you Michael Penn. Can you recommend any linear algebra or abstract algebra book which contains concrete problems such as this? This was very illuminating and helped me understand the subject.
((2 3) (-3 2)) is also a complex number under the usual representation; namely 2+3i. We can "just" raise that to 3030 then map back. Then we map it under T, discard sin component. The eval at 0 is both a matrix, and an operation on complex numbers.
The reason i naturally comes up here is because sine (and cosine) return to themselves after 4 applications of the derivative, and i is a a number which you need to raise to the 4th power to get to 1, i.e. both these objects have a period of 4.
Techniques from linear algebra. greatly complicate the solution of this problem. It 's easier this way: f(x)= e^2x*sin3x =1/(2i)[exp(2+3i)*x -exp(2-3i)x]. g(x)=d(3030)f(x)/d(3030)x =1/(2i)[(2+3i)^3030 *exp(2+3i)*x -(2-3i)^3030*exp(2-3i)x]. g(0)=1/(2i)[(2+3i)^3030 -(2-3i)^3030*], which in the end was received on the board. But you can continue the transformation by going to the trigonometric form of a complex number and using de Moivre's formula , g(0)=(13)^1515 * sin(3030*φ), where φ= atan(3/2).
If you're used to doing linear algebra but not used to dealing with complex exponentials, I can see how Michael's method could actually seem more natural. But indeed yours is much more direct. P.S. You mean tan(φ) = 3/2, or φ = atan(3/2).
This only works exponentials (including complex ones for the trig) multiplayed by polynomials because if you look at tye jordan decomposition of the matrix on R as a differential equation it becomes a very specific type which has solutions in the form of cx^ne^xg where g is an eigenvalue
you could just have used the set of the complex function to calculate the 3030 derivative f(x)=Im(e^((2+3i)x)) f(3030)(x)=Im((2+3i)^3030*e^((2+3i)x)) f(3030)(0)=Im((2+3i)^3030)
@@bilzebor8457 Well you can write the result as 13^1515 × sin(3030*atan(3/2)) which might be a little easier to read. Also makes it more clear it's definitely real
the polar form can only provide a more elegant "closed-form" expression for the 2030th power, but from a computational point of view there is no difference since arctan[3/2]is not a "nice" angle
beta can be written as Im((2+3i)^3030) since 2-3i is complex conjugate of 2+3i and the same holds for the respective exponents. The result isn't too surprising, since f(x) can be written as Im(exp{(2+3i)*x}). The method of going through a vectorspace of functions is more powerful though, e.g. considering a product of exp(2x) with some polynomial of grade n. The vectorspace span({exp(2x), 1/1! * exp(2x)*x... ,1/n! * exp(2x)*x^n}) would then cover all possible derivatives and d/dx would be represented by a matrix with 2 on the diagonal and 1 on the first upper offdiagonal (i.e. Jordan normal form).
Hey Good Place To Stop! I was wondering if you could help with something. Something about this video is very perplexing. How did differentiating a real function 3030 times lead to a complex value? Like the work done is logical but it does not seem intuitive to have a complex value when differentiating a real function...
@@antoniohaddad8816 If you evaluate the final answer (if you take the 3030th power of the stuff in the parentheses), you'll see that the end result is a real number.
@@antoniohaddad8816 Also worth noting that the result is a complex number added to it's complex conjugate, which can be seen without raising the complex numbers involved to the power 3030. Such a number is guaranteed to be real. :)
I think it's better to consider the span of the 2 functions over C since we had to diagonolize the matrix D giving us complex number in the matrices P and P-1, this assumes that P-1 is representation of a lineair transformation going frome C^2. the method and the diagram are the same we just chage the field from R to C and those R2 to C2
Eh, it's personal preference. When a matrix over a field _F_ doesn't diagonalize over _F_ , only then do you need to enlarge to the algebraic closure of _F_ / an algebraic extension of _F_ (and there's nothing to it when making this enlargement, you literally just say "In _F_ bar:"). Since algebraic closures of fields other than *R* can be hard to describe, in general it feels slightly "inefficient" to go to the algebraic closure when a lot of the time you didn't actually need to
I think the key point, which wasn't mentioned is that all derivatives of the original function are a linear combination of two functions, this will motivate the definition of the vector space V.
I appreciate the effort to show the usefulness of linear algebra. However, when dealing with exponentials and trig functions, it is usually always best to transform everything into exponentials first, as they will always lead to a basis wrt which the matrix is already diagonal, so we don't need any computations of eigenvalues and -vectors. Here: after converting sin(3x) to (e^{3xi}-e^{-3xi})/2i, we get f(x)=1/2i*(e^{2x+3xi}-e^{2x-3xi}). Using the basis v_1=e^{2x+3xi} and v_2=e^{2x-3xi} we get v_1'=(2+3i)v_1, v_2'=(2-3i)v_2 (so, we already have a basis of eigenvectors) This means f^(3030)=1/2i*(v_1^(3030)-v_2^(3030))=1/2i*((2+3i)^(3030)v_1-(2-3i)^(3030)v_2) And with v_1(0)=1, v_2(0)=1, we have f^(3030)(0)=1/(2i)*((2+3i)^(3030)-(2-3i)^(3030)) (To simplify this further, we might want to use polar coordinates, then the expressions are just R^(3030)*e^(it*3030) and R^(3030)*e^(-it*3030), which means that we get R^(3030)*sin(t*3030) as our final result, where R,t are the polar coordinates of 3+2i) The underlying reason is the following: in the (infinite dimensional) vector space of all differentiable functions R->C, the eigenvectors of the differential are precisely the functions c*e^zt with constants c and z. Of course, in order to understand the usefulness of this method, it does help to see the version with trigs first.
