Arriving at the fact that the polynomial is having infinitely many roots which is a contradiction was easy and which occurs a lot in many other problems. But ur way of handling the complex numbers was cool.
Here’s the naive way in which I found the solution. Take x=1, you then have f(1)*f(2)=f(3) or f(2)*(f(1)+1)=125. Now, in light of the given equalities, make the educated guesses that the function is growing with x and that coeficients are integers (big assumptions, I know). If you factorize 125, that indicates that f(1) should be 4 and f(2) should be 25. And then f(3) is 4*25=100. So, with f(0)=1, f(1)=4, f(2)=25, f(3)=100, it is evident to extrapolate that f(x)=(x^2+1)^2 and f(5)=676. Again, this was very unscientific, but it worked.
To show r=+-i, it is easier to write it as e^i×t1 (after showing |r|=1) then notice that r(2r^2+1) is also a root and should be in the form e^i×t2. We then get the equality : e^i×t2=e^i×t1(2e^2i×t1+1) Which gives 2e^2i×t1+1=e^i×(t2-t1) Right hand side has modulus = 1 For left hand side to have modulus 1, we must have 2t1 = pi +2n×pi Then we get t1=pi/2 (r=i) or t1=3pi/2 (r=-i)
I got stumped by this one, because I couldn't figure out the degree of the polynomial. I now see that that was probably just a bad way to try to solve this problem.
|2r^3 + r| > r is false - there is equality when r=0. So your argument rules out the existence of non-zero real roots. However, f(0) = 1, so 0 is not a root, so your original result of f having no real roots still holds. Additionally at 12:00, some clarification is needed. The fact the roots are +- i means that f(x) = (x+i)^p * (x-i)^q. This can take two forms: f(x) = (x^2+1)^a * (x+i)^b OR f(x) = (x^2+1)^a*(x-i)^b. Now using the condition that f(x) has real coefficients means f(x) is real for any real x. This means that in either case b=0, which is only a slight challenge to show. This then gives you f(x) = (x^2+1)^n for some n. Lovely work though, I just thought I would clarify some results :D
For any polynomial with real coefficients, its complex roots (if any exist) always come in conjugate pairs. That is enough to show that p = q = n. While that fact is unstated in the video, I think it's well known enough to be assumed for an audience who is familiar with complex roots.
This is how I did this: f(1)f(2)=f(3) f(1)f(2)+f(2)=f(3)+f(2) f(2)(f(1)+1)=125=5² * 5 Lets assume that f(2) > f(1), then we have f(2)=5² and f(1)=4=2² For f(3) we have: f(3)=125-f(2)=100=10² We already know that f(0)=1=1² Notice that for any x in f(x), our answer is of the form (g(x))² . Notice that the function g(x) is changing like this: 1 -> 2 -> 5 -> 10 So differences are: g(1)-g(0) = 1 g(2)-g(1) = 3 g(3)-g(2) = 5 Which means that each step, the change in difference is increasing like this: +2, +4,... compared to the first difference which is 1. So this means that the derivative of g(x) is 2x, so g’(x)=2x, which means that its a polynomial of the form ax²+bx+c, where a=1 and b=0. Now we have g(x)=x²+C, in order for g to satisfy the previous values, C has to be 1. So we have for the whole function: f(x)=(x²+1)² , and now for f(5): f(5)=(26)²=676
I think there was some more explaining that could have been done when you said s being a complex root implies 2s^3+s is also a root. We know that identity holds for real roots (if they existed) but why should it hold for complex roots? The reason is because since the identity holds for all real numbers, we know f(x)f(2x^2)=f(2x^3+x) AS POLYNOMIALS (as in all the coefficients must be the same), rather than just at individual x inputs. So the identity also holds in C[x] rather than just in R[x].
I agree that this was missing from the proof. Your claim, as I understand it, is: "given real value poly f, with for all x in R, f(x)f(2x^2)=f(2x^3+x), then f(X)*f(2X^2)=f(2X^3+X), with X as a placeholder rather than variable. i.e. that the LHS and RHS are polys with same coefficients." How do you prove this?
@@IsaacLevy Yes, so it turns out that if two polynomials agree on enough inputs, then they need to be the same polynomial. This is because if two polynomials f and g are both of degree at most n, and if they agree on n+1 inputs, then f-g is of degree at most n, yet has at least n+1 roots. Therefore, f-g must be the 0 polynomial.
