Watch this next: the last question on my calc 2 final: kzitem.info/news/bejne/sYx6n5ycj6xnmG0
@diegoalejandroordonezcastr5963
Жыл бұрын
You can solve the integral of x/tan(x) using the polilogarithm please😅
@AT-zr9tv
Жыл бұрын
I really enjoy when a calculus exercise is given a geometric context that makes the answer obvious. This was elegant, fun to watch.
@Sphinxycatto
Жыл бұрын
I remember watching him while I was in 8th grade in summer holidays
@BHNOW100
Жыл бұрын
That's sad bro go outside
@ren695
Жыл бұрын
@@BHNOW100 what 💀☠️☠️
@Sphinxycatto
Жыл бұрын
@@BHNOW100 after pondering for 120 seconds i understood the joke 🤣🤣
@extreme4180
Жыл бұрын
@@Sphinxycatto and now ur in which grade? Indian ho kya?
@Sphinxycatto
Жыл бұрын
@@extreme4180 name me hi toh hai Me 11th me hu
@drpeyam
Жыл бұрын
How fun!! I love fixed points! 😁
@infernape716
Жыл бұрын
ok
@The-Devils-Advocate
Жыл бұрын
What is a fixed point?
@pwmiles56
Жыл бұрын
@@The-Devils-Advocate A fixed point is a point which maps to itself under the given rational function. (ax + b) / (cx + d) = x Multiplying out ax + b = cx^2 + dx Make a quadratic in x cx^2 + (d - a)x - b = 0 x = (a - d +/- sqrt((a - d)^2 + 4bc))/(2c) So in general there are two fixed points. However, it can happen that the fixed points are not real. As an example: take f(x) = 1/x We have a=0, b=1, c=1, d = 0 The fixed points are x = +/- sqrt(4)/2 x = +/- 1 This is called the hyperbolic case. But if f(x) = -1/x then a=0, b=1, c=-1, d=0 x = +/- sqrt(-4)/2 x = +/- i This is called the elliptic case
@The-Devils-Advocate
Жыл бұрын
@@pwmiles56 ah, thanks/
@aaron9262
Жыл бұрын
nice
@tobybartels8426
Жыл бұрын
The c = 0 case gives you the linear involutions: f(x) = B − x and (when also d = 0) f(x) = x.
@lawrencejelsma8118
Жыл бұрын
On the mathematics of translational, rotational and mirror images you were interestingly getting into showing that the y= f(x) = (1)(x) and y= f(-x) = g(x) = k/x has g(x) a mirror image over f(x). These are really important in vector space mathematics where we define rotational and translational matrices to change a matrix in vector space. Video game designers as well as engineering projects rely on defining those matrices.
@Ninja20704
Жыл бұрын
I also had questions like this in my homework and tests. They called them self-inverse functions. An interesting follow on question from it is to find like say f^2023(5), where f^n(x) means composing f with itself n times. Knowing this self-inverse property will get you to the answer pretty fast.
@icebeargt5142
Жыл бұрын
h2 maths?
@goodguyamr6996
Жыл бұрын
I can already tell this man's personality is immaculate by watching his videos
@idonthaveusername9907
Жыл бұрын
easily one of my most favourite channels ever! this guy explains so well
@fizixx
Жыл бұрын
Great vid! I've not heard of the Involution function before. It's quite fascinating and will have to do some reading. Thanks BPRP!
@blackpenredpen
Жыл бұрын
Glad you enjoyed it!
@josephparrish7625
Жыл бұрын
I love watching you teach!!!
@pwmiles56
Жыл бұрын
Here's a fun theorem about involutive rational functions. We work in the complex number field so these are Moebius functions, but defined in the same way f(z) = (az + b) / (cz + d) with ad - bc non-zero Show that if f(x) is involutive but not the identity function; w1 = f(z1), w2 = f(z2); the lines z1z2 and w1w2 intersect at w3; and the lines z1w2 and w1z2 intersect at z3 then w3 = f(z3) PS I think I discovered this. I call it the "triangle involution". z1, z2, and z3 form a triangle and w1, w2, w3 fall on a line. The figure is the so-called complete quadrilateral, i.e. all the point intersections of four lines, namely z1z2w3; z2z3w1; z3z1w2; w1w2w3
@fantiscious
Жыл бұрын
Fun fact: If f(x) is an involution, and g(x) is an invertible function, then h(x) = g(f(g^-1(x)) is also an involution! Example: f(x) = C - x, g(x) = e^x where C is an arbitrary constant g^-1(x) = ln(x) h(x) = g(f(g^-1(x)) = e^(C - lnx) = e^C/x = C/x and h(h(x)) = C/(C/x) = x so it does work 😊
@chessematics
Жыл бұрын
Yeah the inverse is g(f⁻¹(g⁻¹(x))) and then you use the fact that f = f⁻¹. Good one.