You can can the same answer in three lines by writing down the Taylor series in the complex plane. Still interesting and this gives a more general result than just evaluating at zero
Coming from electrical engineering, I implicitly saw e^2xSin3x as Ae^{(2+3j)x} + Ae^{(2-3j)x}, whose nth derivative is thus A(2+3j)^n + A(2-3j)^ne^{(2-3j)x}. Arguably that's the same approach because complex exponentials are eigenfunctions of the derivative operator. I guess the quick answer is (13^1515)e^2xSin(3x+phase).
I got it down to 13¹⁵¹⁵sin(3030arctan(1.5)) which is, let's be honest, a pretty decent sized integer. Curiously, the sine part of that is not an integer. But it is rational, and so its denominator must be a power of 13. Would be interesting to try to get an answer in a more closed form. For example, sin(2arctan(1.5)) = 12/13, and sin(4arctan(1.5)) = -120/169.
2:44 I dont get it, isnt V just a subset of R? Like isnt the dimension of V just 1? And how can we be sure that the representation of an element in V is unique (resolved if dim = 2)? Multiple representations of an element surely should create problems when multiplying a matrix
You gotta consider that x is a variable, and can take on multiple different values. V is the space of all functions f(x) = a*e^(2x)*sin(3x) + b*e^(2x)*cos(3x), where a and b are free. Even though this is a real-valued function, each individual element can’t really be considered to be a real number itself.
The elements of V are not real numbers. They are R->R functions. In fact, vectors are almost always functions. Consider [5, 8, 6] in R^3. The first component is 5, the second component is 8, and the third component is 6. In other words, this vector is the {1,2,3} -> R function mapping 1 to 5, 2 to 8, and 3 to 6. The vectors in V are similar except instead of domain {1,2,3} the domain is all real numbers. V is a subspace of R^R
That's what he did. Except what you call C is what he had unnamed and is the matrix with the eigenvalues on the diagonal D = PCP⁻¹ P⁻¹(D)P = P⁻¹(PCP⁻¹)P = C
I am going through Linear Algebra 2 at the moment and we are building up the theory for diagonalization and trigonalization etc. and this gives a nice perspective on how powerful this theory can be in application. Thx a lot.
that's just because the final answer is a sum of a complex number and its conjugate (you can see this using basic properties of the conjugate operation), and you have (a+bi) + (a-bi) = 2a which is real.
It is a real number. One way to see this is to replace 'i' by '-i' and observe that the answer does not change. This means the imaginary part must be 0.
@@tdchayes alternatively (very similar but a tiny bit easier in my opinion), observe that the expression is a complex number plus its conjugate. this moreover tells us that the real number Beta equals is twice the real part of either of the two terms (since (a+bi) + (a-bi) = 2a).
@@schweinmachtbree1013 I can almost see the complex conjugate here in this example. But here is another solution that bprp posted that is less obvious. i ( ln ( x+i ) - ln (x - i) )
@@tdchayes that example would remain true if ln were replaced by any other real function which has a natural extension to *C* (exp, sin, tan, arcsin, tanh, Gamma, Zeta, etc.) - it is just using the fact that complex conjugation commutes with not only finite sums but also infinite sums (here, Taylor series).
@@schweinmachtbree1013 And are you trying to make this so complicated for the average viewer that they can't understand it? If you replace 'i'' by ''-i' the function is the same is a very understandable thing. What is your point? Is what you say helping the average viewer? I'm not sure why this is highlighted.
I'm not sure whether my idea works out or not😅, but if you write that matrix as √(2^2+3^2)*R, then R is an orthogonal matrix, and it's equivalent to a rotation of an angel phi. So if one raises it to the nth power, then it's a rotation of an angel phi*n. Then one raises the coefficient to the nth power, multiply them together, and you have the original matrix raised to the nth power. I don't know linear algebra that well, so maybe there's a mistake in my solution, so please let me know if it's right or wrong. Anyway, nice video as usual. Thank you!
Hmm I suppose so yeah. Writing the map from V to R^2 as φ, you'd "pull back" the dot product in R^2 to an inner product in V defined by v . w = φ(v) · φ(w)
@@schweinmachtbree1013 Can you make that specific? I mean, you now just transformed into the vectorspace R^2 and carried out the straight forward scalar product, didn't you?
