Solving a Functional Equation by Elimination

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PLAYLISTS 🎵 :

[michaelempeigne]

2x^2 - 3x + 1 and x^2 - 2x + 1 have a common factor

[sohelzibara]

Thks for the vid. The sought function is in fact a piecewise defined function which is equal to 1/2 for x = 1 and equal to (2x – 1)/(x – 1) for x # 1.

[gianlucatramutola]

The solution of this formula is ok, but it is not defined in x=1. However, if we substitute the value 1 in the original formula, we have f(1) = 1/2. Therefore, in my opinion, the solution must be completely defined by adding {f(x) = 1/2 if x=1}

[theuserings]

You can simplify the result more
2x^2 - 3x + 1 = (x - 1)(2x - 1)
and
x^2 - 2x + 1 = (x - 1)(x - 1)

cancel the x-1 terms and we have
f(x) = (2x - 1)/(x - 1)

[jmart]

Nice problem. The result can be simplified to (2x-1)/(x-1)

[gianfrancocazzaro]

Hello everyone, maybe the denominator could be written in a simpler way as (1-x)(x-1) so that Thanks SyberMath for your interesting work!

[nirajkumarverma]

Elegantly explained a beautiful functional problem...I really enjoyed it.👍

[felaxwindwalker]

I took a slightly different approach. Like you, I substituted x/(2x-1) into the original equation to get a second equation, but then I subtracted the second equation for the first and got f(x/(2x-1)) in terms of f(x). Substituting that into the original equation and solving for f(x), I got that f(x)=(2x-1)/(x-1), which is the simplified version of what you got. I did notice in there that the domain had shifted from being undefined at x=1/2 to being undefined at x=1. Is this related to the fact that f(1) by the original equation is equal to 1/2, while f(1/2) by the solution is equal to 0 and f(0) by either equation is equal to 1?

[adelsmia]

we can semplify f(x) and it can be 2x-1/x-1

[pwmiles]

A clue is that in t(x) = x/(2x-1) the coefficient of x in the numerator and the constant in the denominator sum to 0. If this happens the function is involutive i.e. t(t(x)) = x

[SuperYoonHo]

Please do a homemade video! Those are kinda interesting and very nice to watch:)

[misterdubity]

If you test x=1 in the equation you get f(1) + 1*f(1) = 1, or f(1) = 1/2. But f(x) is (2x-1)/(x-1), so f(1) is undefined. Why did this happen?

[angelmendez-rivera]

The answer given in the video is incorrect. In actuality, the correct answer is that f is a function R\{1/2} —> R such that f(x) = (2·x – 1)/(x – 1) everywhere EXCEPT at 1, while f(1) = 1/2. The answer in the video is false, because it treats the domain of f as R\{1}, and it completely ignores f(1) = 1/2, while treating 1/2 as part of the domain.

Where did the video go wrong? At 7:08 the video achieves the critical equation of this problem, the equation f(x)·(2·x – – x^2)/(2·x – 1) = 1 – x, which holds everywhere (meaning for all x in the domain). However, the video proceeded to divide both sides of the equation by (2·x – 1 – x^2)/(2·x – 1), and this step, as is, is invalid. Why? Because you are not taking into account the division by 0 that can occur when you do this. To clarify this, we should first rewrite the equation. Notice that 2·x – 1 – x^2 = –(x – 1)^2, so the equation can just be rewritten as f(x)·(x – 1)^2/(2·x – 1) = x – 1. You cannot just divide both sides by (x – 1)^2, unless you look out for the division by 0 that happens. Notice that if x = 1, then f(1)·(1 – 1)/(2·1 – 1) = 1 – 1 = 0 = f(1)·0 = 0. Here, f(1) = 1/2, and you can compute this directly from the original functional equation: f(1) + 1·f(1/(2·1 – 1)) = f(1) + f(1) = 2·f(1) = 1, so f(1) = 1/2. So, f(x)·(x – 1)^2/(2·x – 1) = x – 1 implies not f(x)·(x – 1)/(2·x – 1) = 1 everywhere, but it implies f(x)·(x – 1)/(2·x – 1) = 1 everywhere EXCEPT at 1. Once you this is noted, _now_ you can divide, since division by 0 has been removed from the equation. As such, f(x) = (2·x – 1)/(x – 1) everywhere except at 1, and f(1) = 1/2. This solves the functional equation properly.

[mathaca]

How do we take f(1)=1/2? When x is not 0, f(x)=(2-1)/(x-1) is solved.

[frankchen]

According to the original equation, x=1/2 should be excluded from the domain.

[caesar_cipher]

also needs to be stated : f(x) is not defined at x=1 and x=1/2.
This is a great channel but for first time I give dislike because of obvious errors