A Cool Functional Equation
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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
Good point!
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).
Nice!
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.
Can you tell me please how you did to solve it
So clever
@@chaimaafdi
this is how i did it:
original equation:
3f(-x)+f(1/x)+f(x)=x -- (i)
replacing x with -x, we get:
3f(x)+f(-1/x)+f(-x)=-x -- (ii)
Adding these 2 equations together, we get:
4f(x)+4f(-x)+f(1/x)+f(-1/x)=0
i.e., [f(1/x)+f(-1/x)]= -4[f(x)+f(-x)] -- (iii)
put x=1/x in (iii), and we get:
[f(x)+f(-x)]=-4[f(1/x)+f(-1/x)] -- (iv)
Using (iii) in (iv), we get:
[f(x)+f(-x)]=-4[-4 {f(x)+f(-x)}]=16[f(x)+f(-x)] -- (v)
i.e, -15[f(x)+f(-x)]=0 => f(x)+f(-x)=0
i.e., f(x)=-f(x) => f(x) is an odd function.
Thus, equation (i) now becomes:
-3f(x)+f(1/x)+f(x)=x -- (vi)
i.e., -2f(x)+f(1/x)=x -- (vii)
in this equation, we replace x with 1/x, and we get:
-2f(1/x)+f(x)=1/x -- (viii)
multiplying (vii) by 2 and adding to (viii), we get:
-3f(x)=2x+1/x
or f(x)=-(2x+1/x)/3=-(2x²+1)/3x
Nice functional equation, i need to get better at finding more solvable ones lol
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.
Good thinking! My audience is smarter than me! I’m proud of you all! ☺️
@@SyberMath ypu don't kbow that's true though for all x you only know it's true when x equals 1 ..see what i mean..can't you solve finctional equations witbout substitution? I HATE THEM nonone thinks that wayngove me a break.why wpuld anyine think like that
x=1 and x=-1, -f(1)=f(-1) is right. But it can't guarantee that f(x) is odd function for all x(except 0) by the time we solve the equation or figure out the feature of f(x) through other methods. f(x) = x*x-1, -f(1)=f(-1)=0, is f(x) odd function?
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.
It could have been! I have a tendency to complicate things, sometimes! 😄😁😂
@@SyberMath that's weird indeed
@@SyberMath Was there a specific reason you didn't use matrices and Gaussian elimination? Except that tendency of course.
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.
Np. Thank you for the kind words! 🥰
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.
In fact, one only needs x->-x, and x->1/x to figure out the answer.
I agree, it's the easy way to solve the aquation. We have f(1/x) = x + 2. f(x) and f(x) = 1/x + 2.f(1/x) this system leads to f(x) = -(1+x^2)/3.x
Easy way to do
Using x->-x and x->1/x->-1/x we have
f(-x) = -f(x) input that in functional equation leads to
f(1/x) = x + 2. f(x) and f(x) = 1/x + 2.f(1/x) which f(x) = -(1+x^2)/3.x
🤔😉
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.
It may appear easier to use Gauss method instead of calculating 2 determinants
Mistake in 8:08, the right hand side x -1/x cannot be 0
The error was acknowledged and fixed 2 steps later.
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)
Thank you for posting excellent videos for students of all ages!
Np. Thank you for watching! 😊
Clever. You've turned it into a linear algebra problem essentially.
Thank you!
11:10 Well I can't argue with that.
The generalization of a functional equation of this form is that the variable is used to solve the system of equations...
If you start with assumption that f(x) is _Odd_ [ i.e. f(-x)=-f(x)] you immediately get the equation:
-2f(x)+f(1/x)=x _ (1)
eliminate f(1/x)=2f(x)+x
which means, that here on subst. x→1/x
f(x)=2f(1/x)+(1/x)
⇒f((1/x) = ½[f(x)-1/x] _ (2)
replace f(1/x) in eq(1) with that in eq(2)
and presto, you have your f(x)= -⅓[2x^2+1]/x.
_Must admit f(x) could be even. or neither, haven't tested these hypothesis; but have found a solution that works, so that's a nice place to stop_ ▖シ
pre-analysis: f (x) is an odd function (LHS combinaison of f(x) is equal to RHS x (odd function)
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.
