I love this problem!! :-D You can generalize your solution, and, in my opinion, the generalized version is in some ways simpler. Here’s the solution: It’s very easy to show that any function f(x) can be written as the sum of an even function and an odd function: f(x) = O(x) + E(x) where O(-x) = - O(x) and E(-x) = E(x) Substituting into the equation, we get O(x)E(x) = x Now, both sides are odd, since an even function times an odd function is odd (which is good, because else we’d have no solution). Thus, O(x) = x/E(x) And the general family of solutions is f(x) = x/E(x) + E(x) where E(x) is any even function. We obtain your solution by setting E(x)=c. However, f(x) = 1/x + x^2 is also a solution. Want f(x) to be continuous? Choose an E(x) that has no real roots. For example: f(x) = x/(x^2 + 1) + x^2 + 1 is a solution that is continuous everywhere. :-D [Footnote: to show that any function can be written as the sum of an even function and an odd function, consider any function f(x). I’ll use the Michael Penn trick of “multiplying by one and adding zero”: f(x)=(1/2)(f(x) + f(x)) + (1/2)(f(-x) - f(-x)) Which we can rearrange to get f(x)=(1/2)(f(x) + f(-x)) + (1/2)(f(x) - f(-x)) But, (f(x) + f(-x)) is even and (f(x) - f(-x)) is odd, so we can see that f(x) is the sum of even function E(x) = (1/2)(f(x) + f(-x)) and odd function O(x) = (1/2)(f(x) - f(-x)) :-D]
@@anirudh67 Consider any function f(x). I’ll use the Michael Penn trick of “multiplying by one and adding zero”: f(x)=(1/2)(f(x) + f(x)) + (1/2)(f(-x) - f(-x)) Which we can rearrange to get f(x)=(1/2)(f(x) + f(-x)) + (1/2)(f(x) - f(-x)) But, (f(x) + f(-x)) is even and (f(x) - f(-x)) is odd, so we can see that f(x) is the sum of even function E(x) = (1/2)(f(x) + f(-x)) and odd function O(x) = (1/2)(f(x) - f(-x)) :-D
That's exactly what I did. The f(-x) is a clear indicator that function parity is relevant to this problem. Just one more thing: Given that f²(x) - f²(-x) = 4x, if f(x) was a purely even function or a purely odd function, we would get 0 = 4x, which is false. Therefore, f(x) must have an even and an odd component. Also, f(x) = cos(x) + x*sec(x) is also a solution.
Nice... but actually infinite solutions if all powers of x are assumed f(x) = E(x) + x/E(x) or f(x) = O(x) + x/O(x) Where O and E are odd and even functions
Solution has flaws f(x)^2 -f(-x)^2 has x^2, x^4 ... terms 0 but doesn't necessarily mean cofficients in f(x) are also zero. Coefficient of expression a1a2+a0a3 = 0 doesn't mean a2=0 and a3=0. You have touched upon just one solution. It rather means a2=-a0^2*a3. That gives another solution.
It can be done in an easier way by just observing the eqn. It is of the form of p2 - q2 = 4t Now, we know that (a+b)2 - (a-b)2 = 4ab So, a+b = f(x), a-b = f(-x), ab =x Now put b= x/a a+x/a = f(x), a-x/a = f(-x) So, f(x) = a+x/a with a not equal to 0.
To get all the solutions (including non-polynomials), we can use the fact that all functions can be represented as the sum of an even function and an odd function, f(x)=e(x)+o(x). Then substituting this sum into the equation results in 4e(x)o(x)=4x. Thus, e(x)o(x)=x. The most obvious functions that satisfy this relation are e(x)=c*x^(-2n) and o(x)=1/c*x^(2n+1), where n is an integer. The solution in the video is this at n=0. But other solutions include e(x)=c*x^(2n/(2n+1)), o(x)=1/c*x^(1/(2n+1) and e(x)=c*cos^(a)(bx), o(x)=1/c*x/cos^(a)(bx). Most, if not all, of these extra solutions will be discontinuous at x=0.
I saw a video by Math Elite and he solved such an equation for general case. He showed that f(x) can not be odd or even. Then he uses a theorm that every function can be written as a sum of even and odd function. Then f(x)=O(x) + x/O(x) [O(x) is an odd function] OR f(x)=E(x) + x/E(x). [E(x) is an even function]. In the polinomial case, y=c is an even function.
Determination of the coefficients a's a_0 =a , a_1= 1/a a_0*a_3 + a_1*a_2 = 0 so a_2 = -a_3*a*a etc . f(x).= a+x/a is a feasible solution for any non zero a (in complex plane)
I haven't gone through all the comments, and so I'm not sure whether anyone has given this answer. The given functional equation has infinitely many solutions of the form f(x) = u(x) + x/u(x) where u(x) is any even function (i.e, a function such that u(x) = u(-x) for example: u(x) = cos(x), x^2, x^4, and so on). To obtain the solution you've arrived at, set u(x) = c.
Very nice! It me a minute to figure out how that fits into my general solution (that is, f(x) = x/E(x) + E(x) where E(x) is any even function). We get your solution if we set E(x) = sqrt(|x|). Then, f(x) = x/sqrt(|x|) + sqrt(|x|) = 2 sqrt(x) for x>0 and = 0 for x
@@SyberMath you should fix your solution. I mean that general solution is in form proposed by Leick Robinson. And your solution has only polynomial form.