@@digxx yes that's exactly right, it's as simple as that. this kind of thing is called transport of structure - it's where you have some kind of structure in B and you have an isomorphism/homomorphism from A to B, and you use the map to "transport the structure to A".
the unbounded love for useless calculations really obscures the beautiful connection between calculus (or rather, D-finite power series) and linear algebra :(
either works fine (although we do need the complex numbers later into the video), so the reason is the very superficial/arbitrary one that the question uses the notation _x_ for the variable and not _z_
Maybe we can use Demouvre’s Theorem to keep this going, just to get rid of the imaginary bits, just feels weird having i’s in the final result, when the answer is real.
but- doesn't the D matrix simply correspond to the number 2-3i ? then if you take the exponent form of 2-3i directly you can get the 3030th power easily, and just plug in the values back into a matrix form, then you just skipped a quarter of the video-
you would have to take more than one course in linear algebra to get a somewhat comprehensive account of the applications of linear algebra to mathematics alone! To be honest though, linear algebra is probably the easiest main branch of higher mathematics, so you can just pick linear algebra stuff up as you come across it in your studies.
I have a feeling it might be a meme on math contest problems often having the year number in the problem, so instead of 2020 the person who wrote the problem thought it would be funny to put 3030
In fact, drawing an arrow and writing P is equivalent to drawing the arrow in the opposite direction and writing P^-1. The fact that he chose to draw the arrow from top to bottom mainly comes from some habits inherited by the use of categories.
okay but like…what do you mean this is a good place to stop, you have a question that only uses real numbers and end up with a crazy complex expression?
So, I simplified the answer into a real number in my head as 13¹⁵¹⁵sin(3030arctan(³/₂)), which WolframAlpha agrees with. Mind bogglingly, it's a 1688-digit integer. I think you really dropped the ball by not mentioning that in the video.
Anyhow, a very good example of exercise oriented to the consolidation of a wide range of theoretical backgrounds instead of vain self-referential calcolus virtuosims.... Jesus, if my University teachers would have been like Mike....
@Angel Mendez-Rivera By changing i to -i you change a complex number z to its complex conjugate, and z =the complex conjugate of z iff z is purely real. Thanks.
I don't think you need to have seen commutative diagrams before to follow this video - they are used in a very basic way and Michael explains what it means for a diagram to commute.
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Thank you very much for giving a more formal shape to this method, in order to explain it rigorously. So elegant.
we can solve this using complexification technique.
we should consider the function g(x)=e^(2x) cos(3x) +i e^(2x) sin(3x)=e^(2x+3ix)
now we can calculate the n th derivative of g. it is easy.
but still i like linear algebraic method. many students will benifit by understanding the solution of this problem. linear algebra is a powerful subject used in science and engineering. they will learn so many things if understand this solution.
I’ll say this, it’s useful, but like markedly unnecessary to do this much to arrive at an answer that is otherwise pretty straightforward. But still cool 😎
Yeah, but the method in the video is more general and you can be helped by a calculator that can do matrix operations.
Very hacky solution.
For those looking for how it turns into a real number, note that sin(x)=(e^(ix)-e^(-ix))/(2i)
This means that if you write 2+3i in polar form, you can turn beta into 13^1515×sin(3030×arctan(3/2)) since the magnitude is the square root of 13, and the angle is arctan(3/2)
Nice. Even easier is to notice that the result is equal to its complex conjugate.
@@celkatThank you for this observation. Without it, I had felt sure something had gone wrong.
Interesting method. 22:14 Personally, I think *β = 13¹⁵¹⁵ sin(3030 · arctan(3/2))* is even better. The original task has sine, and this form has the advantage that it uses only real numbers. You can get my expression from (-i/2)(2+3i)³⁰³⁰ + (i/2)(2-3i)³⁰³⁰ using polar form or de Moivre.
The disadvantage of this way it's not easy to see that the answer is in fact an integer. You could argue that it being a real number is closer to it being an integer, but to show that the answer is in fact the imaginary part of (2+3i)³⁰³⁰ seems a lot easier than to show that sin(3030 · arctan(3/2)) is rational (and in fact a non-positive power of 13, since the first factor is 13¹⁵¹⁵).
@@barutjeh It's pretty straightforward to show that cos(atan(x)) = 1/sqrt(1 + x^2) and sin(atan(x)) = x/sqrt(1 + x^2) (just think of a right triangle with shorter sides having lengths of 1 and x). From this, we have sin(atan(3/2)) = 3 sqrt(13)/13 and cos(atan(3/2)) = 2 sqrt(13)/13. Using the angle sum formulae for sine and cosine, we get sin(2 atan(3/2)) = 12/13 and cos(2 atan(3/2)) = - 5/13. Since the angle sum formulae are polynomials in sin(x) and cos(x), if you have rational values for sin(x) and cos(x), you will also have rational values for sin(k x) and cos(k x), for any positive integer k. Further, it's easy to establish by induction that sin(2 k atan(3/2)) = m/13^k, for some integer m.
Thank you for β. :)
My solution was Im[ (2+3i)^3030 ] and I was hoping for Michael P. to get some formulas/simplification for (2+3i)³⁰³⁰ from Jordan decomposition...
At 15:00, I think the third layer of the commutative diagram should be C² -> C², and there should be some discussion why PinvDP maps R² to R² despite taking a detour into C².