@rolflangius1119
Жыл бұрын
If a=d, then c and b have to be zero. This basically gives f(x)=x as a solution
@rolflangius1119
Жыл бұрын
Also, when a=d=0, then b and c can be any value (except for c=0), which gives f(x)=D/x and f(x)=0 as answers
@rolflangius1119
Жыл бұрын
All these possible answers are represented by rewriting as f(x)=(Ax+B)/(Cx+A)
@codahighland
Жыл бұрын
@Rolf Langius f(x) = 0 doesn't satisfy the condition that f(f(x)) = x, though, because a constant function isn't invertible.
@alanaduarte_
Жыл бұрын
Watching this during summer it’s very entertaining!!! 😊😊
@sarithasaritha.t.r147
Жыл бұрын
When graphed, it looks like the function y=1/x
@codahighland
Жыл бұрын
My favorite is the identity function.
@armanavagyan1876
Жыл бұрын
Thanks PROF a good one)
@electroquests
Жыл бұрын
Thanks for the explanation!
@topolojack
Жыл бұрын
we love a fractional linear transformation! i kept wondering if you were going to mention PSL(2) or the hyperbolic plane at any point :)
@arsalmathacademy
Жыл бұрын
Very intlectual person. Good informative lecture
@Denis-bu4ri
Жыл бұрын
make more videos, you are very nice to watch
@shaikshahid1512
Жыл бұрын
Bro forgot his grenade
@guilhermerocha2832
Жыл бұрын
Wow this is so cool. Sugestion: do another video on other types of involution functions
@mnmsean
Жыл бұрын
This is a problem from chapter 3 of spivaks calculus book . There are some brutal exercises in that section if you aren’t used to that kind of math.
@davidgillies620
Жыл бұрын
Möbius transformations are one of those things that look simple but have so many ramifications and pop up in loads of different places.
@sebasFS
Жыл бұрын
GREAT!
@ikvangalen6101
Жыл бұрын
For dramatic f(x) 😂
@mathboy8188
Жыл бұрын
The second approach was a nice observation. Another way to see that a = -d is at least a necessary condition for it to be an involution, in the c not 0 case, is to consider the real number line wrapped up into a circle by adding a point at infinity (call it INF) for both the positive and negative infinite open endpoints of the line. Then think of f as being extended to a map of the circle to itself. If f is as an involution, then f sends some point P to INF, and so must send INF to P (to have that INF = f(f(INF)) = f(P) and that P = f(f(P)) = f(INF)). Often will have P = INF, but in this case P is a "normal" point: If f(x) = (ax + b)/(cx + d) with c not 0, then lim{ x -> - infinity } f(x) = lim{ x -> infinity } f(x) = a/c. Thus on this circle, it's correct to say that lim{ x -> INF } f(x) = a/c = P. (Note that this is actually continuous, with a basis for open neighborhoods about the point INF on the circle corresponding to the points { x in R : |x| > N } on the real number line.) Also have lim{ x -> -d/c } f(x) = INF. (Choose -d/c to make the denominator 0, i.e. solution to cx + d = 0). Thus P = a/c AND P = -d/c. Thus a = -d.
@lanceslance2930
Жыл бұрын
Yo since when did bprp have the swagger watch?
@k_wl
Жыл бұрын
i noticed a ring on his finger lol
@lackethh8179
Жыл бұрын
This was so interesting to watch.