That degree 4 simultaneous equations was definitely a degree 3 in disguise.
We could have found the four unknowns and therefore directly the expression of f(x) by solving the system of four equations with four unknowns using the determinant method. What do you think ?
You should also ensure the solution is unique, by solving the homogeneous system. The reasoning going if another solution f_1 exists then f-f_1 (x) should satisfy the homogeneous system.
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.
Thank you very much!
@@SyberMath: You're welcome. Thanks you for all these great problems. Keep them coming!
I solved equations using Cramer's Rule and it was much easier.
What's that?
Using determinant
This is my first time to see an equation with functions.
Could you elaborate more on its usage in real life applications.?
Bring more functional equations! :D
Or you could guess that f(x) must have the form f(x) =Ax + B/x, put this into the expression on the left, set the coefficient of x to 1 and the coefficient of 1/x to 0 to solve for A and B.
A second way is to make the sub x -> 1/x and get
3f(-1/x) + f(x) + f(1/x) = 1/x.
Since the 2nd and 3rd terms in this equation also occur in the original equation, subtracting this eq. from that one gives
3f(-x) - 3f(-1/x) = x - 1/x, or
f(-x) - f(-1/x) = (x - 1/x)/3.
Then with x -> -x this becomes
f(x) - f(1/x) = -(x - 1/x)/3.
From the first eq. we have
f(x) + f(1/x) = x - 3f(-x).
Add these to get rid of the f(1/x), combine terms on the right and divide by 2 to get
f(x)= x/3 + 1/(6x) - 3f(-x)/2.
Now substitute x -> -x in this equation to get
f(-x) = -x/3 - 1/(6x) - 3f(x)/2.
Substituting this expression for f(-x) into the equation above it, you can solve for f(x): f(x) = -2x/3 - 1/(3x)
I proved that f is odd then I found f more easily. Nice exercise !
Thanks. Nice work
thank you so much this is the first time I understand.
Glad it helped!
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!
I think your way is by far easier than the whole calculation presented here !!!!
You’re welcome. Glad to hear that!
@@fred11298 quite possible 😁
Circle method strikes again ;-)
Nice video!
Thanks 😅
I think I used fewer steps.
Replace x -> -x and add the result to the original equation, to get 3g(x) + g(1/x) + g(x) = 0, where g(x) = f(x) + f(-x)
Then g(1/x) = -4g(x) = 16g(1/x), meaning g(x) = 0, so f(-x) = -f(x)
Now the original reduces to:
f(1/x) = x + 2f(x), so
f(x) = 1/x + 2(x + 2f(x))
And the rest follows.
Easy way to do
Using x->-x and x->1/x->-1/x we have
f(-x) = -f(x) input that in functional equation leads to
f(1/x) = x + 2. f(x) and f(x) = 1/x + 2.f(1/x) which f(x) = -(1+x^2)/3.x
🤔😉
Very good.
And so good to see a functional equation that's not a constant or linear.
Glad it was helpful!
It's too long and it has so many steps and details , however that's a nice work that shows your ability in solving the hardest mathematical equations👌
Thank you very much!
Very nice explanation !! Would be interesting if you made some problems of variation's calculus
very nice thanks for these joyful moments! sybermath!
You could take 4*eq1-eq2-eq3-eq4
It's a nice problem, the approach with system of equations is cool
So many steps to solve this, but I interested with that.
yess
We need more fonctional equation
At 8.23 mark how can x-1/x be zero? Am I missing something?
I fix it at 8:37
Very nice problem where algebra is mixed with functions. Really a good problem and solution is also very elegant.
💖
First, take f as even function -> f(-x) = f(x); On solving we find f(x) comes out to be odd => Contradiction
Then, when you take f as an odd function -> f(-x) = -f(x), we get the same answer as above.
Much less complex and time-consuming.
But most functions are neither even nor odd. You haven't dealt with that possibility.
@@andrewrao634 Yes, however you can represent f as the sum of its even and odd parts, which quickly gives you that its even part is zero, so f is odd.