In fact, since this equation only gives the relation between f(a) and f(-a) for each a, for a>=0 we may let f(a) be anything such that f(a)^2>=4a and then we may set f(-a)=+-sqrt(f(a)^2-4a) to make f satisfies the equation.
@@SyberMath I got as far as noticing x plus 1 works as a solution and x plus one half of you multiply the x by 2..don't you think that counts as most of the correct answer anyway? I just forgot to try to generalize further.
Silly question: should coefficients 2*a0*a1 = 1, instead of just a0*a1 = 1? There are 2 such terms by distributing the multiplication across terms in the final step. Or am I missing something obvious? Thanks!
What about this : h (x) = ( f(x) + f(-x) )/2 ; g(x) = ( f(x) - f(-x) )/2. The equation becomes: h g = x; g is odd, h is even, so g has zeros; h is even and need not have zeros. So: g(x) = x / h(x) for h even and nowhere zero. Then f (x) = h(x) + g(x) = h(x) + x/ h(x) is defined (and a solution) for any h even and nowhere zero, without further hypothesis on the regularity of h.
why would you assume f must be a polynomial? For example, one solution is f(x)=sqrt(3x^2+2x+1). Indeed, if f(x)=sqrt(ax^2 + 2x + c) where a>0 and c is such that ax^2+2x+c is always nonnegative, then this would satisfy the equation. In fact, more generally, f(x)=sqrt(2x+ g(x)) where g(x) is any even function and g(x)+2x is always nonnegative, will work.
It is easier to solve: we denote f (x) = a, f (-x) = b. Then (a + b) (a-b) = 4x. a + b - even function -> a + b = 2; a-b - odd function -> a-b = 2x, whence f (x)=1-x.
I did it in a similar way. Given that [f(x) + f(-x)]*[f(x) + f(-x)] = 4x. Now assume that 4x represents product of 1*4x, or 2*2x, or 4*x. Then I (wrongly) assumed that [f(x) + f(-x)] is equal to 4x or 2x or x, and [f(x) - f(-x)] is equal to 1 or 2 or 4, but I still somehow got the right answer: c*x + 1/c.
Syber, I went on about solving this in a similar fashion, but considered instead the power series for f^2(x). This led me to a whole family of solutions, namely f^2(x) must be of the form a_0 + 2x + a_2x^2 + a_4x^4 + ... + a_(2m)*2x^(2m) + a_(2m + 2)*x^(2m + 2) + ... And from here you could cut of at any point, or if you're feeling bold, you can even go with an infinite power series. So say, a function like f(x) = sqrt(2x + cos(x)) should work as well. Did I miss something in the definition of the function f(x) or is this solution acceptable?
One issue I see is that there is an implicit assumption that we are looking for real-valued functional solutions. But, f(x) = sqrt(2x + cos(x)) is not real-valued for sufficiently large negative values of x.
@@leickrobinson5186 you still could keep summing as many powers as you want. sqrt(x^4 + x^2 + 2x + 1) is an example of such function which works for all real values of x.
Just split f into its even and uneven part e(x) and o(x), with f(x) = e(x) + o(x) and e(x) = e(-x) and o(x) = -o(-x). You get e(x).o(x) = x as only condition. you'll also find solutions like e(x) = c (x^(2m))^(1/(2n+1)) o(x) = 1/c (x^(1-2n+2m))^(1/(2n+1)) f(x) = c (x^(2m))^(1/(2n+1)) + 1/c (x^(1-2n+2m))^(1/(2n+1))
[f(x) - f(-x)][f(x) + f(-x)] = 4x = O(x)*E(x) -- the product of the odd and even parts of f(x). Here's a solution: E(x) = 2*cosh(x), and O(x) = 2x/cosh(x). This means f(x) = cosh(x) + x/cosh(x). We can check: [f(x)]² = cosh²(x) + 2x + x²/cosh²(x) and [f(-x)]² = cosh²(x) - 2x + x²/cosh²(x). Subtract them and you get 4x. The solution's not unique. I also tried E(x) = 2(1 + x²). EDIT: Okay, what I called the odd and even parts of f(x) are twice the standard parts.
Let e(x) and o(x) be the even and odd parts of f(x) respectively, i.e. e(x) = (f(x)+f(-x))/2 and o(x) = (f(x)-f(-x))/2. Then the equation reduces to 4 e(x) o(x) = 4x, i.e. e(x) o(x) = x. So just take any couple of even and odd functions e(x) and o(x) such that e(x) o(x) = x. Then f(x) = e(x) + o(x) satisfies the functional equation. One can start with choosing e(x) such that e(x) at most vanishes for x=0. Then set o(x) = x/e(x), with o(0) arbitrary if e(0) = 0. Example 1: e(x) = 1+x², o(x) = x/(1+x²). Example 2: e(x) = x², o(x) = 1/x with o(0) = 42.
Another example: f(x) = e(x) + o(x), where e(x) = 1 if x is rational, e(x) = 2 if x is irrational, and o(x) = x if x is rational, o(x) = x/2 if x is irrational. This is a discontinuous solution.
