I asked my teacher about this and she said it wasn't possible. Gonna annoy her tomorrow 😂😂

I did not use linear algebra. Me solution uses complex numbers. Here it is: In order to find a mapping transformation for a mirror placed on y=2x i started by making a point in the complex plane such that its angel with the "real" number line is that of the line y=2x and its magnetude is one. Then i made a second, general point that the transformation would act onto. Now, my transformation is the following: to get the output point i multiply the input point by the square of the ratio ((special point)/(input point divided by its magnitude)) In the end i can take the general coordinates of my output complex number and they would be the answer.

what an absolute banger of a video

Yo thumbnail is on point again, Dr. P!! Love it!

@drpeyam

5 жыл бұрын

Thank you!!!!!! 😄

@drpeyam

5 жыл бұрын

That one took an hour to make, I totally understand you now!

@blackpenredpen

5 жыл бұрын

exactly!

@mfadhilal-fatih1427

3 жыл бұрын

Hello mathman

You could easily do it for any linear equation y=mx+n and a point P(a,b) by finding the perpendicular line to the linear function that also passes through the point, which will be y=((a-x)/m)+b, and then you can find the intersection, which is the midpoint of P and P', so you can find P'

Woah! I used linear algebra on a school assignment once. The final for my precal class was to draw a picture using graphs. I used linear transformations to rotate functions to what I wanted. I also used calculus to come up with really accurate features. (What I did was plot the points I wanted to connect with my equation, then found the slope between the points and created a derivative graph. Then used regression and integration to smooth things out)

@josuequintero-td5gh

3 ай бұрын

Integration in precalc? That's way overkill

Thank you so much about your knowledge sharing

Very cool!

I've found the same matrix by simply solving a system of equations { T(1, 2) = (1, 2) , A(-2, 1) = (2, 1) }, where T is a linear operator with matrix { { a , b } , { c , d } } in Cartesian coordinate system.

Dr Peyam! Two days ago I just had an important exam which had an exercise about this. If only you would have uploaded this video earlier 😢

@drpeyam

5 жыл бұрын

Awwwww!!! It was on my playlists though!

Not completely related to the video, but I found a function that describes part of the graph e^x flipped across y = 2x. It has the Lambert W function in it, here it is [For x>0.7726, y

The truth I love your videos I am your fan and I would like you to share more math topics that this matery I love I sleep two hours to study it, to be a great mathematician like you and I would like you to share topics, only basic topics like theorem of pick and others I hope you share more topics: '(

You could get the same results with 3 simple tranformations? Rotation[[cos(tan^-1(.5)), sin(tan^-1(.5))],[-sin(tan^-1(.5)),cos(tan^-1(.5))] Reflection[[-1,0],[0,1]] Rotation[[cos(-tan^-1(.5)),sin(-tan^-1(.5))],[-cos(-tan^-1(.5)),sin(-tan^-1(.5))]] Rotate your axis of reflection so that it becomes the y axis, reflect around the y-axis, then rotate back.

It really do be like that sometimes

The Linear Algebra in this is too hard for me :/ Is there a recommendation of where to learn this?

@drpeyam

5 жыл бұрын

All my playlists!

@benjaminbrady2385

5 жыл бұрын

@@drpeyam Thanks but which one should I start with?

@drpeyam

5 жыл бұрын

Start with Linear Equations (Lay Chapter 1), finish all the Lay ones, then do all the Friedberg ones

Woahhh, its beautiful

T is much nicer if you use y=(tanθ)x instead of y=mx

Half of that kinda flew over my head because I dont know all of the stuff that this builds off of but I think I understand

@drpeyam

4 жыл бұрын

Check out my Linear Transformations Playlist: kzread.info/head/PLJb1qAQIrmmCGJgXVb1gt0q1weM_Bt6aE

I derived the formula for reflection across y=mx using vector projections! kzread.info/dash/bejne/fHyF1qNvgprMicY.html Sadly I never learned the basis stuff that you use in this video, but I have my own methods!