The aproximate answer is -1.5 × 10^1687
And that is the calculated exact answer:
-15018255505063763257087672670709830706621447608990211783269486428013011453559304704180313050291715215424340044168893744476500362245682528403047785586557182697833303424319495343261638479908442941457950507062295565330841752403362771528128679101816061530173846075950314137424690962115852949501512302859651778278255235726892360166045065905930096733842057581718458064428985516626392077710287235350664155415391405587168171614324047854339887473862335585608322420104069884939883199181004805433050290925448965257738757988699891524693275967481854222285195323271231879995590409838311205965156178286673620939661452988891623608690677839262067219938754701843412720840233261915661138964590911610820585782720719584900311258982941391373884926843974520216534592154744662052176911839876334494052258307963557527351037665864714160533041099932114971026737491595926803316049215661121083103021154948754029788416261318828197875321936936871779964935291799071394815848503041933918322238783514127555366715175685601546635563264091592043846459254910298783135358287725838400612526839046008782349039266146804411247048515286420816714176869731412760896663688772529612955695633724241060095697903486439036905058906251017157927740734180504264514938889152001938614369429576967228579148023156450659595744505904648991760099816458992262378134883392248415663225704452218662928910498786161269399801517100791471862979570551973287054167920129507370948135710472897395347181181191698692712930025905527486041017261069074793847913444927870735475169153383155390389057721273891657640145726526485493745070698712980097514216809556189562905070539795930536026820218060858834753826417565912950898819384331586023746036852071822726519739024539868
This was absolutely beautiful, using concepts and techniques from one field to find a solution in another.
Interesting, also because in quantum mechanics switching from calculus to linear algebra avoids solving difficult differential equations and integrals. It's not just a math trick its a powerful instrument.
Other expression: The value f^(3030)(0) can be written as a term of a generalized Lucas sequence.
Note that f(x)=e^(2x)sin(3x) is a solution of an ODE: f''(x)-4f'(x)+13f(x) = 0. -- (☆)
For particular values, f(0) = 0 and f'(0) = 3 since f'(x) = 2e^(2x)sin(3x)+3e^(2x)cos(3x).
Next, differentiate (☆) n times and substitute x=0, then f^(n+2)(0)-4f^(n+1)(0)+13f^(n)(0) = 0.
Define a sequence a_{n} = f^(n)(0), and then we get "a_{n+2} = 4a_{n+1}-13a_{n}, a_{0}=0, a_{1}=3".
By definition of Lucas sequence of first kind, a_{n} = 3U_{n}(4, 13).
※ U_{n}(4, 13) = [(2+3i)^{n}-(2-3i)^{n}]/(6i)
Therefore, f^(3030)(0) = 3U_{3030}(4, 13).
6i must be inside grouping symbols, such as (6i).
@@robertveith6383 Oh, thanks. I added () mark at the denominator part.
how fun! your channels have given me so much insight into the beauty of math that i totally missed in school. I always saw math just as a convenient tool for physics, but I am starting to get a glimpse of the deep and fascinating structure of mathematics. Thank you!
my first instinct on seeing this was to use cauchys integral formula for derivatives and then the residue theorem, which is a pretty efficient solution imo. love this linear algebra approach too, using the derivative as a linear mapping is great
edit: a note connecting these two solutions- we find D to be (2 -3; 3 2) which is the matrix representation of the number 2+3i. thus taking the derivative is the same as multiplying by 2+3i. in the complex solution we have something looking like exp((2+3i)z) and indeed as we know of the exponential function taking the derivative each time will have the effect of multiplying an additional 2+3i. isn’t math so cool?
defining (2+3i)^n=a_n+i*b_n with two integer sequences a_n and b_n, we can infer that a_(n+2)=4a_(n+1)-13a_n and the same relation for b_n. the initial values are a_0=1, a_1=2, b_0=0, b_1=3. note that trying to find a solution using the characteristic polynomial will only result in the corresponding formulas for the real and imaginary part of (2+3i)^n. instead we can try to work out the elements of the sequence using generating functions. after some manipulations you can find A(x)=(1-2x)/(1-4x+13x²) and B(x)=3x/(1-4x+13x²) for a_n and b_n respectively. you can work out a formula for the two sequences using partial fractions and the geometric series. unlike the method using the polar form for (2+3i)^n, we get a algebraic solution. the value we want to find is b_3030 for the solution of the problem. the actual correct value turns out to be
-15018255505063763257087672670709830706621447608990211783269486428013011453559304704180313050291715215424340044168893744476500362245682528403047785586557182697833303424319495343261638479908442941457950507062295565330841752403362771528128679101816061530173846075950314137424690962115852949501512302859651778278255235726892360166045065905930096733842057581718458064428985516626392077710287235350664155415391405587168171614324047854339887473862335585608322420104069884939883199181004805433050290925448965257738757988699891524693275967481854222285195323271231879995590409838311205965156178286673620939661452988891623608690677839262067219938754701843412720840233261915661138964590911610820585782720719584900311258982941391373884926843974520216534592154744662052176911839876334494052258307963557527351037665864714160533041099932114971026737491595926803316049215661121083103021154948754029788416261318828197875321936936871779964935291799071394815848503041933918322238783514127555366715175685601546635563264091592043846459254910298783135358287725838400612526839046008782349039266146804411247048515286420816714176869731412760896663688772529612955695633724241060095697903486439036905058906251017157927740734180504264514938889152001938614369429576967228579148023156450659595744505904648991760099816458992262378134883392248415663225704452218662928910498786161269399801517100791471862979570551973287054167920129507370948135710472897395347181181191698692712930025905527486041017261069074793847913444927870735475169153383155390389057721273891657640145726526485493745070698712980097514216809556189562905070539795930536026820218060858834753826417565912950898819384331586023746036852071822726519739024539868
which is approximately -1.5*10^1687
having matrix
[2 -3
3 2]
we already have a representation of complex number. Say, 2+3i.