Mathematics is the science that deals with the logic of shape, quantity and arrangement. Math is all around us, in everything we do. ... The needs of math arose based on the wants of society. The more complex a society, the more complex the mathematical needs.
at 5:50 how do you go from a1 a2 + a0 a3 = 0 to concluding that a2=a4=a6=...=0? could it not be that just a1 a2 = -a0 a3? and likewise for the coefficients of higher degrees
I agree with you. If a3=d then a2=-d.c^2 will make the x^2, coefficient zero. It will get a lot more complicated with higher order coefficients. I think the linear function is only one of the solutions. You can certainly make the higher order polynomials work by choosing proper coefficients.
This was easy. I had the solution at the first look. Because the difference: f(x)² - f(-x)² = some linear function (x) --> positive 'x' raises more than negative 'x' in a linear way (1st order) --> conclusion: f(x)² = not balanced = shifted parabola. --> f(x) = some linear funktion. And with that in mind, you can easily solve it.
Not 100% proven as a1a2+a0a3 = 0 does not require a2 and a3 to be zero. They could be a2=a0 and a3=-a1. However that had implications on subsequent terms.
Consider E(x) = (f(x) + f(-x))/2. Note that E(-x) = (f(-x) + f(x))/2 = E(x). Hence, E(x) is even. Next, consider O(x) = (f(x) - f(-x))/2. Note that O(-x) = (f(-x) - f(x))/2 = -O(x). Hence, O(x) is odd. Next, note that O(x) + E(x) = (f(x) - f(-x))/2 + (f(x) + f(-x))/2 = 2f(x)/2 = f(x). So, f(x) = O(x) + E(x). Also note that E(x) - O(x) = (f(x) + f(-x) - f(x) + f(-x))/2 = 2f(-x)/2 = f(-x). So, f(-x) = E(x) - O(x) Now, consider [f(x)]^2 - [f(-x)]^2 = 4x. Substitute in f(x) = O(x) + E(x), and f(-x) = E(x) - O(x), giving us [O(x) + E(x)]^2 - [E(x) - O(x)]^2 = 4x. Simplifying the left hand side(and skipping a few steps), we get 4O(x)E(x) = 4x O(x)E(x) = x O(x) = x/E(x) or E(x) = x/O(x) So, for example, sin(x) is odd. So O(x) = sin x, E(x) = x/sin(x) gives us f(x) = sin x + x/(sin x). In fact, as a general solution, let g(x) be any non-even function. Then f(x) = g(x) + g(-x) + x/(g(x) + g(-x)) will solve the given quadratic.
So, when you said you would be solving for 'f of x' what you really intended to do was find f (i.e. the definition of f), and the initial equation was an identity, not just an equation. I was confused until the very end.
I dont get why a2,a3 and so on need to equal 0. Just because the sums of products of the coefficients equal 0, doesn't mean all a2, a3, ... need to equal zero.
we have proven that a2=a3=....=0 gives.one polynomial solution. But how do we know that this is the only polynomial solution? Similarly, we have proven that a linear polynomial is a solution. How do we prove that it is the only solution?
@@mohamedramadan-qt9yl Nope. That's like saying that 3 + 0.1 + 0.04 + 0.001 + ... is an appropiximation for pi. Chopping it off early makes it an approximation, but the infinite series is equal.
@@chaosredefined3834 He should say that f(x) is a polynomial, if he doesn't say that he must mention that the function satisfies Taylor expansion conditions. Otherwise the proof is not valid for every function f(x)
@@SyberMath I tried using the same method and got the same value for f(x) I got:(assuming that f(x) is not a polynomial) f(x)=a+b/x and that ab=x^2 So if b=(x^2)/a f(x)=a+(x^2) × 1 a x I.e f(x)=a+x/a
@@aaryangre7809 The general family of solutions is f(x) = x/E(x) + E(x) where E(x) is any even function. See my main comment for the derivation details. :-D
If we’re assuming that f is a polynomial of degree n, then when you expand everything out you have a polynomial of degree 2n on the left and a constant value on the right, which only works if the polynomial on the left is also a constant. This breaks down if f is an arbitrary power series, which leads to the alternative solutions some people are discussing in the comments.
If the polynomial has finite degree, meaning the polynomial has a largest exponent, then all the coefficients other than a₀ and a₁ do have to be zero. On the other hand, you could have what amounts to an infinite polynomial, as happens with a Taylor Series. Solutions to this problem with an infinite polynomial are possible, and I have no doubt all the alternate solutions given in other comments here fall into this category.
Thank you for the compliment but BlackPenRedPen is a college professor with lots of knowledge and good content! I have respect and admiration for his work! I'm just trying to share my passion for problem solving.
Actually, there are *many* more functions that satisfy this equation. The general family of solutions is f(x) = x/E(x) + E(x) where E(x) is any even function. (See my main comment for a detailed derivation of the solution.) :-D
@@MathElite This is actually included in my family of solutions! Since the product of two odd functions is an even function, we can choose E(x) = x/O(x) Substituting, we get f(x) = x/(x/O(x)) + x/O(x) = O(x) + x/O(x) :-D
f(x) = h(x) + g(x) h is event g is odd, so : f(x) = h(x)+x/h(x) chose h(x) when h for any x never equal 0 f(x)² - f(-x)² = 4 x (h(x)+g(x))²-(h(x)-g(x))² = 4x 4 h(x) g(x) = 4x sp : g(x) = x/ h(x) so f(-x) = h(x) -x / h(x) ...