You proved: If r is root of f(x) => 2r^3+r is root of f(x). However |2r^3 + r| = |r| |2r^2 + 1| >= |r|. Notice that r = 0 is possible, except that we know f(0) = 1, so r != 0. (This is an important note, because otherwise, the infinite sequence of roots would not hold.) Second, note that if r < 0, then r(2r^2+1) < 0, and if r > 0, then r(2r^2 + 1) > 0, so we have either a strictly decreasing, or else a strictly increasing sequence of numbers, which leads to an infinite number of roots.
It seems sort of “out of the box” to start with the root of the function f, when we are supposed to calculate its value at a given point. Any other approach to go about this?
Hmm, I don't think it's that "out of the box". In order to find f(5) you could either use some clever manipulations of f provided in the question, or you could just find f. As you know f is a polynomial you can try and determine the roots of f, as these define the polynomial. Seems sensible to me. What ideas do you have for an alternative approach out of interest?
I don’t have any at the moment. However, i think I understand the reason to look at roots of the polynomial. To determine a polynomial explicitly, it suffices to determine its roots along with the multiplicity. The multiplicity in this case is found by the value of f(2)+f(3) as shown here.
Here is a naive solution: (1) Let n be the degree of f(x); then the coefficient of x^n is a_n = lim_{x-->infinity} f(x)/x^n = lim_{x-->infinity} f(2x^3+x)/(f(2x^2)*x^n) = 1. (2) Since f(x)/f(2x^3+x) is an even function, it's reasonable to assume that f(x) is even or odd; the latter possibility can be discarded because f(0) is nonzero; therefore, f(x) = 1 + a_2*x^2 + ... + a_{2n-2}*x^{2n-2} + x^{2n}. (3) Now, let's assume --- without justification! --- that a_2, ... , a_{2n-2} are nonnegative; then 2^{2n}+3^{2n} < f(2)+f(3) = 125, which implies n
Your approach is completely incorrect and has several miscalculations. First: For any two polynomials g(x) and h(x), the degree of g(x)h(x) is the sum of the two degrees, NOT the product. Your degree calculations of degree((f'(x)f(2x²)) = (n - 1)2n = 2n² - 2n and degree(4xf(x)f'(2x²)) = n2(n-1) + 1 = 2n² - 2n + 1 are incorrect, and the correct values are both at 3n - 1. So you can not establish an equation of n to solve for n. Second: The two roots of 2n² - 5n + 2 = 0 are 2 and 1/2, NOT 4 and 1, since the denominator used in the quadratic formula should be 4, NOT 2.
I’d be surprised if more than a handful could do this problem. Why would you look at the roots of a polynomial when calculating values of it for a value of x?
In this case he already had the intuition that the polynomial must be a very special one and analyzing its roots reduced the solution space quite a bit. Functional equations can often be solved by plugging in values and learning bit by bit more about the function's properties, which is also used here. But sometimes this technique may leave too many possible choices for the right polynomial (I tried and didn‘t succeed applying the other conditions given). So often these problems want you to use simple properties of the mathematical objects at hand but in a very clever way. And for polynomials, roots are one of the first things you learn about them!
When dealing with polynomials its often times useful to find its roots and then write it is c(x-r_1)....(x-r_n) because the factored form is very nice to deal with
Think about the complex numbers and the goniometric circle: the complex numbers having modulus 1 are exactly those on the goniometric circle (cf. polar coordinates). Now also think back about what happens when complex numbers z_1 = e^{ theta_1 i } and z_2 = e^{ theta_2 i } are multiplied, we get as result z_1 z_2 = e^ { ( theta_1 + theta_2 ) i }. So, one just has to add the angles; it’s easy and visual. Finally, remember that the square root of a number is a number that, multiplied by itself, gives the original number. I hope this helps.
My complaint is about fast verbosity, hard to distinguish his ,r,2,3 in step-wise narrative arriving at the final lines of the solution. Why can't he explain in simple steps the strategy for solution after opening his toolbox?His style to claim separate space for solution(s)adds to the confusion for learners,(copying each step on paper) , who are used to traditional algebraic protocols of textbooks on Higher Algebra,such as authored by Bernard and Child.
