There are unknown way to visualize subspace, or vector spaces. You can stretching the width of the x axis, for example, in the right line of a 3d stereo image, and also get depth, as shown below. L R |____| |______| TIP: To get the 3d depth, close one eye and focus on either left or right line, and then open it. This because the z axis uses x to get depth. Which means that you can get double depth to the image.... 4d depth??? :O It's possible to add 4d depth to a isometric cube, by moving the skew square side along the x axis of the right cube, individually, and leave the left cube unchanged. It's even possible to create an isometric/true perspective hybrid, by leaving the north west/south east line unchanged, while adding perspective to the north east/south west line. Would be awesome, if someone make a game (even KZbin video) that way! Nice video! :-) p.s You're good teacher!
@mhoefnagel924 ай бұрын
Thanks! Interesting idea - would be interesting to see how 4D space could be visualised in this way.
@cynthiastreet50735 ай бұрын
why do you have so many ads🥲
@mhoefnagel925 ай бұрын
Sadly, its not my choice to have ads -- I would prefer not to.
@MikeMagTech5 ай бұрын
Excellent video. Thank you.
@mhoefnagel925 ай бұрын
Thanks!
@mikehoef6 ай бұрын
Mistake at 29:35, the last entry of the first column of Q should be -1/sqrt(2).
@mhoefnagel926 ай бұрын
Mistake at 08:45, the c's in the second equation should be x's.
@quantumgaming91806 ай бұрын
If only there were more professors like you in the world, the world would be a better spacr
@mhoefnagel926 ай бұрын
Thanks - I will be doing another video on vector spaces soon.
@mhoefnagel926 ай бұрын
Thank you for the generous complement!
@quantumgaming91806 ай бұрын
Amazing video. Hope you continue the series
@goldenblade31796 ай бұрын
Timestamp would be great
@mhoefnagel925 ай бұрын
Thanks --- I agree. Let there be time stamps.
@mhoefnagel926 ай бұрын
Mistake at 10:13 (the table is wrong). The columns after the first row should swap around, and then the right column first row should read "f(t)"
@fortified0017 ай бұрын
Thanks
@peradies70447 ай бұрын
The Derivatives are in the form of Multivariable functions, yet the first solution seems like a singular variable function?
@mhoefnagel925 ай бұрын
I missed this comment. In the theory of 'ordinary differential equations,' which is what we are doing here, we always assume that y is a function of a single variable x. So when we are given a DE like x dx + y dy = 0, we assume that y is a function that satisfies the equation above. Of course, we then want to solve the equation (find all possible y). Solutions are given in two ways: either by writing y as a function of x, or implicitly by writing y as a solution to a certain two-variable equation.
@joshuakotzee80297 ай бұрын
Hate to be that guy, but your cubic had a tiny error. Roots were (r-1)^2(r-2), not (r-1)(r-2)^2 :) Brilliant video, thank you for making these! They really help a lot.
@mhoefnagel927 ай бұрын
Thanks for this comment --- it is appreciated! It is indeed an error (which I should have picked up). Glad to hear the videos help.
@peradies70447 ай бұрын
So any combination of solutions multiplied by a constant is itself a solution? Wouldn't you then only need one of these solutions if its already two solutions multiplied?
@mhoefnagel927 ай бұрын
Yes - any linear combination of solutions to a homogeneous differential equation (with 0 on the right) is itself a solution. Solving such a differential equation means finding all possible solutions. For this reason, finding one solution does not solve the differential equation. However, if we are given initial values, then we can single out one solution - the unique solution. For example, solving the equation x^2 - 1 = 0 means to find all possible x that satisfy the equation (x = 1, -1). For this reason, x=1 is not the solution; it is 'a' solution.
@peradies70447 ай бұрын
ahhh thank you I understand, now it makes sense why linear independance matters!
@mhoefnagel927 ай бұрын
@@peradies7044 I am glad!
@hastyroehling57538 ай бұрын
P R O M O S M
@Rockyroadpool8 ай бұрын
Hi sir. Not sure if this is the best way to ask but for the last example- why does it differentiate to n! ?
@mhoefnagel928 ай бұрын
Which part?
@philipc44907 ай бұрын
he's asking about 28:50@@mhoefnagel92
@peradies70448 ай бұрын
Very helpful when you've been sick and missed the in person lectures - Some of these topics are very hard to grasp by self study alone
@mhoefnagel928 ай бұрын
Thanks, I am glad they help
@mikehoef8 ай бұрын
Small error at 47 (second equality in the third line should be dv/dt instead of dx/dt)
@clayton97330 Жыл бұрын
Differential equations, the only equations that matter.
@NurHadi-qf9kl Жыл бұрын
Subs y=uv, yg adalah fungsi2 dg argumen x. Maka dy/dx=udv/dx+vdu/dx .udv/dx+vdu/dx+uv-u^2.v^2=0 .udv/dx=0
@NurHadi-qf9kl Жыл бұрын
.dy/dx+y=y^2
@mathunt1130 Жыл бұрын
Your first example would have been simple if you had taken limits as t->\infty, then you wouldn't be so caught up on your two cases, as s is FIXED.