@@SyberMath Thanks for the hint! I approached the equation from a slightly different angle than in the other comment: Because -(-x)=x, our equation includes infinitely many equations for (x, -x) pairs, which are independant of each other. For example, if you solved the equation for x=1 and x=-1, then these two values are independant of every other x value. So for a given value of x, we'd normally have two equations for the two values f(x) and f(-x). (The first equation is the original one, the second is given by replacing x with -x). Now there's something special here, because the equation does not change by replacing x with -x. So in fact, we only have one equation for two values, which is the very reason why we have infinitely many solutions! Now I solved this by just solving for f(x): f(x)=+/-sqrt(4x+f(-x)²), which means that for any arbitrary function g: ]-inf; 0] --> IR, the function f: IR --> IR given by f(x)=... ...g(x) for x0 (Note that the sqrt always exists!) solves the functional equation. The +/- can be chosen seperately for every value of x>0. This solution should be equivalent to the solution in the other comment f(x)=x/E(x)+E(x) with an arbitrary even function E.
sir, with all due respect, you appear to be clueless as to what a functional equation is, you didn't even define the domain and codomain of the function which makes your equation ill-defined and your assumption that the function is a polynomial literally gave me a hemmorhage
I love this problem!! :-D
You can generalize your solution, and, in my opinion, the generalized version is in some ways simpler. Here’s the solution:
It’s very easy to show that any function f(x) can be written as the sum of an even function and an odd function:
f(x) = O(x) + E(x)
where O(-x) = - O(x)
and E(-x) = E(x)
Substituting into the equation, we get
O(x)E(x) = x
Now, both sides are odd, since an even function times an odd function is odd (which is good, because else we’d have no solution).
Thus,
O(x) = x/E(x)
And the general family of solutions is
f(x) = x/E(x) + E(x)
where E(x) is any even function.
We obtain your solution by setting E(x)=c.
However,
f(x) = 1/x + x^2
is also a solution.
Want f(x) to be continuous? Choose an E(x) that has no real roots. For example:
f(x) = x/(x^2 + 1) + x^2 + 1
is a solution that is continuous everywhere.
:-D
[Footnote: to show that any function can be written as the sum of an even function and an odd function, consider any function f(x). I’ll use the Michael Penn trick of “multiplying by one and adding zero”:
f(x)=(1/2)(f(x) + f(x)) + (1/2)(f(-x) - f(-x))
Which we can rearrange to get
f(x)=(1/2)(f(x) + f(-x)) + (1/2)(f(x) - f(-x))
But, (f(x) + f(-x)) is even and (f(x) - f(-x)) is odd, so we can see that f(x) is the sum of even function
E(x) = (1/2)(f(x) + f(-x))
and odd function
O(x) = (1/2)(f(x) - f(-x))
:-D]
How can you say that any function can be written as a sum of odd and even functions? Exponential and logarithms are neither odd nor even
@@anirudh67 exponential functions can be written as a sum of their odd and even parts still. E.g.
e^x = (e^x+e^-x)/2 + (e^x-e^-x)/2
@@anirudh67 Consider any function f(x). I’ll use the Michael Penn trick of “multiplying by one and adding zero”:
f(x)=(1/2)(f(x) + f(x)) + (1/2)(f(-x) - f(-x))
Which we can rearrange to get
f(x)=(1/2)(f(x) + f(-x)) + (1/2)(f(x) - f(-x))
But, (f(x) + f(-x)) is even and (f(x) - f(-x)) is odd, so we can see that f(x) is the sum of even function
E(x) = (1/2)(f(x) + f(-x))
and odd function
O(x) = (1/2)(f(x) - f(-x))
:-D
@@leickrobinson5186 great thanks :D
That's exactly what I did. The f(-x) is a clear indicator that function parity is relevant to this problem. Just one more thing:
Given that f²(x) - f²(-x) = 4x, if f(x) was a purely even function or a purely odd function, we would get 0 = 4x, which is false. Therefore, f(x) must have an even and an odd component.
Also, f(x) = cos(x) + x*sec(x) is also a solution.
Nice... but actually infinite solutions if all powers of x are assumed
f(x) = E(x) + x/E(x)
or
f(x) = O(x) + x/O(x)
Where O and E are odd and even functions
There are non-polynomial solutions such as f(x)= c x^n +(1/c) x^(1-n)
Solution has flaws
f(x)^2 -f(-x)^2 has x^2, x^4 ... terms 0 but doesn't necessarily mean cofficients in f(x) are also zero.
Coefficient of expression a1a2+a0a3 = 0 doesn't mean a2=0 and a3=0. You have touched upon just one solution.
It rather means a2=-a0^2*a3. That gives another solution.
Exactly, I don't know how no one else pointed this out. The logic at 5:55 is flawed
@@bscutajar I did
Edit: please don't post solutions until the premiere starts
This channel is wonderful, made me like solving equations again
Thank you! 💖
^^^ You make great videos! Dear friends, check it out! ^^^
It can be done in an easier way by just observing the eqn.
It is of the form of p2 - q2 = 4t
Now, we know that (a+b)2 - (a-b)2 = 4ab
So, a+b = f(x), a-b = f(-x), ab =x
Now put b= x/a
a+x/a = f(x), a-x/a = f(-x)
So, f(x) = a+x/a with a not equal to 0.