The product of all the roots has modulus 1 (because f(0)=1) and no root can have modulus greater than 1. So, if a root has modulus < 1 then there is a contradiction.
@@angelvalera1997 only if the leading coefficient is +-1, the full reasoning is in the responses to too comment Edit: nvm that wasnt the question but i think it is the key observation
Let P(z)=az^3+bz^2+cz+d where a,b,c,d are unimodular complex numbers (|a|=|b|=|c|=|d|=1 ).Show that |P(z)| >=6 for at least one z€C such that |z|=1 Try to solve this problem inone of ur upcoming videos..m
Are you sure? If a=b=c=d=1, P(z) = (z^4 - 1)/(z - 1); if |z| = 1, we can write z = e^{it}, where t is real, so P(z) = (e^{4it} - 1)/(e^{it} - 1) = e^{3it/2}*sin(2t)/sin(t/2), therefore |P(z)| = |sin(2t)/sin(t/2)|
This was quite a hard problem for an AIME ... I guess a lot of people might have just guessed the polynomial and got it that way.
Arriving at the fact that the polynomial is having infinitely many roots which is a contradiction was easy and which occurs a lot in many other problems. But ur way of handling the complex numbers was cool.
Of all variable names he chose "r", the man's got balls! :)
Here’s the naive way in which I found the solution. Take x=1, you then have f(1)*f(2)=f(3) or f(2)*(f(1)+1)=125. Now, in light of the given equalities, make the educated guesses that the function is growing with x and that coeficients are integers (big assumptions, I know). If you factorize 125, that indicates that f(1) should be 4 and f(2) should be 25. And then f(3) is 4*25=100. So, with f(0)=1, f(1)=4, f(2)=25, f(3)=100, it is evident to extrapolate that f(x)=(x^2+1)^2 and f(5)=676. Again, this was very unscientific, but it worked.
To show r=+-i, it is easier to write it as e^i×t1 (after showing |r|=1) then notice that r(2r^2+1) is also a root and should be in the form e^i×t2.
We then get the equality : e^i×t2=e^i×t1(2e^2i×t1+1)
Which gives 2e^2i×t1+1=e^i×(t2-t1)
Right hand side has modulus = 1
For left hand side to have modulus 1, we must have 2t1 = pi +2n×pi
Then we get t1=pi/2 (r=i) or t1=3pi/2 (r=-i)
Amazing problem!! I was not able to solve it but i loved how you did it
I got stumped by this one, because I couldn't figure out the degree of the polynomial. I now see that that was probably just a bad way to try to solve this problem.
|2r^3 + r| > r is false - there is equality when r=0. So your argument rules out the existence of non-zero real roots. However, f(0) = 1, so 0 is not a root, so your original result of f having no real roots still holds.
Additionally at 12:00, some clarification is needed. The fact the roots are +- i means that f(x) = (x+i)^p * (x-i)^q. This can take two forms: f(x) = (x^2+1)^a * (x+i)^b OR f(x) = (x^2+1)^a*(x-i)^b. Now using the condition that f(x) has real coefficients means f(x) is real for any real x. This means that in either case b=0, which is only a slight challenge to show. This then gives you f(x) = (x^2+1)^n for some n.
Lovely work though, I just thought I would clarify some results :D
For any polynomial with real coefficients, its complex roots (if any exist) always come in conjugate pairs. That is enough to show that p = q = n. While that fact is unstated in the video, I think it's well known enough to be assumed for an audience who is familiar with complex roots.
@@n8cantor Of course! I figured I was missing something :)
Outstanding!
This is how I did this:
f(1)f(2)=f(3)
f(1)f(2)+f(2)=f(3)+f(2)
f(2)(f(1)+1)=125=5² * 5
Lets assume that f(2) > f(1), then we have
f(2)=5² and f(1)=4=2²
For f(3) we have:
f(3)=125-f(2)=100=10²
We already know that f(0)=1=1²
Notice that for any x in f(x), our answer is of the form (g(x))² . Notice that the function g(x) is changing like this:
1 -> 2 -> 5 -> 10
So differences are:
g(1)-g(0) = 1
g(2)-g(1) = 3
g(3)-g(2) = 5
Which means that each step, the change in difference is increasing like this: +2, +4,... compared to the first difference which is 1. So this means that the derivative of g(x) is 2x, so g’(x)=2x, which means that its a polynomial of the form ax²+bx+c, where a=1 and b=0. Now we have g(x)=x²+C, in order for g to satisfy the previous values, C has to be 1.