I did this without linear algebra, and was honestly quite easy, but the linear algebra way is very elegant :)

Hello, I am Mexican, I dream some day to be like you, to know a lot of mathematics,

Why not just solve the system of equations using the known points @4:36 ? You have a + 2b = 1 and c + 2d = 2, as well as -2a + b = 2 and -2c + d = -1. This easily becomes 5b = 4 and b = 4/5, and then a = -3/5, and similarly 5d = 3 and d = 3/5, c = 4/5

I think finding the reflection about y = mx + c with algebra only is not too difficult.

Reflect about the line x sin(t) = y cos(t): if we rotate the plane such that the line meets the x axis, then negate, then rotate back, it should work. so let's take the formula: e^-it * [1,0;0,-1] * e^it, e^it being a rotation matrix. The first two multiply to give [cos,sin;sin,-cos], and multiplying that by e^it gives [cos^2-sin^2, 2sincos; 2sincos, sin^2-cos^2] which if i recall is equivalent to the quaternion (cos^2-sin^2)i + (2sincos)k = (1-2sin^2)i + (2sincos)k for any angle input to the sines and cosines. Odd how this is supposed to map from 2d to 2d... is it really guaranteed that this quaternion works?

@MrRyanroberson1

5 жыл бұрын

in this case, the input vector should be formatted as a complex number ix + ky, and multiplying this by ai+bk we get -(ax +by) -(ay+bx)j, which seems like a reasonable linear combination of x and y, though i can't prove the quaternion solution i remain confident on the matrix one. really it should work with the proper map for (x,y) -> [quaternion]

@MrRyanroberson1

5 жыл бұрын

Algebraic solution: take (x,y) -> x+iy = z. c(z) = x-iy, the complex conjugate. z' = c(z * e^-it) * e^it, as the conjugate flips about the x axis, we simply need to rotate into our new basis, since the line of reflection is indeed a line which is conservative in terms of area and even the position of the origin point in this case. If, however, the line is moved H units away from the origin along a perpendicular line (the exact distance can be determined through basic algebra), simply subtract H before conjugating. The linear algebra version of this is just replacing each function with its matrix

@MrRyanroberson1

5 жыл бұрын

it also seems that by your solution sincos = m/(1+m^2) and (1-m^2)/(1+m^2) = cos^2-sin^2, if 1+m^2 = sec^2, then 1-m^2 = 1 - tan^2, yes m = tangent, this makes total sense. tan/(1+tan^2) = tan/(sec^2) = sincos. our formulas totally agree under m = tangent, and your solution operates on the tangent (the point 1,M) whereas mine operates on the unit circle (a,b where aa+bb=1)

It's special orthogonal group 2!

easy to do without linear algebra. For example reflect the point (6,3) about line y = 2x step one: calculate equation of line with (6,3) that is perpendicular to y = 2x it is y = -1/2 x + 6 step two: use systems of equations to solve for point of intersection which is (12/5, 24/5) third: since this is midpoint the reflected point easy to solve for (-6/5, 33/5) Lot easier!

@drpeyam

2 жыл бұрын

But systems of equations *is* linear algebra, that’s the whole point!! I’m just doing a more systematic way of what you’re doing!

Is it bad that I'd rotate and rotate back?

@drpeyam

5 жыл бұрын

Inefficient 😂

This video reflects well on you.

@drpeyam

5 жыл бұрын

Hahaha, pun intended? 😉

@michaelz2270

5 жыл бұрын

@@drpeyam yes :-)

I hope and see it I'm your fan :'(💔

imgur.com/a/QuopTEn To be fair it is easier with simple Cartesian geometry. Always love linear algebra tho

Reflection about y=mx+b ?

@drpeyam

5 жыл бұрын

I mentioned that in the video, you do a translation first

@husklyman

5 жыл бұрын

am whoops my mistake, I ment reflection of a function.

YAY | YAY

@drpeyam

5 жыл бұрын

Oooooon!!!! 😄😄😄

@OonHan

5 жыл бұрын

@@drpeyam What's up???!!!