Nice!
To get all the solutions (including non-polynomials), we can use the fact that all functions can be represented as the sum of an even function and an odd function, f(x)=e(x)+o(x). Then substituting this sum into the equation results in 4e(x)o(x)=4x. Thus, e(x)o(x)=x.
The most obvious functions that satisfy this relation are e(x)=c*x^(-2n) and o(x)=1/c*x^(2n+1), where n is an integer. The solution in the video is this at n=0.
But other solutions include
e(x)=c*x^(2n/(2n+1)), o(x)=1/c*x^(1/(2n+1)
and
e(x)=c*cos^(a)(bx), o(x)=1/c*x/cos^(a)(bx).
Most, if not all, of these extra solutions will be discontinuous at x=0.
Everytime he says "so whats that supposed to mean?" You have to take a shot
I saw a video by Math Elite and he solved such an equation for general case. He showed that f(x) can not be odd or even. Then he uses a theorm that every function can be written as a sum of even and odd function. Then f(x)=O(x) + x/O(x) [O(x) is an odd function] OR f(x)=E(x) + x/E(x). [E(x) is an even function]. In the polinomial case, y=c is an even function.
That's such a beautiful result! He makes great videos!
Determination of the coefficients a's
a_0 =a , a_1= 1/a
a_0*a_3 + a_1*a_2 = 0
so a_2 = -a_3*a*a
etc .
f(x).= a+x/a is a feasible solution
for any non zero a (in complex plane)
I haven't gone through all the comments, and so I'm not sure whether anyone has given this answer. The given functional equation has infinitely many solutions of the form f(x) = u(x) + x/u(x) where u(x) is any even function (i.e, a function such that u(x) = u(-x) for example: u(x) = cos(x), x^2, x^4, and so on). To obtain the solution you've arrived at, set u(x) = c.
Nice!
See the pinned comment.
@@SyberMath OK. Thank you. I'll remove my comments after a day or two. The comment pinned by you has all the details, and a detailed solution as well.
@@PS-mh8ts You don't need to remove it imo
@@SyberMath OK. Thank you. :-)
Are there any other solutions or is this all of them?
F(x) = 2sqrt(x) if x > 0,
F(x) = 0 otherwise.
For x greater 0: F(x)² = 4x.
For x less 0: -F(-x)² = -(2sqrt(-x))²=4x.
Looks like solution to me.
Very nice! It me a minute to figure out how that fits into my general solution
(that is, f(x) = x/E(x) + E(x)
where E(x) is any even function).
We get your solution if we set
E(x) = sqrt(|x|).
Then, f(x) = x/sqrt(|x|) + sqrt(|x|)
= 2 sqrt(x) for x>0 and
= 0 for x
Very nice!
@@SyberMath you should fix your solution.
I mean that general solution is in form proposed by Leick Robinson. And your solution has only polynomial form.
In fact, since this equation only gives the relation between f(a) and f(-a) for each a, for a>=0 we may let f(a) be anything such that f(a)^2>=4a and then we may set f(-a)=+-sqrt(f(a)^2-4a) to make f satisfies the equation.
watching you easily solve these complicated questions is amazing, learning a lot Syber
Glad to hear that! Make sure to check Leick Robinson's comments. They are very insightful!
@@SyberMath I got as far as noticing x plus 1 works as a solution and x plus one half of you multiply the x by 2..don't you think that counts as most of the correct answer anyway? I just forgot to try to generalize further.
@@SyberMath Hope you can respond when you can. Thanks very much.
How do we know that f(x) is a polynomial? It might be for example 1/x.
hmm ye same question here
Does 1/x satisfy the equation? 🤔
The general family of solutions is
f(x) = x/E(x) + E(x)
where E(x) is any even function. See my main comment for the derivation details. :-D
A Laurent series: ...+a(-2)/x^2+a(-1)/x+a(0)+a(1)*x+a(2)*x^2+...+b(1)*log(x)+...
@@vascomanteigas9433 Can you elaborate?
Silly question: should coefficients 2*a0*a1 = 1, instead of just a0*a1 = 1? There are 2 such terms by distributing the multiplication across terms in the final step. Or am I missing something obvious? Thanks!
how about function in complex domain -> z(x)=sqrt(2x)?
[sqrt( 2x)]^2 = 2x
[sqrt(-2x)]^2 = [i*sqrt(2x)]^2 = -2x; the sum is 4x.
What about this : h (x) = ( f(x) + f(-x) )/2 ; g(x) = ( f(x) - f(-x) )/2.
The equation becomes: h g = x;
g is odd, h is even, so g has zeros; h is even and need not have zeros. So: g(x) = x / h(x) for h even and nowhere zero. Then
f (x) = h(x) + g(x) = h(x) + x/ h(x) is defined (and a solution) for any h even and nowhere zero, without further hypothesis on the regularity of h.
why would you assume f must be a polynomial? For example, one solution is f(x)=sqrt(3x^2+2x+1). Indeed, if f(x)=sqrt(ax^2 + 2x + c) where a>0 and c is such that ax^2+2x+c is always nonnegative, then this would satisfy the equation. In fact, more generally, f(x)=sqrt(2x+ g(x)) where g(x) is any even function and g(x)+2x is always nonnegative, will work.