So we have for the whole function:
f(x)=(x²+1)² , and now for f(5):
f(5)=(26)²=676
I think there was some more explaining that could have been done when you said s being a complex root implies 2s^3+s is also a root. We know that identity holds for real roots (if they existed) but why should it hold for complex roots?
The reason is because since the identity holds for all real numbers, we know f(x)f(2x^2)=f(2x^3+x) AS POLYNOMIALS (as in all the coefficients must be the same), rather than just at individual x inputs. So the identity also holds in C[x] rather than just in R[x].
I agree that this was missing from the proof. Your claim, as I understand it, is: "given real value poly f, with for all x in R, f(x)f(2x^2)=f(2x^3+x), then f(X)*f(2X^2)=f(2X^3+X), with X as a placeholder rather than variable. i.e. that the LHS and RHS are polys with same coefficients." How do you prove this?
@@IsaacLevy Yes, so it turns out that if two polynomials agree on enough inputs, then they need to be the same polynomial. This is because if two polynomials f and g are both of degree at most n, and if they agree on n+1 inputs, then f-g is of degree at most n, yet has at least n+1 roots. Therefore, f-g must be the 0 polynomial.
You proved: If r is root of f(x) => 2r^3+r is root of f(x). However |2r^3 + r| = |r| |2r^2 + 1| >= |r|. Notice that r = 0 is possible, except that we know f(0) = 1, so r != 0. (This is an important note, because otherwise, the infinite sequence of roots would not hold.) Second, note that if r < 0, then r(2r^2+1) < 0, and if r > 0, then r(2r^2 + 1) > 0, so we have either a strictly decreasing, or else a strictly increasing sequence of numbers, which leads to an infinite number of roots.
So awesome, keep up the good work!
great video!
Claim there is no real roots. Then find complex roots. Use the condition to get final answer. Good job
Can u make discord server it would be amazing for whole community to interact and share good methods and problems to solve
I’m not terribly familiar with discord but I will try to make one! Heard that discord is interesting!
That would be awesome
Nice solution
It seems sort of “out of the box” to start with the root of the function f, when we are supposed to calculate its value at a given point. Any other approach to go about this?
Hmm, I don't think it's that "out of the box". In order to find f(5) you could either use some clever manipulations of f provided in the question, or you could just find f. As you know f is a polynomial you can try and determine the roots of f, as these define the polynomial. Seems sensible to me. What ideas do you have for an alternative approach out of interest?
I don’t have any at the moment. However, i think I understand the reason to look at roots of the polynomial. To determine a polynomial explicitly, it suffices to determine its roots along with the multiplicity. The multiplicity in this case is found by the value of f(2)+f(3) as shown here.
Here is a naive solution:
(1) Let n be the degree of f(x); then the coefficient of x^n is a_n = lim_{x-->infinity} f(x)/x^n = lim_{x-->infinity} f(2x^3+x)/(f(2x^2)*x^n) = 1.
(2) Since f(x)/f(2x^3+x) is an even function, it's reasonable to assume that f(x) is even or odd; the latter possibility can be discarded because f(0) is nonzero; therefore, f(x) = 1 + a_2*x^2 + ... + a_{2n-2}*x^{2n-2} + x^{2n}.
(3) Now, let's assume --- without justification! --- that a_2, ... , a_{2n-2} are nonnegative; then 2^{2n}+3^{2n} < f(2)+f(3) = 125, which implies n
Maths is beautiful
My solution: f polynomial is derivable. I derive the relation f'(x)f(2x²) + 4xf(x)f'(2x²) = (6x²+ 1)f'(2x^3 +x). now if n is the degree of polynomial function f, using the derived relation, we obtain:
degre(f'(x)f(2x²)) = (n - 1)2n = 2n² - 2n
degree(4xf(x)f'(2x²)) = n2(n-1) + 1 = 2n² - 2n + 1
degree(6x² + 1)f'(2x^3 + x)) = 3(n-1) + 2 = 3n - 1
2n² - 2n + 1 = 3n - 1 => 2n² - 5n + 2 = 0 => n = (5 + 3)/ 2 or (5 - 3)/2 => n = 4 or n = 1
if n = 1, f(x) = ax + b, f(0) = 1 => b = 1, the relation gives (ax + 1)(2ax² + 1) = 2ax^3 + ax + 1 => 2a²x^3 + 2ax² + ax + 1 = 2ax^3 + ax + 1 => a = 0 => f(x) = 1 or f(2) + f(3) = 125 => impossible
if n = 4 f(x) = ax^4 + bx^3 +cx² + dx + e, f(0) = 1 => e = 1. substituting in relation f(x)f(2x²) = f(2x^3 + x) gives finally after a boring computation the values a=1, b = 0, c = 2, d = 0, e = 1 => f(x) = (x² + 1)²
Notice that for n = 4 i do not use f(2) + f(3) = 125 ...