@@actions-speak In this case, it's not difficult to show that f is odd (by other means), but in general it's not safe to assume that any function is the sum of even and odd parts.
@@andrewrao634 it's pretty safe to assume, because every function IS, uniquely, the sum of even and odd parts. f(x) = ½(f(x) + f(-x)) + ½(f(x) - f(-x)) - the first part is even, the second part is odd
@@lexyeevee What are the even and odd parts of f(x) = x^(1/2) ?
-a is equal to c. We have to find the value of c which isf(x). a=(2x^2-1)/3x. Why not we find the value of -a?? Than the answer will be (1-2x^2)/-3x.
understand your logic bro, thanks for sharing
No problem 👍
I like this problem because as you can see in the comments there are many ways to solve this but in the end we all get the same answer😁
That's so true!
Very smart way for solving that equation !!
You should also have checked at the end if the resulting f(x) satisfies the original equation.
That's right!
You are a genius sir 😊
We can use matrix to calculate a, b, c, d.
That's right!
such a nice problem which can show me lights
X ADDED WITH -I/X IS ZERO? GOOD LESSON. ATLEAST AHAVE A MIND TO CORRECT. WELL DONE
Without doubt, this is an excellent problem and I have learnt how to go about solving the problem with suitable substitution..
Wow , very good functional equation .Your method is nice .
Write system of equations:
3f(-x)+f(1/x)+f(x)=x
3f(x)+f(-1/x)+f(-x)=-x
3f(-1/x)+f(x)+f(1/x)=1/x
3f(1/x)+f(-x)+f(-1/x)=-1/x
Matrix form: A*y=b
A=
[1 3 1 0]
[3 1 0 1]
[1 0 1 3]
[0 1 3 1]
y=[f(x) f(-x) f(1/x) f(-1/x)]^T
b=[x -x 1/x -1/x]^T
Solve: y=(A^-1)*b
A^-1=(1/15)*
[-3 7 -3 2]
[7 -3 2 -3]
[-3 2 -3 7]
[2 -3 7 -3]
f(x)=1st coordinate of y
f(x)=(1/15)(-3x+7[-x]-3[1/x]+2[-1/x])
f(x)=-2x/3-1/(3x)
Very nice! 💖
wow syber never seen such a solid explanation!
Great job i keep looking forward seeing your next VDO
Nice equation, substitution is a great method, but I'd like to see other methods once, cuz I don't really know them...
8:04 why is the right hand side zero? x - (1/x) = 0 ?
I LIKE IT.
How 8.20 in equation=0 in my point of view instead of 0 there are x-1overx
f(x)=(16x+5/x)/27.....non so se i calcoli sono esatti....ma non ho voglia di controllare
ok, dipende da te 😄
I use different méthode but we get the same result at the end nice work
In your “second equation,” I think you might have meant to rename your variable to something like u? What you wrote is that f is an odd function, and that wasn’t a fact we were given.
One way to verify the answer for f(x) is the determined condition a=-c or c=-a. And c is f(x) :)
Eu cheguei tão perto... eu quase cobseguiii
I love this . Show me more
Solved!!!
You're a Math God!
~ from the philippines
th-cam.com/video/L3wKzyIN1yk/w-d-xo.html
😂
Nice.
I found it more faster by using the proprietary of impair function
Please tell the name of the book where you have got these beautiful and interesting functional equation problems?
That's a good question! Unfortunately these problems are not from a single book. They are all scattered. Some are my own problems, some are taken from books, journals, and math competitions. Some are adaptations of classical problems. I usually share the resource if I know where the problem comes from.
Nice exercise, indeed. And not so difficult, by the way
Interesting question.
8:05, why is the whole thing equal to 0？
Corrected at 8:31
@@SyberMath Thank you. I kept rewinding before making it to 8:31
How are there four degrees of freedom in the original functional equation?
Is it +1 for the equation, and then +1 for each argument given?
let a=-x, b=1/x, c=-1/x
Cooool , nice problem !!!
Thanks!
I have a fonction équation for you :
find f(x) when f(-x) - f(x) = 2x.
can you solve this in future please?