It is easier to solve:
we denote f (x) = a, f (-x) = b. Then (a + b) (a-b) = 4x. a + b - even function -> a + b = 2; a-b - odd function -> a-b = 2x, whence f (x)=1-x.
I did it in a similar way. Given that [f(x) + f(-x)]*[f(x) + f(-x)] = 4x. Now assume that 4x represents product of 1*4x, or 2*2x, or 4*x. Then I (wrongly) assumed that [f(x) + f(-x)] is equal to 4x or 2x or x, and [f(x) - f(-x)] is equal to 1 or 2 or 4, but I still somehow got the right answer: c*x + 1/c.
Syber, I went on about solving this in a similar fashion, but considered instead the power series for f^2(x). This led me to a whole family of solutions, namely f^2(x) must be of the form a_0 + 2x + a_2x^2 + a_4x^4 + ... + a_(2m)*2x^(2m) + a_(2m + 2)*x^(2m + 2) + ...
And from here you could cut of at any point, or if you're feeling bold, you can even go with an infinite power series. So say, a function like f(x) = sqrt(2x + cos(x)) should work as well. Did I miss something in the definition of the function f(x) or is this solution acceptable?
One issue I see is that there is an implicit assumption that we are looking for real-valued functional solutions.
But, f(x) = sqrt(2x + cos(x)) is not real-valued for sufficiently large negative values of x.
Leaving the problem a little open-ended helped spark some good conversation
😁
@@leickrobinson5186 you still could keep summing as many powers as you want. sqrt(x^4 + x^2 + 2x + 1) is an example of such function which works for all real values of x.
Sqrt( even fuction +2x) is solution
Just split f into its even and uneven part e(x) and o(x), with f(x) = e(x) + o(x) and e(x) = e(-x) and o(x) = -o(-x).
You get e(x).o(x) = x as only condition.
you'll also find solutions like
e(x) = c (x^(2m))^(1/(2n+1))
o(x) = 1/c (x^(1-2n+2m))^(1/(2n+1))
f(x) = c (x^(2m))^(1/(2n+1)) + 1/c (x^(1-2n+2m))^(1/(2n+1))
[f(x) - f(-x)][f(x) + f(-x)] = 4x = O(x)*E(x) -- the product of the odd and even parts of f(x). Here's a solution: E(x) = 2*cosh(x), and O(x) = 2x/cosh(x). This means f(x) = cosh(x) + x/cosh(x).
We can check: [f(x)]² = cosh²(x) + 2x + x²/cosh²(x) and [f(-x)]² = cosh²(x) - 2x + x²/cosh²(x). Subtract them and you get 4x.
The solution's not unique. I also tried E(x) = 2(1 + x²).
EDIT: Okay, what I called the odd and even parts of f(x) are twice the standard parts.
F(x) = (a+x/a) with a as non zero constant
Let e(x) and o(x) be the even and odd parts of f(x) respectively, i.e. e(x) = (f(x)+f(-x))/2 and o(x) = (f(x)-f(-x))/2. Then the equation reduces to 4 e(x) o(x) = 4x, i.e. e(x) o(x) = x. So just take any couple of even and odd functions e(x) and o(x) such that e(x) o(x) = x. Then f(x) = e(x) + o(x) satisfies the functional equation.
One can start with choosing e(x) such that e(x) at most vanishes for x=0. Then set o(x) = x/e(x), with o(0) arbitrary if e(0) = 0.
Example 1: e(x) = 1+x², o(x) = x/(1+x²).
Example 2: e(x) = x², o(x) = 1/x with o(0) = 42.
Another example: f(x) = e(x) + o(x), where e(x) = 1 if x is rational, e(x) = 2 if x is irrational, and o(x) = x if x is rational, o(x) = x/2 if x is irrational. This is a discontinuous solution.
Mathematics is the science that deals with the logic of shape, quantity and arrangement. Math is all around us, in everything we do. ... The needs of math arose based on the wants of society. The more complex a society, the more complex the mathematical needs.
I agree
keep it up bro. well done. very good content
Thanks, will do!
at 5:50 how do you go from a1 a2 + a0 a3 = 0 to concluding that a2=a4=a6=...=0? could it not be that just a1 a2 = -a0 a3? and likewise for the coefficients of higher degrees
I agree with you. If a3=d then a2=-d.c^2 will make the x^2, coefficient zero. It will get a lot more complicated with higher order coefficients. I think the linear function is only one of the solutions. You can certainly make the higher order polynomials work by choosing proper coefficients.
what about f(x)=sqrt(2x)....[sqrt(2x)]^2-[sqrt(-2x)]^2=2x- (-2x)=4x
This was easy. I had the solution at the first look.
Because the difference: f(x)² - f(-x)² = some linear function (x) --> positive 'x' raises more than negative 'x' in a linear way (1st order)
--> conclusion: f(x)² = not balanced = shifted parabola. --> f(x) = some linear funktion. And with that in mind, you can easily solve it.
Nice!
Not 100% proven as a1a2+a0a3 = 0 does not require a2 and a3 to be zero. They could be a2=a0 and a3=-a1. However that had implications on subsequent terms.
How do we prove the solution must be a polynomial tho? That seems to be tbe hardest part of the problem.
It doesn’t have to be a polynomial.