Your approach is completely incorrect and has several miscalculations. First: For any two polynomials g(x) and h(x), the degree of g(x)h(x) is the sum of the two degrees, NOT the product. Your degree calculations of degree((f'(x)f(2x²)) = (n - 1)2n = 2n² - 2n and degree(4xf(x)f'(2x²)) = n2(n-1) + 1 = 2n² - 2n + 1 are incorrect, and the correct values are both at 3n - 1. So you can not establish an equation of n to solve for n. Second: The two roots of 2n² - 5n + 2 = 0 are 2 and 1/2, NOT 4 and 1, since the denominator used in the quadratic formula should be 4, NOT 2.
I’d be surprised if more than a handful could do this problem. Why would you look at the roots of a polynomial when calculating values of it for a value of x?
In this case he already had the intuition that the polynomial must be a very special one and analyzing its roots reduced the solution space quite a bit.
Functional equations can often be solved by plugging in values and learning bit by bit more about the function's properties, which is also used here. But sometimes this technique may leave too many possible choices for the right polynomial (I tried and didn‘t succeed applying the other conditions given). So often these problems want you to use simple properties of the mathematical objects at hand but in a very clever way. And for polynomials, roots are one of the first things you learn about them!
When dealing with polynomials its often times useful to find its roots and then write it is c(x-r_1)....(x-r_n) because the factored form is very nice to deal with
But f should be monic for the product the roots is 1.
Why modulus of √r is 1??
Think about the complex numbers and the goniometric circle: the complex numbers having modulus 1 are exactly those on the goniometric circle (cf. polar coordinates).
Now also think back about what happens when complex numbers z_1 = e^{ theta_1 i } and z_2 = e^{ theta_2 i } are multiplied, we get as result z_1 z_2 = e^ { ( theta_1 + theta_2 ) i }. So, one just has to add the angles; it’s easy and visual.
Finally, remember that the square root of a number is a number that, multiplied by itself, gives the original number.
I hope this helps.
My complaint is about fast verbosity, hard to distinguish his ,r,2,3 in step-wise narrative arriving at the final lines of the solution. Why can't he explain in simple steps the strategy for solution after opening his toolbox?His style to claim separate space for solution(s)adds to the confusion for learners,(copying each step on paper) , who are used to traditional algebraic protocols of textbooks on Higher Algebra,such as authored by Bernard and Child.
Here f has a single real root and other roots are imaginary. We know how f looks.
How do we know that the modulus of all the roots is exactly 1?
The product of all the roots has modulus 1 (because f(0)=1) and no root can have modulus greater than 1. So, if a root has modulus < 1 then there is a contradiction.
@@angelvalera1997 thank you
@@angelvalera1997 only if the leading coefficient is +-1, the full reasoning is in the responses to too comment
Edit: nvm that wasnt the question but i think it is the key observation
.
Let P(z)=az^3+bz^2+cz+d where a,b,c,d are unimodular complex numbers (|a|=|b|=|c|=|d|=1 ).Show that |P(z)| >=6 for at least one z€C such that |z|=1
Try to solve this problem inone of ur upcoming videos..m
Are you sure? If a=b=c=d=1, P(z) = (z^4 - 1)/(z - 1); if |z| = 1, we can write z = e^{it}, where t is real, so P(z) = (e^{4it} - 1)/(e^{it} - 1) = e^{3it/2}*sin(2t)/sin(t/2), therefore |P(z)| = |sin(2t)/sin(t/2)|
is this from arthur engel book?
I don't think so but there are many similar question in it