@michaelbaum6796
Жыл бұрын
Thanks a lot for this nice video👍
@tonyhaddad1394
Жыл бұрын
The easiest way in this case (multiple choice) if you calculat f(0) = 23 So f^-1(23) = 0 then choice b is the correct answer
@tonyhaddad1394
Жыл бұрын
Or the second from the top 😅
@mairc9228
Жыл бұрын
tecnically speaking in the f(x)=(tx-t^2+k)/(x-t) case you still have a=-d; they just both happen to be t. If you have t=0 you go back to the case where f(x)=k/x and a=d=0; you can do that because you divifed by neither of those variables in the proof.
@MyOneFiftiethOfADollar
Жыл бұрын
Involutory functions, f(f(x)) = x work great with functional equations Consider 3g(f(x)) + 5 = x with the instructions "solve for g(x)" substituting f(x) for x in the above functional equation, We get 3g(x) + 5 = f(x) and we see g(x) = (f(x) - 5)/3 So if f(x)=(23-x)/(1+4x) g(x) = ((23-x)/(1+4x) - 5)/3 = ((23-x) - 5 - 20x)/(3+12x) = (18-21x)/(3+12x) = (6 - 7x)/(1+4x)
@scarletevans4474
11 ай бұрын
12:40 Not exactly they can be "anything", you missed one extra, tiny thing: BC≠−1 Because, we then get that B=−1/C, thus (−x+B)/(Cx+1) =−(x+1/C)/(Cx+1) = −1/C (Cx+1)/(Cx+1) = −1/C and this is just a constant 👍
@youtubeuserdan4017
Жыл бұрын
Bro straight up trolled his students. Respect.
@akf2000
Жыл бұрын
I'm just fixated on the pen switching
@gregebert5544
Жыл бұрын
I swear I would have been a math major instead of an engineer if he was my teacher.
@kevinstreeter6943
Жыл бұрын
I majored in math. You made the right choice.
@gregebert5544
Жыл бұрын
@@kevinstreeter6943 Perhaps. I'm retiring in 2 weeks, having finally burned-out as an engineer for 38 years. It was great for about 20 years, but I allowed myself to get pigeonholed into dull-ish, though job-secure, work at a large company. I blame Wall Street for killing-off many of the great theoretical think-tanks we had in the past that hired PhD's in math, chem, and physics (Bell Labs, IBM, HP, Hughes, TRW, EG&G, and many more) to do cutting-edge R&D, which resulted in many new technologies. People who enjoy, or love, math should be rewarded with high-paying and secure jobs.
@LuigiElettrico
Жыл бұрын
Cool and simple.
@aguyontheinternet8436
Жыл бұрын
That's pre-cal? That looks ez!
@xX_SushiRoll_Xx
Жыл бұрын
I love watching this while high
@marwachayma4694
Жыл бұрын
The best teacher in the world 🌎❤
@MrConverse
Жыл бұрын
8:15, how did you know that something was wrong? Impressive.
@rylanbuck1332
Жыл бұрын
interesting enough, it also works with irrational numbers as well!
@diegoalejandroordonezcastr5963
Жыл бұрын
You can solve the integral of x/tan(x) using the polilogarithm please😅
@sparky2141
9 ай бұрын
Just My way of approaching, solely to quickly solve questions like this should the situation arise, Inputting 23 in the original function, gives 0 So inputting 0 in the Inverse Should give 23 That by a glance eliminates option A and D And notice how -1/4 is not in the domain of the original function, due to zero in the denominator Hence it should not be in the domain of the inverse function too, And that by quick inspection gives the right answer which is the function itself Although I do know that solving this question was not the intention of this video, I thought I would share what I did after seeing the thumbnail
@PunmasterSTP
Жыл бұрын
Involution function? More like "Interesting information; thanks a ton!"
@tomctutor
Жыл бұрын
I postulate: There is no proper real Rational Polynomial f(x) = P(x)/Q(x) such that f⁻¹(x)= [f(x)]⁻¹= 1/f(x). by proper real I mean non-trivial e.g. f(x) ≠ constant and x∊ℝ PS> I know that w = (1+iz)/(z + i) in complex field has such property, If someone posts a counter example then I will eat my hat! ( I have done similar algebraic manipulation as BPRP's and came up with contradictions in all possible cases).🧐
@pwmiles56
Жыл бұрын
It's fairly easy to prove if P and Q are linear or constant. Suppose f(x) = (ax + b)/(cx +d) Put x = x1/x2, f(x) = f1/f2 (so-called homogeneous coordinates). The function is then the linear operation [f1] = [a b] [x1] [f2] [c d] [x2] The proposition amounts to [ d -b] = k [c d] [-c a] [a b] with k an unknown constant. With a bit of work you get k^4 = 1 The real roots, k=1 and k=-1, both result in singular matrices, such that f(x)= -1=constant, so it doesn't have an inverse. The imaginary roots , k=i and k= -i, give results like the one you show :-))
@lychenus
Жыл бұрын
when going through question, more like i want to see how student struggle and what they are failing
@krish-502
Жыл бұрын
Can you find the conjugate of the quantity (1+i)^(1+i)? Thank you!
