An amazingly powerful tool: solving a Calculus problem using linear algebra 

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.

Thank you very much for giving a more formal shape to this method, in order to explain it rigorously. So elegant.

The aproximate answer is -1.5 × 10^1687 And that is the calculated exact answer: -15018255505063763257087672670709830706621447608990211783269486428013011453559304704180313050291715215424340044168893744476500362245682528403047785586557182697833303424319495343261638479908442941457950507062295565330841752403362771528128679101816061530173846075950314137424690962115852949501512302859651778278255235726892360166045065905930096733842057581718458064428985516626392077710287235350664155415391405587168171614324047854339887473862335585608322420104069884939883199181004805433050290925448965257738757988699891524693275967481854222285195323271231879995590409838311205965156178286673620939661452988891623608690677839262067219938754701843412720840233261915661138964590911610820585782720719584900311258982941391373884926843974520216534592154744662052176911839876334494052258307963557527351037665864714160533041099932114971026737491595926803316049215661121083103021154948754029788416261318828197875321936936871779964935291799071394815848503041933918322238783514127555366715175685601546635563264091592043846459254910298783135358287725838400612526839046008782349039266146804411247048515286420816714176869731412760896663688772529612955695633724241060095697903486439036905058906251017157927740734180504264514938889152001938614369429576967228579148023156450659595744505904648991760099816458992262378134883392248415663225704452218662928910498786161269399801517100791471862979570551973287054167920129507370948135710472897395347181181191698692712930025905527486041017261069074793847913444927870735475169153383155390389057721273891657640145726526485493745070698712980097514216809556189562905070539795930536026820218060858834753826417565912950898819384331586023746036852071822726519739024539868

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.

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.

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.

The final result has a total of 1688 decimal digits (aprox. -15.018*10^1686). That's a good place to stop.

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).

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?

This was absolutely beautiful, using concepts and techniques from one field to find a solution in another.

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...

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

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!

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)]

This is beautiful !! Many thanks for sharing.

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.

Definitely could have simplified further by writing the complex number in its polar form

This was so cool, linear algebra is actually really satisfying

This is amazing! I really like linear algebra, would've never thought it can be used this way tho.

I showed your previous video about this topic to my teacher, she held a course for 11th year students about matrices and vectors

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).

maybe you could do videos about thinking with commutative diagrams? it might really help me (and maybe others) apply algebra knowledge. just an idea.

15:08 those should be C, not R, given that P pushes them into the complex numbers

really enjoyed this one!

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)

22:20

Thx for the clear explanation!

Very cool. Learned something new.

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.

Wow. A very fun problem solution. Thanks.

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.

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.

((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 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

it is beautiful . love it

Wow that was an awesome problem. I'm gonna have to watch this video again and really process it.

Linear algebra on main channel...? WE GET A NEW LINEAR ALGEBRA MATHMAJOR COURSE VIDEO TODAYYYYYYYYY HELL YEAH LETS GOOOOOO

I thought of solving it by rewriting sin3x using Euler's formula

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).

Good. Exponential function is also a good way. e^(2+3i)x

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.

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.

Nice commutative diagram :D

تمرين رائع.سفر في الذكريات بين الخاصيات والعمليات.

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.

an amazingly powerful tool: building a sand castle with concrete and power tools

I can't wait to get this problem on my calculus test tomorrow.

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 like how we don't even get exactly the answer

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

Oooh, so close to Laplace transforms (which it's equivalent to). But the category theory flavor of this approach was nice.

16:54 to 17:20 me at my math exam having no idea what I'm doing

so good

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

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.

Nth derivative of e^(ax) sinbx is (a^2+b^2)^(n/2) e^(ax) sin(bx + n(arctan b/a))

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.

I solved a similar problem with complex analysis using the cauchy formula, but this is definitely better since it's easier to understand

This take five minutes with complex exponentials. Just take the imagebery part of e^x(2+3i)=f(x)

LOl, that's genius i like it

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!

the unbounded love for useless calculations really obscures the beautiful connection between calculus (or rather, D-finite power series) and linear algebra :(

Around 15:00 he finishes the diagram with two arrows pointing downwards with labels P-1. I think the correct labels are P.

Can the same technique be applied to multivariable functions though?

Why would you not normalize P so it's a unitary matrix?

wow linear algebra is really powerful! kinda wish i learned all its usefulness in class :(

It reminds me the Fourier transform, if we try to use this to solve an ODE like f''(x) + f'(x) - f(x) = 0.

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.

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.

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

I always smell chalk dust and have to brush on my fingers when I watch these videos.

Beta should actually end up being a real number, right?

Couldn’t you diagonalize it using the PCP^-1 version instead since it has complex eigenvalues?

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?

Wooooow

Are you a future traveller who have stolen this problem from 3030 math contest lol

Now we need the proof that the answer is real

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....

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?

Cool

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-

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

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

D is in fact the JACOBIAN matrix right?

Why is a,b a real number in the linear combination. Why not a complex number?

And if we take the second derivative of R2, we get R2-D2.

Now that we have a Vectorspace in R^2, can we make sense of the scalar product back in V?

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?

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?

Or we can write the function in polar form and take all the derivatives. Easy peasey.

What’s the significance of 3030? Please tell me it’s a Deltron reference

Last steps were rushed. What is T inverse?

You really should specify whether your matrix is a real one (since V you observe over R) or a complex one (since you diagonalize over C). You can't have both.

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?..

For beeing in R² there are a lot of imaginary numbers i.. o.O

Can't we just make it a complex function? I guess its just a demonstration of the lineat method.

now how d'U show that the result is real?

Are all finite dimensional linear dynamics represented in differentiation?

I guess the follow up question would be how could we compute the final value into a single form. But that is very much a video for another time.