Conversion of Irrational roots for cubic functions and higher

[WhiteTrout]

I have been seeing a few during in my practice questions which leaves me worrying.
If it is a quadratic function, the irrational numbers can be easily obtained using the equation.

However, I got a question today which eventually took this form:

28D³+36D²-41D²+4 = 0
(I reevaluated everything a few times and I think it is correct)

And it reminded me of various other instances where this also happened.

I punched the numbers into my calculator (still using a Casio fx-570MS) but all hell broke loose when all three roots are irrational.

So, I would like to ask if there is any way to convert an irrational root into a number in terms of its squareroot? Lecturer assured there shouldn't be many of this kind of questions around, but at least it is good to be informed.

The easier way is to get a more up to date calculator. I recall there was one or two capable of converting to squareroots.

 

[ehild]

Is not it

28D³+36D²-41D¹+4 = 0?

Before deciding that all roots are irrational (it is impossible for a third-order equation) try to find the extrema and the limits both at -∞ and ∞. If a continuous function is negative somewhere and positive somewhere else, it has to be zero in a point in between.

ehild

 

[HallsofIvy]

You can use the "rational root theorem". Any rational root of that equation must be of the form 1/p where p is an integer factor of 28. Those are, of course, \pm 1, \pm 1/2, \pm 1/4, \pm 1/7, \pm 1/14, \pm 1/28.

 

[LCKurtz]

28D³+36D²-41D¹+4 = 0?

Before deciding that all roots are irrational (it is impossible for a third-order equation) try to find the extrema and the limits both at -∞ and ∞.

Presumably you mean not all the roots are complex.

 

[HallsofIvy]

I wondered that myself but I think he meant that, for an equation like this, with integer coefficients, all three roots cannot be real, irrational numbers (the irrational numbers being a subset of the rational numbers). If \alpha, \beta, and \gamma, are roots, then the equation is of the form a(x- \alpha)(x- \beta)(x- \gamma)= ax^3+ a(\alpha+\beta+ \gamma)x^2+ a(\alpha\beta+ \alpha\gamma+ \beta\gamma)x+ a\alpha\beta\gamma= 0. If all three of \alpha, \beta, and \gamma are irrational, the coefficients cannot all be integer.

 

[LCKurtz]

I wondered that myself but I think he meant that, for an equation like this, with integer coefficients, all three roots cannot be real, irrational numbers (the irrational numbers being a subset of the rational numbers). If \alpha, \beta, and \gamma, are roots, then the equation is of the form a(x- \alpha)(x- \beta)(x- \gamma)= ax^3+ a(\alpha+\beta+ \gamma)x^2+ a(\alpha\beta+ \alpha\gamma+ \beta\gamma)x+ a\alpha\beta\gamma= 0. If all three of \alpha, \beta, and \gamma are irrational, the coefficients cannot all be integer.

Yes, I agree that may very well be what he meant.

 

[ehild]

Is not it

28D³+36D²-41D¹+4 = 0?

Before deciding that all roots are [STRIKE]irrational [/STRIKE]

I meant complex instead of irrational, silly me.
ehild

 

[Char. Limit]

I meant complex instead of irrational, silly me.
ehild

The funny thing is, Halls of Ivy, it seems to me, proved your original statement correct.

 

[ehild]

The funny thing is, Halls of Ivy, it seems to me, proved your original statement correct.

All roots looks irrational solved by Mapple {x = -2.038433315}, {x = .1088438114}, {x = .6438752182}. From the possible rational roots p/q (p, q are relative primes, p is integer divisor of 4 and q is that of 28) none is close to these roots.

ehild

 

[Char. Limit]

All roots looks irrational solved by Mapple {x = -2.038433315}, {x = .1088438114}, {x = .6438752182}. From the possible rational roots p/q (p, q are relative primes, p is integer divisor of 4 and q is that of 28) none is close to these roots.

ehild

Ahhh... then what's wrong with HallsOfIvy's statement? Because if there's no rational roots, his statement must be wrong.

 

[ehild]

He did not prove it, just said that my wrong statement might have been be true which was not.

ehild

 

[HallsofIvy]

What I showed was that if all three roots of a cubic equation are real and irrational then at least one coefficient must be irrrational also- so a cubic equation like this, with integer coefficients, canot have all three roots irrational.

 

[I like Serena]

@HallsofIvy: Interesting. But I don't quite get it.
You have 3 expressions for a₀, a₁, and a₂, with 3 possibly irrational unknown roots.
Why would there not be a solution for that with integer coefficients?