Thus, power of matrix actually consists from real and imag parts of (2+3*i)^3030.
Latter can be evaulated using Muavre formula:
norm = sqrt(2^2+3^2)=sqrt(13)
arg = atan(3/2)
(2+3*i)^3030 = norm^3030*exp(i*3030*arg)
Thus power of matrix to be
norm^3030*
[cos(arg*3030) -sin(arg*3030)
sin(arg*3030) cos(arg*3030)]
That's a really smart shortcut.
We can also get rid of linear algebra here and jump right into complex analysis using the following decomposition of sin(x):
sin(x) = (e^(ix) - e^(-ix)) / 2i, thus
e^(2x) * sin(3x) = (e^((2 + 3i)x) - e^((2 - 3i)x)) / 2i
Differentiating the rhs 3030 times we get:
((2 + 3i)^3030 * e^((2 + 3i)x) - (2 - 3i)^3030 * e^((2 - 3i)x)) / 2i
The only issue with this approach is that you need to combine the result to get back to real-valued functions
@@oleg67664 Or we can take the 3030th derivative of e^((2+3i)x) and take the imaginary part.
The final result has a total of 1688 decimal digits (aprox. -15.018*10^1686). That's a good place to stop.
From the moment I saw the 3030th derivative in combination with Linear Algebra I just KNEW this is going to be about Diagonalization. It is absolutely a marvelous method. I have introductory Linear Algebra students and Diagonalization is towards the end. Unfortunately this application is just too far fetched for my students, but I am going to try this out on a similar problem: An epower with a cosine term...
This was so cool, linear algebra is actually really satisfying
15:08 those should be C, not R, given that P pushes them into the complex numbers
If you squint then can interpret Michael's ( *R* ^ 2)bar's as ( *R* bar)^2's where *R* bar ≅ *C* is the algebraic closure of *R* :p
I'd prefer to use the 3030th derivative of 1/2i(e^(2+3i)z-e^(2-3i)z) because mathmajor is further along in Complex Analysis than Linear Algebra right now.
I showed your previous video about this topic to my teacher, she held a course for 11th year students about matrices and vectors
Maybe explore -> take derivative once to see the two set of functions before jumping to linear algebra would help tremendously why we are doing it this way.
agreed!
maybe you could do videos about thinking with commutative diagrams? it might really help me (and maybe others) apply algebra knowledge. just an idea.
I found this really helpful to understand the usefulness of linear algebra. Thank you Michael Penn. Can you recommend any linear algebra or abstract algebra book which contains concrete problems such as this? This was very illuminating and helped me understand the subject.
((2 3) (-3 2)) is also a complex number under the usual representation; namely 2+3i. We can "just" raise that to 3030 then map back. Then we map it under T, discard sin component.
The eval at 0 is both a matrix, and an operation on complex numbers.
This is beautiful !! Many thanks for sharing.
The reason i naturally comes up here is because sine (and cosine) return to themselves after 4 applications of the derivative, and i is a a number which you need to raise to the 4th power to get to 1, i.e. both these objects have a period of 4.
Techniques from linear algebra. greatly complicate the solution of this problem. It 's easier this way:
f(x)= e^2x*sin3x =1/(2i)[exp(2+3i)*x -exp(2-3i)x].
g(x)=d(3030)f(x)/d(3030)x =1/(2i)[(2+3i)^3030 *exp(2+3i)*x -(2-3i)^3030*exp(2-3i)x].
g(0)=1/(2i)[(2+3i)^3030 -(2-3i)^3030*], which in the end was received on the board.
But you can continue the transformation by going to the trigonometric form of a complex number and using de Moivre's formula ,
g(0)=(13)^1515 * sin(3030*φ), where φ= atan(3/2).
If you're used to doing linear algebra but not used to dealing with complex exponentials, I can see how Michael's method could actually seem more natural. But indeed yours is much more direct.
P.S. You mean tan(φ) = 3/2, or φ = atan(3/2).
@@theadamabrams φ = atan(3/2). Thank you for pointing out the typo.
This only works exponentials (including complex ones for the trig) multiplayed by polynomials because if you look at tye jordan decomposition of the matrix on R as a differential equation it becomes a very specific type which has solutions in the form of cx^ne^xg where g is an eigenvalue
you could just have used the set of the complex function to calculate the 3030 derivative f(x)=Im(e^((2+3i)x))
f(3030)(x)=Im((2+3i)^3030*e^((2+3i)x))
f(3030)(0)=Im((2+3i)^3030)
Definitely could have simplified further by writing the complex number in its polar form
Definitely
not really, to get the polar form you'd need to simplify arctan(3/2) to get the angle of 2+3i, but I don't think that's doable
There just isnt a nice value for the angle
@@bilzebor8457 Well you can write the result as 13^1515 × sin(3030*atan(3/2)) which might be a little easier to read. Also makes it more clear it's definitely real
the polar form can only provide a more elegant "closed-form" expression for the 2030th power, but from a computational point of view there is no difference since arctan[3/2]is not a "nice" angle
Very cool. Learned something new.
beta can be written as Im((2+3i)^3030) since 2-3i is complex conjugate of 2+3i and the same holds for the respective exponents. The result isn't too surprising, since f(x) can be written as Im(exp{(2+3i)*x}).