@tomctutor
Жыл бұрын
Using DeMoivre, know 1+i ≡ √2 e^(iπ/4) simply raise this quantity to (i+i) to get = √2 ((√2 )^i) e^(iπ/4) e^(-π/4) noting that (√2 )^i ≡ e^i(½ln2) = cos(½ln2)+ i sin(½ln2) giving = e^(-π/4) [cos(½ln2)+ i sin(½ln2)] (1+i ) polar form (r, θ): r =√2 e^(-π/4) , θ = 1.13197 conjugate is therefore r =√2 e^(-π/4) , θ = - 1.13197 😲
@anupamamehra6068
Жыл бұрын
Hi, blackpenredpen! This is Shiv, and I have a challenge for you - find the general answer/expression for (n/2) factorial or (n/2) ! where n is an odd positive integer. All the best!
@apleb7605
Жыл бұрын
But if you close your eyes…. Does it almost feel like nothing changed at all?
@creativename.
11 ай бұрын
Gotta be a right (before watching video)
@romanbykov5922
Жыл бұрын
7:12 Shouldn't it be "bd" rather than "bx"?
@G_4J
Жыл бұрын
he changes it at 8:11
@thewhat2
Жыл бұрын
2:09 "HIV" in captions 💀
@FIN2827
Жыл бұрын
What is the intégral from 0 to pi÷2 of: (sinx)(cosx)÷[(tanx)^2+(cotanx)^2
@tomctutor
Жыл бұрын
Go Wolfram my friend: integrate (sinx cosx)/((tanx)^2+(cotx)^2) dx from x=0 to x=pi/2 gives ⅛ (π-2) ~ 0.14270 🙄
@wolfiegames1572
Жыл бұрын
This is EPIC
@neutronenstern.
Жыл бұрын
looks like möbius transform
@Ghi102
Жыл бұрын
I'm trying to see how we can get the first function from the g(x) = tx - (t^2) + k / x - t. If we divide everything by -t (and add t != 0), we can get g(x) = -x + t + k / (x/t) + 1. Since both t and k are constant, we can replace them with another constant and say B = t + k. We can then say C = (1/t) since t is constant and we can get g(x) = -x + B / Cx + 1. Could anybody confirm if I skipped a step or did anything wrong?
@kostantinos2297
Жыл бұрын
The reasoning is correct, just forgot to divide k by -t. You get: g(x) = (tx - t^2 +k)/(x-t) = (-x + t - k/t)/(1 - x/t) = (-x+B)/(Cx+1) Therefore C = -1/t, B = t - k/t
@ur2moon
Жыл бұрын
Sir how can I remeber the formulas I've learnt for longer time or rather say for my whole life ?
@Sanatan_saarthi_1729
Жыл бұрын
One second approach:- f(0)=23 so f(23)=0 check it's option c in thumbnail 😂😂
@tiffaz84
Жыл бұрын
Sir, at 7:14 there is an error in the expansion. Should be bd not bx
@G_4J
Жыл бұрын
he changes it at 8:11
@oggermcduckling3274
Жыл бұрын
Not relevant to the video but I've confused myself. Sqrt(x)=-5 has no solution.. but.. What if x=i²i²5²=25. Then sqrt(x)=sqrt(i²i²5²)=ii5=i²5=-5 ?? Can someone help explain what is wrong with this?
@pwmiles56
Жыл бұрын
-5 actually is a square root of 25. (-5)(-5)=25. Insisting on the positive square root is just a convention, you often have to look out for the negative one
@mayankshekhar9631
Жыл бұрын
Aah mobius transfomrations
@mostafamxs8554
Жыл бұрын
خويه انت شدسوي بينة
@YTBRSosyalEmre
Жыл бұрын
only real people know that the old title was " when the answer is same as the question..."
