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cubic one root proof

For the cubic y = x^3 - Nx^2 +2Nx -4N
I was trying to prove that it has at most has one real root.

From my understanding a cubic can only have one real root if it
a) only has one turning point so the derivative only has one real solution - an increasing function or b) has no turning points which means its an increasing function so only hits the xaxis once
or c) has 2 turning points but theyre either both above or below the xaxis.

the hint for c was Prove that if f'(p) = f'(q) = 0 (for distinct reals p, q), then either f(p), f(q) >= 0 or f(p), f(q) <= 0

but im pretty stuck on part c - if anyone could help out that would be greatly appreciated.

(edited 9 months ago)

Working:

Original post by mosaurlodon

Working:

Where is the question from? I can possibly see a couple of ways of going, one doing a couple of transformations and a few arguments about the order of stationary points/adding polynomials etc and the other algebra heavy.

(edited 9 months ago)

Reply 3

mosaurlodon

it comes from this video https://youtu.be/xVtbj5U6KJg?si=HMb1NlUaSuiiskkx
Around 10:00
They mention it as an algebra exercise so I’m assuming they mean to do it using the algebra way

Reply 4

mqb2766

Original post by mosaurlodon

it comes from this video https://youtu.be/xVtbj5U6KJg?si=HMb1NlUaSuiiskkx
Around 10:00
They mention it as an algebra exercise so I’m assuming they mean to do it using the algebra way

Without watching it fully, it looks like they want you to do a cubic discriminant, so just google that. One thing to notice is that the quadratic, linear and constant coeffs are in a geometric sequence so you could do a transformation x -> -2x to get
-8x^3 - 4N (x^2 + x + 1)
and scale by -4
2x^3 + N(x^2 + x + 1)
or
Mx^3 + (x^2 + x + 1)
which is a bit friendlier to think about as the quadratic is positive etc.

(edited 9 months ago)

Reply 5

mqb2766

Original post by mqb2766

If you go with the last form
Mx^3 + (x^2 + x + 1)
then the discriminant is relatively simple and its easy to show that its negative for all M (not that the original form is that hard to do).

A bit more ad hoc, but assuming the roots vary smoothly as a function of M, then you could check to see if there is a value of M which corresponds to a repeated root and its fairly easy to show that there isnt.

Reply 6

mosaurlodon

I get that this problem is a lot easier to solve with the cubic discriminant way, its just that I was trying to do it the graphical way which is prob the method id go for if I got this question in an interview, in fact in my original post, the problem is just an exercise from 10:17 in the video.

so with Mx^3 + (x^2 + x + 1), its easier to do the discriminant with, and since its negative for all M it proves that theres not a repeated root - atleast I think thats what you mean with the "ad hoc" paragraph? I could definitely be misinterpreting that.

(edited 9 months ago)

Reply 7

mqb2766

Original post by mosaurlodon

I get that this problem is a lot easier to solve with the cubic discriminant way, its just that I was trying to do it the graphical way which is prob the method id go for if I got this question in an interview, in fact in my original post, the problem is just an exercise from 10:17 in the video.
so with Mx^3 + (x^2 + x + 1), its easier to do the discriminant with, and since its negative for all M it proves that theres not a repeated root - atleast I think thats what you mean with the "ad hoc" paragraph? I could definitely be misinterpreting that.

For the bespoke method, if M is small (both positive and negative) then
Mx^3 + (x^2 + x + 1)
must have a local minimum (positive value) roughly where the quadratic is minimum and so the cubic has a single crossing point as trivially the local maximum will also be positive.

So you have one (or two) values of M for which there is a single crossing point. The function changes smoothly as you alter M so reallly you want to show that there is no value of M which corresponds to a repeated root (so the minimum drops down and touches the x-axis). So this would be when
k(x-a)(x-b)^2
is equivalent to the cubic. So expand see if you can determine real a and b which will give the unity coefficients for the quadratic terms. It fairly quickly turns into a quadratic with complex roots so the cubic never touches the x-axis so the minimum (and maximum) is always above.

A bit of playing in desmos
https://www.desmos.com/calculator/fzazo9xxo1
shows that as long as the minimum of the quadratic is sufficiently above the x-axis, this seems to be a fairly general property, but if the minimum is just above the x-axis (small negative discriminant / imaginary values), then the addition of the cubic term can pull it below the x-axis, so its not a simple story and doing a bit of algebra to determine whether it happens is necessary. In the end, youre doing an ad-hoc cubic discriminant.

Im sure there are other ways to do it, but calculating the x values and subbing them back in didnt look like it was going to easily give. But didnt spend lots of time on it either as it looked like an algebra slog.

Reply 8

mosaurlodon

I see so using sum of roots, sum of two roots, product roots etc. I got a and b as complex so it can never have a repeated root.

I guess learning the cubic discriminant might save you a lot of trouble so i might have to get that one into memory.

The transformation of x^3 - Nx^2 +2Nx -4N -> Mx^3 + (x^2 + x + 1) is quite useful and really smart I wouldve never thought of that.
initially I was quite confused since I was like how can you take a function and 'transform' it? Then I realised its literally just the same function but with -2x.
I also didnt get how you can take out 'factors' e.g. -8x^3 - 4N (x^2 + x + 1) -> 2x^3 + N(x^2 + x + 1) but then I also realised you only care about the roots of the function not how it looks like, shape etc.

Thank you very much.

Reply 9

mqb2766

Original post by mosaurlodon

I see so using sum of roots, sum of two roots, product roots etc. I got a and b as complex so it can never have a repeated root.

I guess learning the cubic discriminant might save you a lot of trouble so i might have to get that one into memory.
The transformation of x^3 - Nx^2 +2Nx -4N -> Mx^3 + (x^2 + x + 1) is quite useful and really smart I wouldve never thought of that.
initially I was quite confused since I was like how can you take a function and 'transform' it? Then I realised its literally just the same function but with -2x.
I also didnt get how you can take out 'factors' e.g. -8x^3 - 4N (x^2 + x + 1) -> 2x^3 + N(x^2 + x + 1) but then I also realised you only care about the roots of the function not how it looks like, shape etc.
Thank you very much.

Cant guarantee its what they wanted, but the roots are a locus (function of N or M) so having one point on that locus (M small or N large) and showing that it never goes through the relevant point (touches the x-axis) is one way to crack the problem.

Note that the transformation(s) dont really affect the problem and strictly speaking youd also have to handle the trivial case of N=0 seperately. The one hidden advantage they have is that you think of it as a positive quadratic + varying cubic term and its easier to see what arguments you need to make, so the addition of the cubic term never pulls the positive minimum down to the x-axis. The original quadratic has the same properties, though its a bit harder to decide which arguments to make and the N positive/negative flips the quadratic so that has to be accounted for.

The geometric sequence for the quadratic coefficients so
a x^2 + ar x + ar^2
immediately implies that the roots cannot be real (discriminant, or map to x^2+x+1 as above) and indeed the complex, cubic roots are something like (1+/-sqrt(3)i) which if you chuck either of the cubics (in M or N) into wolfram, youll see something similar in the calculated cubic roots.

Tbh, putting a cubic in depressed form (like completing the square of a quadratic)
https://en.wikipedia.org/wiki/Cubic_equation#Depressed_cubic
and knowing about the cubic discriminants come in handy at times, so its worth having a bit of a read up about them.