A Rational Functional Equation Solved in Two Ways

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#ChallengingMathProblems #FunctionalEquations

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[antonyqueen]

Your 2nd method is using 1st method. Instead of calculating each time separately x to have the value of 1 and -1, use the result of the first method and replace z by 1 and -1 to calculate x.
In nutshell: both methods have in common the step of calculating x as function of z. After that:
1st method consists on finding f(z), and
The second method consists on calculating x for z=*/-1

[loohooi]

When x=1, ax-b/bx-a= -1.
When x=-1, ax-b/bx-a=1.So we can just substitute these values in, and get the value of f(-1)+f(1).

[thatdude_]

Since the equality holds for all a and b, just choose a=1 and b=0. then you get f(-x)=x^2-3x+1, hence f(x)=x^2+3x+1. From this f(x)+f(-x)=2(x^2+1) so the solution is 4.

[Anmol_Sinha]

Just substitute x =1 then substitute x= -1, add them and you're done.

[stevenlitvintchouk]

I did it the same way as your first method, except I used r (for "ratio") instead of z. 🙂 I'm glad you're showing us some easier-type problems that I can actually solve.

[kepler]

Problems like these are so interesting

[shrivardhank.]

I can see some guys commenting about how easy the problem is, I think they should know that if something is easy for you, It is not necessary that Its also easy for others😅

[davidseed]

whats interesting about this problem is not the solution, (which we can do by guesswork evaluating at x=+-1), but the fact that this is a family of functional equations.
if a=0, then f(z)=1/z² +3/z+1
otherwise f(z) = ((z-c)/(cz-1))² -3(z-c)/(cz-1) +1 where c=b/a and cz/=1, x= (z-c)/(cz-1)
so for example c=1, then x=1, , gives f(z)= 1-3+1=-1
c=0 gives x=-z, f(z)=z² +3z +1
c=-1 gives x=-1 f(z)= 1+3+1=5

[cicik]

there is very short way;: f(1) will be if (ax-b)/(bx-a) = 1 => ax-b = bx-a => x = -1
the same. f(-1) will be if (ax-b)/(bx-a) = -1 or x = 1
substitute now: f(1) = (-1)²+3+1 ; f(-1) = 1²-3+1 ; f(1) + f(-1) = 4

[satrajitghosh]

At ax - b = bx - a,
value of x is x = (b-a)/(a-b) = -1
Again at ax - b = -bx +a,
value of x is x = (a+b)/(a+b) = 1
So f(1) + f(-1)
= (1)^2 -3*1 + 1
+ (-1)^2 +3*1 + 1
= 4

[Zarunias]

The problem as stated makes no sense. Let a=b=x=2 and we get: f(1)=-1. Let a=b=x=3 and we get: f(1)=1. As a function cannot have to different values for the same argument, the definition of this function is not valid.
The thing is that it was never stated that a must not equal b, but this is a crucial statement to get a reasonable problem. After looking at your solution it was clear that this detail was missing in the wording, but before that I was pretty confused.

[thierryvieville]

Merci beaucoup pour cette nouvelle très intéressante vidéo :D

Just a small digression regarding z = (a x - b) / (b x - a)

Suprizing that z on x = +- 1 does neither depends on a nor in b, does'nt it ?

In fact the curve only depends on c = a / b not on a and b independently, i.e. z = (c x - 1) / (x - c)

And it is its own inverse function, as any homographic function (i.e., z = (a x - b) / (c x - d))

It is in fact the usual basic hyperbola v = 1 / u.
up to a translation of length c in x and a zoom of ratio (c^2 - 1) followed by translation of length c in z
as visible writing u = (x - c) and z = c + (c^2 - 1) v

The clever point is the choice of the zoom so that all curves intersects in (-1, 1) and (1, -1)
- The reciprocial is true, if an homographic function includes (-1, 1) and (1, -1) it must be of the form (c x - 1) / (x - c) as above (easy to prove: juste set the equations and solve)

What ? Yes ! This is only a few blabla around this @SybermMath nice problem :)
... and my purpose is simply to show that these are always very interesting problems, often beyond juste finding the solution !!!

And here https:// tinyl.io/5v7E is the drawing of a few curves

[writerightmathnation]

When writing, sometimes the closed caption covers what you're writing. You may want to watch out for this in future videos.

[佐藤広-cp]

I think the 2nd method is correct, but if you look at the expression on the left side and try to put 1 or -1 in x, the left side becomes f(-1) or f(1), and the solution came out, I feel it is mathematically incorrect.

[RAG]

I disagree with your judgement here. In method 1 you got a general formula for x in terms of z. All you had to do was find x for each z, i.e -1 for 1 and 1 for -1, and substitute these in for the answer. Method 2 was actually longer because you solved the formula for z for each value separately. You just did it quicker because you sped up. Method 1 was clearly the better one.

[isacnordin]

I feel some assumptions are left out and the question is weird. Here are two assumptions and their conclusions
my own writing:
* f(a, b, x)= f( (ax-b) / (bx-a) )
* g(x)=x^2-3x+1
a, b Real == same as me saying a&b can take any value (given not division by zero)

1)
Assumption: a, b, x Real
->f(a, b, 0)=f(-b/-a)=f(c)=g(0)=1, c=b/a, c is Real since a, b Real. --> f(z)=1 for all z
Error1: f(1)+f(-1)=2, not =4 as said in video
Error2: f(a, a, x) = f( (ax-a)/(ax-a) ) = f(1) = g(x), for all x . but g(0) not same as ex: g(1)

2)
Assumption: a, b are unknown fixed values, x is Real
->f(a, b, 1) = f( (a-b)/(b-a) ) = f(-1) = g(1) = -1 ---> f(-1)=-1
->f(a, b, -1)= f( (-a-b)/(-b-a) ) = f(1) = g(-1) = 5 --> f(1)=5
f(k)=f( (ax-b)/(bx-a) )--> k=(ax-b)/(bx-a) --> x(k) = (ka-b)/(kb-a)-->
f(a, b, x(k)) = f(k) = g( x(k) ) (x=a/b not allowed, division by zero else allowed)

->can't find a contradiction, and see f(1)+f(-1)=4 as said in video
furthermore f(k)= g(x(k)) = [ka-b)/(kb-a)]^2 - 3[(ka-b)/(kb-a)] + 1 same as in video
for all values of k (except for when ka-b = 0 which will depend on the unknown variables a, b (if they are not 0))

[Handelsbilanzdefizit]

Every day a video.
This guy is more productive than euler was.

[sebastianviacava]

i understand second method much better than first please someone can tell me why b minus a.z is equal to a.z minus b and a minus b.z is equal to do b.z minus a? thanks a lot

[tomasadaogoncalves]

Please, tell me the name of this programme.

[zexlagamer]

Bastante fácil ..pero para los nuevos es muy útil...