For example,
f(x) = x/(x^2 + 1) + x^2 + 1
is a solution. (See my main comment for details and derivation.)
@@leickrobinson5186 Thanks, that's very cool
I saw it like power series
Thanks for the problem!
5:53 ...I think the conclusion is erroneous.
The equations are:
x^0: a0 a1 = 1 a0 = 1/a1
x^2: a0 a3 + a1 a2 = 0 a2 = -a0^2 a3
x^4: a0 a5 + a1 a4 + a2 a3 = 0 a4 = a0^3 a3^2 - a0^2 a5
x^6: a0 a7 + a1 a6 + a2 a5 + a3 a4 = 0 a6 = -a0^2 a7 + 2 a0^3 a3 a5 - a0^4 a3^3
...
I seems that for *any* odd function O(x) (free coefficients a1 a3 a5 a7...) there is and even function E(x) (coefficients a0 a2 a4 a6...) such that f(x) = O(x) + E(x) is solution to the equation.
Consider E(x) = (f(x) + f(-x))/2. Note that E(-x) = (f(-x) + f(x))/2 = E(x). Hence, E(x) is even.
Next, consider O(x) = (f(x) - f(-x))/2. Note that O(-x) = (f(-x) - f(x))/2 = -O(x). Hence, O(x) is odd.
Next, note that O(x) + E(x) = (f(x) - f(-x))/2 + (f(x) + f(-x))/2 = 2f(x)/2 = f(x). So, f(x) = O(x) + E(x). Also note that E(x) - O(x) = (f(x) + f(-x) - f(x) + f(-x))/2 = 2f(-x)/2 = f(-x). So, f(-x) = E(x) - O(x)
Now, consider [f(x)]^2 - [f(-x)]^2 = 4x. Substitute in f(x) = O(x) + E(x), and f(-x) = E(x) - O(x), giving us [O(x) + E(x)]^2 - [E(x) - O(x)]^2 = 4x. Simplifying the left hand side(and skipping a few steps), we get
4O(x)E(x) = 4x
O(x)E(x) = x
O(x) = x/E(x) or E(x) = x/O(x)
So, for example, sin(x) is odd. So O(x) = sin x, E(x) = x/sin(x) gives us f(x) = sin x + x/(sin x).
In fact, as a general solution, let g(x) be any non-even function. Then f(x) = g(x) + g(-x) + x/(g(x) + g(-x)) will solve the given quadratic.
So, when you said you would be solving for 'f of x' what you really intended to do was find f (i.e. the definition of f), and the initial equation was an identity, not just an equation. I was confused until the very end.
Onctly i dont know how to solve functionel equation but i will learn from you !!!! 😊
You can do it!
I dont get why a2,a3 and so on need to equal 0. Just because the sums of products of the coefficients equal 0, doesn't mean all a2, a3, ... need to equal zero.
Quadratic functional equation? More like "Cool, tricky problem with a great explanation!" 👍
Thanks!
F(x)=square of(ax2+2x+c) also is solution
So c = c(x) Can be and non-zero, even function, for example just a constant function…. (c not zero).
we have proven that a2=a3=....=0 gives.one polynomial solution. But how do we know that this is the only polynomial solution?
Similarly, we have proven that a linear polynomial is a solution.
How do we prove that it is the only solution?
how do I know that f(x) is polynomial function?
Every function is polynomial, thanks to the taylor series.
@@chaosredefined3834 this is an approximation under some conditions on X, this solution is a restricted solution not in the general case
@@mohamedramadan-qt9yl Nope. That's like saying that 3 + 0.1 + 0.04 + 0.001 + ... is an appropiximation for pi. Chopping it off early makes it an approximation, but the infinite series is equal.
@@chaosredefined3834 He should say that f(x) is a polynomial, if he doesn't say that he must mention that the function satisfies Taylor expansion conditions.
Otherwise the proof is not valid for every function f(x)
What if f is not polynomial?
Great video by the way!
Good question! I challenge you and everyone else to find a non-polynomial solution to this equation!
🙃😉😁🤩
Assume that f are analytic, and place the Laurent series on it...
@@SyberMath Lol, challenge retroactively accepted… and done. See the main comment I just left for details. :-D
@@SyberMath I tried using the same method and got the same value for f(x)
I got:(assuming that f(x) is not a polynomial)
f(x)=a+b/x and that ab=x^2
So if b=(x^2)/a
f(x)=a+(x^2) × 1
a x
I.e f(x)=a+x/a
@@aaryangre7809 The general family of solutions is
f(x) = x/E(x) + E(x)
where E(x) is any even function. See my main comment for the derivation details. :-D
You're a genious!
Aww, thank you for the kind words! 🥰💖
Can we differentiate the two sides?
Why not? 😁
I didn' t understand why the rest of the coefficients must be zero, can someone elaborate please 5:45?
If we’re assuming that f is a polynomial of degree n, then when you expand everything out you have a polynomial of degree 2n on the left and a constant value on the right, which only works if the polynomial on the left is also a constant.
This breaks down if f is an arbitrary power series, which leads to the alternative solutions some people are discussing in the comments.
Creative solutions
Thank you!
Damn you're so welcome.......
Appreciate you
The solution reminds us of the parallelogram rule:
||X+Y||^2 - ||X-Y||^2 = 4X•Y
That's nice!
does f(x) = sqrt(2x) not work?