First, let's take a look at the equation from the symetry that result from it:
f(-x) - f(x) = 2x
By substitution x = -x, we see that:
f(x) - f(-x) = -2x
Since there's a clear symetry, this means that f(-x) = -f(x) so we can write the equation as:
-f(x) - f(x) = 2x
-2f(x) = 2x
f(x) = -x
So f(x) = -x but can be writen more generaly as f(x) = a - x where a is whatever real you want.
@@IkikaeruRaimei thanks
Thanks for your video and work 🤠 i wish you good luck 🙂😉 see you soon 😸
Damn! Lucid explanation.
غاتسطينا!
Bravo to indians
Nice sir. Which software is used to write ?
Thanks. It is Notability!
x + -1/x = 0? What am I missing?
Sorry for bad english :3
I built a system of equations by putting
x -> x
x -> 1/x
x -> -x
x -> -1/x
I set a = f(x) b = f(-x) c = f(1/x) d = f(-1/x)
And realized that a = -b and c = -d because f has to be odd
(I would have solved that with linear combinations of my 4 equations and, x and 1/x are odd so.. the result has to be odd to)
I found f(x) = -1/3x - 2x/3 which is the same thing :)
It's the same method than you but idk if it's allowed to say that f is odd with my argument
you found it with calculations which is a better argument ^^
Yes, that looks right!
Your English is fine! Don't worry about it! 😊
Sorry if stupid but how can you assume the substitutions work? Like how do you know subbing in 1/x will make the sum equal 1/x?
No question is stupid! We are trying to get a system of equations and substitution always works. We're taking advantage of the fact that 1/(1/x)=x and -(-x)=x
@@SyberMaththanks
Ayooooo thats called using the given info till the limit
thank you.
You're welcome
8:03 why is the whole thing equal to 0?
It's not. Correction at 8:31
This was quite easy compared to the IMO ones... some of those equations are ridiculously difficult to solve, I struggled a lot with them
They are hard. I try to pick manageable ones
@@SyberMath yup, I think that'll be beneficial for most of the viewers... although you could explain such a difficult problem, it would be very difficult to answer. :D
Trust me. It would be hard for me, too! 😁
For some reason I feel imo questions easy and these ones difficult lol maybe I am taking a lot of surprisingly correct assumptions in those problems
-(2x^2+1)/3x
Sorry, I’m not as smart as everyone else is, but why is a + c = 0 also @ 6:15?
Late answer, but here you go;
3(a + c) + (a + b + c + d) = 0
We know a + b + c + d is 0, so substitute that into the equation
3(a + c) + 0 = 0
3(a + c) = 0
Solve for a + c, you need to isolate a + c.
Dividing 3 on both sides gives us:
3(a + c) ÷ 3 = 0 ÷ 3
a + c = 0
Thus, a + c = 0
Wow
took me 15 minutes but quite easy
Too many xes. Maybe use a t or an s?
なるほどーーーーーー
おもしろかった Great!😉👍
私はそれを聞いてうれしい！
Brazil. Rio branco/AC
Nice solution. I really enjoyed the way you you simplified it. But, I have a doubt. Are there other functions those satisfy the given condition ? I'm asking this because I somehow thought of another function that satisfies the given condition...
f(x) = (-x/|x|)e^(|ln|x||)
The above function also satisfies the condition given in the question. So it could be a solution too. So, could there be other functions satisfying that condition that I can't think of ?
That's a good question. There should not be imo but I could be wrong. Could your answer be equivalent to the one we found?
@@SyberMath @SyberMath I don't think the functions can be equivalent.
Say,
f(x) = -(2x²+1)/3x = -(2x/3 + 1/3x)
g(x) = (-x/|x|)e^(|ln|x||)
Basically,
g(x) = -x , when x ≤-1
g(x) = -1/x , when -1
@@arkanilpaul9501 Your function simply equates to:
f(x) = -x for |x| >= 1
f(x) = -1/x for 0 < |x| < 1
And it *doesn't* satisfy the original problem for |x| < 1 (try x = 1/2), so it's not a solution.
@@andrewrao634 Thanks. I can see it now 😁
Put 1 and minus 1
I guess solveable