@@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
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.
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.
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
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 😊
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.
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)
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.
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 👍
@@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.
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).🧐
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 :-))
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!
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
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?
-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
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?
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
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
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.
Watch this next: the last question on my calc 2 final: th-cam.com/video/PVD9hfVz6e8/w-d-xo.html
You can solve the integral of x/tan(x) using the polilogarithm please😅
I really enjoy when a calculus exercise is given a geometric context that makes the answer obvious. This was elegant, fun to watch.
How fun!! I love fixed points! 😁
ok
What is a fixed point?
@@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
@@pwmiles56 ah, thanks/
nice
The c = 0 case gives you the linear involutions: f(x) = B − x and (when also d = 0) f(x) = x.
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.
Great vid! I've not heard of the Involution function before. It's quite fascinating and will have to do some reading. Thanks BPRP!
Glad you enjoyed it!
I love watching you teach!!!
easily one of my most favourite channels ever! this guy explains so well
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.
h2 maths?
I remember watching him while I was in 8th grade in summer holidays
That's sad bro go outside
@@BHNOW100 what 💀☠️☠️
@@BHNOW100 after pondering for 120 seconds i understood the joke 🤣🤣
@@Sphinxycatto and now ur in which grade? Indian ho kya?
@@extreme4180 name me hi toh hai
Me 11th me hu
I can already tell this man's personality is immaculate by watching his videos
Watching this during summer it’s very entertaining!!! 😊😊
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
Bro straight up trolled his students. Respect.
My favorite is the identity function.
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 😊
Yeah the inverse is g(f⁻¹(g⁻¹(x))) and then you use the fact that f = f⁻¹. Good one.
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.
Wow this is so cool. Sugestion: do another video on other types of involution functions
Very intlectual person. Good informative lecture
Thanks for the explanation!
If a=d, then c and b have to be zero. This basically gives f(x)=x as a solution
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
All these possible answers are represented by rewriting as f(x)=(Ax+B)/(Cx+A)
@Rolf Langius f(x) = 0 doesn't satisfy the condition that f(f(x)) = x, though, because a constant function isn't invertible.
Möbius transformations are one of those things that look simple but have so many ramifications and pop up in loads of different places.
Yo since when did bprp have the swagger watch?
we love a fractional linear transformation! i kept wondering if you were going to mention PSL(2) or the hyperbolic plane at any point :)
When graphed, it looks like the function y=1/x
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)
Bro forgot his grenade
I'm just fixated on the pen switching
make more videos, you are very nice to watch
Thanks PROF a good one)
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
Or the second from the top 😅
This was so interesting to watch.
Thanks a lot for this nice video👍
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.
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.
The best teacher in the world 🌎❤
For dramatic f(x) 😂
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 👍
I swear I would have been a math major instead of an engineer if he was my teacher.
I majored in math. You made the right choice.
@@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.
i noticed a ring on his finger lol
8:15, how did you know that something was wrong? Impressive.
Can you find the conjugate of the quantity (1+i)^(1+i)? Thank you!
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
😲
You can solve the integral of x/tan(x) using the polilogarithm please😅
GREAT!
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).🧐
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 :-))
interesting enough, it also works with irrational numbers as well!
Cool and simple.
I love watching this while high
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!
7:12 Shouldn't it be "bd" rather than "bx"?
he changes it at 8:11
That's pre-cal? That looks ez!
Involution function? More like "Interesting information; thanks a ton!"
when going through question, more like i want to see how student struggle and what they are failing
Sir, at 7:14 there is an error in the expansion. Should be bd not bx
he changes it at 8:11
But if you close your eyes….
Does it almost feel like nothing changed at all?
What is the intégral from 0 to pi÷2 of:
(sinx)(cosx)÷[(tanx)^2+(cotanx)^2
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 🙄
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
Sir how can I remeber the formulas I've learnt for longer time or rather say for my whole life ?
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?
-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
only real people know that the old title was " when the answer is same as the question..."
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?
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
One second approach:- f(0)=23 so f(23)=0 check it's option c in thumbnail 😂😂
Gotta be a right (before watching video)
Could you suggest books for calculus beginners?
G Tewani
@@gamingzo888 bruh everyone is not in India like you also g tewani is spoon feeding book
thomas calculus
2:09 "HIV" in captions 💀
This is EPIC
Is the answer A?
Its not a
A = correct answer
looks like möbius transform
PROF i think better 7 hour is UR style better)
خويه انت شدسوي بينة
2:08
"Deja vu" has been mistyped as "HIV".
Hi bprp can you solve this:
((x^2-4x+4)^2)/(|x-2|)=0
I hope you can see this comment
Aah mobius transfomrations
The V mudra is too generic. Show us triad claw.
Fact : You'll find the fact when you find out the inverse of function given by:
f(x) = (4x+3)/(6x-4)
Why your long beard sometime appears and then again disappears? 😑 Yes, I am serious! Please ANSWER....My curiosity is rising obove my head.
1st
Heh
Too late bro
@@Sphinxycatto check again 😏
@@j.o.k.e7864 there is about 1 min diff
@@Sphinxycatto Thanks for the confirmation that I'm the first 😏
@@j.o.k.e7864 ay it's ok 👍
Atleast I was smart
Third.
Nice sloppy notation, you wrote y = x lol.
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
@@xXJ4FARGAMERXx May Allah SWT reward you akhi 🤲🏽♥️
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.
This was so interesting to watch.