A very nice lemma from the method of differences: f(x)-4f(x+1)+6f(x+2)-4f(x+3)+f(x+4)=0 Holds for any cubic f. Indeed, you might notice the similarity to the binomial theorem, and I have proven (though RUclips won't let me link it) that similar relations hold whenever the degree of the corresponding binomial is higher than the degree of the polynomial. Needless to say, the particular statement I quoted destroys this question in two lines.
My approach to formalize how f(x) is composed is the following: Let be g(x)=f(x)-12 and h(x)=f(x)+12. We know that g(x)h(x)=C(x-1)(x-2)..(x-7). This tells us that 1) g and h have the same coefficients except for the constant term, and 2) C=k^2 for some k since C is the product of the leading terms of g and h (which are equal). Looking up to the x^2 term for both g and h, they must be equal, and this means we have to separate (x-1),(x-2),.., (x-7) in two groups such that it is constant in each group the sum of the constant terms (since in both side the coefficient of x^2 must be 12k). This lead to a unique possibility: 2,3,7 on a side and 1,5,6 on the other. Then the same reasoning.
I noticed the symmetry about point (4,0), so g(x)=f(x+4) has to be an odd function: g(x)=ax^3+bx, where g(1)=g(2)=-12, g(3)=12. This solves to a=2, b=-14. Now f(x)=g(x-4)=2(x-4)^3-14(x-4)=2x^3-24x^2+82x-72. And |f(0)|=72.
Perfect start! Then we can skip the closed form of f(x). ... This solves to a=2, b=-14. So g(x) = 2x(x^2-7). Now f(x)=g(x-4), So f(0) = g(-4) = 2(-4)((-4)^2-7) = -72. And |f(0)|=72
My solution was more tedious: I inputted all of the x-values into the standard cubic polynomial ax^3 + bx^2 + cx + d, and then solved the system of equations, you get four solutions. I then reconstructed the polynomials with those coefficients and was able to eliminate two solutions because they result in a constant that is not equal to zero, and the only solutions left were those that when you input zero, you get 72 and -72.
I did pretty much the same thing except I first horizontally translated the function to the left by 4, so I ended up with an odd function and eliminated d. That way I only needed to solve one system of equations, and after solving it I realised I could’ve eliminated bx^2 as well.
Wow! What a neat solution! I tried to solve it algebraically, trying to obtain additional equation by using the property "Every cubic function is symmetric with respect to its inflection point" with the point (4,0) being the inflection point by the symmetry of the |f(x)| values. But your approach is way simpler!
My solution was to notice that this is (anti)symmetric around x=4, so I introduced g(x)=f(x+4), which is an odd function and |f(0)|=|g(±4)|. Since g(x)=-g(-x), g(x) has form g(x)=ax+bx³. Using this and g(1)=g(2)=±12, one gets g(x)=±2x(7-x²). One can also use g(3)=∓12, just to check, but that part is redundant. Finally, using this, one gets |f(0)|=|g(±4)|=72 and this can all be easily done in head.
I did a system of equations, with 4 of the 6 inputs provided and got the coefficients for the polinomial (didnt need to go that far, but might as well) Got the “d” term, and was -72 for a>0, so yeah
i got as far as the figure and that f(1)=f(5)=f(6)=12 (w.lo.g ) but did not see this trick f(x)-12 has these as solutions .. massive and really neat shortcut to the solution.
@@lgooch yea but this is about a polynomial so that's not something that would.occur to anyone..we don't even know if we have any negstivr tersm here..of ckirse I know we swuare numbers in general tonget rid of negstives but not here..
This is not a correct answer, because if a = 0, then f(x) is not a cubic polynomial in x, by definition. A cubic polynomial in x over a ring R is defined as ax^3 + bx^2 + cx + d, where a, b, c, d are elements of R, and a is nonzero.
A very nice lemma from the method of differences:
f(x)-4f(x+1)+6f(x+2)-4f(x+3)+f(x+4)=0
Holds for any cubic f. Indeed, you might notice the similarity to the binomial theorem, and I have proven (though RUclips won't let me link it) that similar relations hold whenever the degree of the corresponding binomial is higher than the degree of the polynomial.
Needless to say, the particular statement I quoted destroys this question in two lines.
If you dont mind, can you name the channel who proofed that?
My approach to formalize how f(x) is composed is the following:
Let be g(x)=f(x)-12 and h(x)=f(x)+12.
We know that g(x)h(x)=C(x-1)(x-2)..(x-7). This tells us that 1) g and h have the same coefficients except for the constant term, and 2) C=k^2 for some k since C is the product of the leading terms of g and h (which are equal).
