A Cool Functional Equation

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#ChallengingMathProblems #FunctionalEquations
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[Wurfenkopf]

Actually the video required one more step to be completed, because a priori you don't know if a solution exists.
We know that, IF a solution exists, then it must coincide with that formula, but to state that it IS a solution we have to substitute it in the first equation and see that there is no contradiction

[adandap]

Nice. I wrote the four equations in matrix form and then used row operations on the coefficient matrix ((1, 3, 1, 0), (3, -1, 0, 1), (1, 0, 1, 3), (0, 1, 3, -1)) to get f(x).

[baptiste]

actually, you can avoid this complex system of equations if instead you remove little by little the terms which are not f(x) using substitution. I did this and end up with the same answer in a way I personnaly find less complex.

[MathElite]

Nice functional equation, i need to get better at finding more solvable ones lol

[musicmakelightning]

I wish you had been my math teacher. I've been out of school for a long long while now (probably since before you were born), but I thoroughly enjoy watching you solve equations and I often tune into your channel when the events of the day are getting me down and I need something positive. Thanks for your work.

[sunnysidechrome]

This particular equation can be solved by guessing the functional form of the solution. The presence of the x on the RHS suggests a term proportional to x and the presence of f(1/x) suggests the inclusion of a term proportional to 1/x. Writing f(x) = ax + b/x, we see that f(1/x) = a/x + bx, that we can eliminate 1/x on the LHS. Simultaneous equations in a and b result from comparing cofficients and these yeild the given answer.

[arbenozturk]

Actually, if we write this equation for x=1 and x=-1 we clearly can see f(x) is an odd function and thus f(-x)=-f(x). Then we can arange the equation like " f(1/x) -2f(x) = x ". Also we can use " f(x) -2f(1/x) = 1/x ". In this way we can find f(x) in a less compliated way. Once we realize that f(x) is an odd function the question gets really easier.

[dodokgp]

I think that a lot of steps could have been avoided because our goal is just to solve for "c" which is equal to f(x). That should have been the focus after obtaining the set of 4 equations.

[MathNotationsVids]

Thank you for posting excellent videos for students of all ages!

[roberttelarket]

It's ingenious and beautiful! Further it's wonderful that f(x) isn't equal to the usual identity function for these types of problems.

[joaquingutierrez]

I have not seen the video yet. Here's my approach. Change x to -x, 1/x and -1/x. We then have a system of 4 equations. Let f(x) = u1, f(-x) = u2, f(1/x) = u3 and f(-1/x) = u4. We just need to solve the system.

I think that the best way to solve it is finding the inverse of the system matrix. Maybe Cramer's rule is easier.

[NoNameFound--years-ago]

This is my first time to see an equation with functions.
Could you elaborate more on its usage in real life applications.?

[ethanwinters]

Really liked this problem! : ) I got to the same system of equations as you, but I’m not so adept at manipulating these sorts of big systems efficiently and I **definitely** didn’t want to invert that 4x4 matrix by hand, lol. But a nice work-around I found was noticing that if I did do the inversion, I’d find that f(x) was a linear combination of x, -x, 1/x, and -1/x, which would simplify to a linear combination of just x and 1/x. Thus I could sub in f(x) = px + q/x into the original equation and solve for p and q much more easily.

Thanks for the brain-tickler, it was pretty satisfying to solve!

[MushookieMan]

I wrote three equations in four unknowns. Then, I was able to eliminate two unknowns. Then I performed the same substitution again, x->1/x, giving me two equations in two unknowns. And solved it with Cramer's rule. It seems as though making the same substitution multiple times doesn't cause a linearly dependent system.

[WahranRai]

pre-analysis: f (x) is an odd function (LHS combinaison of f(x) is equal to RHS x (odd function)

[chaosredefined]

Given:
3a + 1b + 1c + 0d = x (1)
1a + 0b + 3c + 1d = -x (2)
0a + 1b + 1c + 3d = 1/x (3)
1a + 3b + 0c + 1d = -1/x (4)

We want to find c:
(1) - 3*(4) => -8b + 1c -3d = x + 3/x (5)
(2) - (4) => -3b + 3c = 1/x - x (6)

At this point, we have three equations without "a": (3, 5, 6). Next, as (6) doesn't contain "d", we should aim to eliminate that:

(5) + (3) => -7b + 2c = x + 4/x (7)

Now, we can cancel the "b"s in (6) and (7)
7*(6) - 3*(7) => 15c = -5/x - 10x

Simplifying this, we get:
c = (-2x^2 - 1)/(3x)

We defined c as f(x), so f(x) = (-2x^2 - 1)/(3x)

[DynestiGTI]

That degree 4 simultaneous equations was definitely a degree 3 in disguise.

[siyumitcgonlnkn]

In fact, one only needs x->-x, and x->1/x to figure out the answer.

[mcwulf]

Very good.
And so good to see a functional equation that's not a constant or linear.

[michaelyap]

Mistake in 8:08, the right hand side x -1/x cannot be 0