It works
Et si f n'est pas polynomiale ?
Why if a1a2+a0a3 is 0 are you sure a2 and a3 are 0? Thanks...
From a1a2 + a0a3 = 0, why is a2 and a3 equal 0? Could be that a2 = a0 and a3 = - a1.
All the sums are 0 not just that one.
If the polynomial has finite degree, meaning the polynomial has a largest exponent, then all the coefficients other than a₀ and a₁ do have to be zero. On the other hand, you could have what amounts to an infinite polynomial, as happens with a Taylor Series. Solutions to this problem with an infinite polynomial are possible, and I have no doubt all the alternate solutions given in other comments here fall into this category.
@@SyberMath yeah so?
What happens if this function is not polynomial
Very well
Thanks
U are helping me for olympiad , thnxxxxxx
Can i ask , is u a college student or completed phd
Neither 😁
@@SyberMath then?? A school student??
@@deepjyoti5610 A former math teacher 😁
One extra like 👍
Thanks 👍
problem？ why cant it be x^-1 or x^1/2
You are even better than BlackPenRedPen
I’m not so sure I can decide who is better, but BRPR certainly has a new contender! :-D
Thank you for the compliment but BlackPenRedPen is a college professor with lots of knowledge and good content! I have respect and admiration for his work!
I'm just trying to share my passion for problem solving.
Two words Fourier transform
1st step - Floor function ISN'T additive:
[ f(x) ] - [ f(-x) ] != [ f(x) - f(-x) ]
This is a wrong approach
Those are brackets not the floor function symbols
This is the functional equation of c+x/c
😃good one
Thanks 😅
Super
Thank you!!!
Why can you assume it's polynomial? In fact you can't!
But how can we prove there aren t any more functions that satisfy this equation?
Actually, there are *many* more functions that satisfy this equation.
The general family of solutions is
f(x) = x/E(x) + E(x)
where E(x) is any even function.
(See my main comment for a detailed derivation of the solution.) :-D
@@leickrobinson5186 also f(x) = x/O(x) + O(x) works too :DD
O(x) is an odd function
@@MathElite This is actually included in my family of solutions!
Since the product of two odd functions is an even function, we can choose
E(x) = x/O(x)
Substituting, we get
f(x) = x/(x/O(x)) + x/O(x)
= O(x) + x/O(x)
:-D
@@leickrobinson5186 ah nice
@@leickrobinson5186 Which means that the 2 solutions are equivalent.
Why a2=a3=...=an=0?
I think [ ] is symbol of Gauss.
Gauss?
👍
Sir! d/ dx( n f(x) = n. d/ dx( f(x))
Int. ( n. f(x)) = n Int of f(x)
Σ(nx^2) = n. Σ( x^2)
But f(nx) # nf(x) can't.
What if the function is not polynomial?
Every function is polynomial, thanks to the taylor series.
@@chaosredefined3834 you mean derivable at any degree?
👍👍👍
You did not really show that a2=a3=a4=....=0
I thought that was the floor of f(x) for a second
😁
f(x) = h(x) + g(x) h is event g is odd,
so : f(x) = h(x)+x/h(x)
chose h(x) when h for any x never equal 0
f(x)² - f(-x)² = 4 x
(h(x)+g(x))²-(h(x)-g(x))² = 4x
4 h(x) g(x) = 4x
sp : g(x) = x/ h(x)
so f(-x) = h(x) -x / h(x) ...
❤
🥰
More like, an add on to Leick's solution.
f(x) = x/g(x) + g(x)
for any non-zero g(x). try
f(x)= x/e^x + e^x
f(x) = x + x
f(x) = sin(x)
Um... where's the proof that f(x) has to be polynomial?!
Does not have to be. Check the comments for a more general solution
@@SyberMath Thanks for the hint!
I approached the equation from a slightly different angle than in the other comment:
Because -(-x)=x, our equation includes infinitely many equations for (x, -x) pairs, which are independant of each other. For example, if you solved the equation for x=1 and x=-1, then these two values are independant of every other x value.
So for a given value of x, we'd normally have two equations for the two values f(x) and f(-x). (The first equation is the original one, the second is given by replacing x with -x). Now there's something special here, because the equation does not change by replacing x with -x. So in fact, we only have one equation for two values, which is the very reason why we have infinitely many solutions!
Now I solved this by just solving for f(x):
f(x)=+/-sqrt(4x+f(-x)²),
which means that for any arbitrary function g: ]-inf; 0] --> IR, the function f: IR --> IR given by f(x)=...
...g(x) for x0 (Note that the sqrt always exists!)
solves the functional equation. The +/- can be chosen seperately for every value of x>0.
This solution should be equivalent to the solution in the other comment
f(x)=x/E(x)+E(x) with an arbitrary even function E.
@@dustinbachstein Wow!!! 🤩
No proof. I just found the polynomial solutions 😁
Lol dude you cheated with the assumption that f(x) is a polynomial. Where is the part where you discuss the other case?
It's in the comments section. 😁
I found a specific set of solutions
sir, with all due respect, you appear to be clueless as to what a functional equation is, you didn't even define the domain and codomain of the function which makes your equation ill-defined and your assumption that the function is a polynomial literally gave me a hemmorhage