The method of going through a vectorspace of functions is more powerful though, e.g. considering a product of exp(2x) with some polynomial of grade n. The vectorspace span({exp(2x), 1/1! * exp(2x)*x... ,1/n! * exp(2x)*x^n}) would then cover all possible derivatives and d/dx would be represented by a matrix with 2 on the diagonal and 1 on the first upper offdiagonal (i.e. Jordan normal form).
22:20
You need to be stopped (but please don't)
Hey Good Place To Stop! I was wondering if you could help with something. Something about this video is very perplexing. How did differentiating a real function 3030 times lead to a complex value? Like the work done is logical but it does not seem intuitive to have a complex value when differentiating a real function...
@@antoniohaddad8816 If you evaluate the final answer (if you take the 3030th power of the stuff in the parentheses), you'll see that the end result is a real number.
@@antoniohaddad8816 Also worth noting that the result is a complex number added to it's complex conjugate, which can be seen without raising the complex numbers involved to the power 3030. Such a number is guaranteed to be real. :)
@@seanmacfoy5326 I think this guy is eautistic 😂
I've been out of undergrad for 6 years or so now, but this is starting to remind me of the diagrams from my differential geometry class.
really enjoyed this one!
Linear algebra on main channel...? WE GET A NEW LINEAR ALGEBRA MATHMAJOR COURSE VIDEO TODAYYYYYYYYY HELL YEAH LETS GOOOOOO
Good. Exponential function is also a good way. e^(2+3i)x
Wow that was an awesome problem. I'm gonna have to watch this video again and really process it.
I think it's better to consider the span of the 2 functions over C since we had to diagonolize the matrix D giving us complex number in the matrices P and P-1, this assumes that P-1 is representation of a lineair transformation going frome C^2.
the method and the diagram are the same we just chage the field from R to C and those R2 to C2
Eh, it's personal preference. When a matrix over a field _F_ doesn't diagonalize over _F_ , only then do you need to enlarge to the algebraic closure of _F_ / an algebraic extension of _F_ (and there's nothing to it when making this enlargement, you literally just say "In _F_ bar:"). Since algebraic closures of fields other than *R* can be hard to describe, in general it feels slightly "inefficient" to go to the algebraic closure when a lot of the time you didn't actually need to
Thx for the clear explanation!
I think the key point, which wasn't mentioned is that all derivatives of the original function are a linear combination of two functions, this will motivate the definition of the vector space V.
derivative operator and matrix both are linear operator, this is why you can get commutative diagram.
@@mathtutorial2761 I get that, and I have nothing against the actual talk, but the actual motivation was a bit mysterious.
@@mathunt1130 The solution is non-obvious but it does show you what does it mean by "think outside of a box"
@@mathtutorial2761 The form of the solution is trivially obvious.
I thought of solving it by rewriting sin3x using Euler's formula
Wow. A very fun problem solution. Thanks.
an amazingly powerful tool: building a sand castle with concrete and power tools
16:54 to 17:20 me at my math exam having no idea what I'm doing
I appreciate the effort to show the usefulness of linear algebra. However, when dealing with exponentials and trig functions, it is usually always best to transform everything into exponentials first, as they will always lead to a basis wrt which the matrix is already diagonal, so we don't need any computations of eigenvalues and -vectors.
Here: after converting sin(3x) to (e^{3xi}-e^{-3xi})/2i, we get f(x)=1/2i*(e^{2x+3xi}-e^{2x-3xi}). Using the basis v_1=e^{2x+3xi} and v_2=e^{2x-3xi} we get v_1'=(2+3i)v_1, v_2'=(2-3i)v_2 (so, we already have a basis of eigenvectors)
This means f^(3030)=1/2i*(v_1^(3030)-v_2^(3030))=1/2i*((2+3i)^(3030)v_1-(2-3i)^(3030)v_2)
And with v_1(0)=1, v_2(0)=1, we have f^(3030)(0)=1/(2i)*((2+3i)^(3030)-(2-3i)^(3030))
(To simplify this further, we might want to use polar coordinates, then the expressions are just R^(3030)*e^(it*3030) and R^(3030)*e^(-it*3030), which means that we get R^(3030)*sin(t*3030) as our final result, where R,t are the polar coordinates of 3+2i)
The underlying reason is the following: in the (infinite dimensional) vector space of all differentiable functions R->C, the eigenvectors of the differential are precisely the functions c*e^zt with constants c and z.
Of course, in order to understand the usefulness of this method, it does help to see the version with trigs first.
You can can the same answer in three lines by writing down the Taylor series in the complex plane. Still interesting and this gives a more general result than just evaluating at zero
I can't wait to get this problem on my calculus test tomorrow.
I like how we don't even get exactly the answer
Around 15:00 he finishes the diagram with two arrows pointing downwards with labels P-1. I think the correct labels are P.
