@@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.
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 😊
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
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 :-))
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)
@@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.
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 👍
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
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
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!
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.
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.
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.
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