Looking up to the x^2 term for both g and h, they must be equal, and this means we have to separate (x-1),(x-2),.., (x-7) in two groups such that it is constant in each group the sum of the constant terms (since in both side the coefficient of x^2 must be 12k).
This lead to a unique possibility: 2,3,7 on a side and 1,5,6 on the other.
Then the same reasoning.
I noticed the symmetry about point (4,0), so g(x)=f(x+4) has to be an odd function: g(x)=ax^3+bx, where g(1)=g(2)=-12, g(3)=12. This solves to a=2, b=-14. Now f(x)=g(x-4)=2(x-4)^3-14(x-4)=2x^3-24x^2+82x-72. And |f(0)|=72.
Perfect start! Then we can skip the closed form of f(x).
... This solves to a=2, b=-14. So g(x) = 2x(x^2-7). Now f(x)=g(x-4), So f(0) = g(-4) = 2(-4)((-4)^2-7) = -72. And |f(0)|=72
My solution was more tedious: I inputted all of the x-values into the standard cubic polynomial ax^3 + bx^2 + cx + d, and then solved the system of equations, you get four solutions. I then reconstructed the polynomials with those coefficients and was able to eliminate two solutions because they result in a constant that is not equal to zero, and the only solutions left were those that when you input zero, you get 72 and -72.
I did pretty much the same thing except I first horizontally translated the function to the left by 4, so I ended up with an odd function and eliminated d. That way I only needed to solve one system of equations, and after solving it I realised I could’ve eliminated bx^2 as well.
I did that and got very easily that
f(x) = +/- { -2(x-4)^3 + 14(x-3) }.
I did this too..isn't this what most ppl would do..WHY ON EARTH WOULD YOU SQUARE AT ALL??
Wow! What a neat solution!
I tried to solve it algebraically, trying to obtain additional equation by using the property "Every cubic function is symmetric with respect to its inflection point" with the point (4,0) being the inflection point by the symmetry of the |f(x)| values. But your approach is way simpler!
Yes, the video's initial approach combined with f(4) = 0 lets you solve for that C where he got "stuck" and shifted gears.
My solution was to notice that this is (anti)symmetric around x=4, so I introduced g(x)=f(x+4), which is an odd function and |f(0)|=|g(±4)|. Since g(x)=-g(-x), g(x) has form g(x)=ax+bx³. Using this and g(1)=g(2)=±12, one gets g(x)=±2x(7-x²). One can also use g(3)=∓12, just to check, but that part is redundant. Finally, using this, one gets |f(0)|=|g(±4)|=72 and this can all be easily done in head.
I did a system of equations, with 4 of the 6 inputs provided and got the coefficients for the polinomial (didnt need to go that far, but might as well)
Got the “d” term, and was -72 for a>0, so yeah
Beautiful problem.
This was an awesome question.
Wonderful solution :D! Thank you :D!
You are from Hong Kong :)
This was a pretty good question 👍👍👍
i got as far as the figure and that f(1)=f(5)=f(6)=12 (w.lo.g ) but did not see this trick f(x)-12 has these as solutions .. massive and really neat shortcut to the solution.
Very interesting problem....
I've been thinking of solving the linear equation raised by the coefficients of f(x)^2 the whole time...
Amazing question
very nice
Why do you write solution/answer like an essay?
Neat solution
Hey iam following you ☺️☺️☺️☺️ upload video regular please based on number theory
Really nice problem
Very good
🔥
uncle roger do mathematic
WHY ON EARTH would anyone think of squaring the function..I don't think anyone would think of that..
We square to ignore the absolute value, since a real number squared is always positive
@@lgooch yea but this is about a polynomial so that's not something that would.occur to anyone..we don't even know if we have any negstivr tersm here..of ckirse I know we swuare numbers in general tonget rid of negstives but not here..
@@leif1075 we can’t assume there isn’t any negative terms so it makes it easier to just square the function
liky
ez
Why did he claim the roots to be: 1,2,3,5,6,7?
Can anyone give an explanation, please?
These are roots for g(x) = f(x)^2-144.
And that's because by hypothesis g(1)=g(2)=..g(7)=0
Does that have something to do with the degree of polynomial?
@@AdityaKarwa yes, a polynomial has exactly as many (complex) roots as its degree. This is called the Fundamental Theorem of Algebra.
12. Using a=b=c=0 and d=12, it holds. Since it does not specify anything else, if it is anything, it is 12.
This is not a correct answer, because if a = 0, then f(x) is not a cubic polynomial in x, by definition. A cubic polynomial in x over a ring R is defined as ax^3 + bx^2 + cx + d, where a, b, c, d are elements of R, and a is nonzero.