@mhoefnagel92 Жыл бұрын
The limit exists or not, depending on the value of s, hence the cases.
@elara6227 Жыл бұрын
Awesome video! Thank you, your videos have really helped me for my upcoming linear algebra exam :)
@mhoefnagel92 Жыл бұрын
I’m glad to hear it :)
@DanicaCapazorio Жыл бұрын
Great Video - really enjoyed it!
@mhoefnagel92 Жыл бұрын
Glad to hear it!
@christianshaw2892 Жыл бұрын
Promo*SM
@AzeddineHamich Жыл бұрын
Matrix to the power
@DavidFMayerPhD Жыл бұрын
Please do a similar video on the Fourier Transform.
@mhoefnagel92 Жыл бұрын
I do aim to do that at some point
@DavidFMayerPhD Жыл бұрын
@@mhoefnagel92 I have NEVER seen a better intuitive explanation of the Laplace transform. Congrats.
@mhoefnagel92 Жыл бұрын
@@DavidFMayerPhD many thanks!
@DavidFMayerPhD Жыл бұрын
@@mhoefnagel92 By the way, my PhD IS in Mathematics. When I went to school, there were NO SUCH VIDEOS. I would have learned more if my books had been supplemented by videos such as yours.
@mhoefnagel92 Жыл бұрын
@@DavidFMayerPhD thanks for these generous remarks, I’m glad the receive them.
@Vykng Жыл бұрын
what would you do if the laplace transform was used on sqrt(t)? would you have to make n=1/2 and use possibly the gamma function to solve?
@mhoefnagel92 Жыл бұрын
This is not an easy Laplace transform. There are a few standard tricks that I have seen --- using gamma function, as you say, but also using Laplace transform of 1/sqrt(t). There are a few videos on KZbin explain these and other approaches.
@lih3391 Жыл бұрын
Yeah, pretty much right, things do turn out suprisingly well when you replace factorials with gamma functions
@EmileCluyse Жыл бұрын
Very helpful!
@roberthuff3122 Жыл бұрын
Thank you, but why does it work?
@mhoefnagel92 Жыл бұрын
We apply the Laplace transform to an IVP, it becomes an algebraic equation, which can then be manipulated (often using partial fraction decompositions) into a form which we can apply the inverse Laplace transform (giving a solution to the IVP).
@lordkelvin100thompson8 Жыл бұрын
Consider this function: y= e^(Ax). To take the derivative, take the derivative of the outer function, which is e^(Ax) since the function is its own derivative, and multiply by the derivative of the inner function: Ax, or simply A and multiply: y'= Ae^(Ax). To find the antiderivative, integrate the outer function, again, e^(Ax) and divide by the derivative of the inner function so: Se^(Ax)dx= (1/A)e^(Ax). This means that taking the derivative is simply multiplication, and integration is simply division. It would be nice to be able to do that for general functions: y= f(x), or something like: y= sin(x). It's not generally the case that the derivatives and integrals are related by constants. Therefore, make your general function a function of e^(Ax) by multiplying with e^(Ax): e^(Ax)f(x). For this, make the substitutions: e^(-sx)f(x) where s is a new variable, and may be real of complex. Next, get rid of the 'x' term by integrating from zero to INF. If the integral converges, you have a new function that's a function of s only. This is your LaPlace Transform. If you can do integration by parts, you can integrate: S[f'(x)e^(-sx)dx and see that F'(s)= sF(s) - F(0) (F(0) takes care of initial conditions).
@roberthuff3122 Жыл бұрын
@@lordkelvin100thompson8 Thank you.
@fasterfaster Жыл бұрын
Nice❤👍
@ShanBojack Жыл бұрын
Subscribed
@Abdalrhman_Kilesee Жыл бұрын
Can u tell me how to make these nice thumbnails? What font u use for the mathematical script
@mhoefnagel92 Жыл бұрын
Sure - I use a Python library for mathematical animations, which is called "manim", short for "Mathematical Animation engine". The mathematical formulas are all in LaTex (rendered with a short manim script).
@Abdalrhman_Kilesee Жыл бұрын
Amazing 😻 im the 135th subscriber , so remember me when u get famous
@bored_as_fuck_890 Жыл бұрын
235th
@larzcaetano Жыл бұрын
Amazing video!!! Would be amazing if you could make a video on how to derive it :D
@mhoefnagel92 Жыл бұрын
Thank you!
@ShanBojack Жыл бұрын
@@mhoefnagel92 yes true you should
@mhoefnagel92 Жыл бұрын
Apologies for the bad sound!
@graeme8866 Жыл бұрын
Weewoo
@rubenmartikyan9266 Жыл бұрын
Keep up the good work bro.
@mhoefnagel92 Жыл бұрын
At 20:22, should be e^(-x) after the (10D - 2C) term.
@jordiswartz6775 Жыл бұрын
a mad one STILL, thanks boss
@mhoefnagel92 Жыл бұрын
Small error at 18:40. After the last equality, the first term should be e^(x) not e^(x^2).