Oooh, so close to Laplace transforms (which it's equivalent to). But the category theory flavor of this approach was nice.
Coming from electrical engineering, I implicitly saw e^2xSin3x as Ae^{(2+3j)x} + Ae^{(2-3j)x}, whose nth derivative is thus A(2+3j)^n + A(2-3j)^ne^{(2-3j)x}. Arguably that's the same approach because complex exponentials are eigenfunctions of the derivative operator. I guess the quick answer is (13^1515)e^2xSin(3x+phase).
Nice commutative diagram :D
I got it down to 13¹⁵¹⁵sin(3030arctan(1.5)) which is, let's be honest, a pretty decent sized integer. Curiously, the sine part of that is not an integer. But it is rational, and so its denominator must be a power of 13. Would be interesting to try to get an answer in a more closed form. For example, sin(2arctan(1.5)) = 12/13, and sin(4arctan(1.5)) = -120/169.
It would be interesting if we could find its prime factorization!
Are you a future traveller who have stolen this problem from 3030 math contest lol
2:44 I dont get it, isnt V just a subset of R? Like isnt the dimension of V just 1? And how can we be sure that the representation of an element in V is unique (resolved if dim = 2)? Multiple representations of an element surely should create problems when multiplying a matrix
You gotta consider that x is a variable, and can take on multiple different values.
V is the space of all functions f(x) = a*e^(2x)*sin(3x) + b*e^(2x)*cos(3x), where a and b are free.
Even though this is a real-valued function, each individual element can’t really be considered to be a real number itself.
The elements of V are not real numbers. They are R->R functions.
In fact, vectors are almost always functions. Consider [5, 8, 6] in R^3. The first component is 5, the second component is 8, and the third component is 6. In other words, this vector is the {1,2,3} -> R function mapping 1 to 5, 2 to 8, and 3 to 6. The vectors in V are similar except instead of domain {1,2,3} the domain is all real numbers. V is a subspace of R^R
it is beautiful . love it
تمرين رائع.سفر في الذكريات بين الخاصيات والعمليات.
Why would you not normalize P so it's a unitary matrix?
Nth derivative of e^(ax) sinbx is (a^2+b^2)^(n/2) e^(ax) sin(bx + n(arctan b/a))
Couldn’t you diagonalize it using the PCP^-1 version instead since it has complex eigenvalues?
That's what he did. Except what you call C is what he had unnamed and is the matrix with the eigenvalues on the diagonal
D = PCP⁻¹
P⁻¹(D)P = P⁻¹(PCP⁻¹)P = C
I am going through Linear Algebra 2 at the moment and we are building up the theory for diagonalization and trigonalization etc. and this gives a nice perspective on how powerful this theory can be in application. Thx a lot.
Can the same technique be applied to multivariable functions though?
Now we need the proof that the answer is real
that's just because the final answer is a sum of a complex number and its conjugate (you can see this using basic properties of the conjugate operation), and you have (a+bi) + (a-bi) = 2a which is real.
Beta should actually end up being a real number, right?
It is a real number. One way to see this is to replace 'i' by '-i' and observe that the answer does not change. This means the imaginary part must be 0.
@@tdchayes alternatively (very similar but a tiny bit easier in my opinion), observe that the expression is a complex number plus its conjugate. this moreover tells us that the real number Beta equals is twice the real part of either of the two terms (since (a+bi) + (a-bi) = 2a).
@@schweinmachtbree1013 I can almost see the complex conjugate here in this example. But here is another solution that bprp posted that is less obvious. i ( ln ( x+i ) - ln (x - i) )
@@tdchayes that example would remain true if ln were replaced by any other real function which has a natural extension to *C* (exp, sin, tan, arcsin, tanh, Gamma, Zeta, etc.) - it is just using the fact that complex conjugation commutes with not only finite sums but also infinite sums (here, Taylor series).
@@schweinmachtbree1013 And are you trying to make this so complicated for the average viewer that they can't understand it? If you replace 'i'' by ''-i' the function is the same is a very understandable thing. What is your point? Is what you say helping the average viewer? I'm not sure why this is highlighted.
I'm not sure whether my idea works out or not😅, but if you write that matrix as √(2^2+3^2)*R, then R is an orthogonal matrix, and it's equivalent to a rotation of an angel phi. So if one raises it to the nth power, then it's a rotation of an angel phi*n. Then one raises the coefficient to the nth power, multiply them together, and you have the original matrix raised to the nth power. I don't know linear algebra that well, so maybe there's a mistake in my solution, so please let me know if it's right or wrong. Anyway, nice video as usual. Thank you!
This take five minutes with complex exponentials. Just take the imagebery part of e^x(2+3i)=f(x)
It reminds me the Fourier transform, if we try to use this to solve an ODE like f''(x) + f'(x) - f(x) = 0.
Now that we have a Vectorspace in R^2, can we make sense of the scalar product back in V?
Hmm I suppose so yeah. Writing the map from V to R^2 as φ, you'd "pull back" the dot product in R^2 to an inner product in V defined by v . w = φ(v) · φ(w)
@@schweinmachtbree1013 Can you make that specific? I mean, you now just transformed into the vectorspace R^2 and carried out the straight forward scalar product, didn't you?