@shadmanhasan4205
Жыл бұрын
A = correct answer
@pranavgaikwad437
Жыл бұрын
Could you suggest books for calculus beginners?
@gamingzo888
Жыл бұрын
G Tewani
@sguptzz
Жыл бұрын
@@gamingzo888 bruh everyone is not in India like you also g tewani is spoon feeding book
@sguptzz
Жыл бұрын
thomas calculus
@andersonseecharan2447
11 ай бұрын
Is the answer A?
@andersonseecharan2447
11 ай бұрын
Its not a
@armanavagyan1876
Жыл бұрын
PROF i think better 7 hour is UR style better)
@vaibhavsrivastva1253
9 ай бұрын
2:08 "Deja vu" has been mistyped as "HIV".
@Flemenjo
Жыл бұрын
Hi bprp can you solve this: ((x^2-4x+4)^2)/(|x-2|)=0 I hope you can see this comment
@guy_with_infinite_power
Жыл бұрын
Fact : You'll find the fact when you find out the inverse of function given by: f(x) = (4x+3)/(6x-4)
@MaximusAurelius1987
Жыл бұрын
The V mudra is too generic. Show us triad claw.
@j.o.k.e7864
Жыл бұрын
1st
@Sphinxycatto
Жыл бұрын
Heh Too late bro
@j.o.k.e7864
Жыл бұрын
@@Sphinxycatto check again 😏
@Sphinxycatto
Жыл бұрын
@@j.o.k.e7864 there is about 1 min diff
@j.o.k.e7864
Жыл бұрын
@@Sphinxycatto Thanks for the confirmation that I'm the first 😏
@Sphinxycatto
Жыл бұрын
@@j.o.k.e7864 ay it's ok 👍 Atleast I was smart
@navamgarg
Жыл бұрын
Why your long beard sometime appears and then again disappears? 😑 Yes, I am serious! Please ANSWER....My curiosity is rising obove my head.
@adityaagarwal636
Жыл бұрын
Third.
@donwald3436
Жыл бұрын
Nice sloppy notation, you wrote y = x lol.
@xXJ4FARGAMERXx
Жыл бұрын
I prefer replacing x with f⁻¹(x), and then solving from there. f(x) = ax + b f(f⁻¹(x)) = af⁻¹(x) + b x = af⁻¹(x) + b x - b = af⁻¹(x) (x - b)/a = f⁻¹(x) f⁻¹(x) = (x - b)/a
@prxject1
Жыл бұрын
@@xXJ4FARGAMERXx May Allah SWT reward you akhi 🤲🏽♥️
@jumpman8282
Жыл бұрын
When deriving the inverse to 𝑓(𝑥) = (𝑎𝑥 + 𝑏) ∕ (𝑐𝑥 + 𝑑), I wonder why you didn't do it the same way that you found the inverse to (23 − 𝑥) ∕ (1 + 4𝑥), because doing so you would quickly arrive at 𝑓⁻¹(𝑥) = (−𝑑𝑥 + 𝑏)/(𝑐𝑥 − 𝑎) from where it is obvious that 𝑎 = −𝑑 ⇒ 𝑓(𝑥) = 𝑓⁻¹(𝑥), regardless of what values we choose for 𝑏 and 𝑐 (including 0). If we want to be thorough we can then set (𝑎𝑥 + 𝑏) ∕ (𝑐𝑥 + 𝑑) = (−𝑑𝑥 + 𝑏)/(𝑐𝑥 − 𝑎), which gives us the quadratic equation (𝑎 + 𝑑)𝑐𝑥² + (𝑎 + 𝑑)(𝑑 − 𝑎)𝑥 − (𝑎 + 𝑑)𝑏 = 0. Since we have already covered the case 𝑎 = −𝑑, we can divide both sides by (𝑎 + 𝑑) to get 𝑐𝑥² + (𝑑 − 𝑎)𝑥 − 𝑏 = 0, which tells us 𝑏 = 𝑐 = 0 and 𝑎 = 𝑑, i.e., 𝑓(𝑥) = 𝑥.
Пікірлер: 149