@@digxx yes that's exactly right, it's as simple as that. this kind of thing is called transport of structure - it's where you have some kind of structure in B and you have an isomorphism/homomorphism from A to B, and you use the map to "transport the structure to A".
do you do like, study of functions in this channel? i stumbled upon an odd one and can't figure why it is the way it is
Using Re(𝑧) = ½ (𝑧 + 𝑧̅) we can simplify the result to Re(𝑖 (2−3𝑖)³⁰³⁰) or further to −Im((2−3𝑖)³⁰³⁰),can't we?
I always smell chalk dust and have to brush on my fingers when I watch these videos.
the unbounded love for useless calculations really obscures the beautiful connection between calculus (or rather, D-finite power series) and linear algebra :(
I did not get the last step where T^-1 acting on (alpha, Beta)=alpha*e^(2x) sin(3x)+beta*e^(2x) cos(3x). How can we get the inverse of T?
(Alpha,beta) is alpha•(1,0)+beta•(0,1) while T^(-1) is a linear operator which maps (1,0) back to the function e^(2x) sin3x.
@@pedroteran5885 Thanks
Why is a,b a real number in the linear combination. Why not a complex number?
either works fine (although we do need the complex numbers later into the video), so the reason is the very superficial/arbitrary one that the question uses the notation _x_ for the variable and not _z_
Maybe we can use Demouvre’s Theorem to keep this going, just to get rid of the imaginary bits, just feels weird having i’s in the final result, when the answer is real.
but- doesn't the D matrix simply correspond to the number 2-3i ?
then if you take the exponent form of 2-3i directly you can get the 3030th power easily, and just plug in the values back into a matrix form, then you just skipped a quarter of the video-
LOl, that's genius
i like it
wow linear algebra is really powerful! kinda wish i learned all its usefulness in class :(
you would have to take more than one course in linear algebra to get a somewhat comprehensive account of the applications of linear algebra to mathematics alone! To be honest though, linear algebra is probably the easiest main branch of higher mathematics, so you can just pick linear algebra stuff up as you come across it in your studies.
I understood everything, except for why he used that other function to begin with. Can someone please explain to me the idea of why he chose it?
What’s the significance of 3030? Please tell me it’s a Deltron reference
I have a feeling it might be a meme on math contest problems often having the year number in the problem, so instead of 2020 the person who wrote the problem thought it would be funny to put 3030
@@schweinmachtbree1013 yeah I was assuming it was some variation of that
from R² bar to R²（from down to up）it should do P ^ -1，but the arrow(from up to down)，i think should write P，not P ^ -1
In fact, drawing an arrow and writing P is equivalent to drawing the arrow in the opposite direction and writing P^-1. The fact that he chose to draw the arrow from top to bottom mainly comes from some habits inherited by the use of categories.
@@becomepostal @becomepostal if Communication diagram is wrong，it would go to another ans→ (T^-1)(p^-1)(Λ^3030)PTf(x)
so good
I solved a similar problem with complex analysis using the cauchy formula, but this is definitely better since it's easier to understand
D is in fact the JACOBIAN matrix right?
Last steps were rushed. What is T inverse?
The inverse map of T
okay but like…what do you mean this is a good place to stop, you have a question that only uses real numbers and end up with a crazy complex expression?
now how d'U show that the result is real?
the result is the sum of a complex number and its conjugate which means it's real
So, I simplified the answer into a real number in my head as 13¹⁵¹⁵sin(3030arctan(³/₂)), which WolframAlpha agrees with. Mind bogglingly, it's a 1688-digit integer. I think you really dropped the ball by not mentioning that in the video.
Isn't it a bit odd that you took a derivative (albeit of a very high degree) of a real function, then substituted x=0 and got a complex number?..
And if we take the second derivative of R2, we get R2-D2.
Anyhow, a very good example of exercise oriented to the consolidation of a wide range of theoretical backgrounds instead of vain self-referential calcolus virtuosims....
Jesus, if my University teachers would have been like Mike....
Or we can write the function in polar form and take all the derivatives. Easy peasey.
Are all finite dimensional linear dynamics represented in differentiation?
Shouldn't f_(3030) evaluated at zero be purely real number, but it seems Mike's result should have an imaginary part? ☹
@@angelmendez-rivera351
@@angelmendez-rivera351 Good thinking...I'll take another look at the answer. Thanks!
@Angel Mendez-Rivera By golly. You are right!☀
@Angel Mendez-Rivera By changing i to -i you change a complex number z to its complex conjugate, and z =the complex conjugate of z iff z is purely real.
Thanks.
Who is the target audience of this video that (1) doesn't know span but (2) knows about vector spaces, isomorphism, and commutative diagrams?
I don't think you need to have seen commutative diagrams before to follow this video - they are used in a very basic way and Michael explains what it means for a diagram to commute.
Can't we just make it a complex function? I guess its just a demonstration of the lineat method.
For beeing in R² there are a lot of imaginary numbers i.. o.O
Isn't it quicker to use desmos?
What? I don't like the final answer. My Calc. students would be expect a real value for the answer. I guess I will just work it out myself.
the final answer is the sum of a complex number and its conjugate which means it's real
Wooooow