Especially since the problem at hand appears to have 3 irrational roots.

Actually, I thought you proved that all roots had to be irrational.
Because if any root were rational, it would have to be one of the fractions you mentioned, but none of those satisfy the equation.
At least, that is what I got from your previous explanation and from the wiki page: http://en.wikipedia.org/wiki/Rational_root_theorem.

@OP: Sorry to be diverging from your problem.

 

[gb7nash]

If the rational root test fails, this only means that there are no rational solutions. You can still have irrational or complex-valued solutions.

 

[WhiteTrout]

@OP: Sorry to be diverging from your problem.

No worries, the resulting discussion is interesting to read actually; in a way it answers the question pretty well.
Please do carry on.

Otherwise I was just speculating if the answer be in terms of
[[a/\sqrt[3]{n}];[b/\sqrt[3]{n}];[c/\sqrt[3]{n}]
Or anything similar.

Uhm. But referring to previous posts... Probably not that simple.
Didn't think that one out too well did I.

 

[uart]

Hi WhiteTrout, you still haven't answered if the equation you're considering is,

28D³-5D²+4 = 0?

or

28D³+36D²-41D+4 = 0?

 

[HallsofIvy]

If \alpha, \beta, and \gamma are roots of a cubic, it must be of the form
a(x- \alpha)(x- \beta)(x- \gamma)= ax^3- a(\alpha+\beta+ \gamma)x+ (\alpha\beta+ a\alpha\gamma+ \beta\gamma)x+ a\alpha\beta\gamma= 0

Now, is it possible for \alpha, \beta, and \gamma to be irrational if a, a(\alpha+ \beta+ \gamma), a(\alpha\beta+ \alpha\gamma+ \beta\gamma, and a\alpha\beta\gamma are all rational?

 

[WhiteTrout]

@uart: Quite sure it is the one in the OP.

@HallsofIvy:
Thank you for reminding me about that post too.
From referring to the thread itself, it seems unlikely for α, β, and γ to hold irrational numbers when the coefficient themselves are rational.

But in that case, the resultant answers from various calculators still leaves me baffled.
It almost seem like the formulation of the question was fired blindly, or perhaps some sort of typo; Which could probably say quite a lot about this.

If it helps the original question looks like this
28y'''' - 20y''' -113y'' +82y' +4y' -8y =0

For some reason, 82y' and 4y' are separated in the question. I unwittingly combined them because they are of the same term.
Converting it into a quartic function in terms of D, the first solution is D= 2
Synthetic division reduces it into

28D³+36D²-41D¹+4 =0

Otherwise, (82+4)y' could seem like the problem, then maybe it did not correctly belong there?

 

[uart]

@uart: Quite sure it is the one in the OP.

Then please read your OP again. The one in the OP is :

28D³+36D²-41D²+4 = 28D³-5D²+4

 

[vela]

If the original problem was 28y^{(5)} - 20y^{(4)} -113y^{(3)} +82y^{(2)} +4y^{(1)} -8y =0 the roots are all rational, so I think the original problem suffers from typos.

 

[LCKurtz]

If \alpha, \beta, and \gamma are roots of a cubic, it must be of the form
a(x- \alpha)(x- \beta)(x- \gamma)= ax^3- a(\alpha+\beta+ \gamma)x+ (\alpha\beta+ a\alpha\gamma+ \beta\gamma)x+ a\alpha\beta\gamma= 0

Now, is it possible for \alpha, \beta, and \gamma to be irrational if a, a(\alpha+ \beta+ \gamma), a(\alpha\beta+ \alpha\gamma+ \beta\gamma, and a\alpha\beta\gamma are all rational?

Apparently, yes. The cubic in question has integer coefficients:

f(x)=28x³+36x²-41x+4

And since

f(-3)=-305, f(-1)=53, f(1/2)=-4, f(1)=27

it has three real roots. None of the candidates given by the rational root theorem are roots so all three roots are irrational.

 

[SammyS]

Leave it to LCKurtz to give a nice direct method to show that indeed the roots of f(x) = 28x³+36x²-41x+4 are irrational; notwithstanding that the coefficients are integers.

I've spent some time on the following polynomial and used WolframAlpha to find its roots and then express them without using ,i, the imaginary unit .

The polynomial has the following zeros:

1-\sqrt{6} \sin((1/3) \tan^{-1}(\sqrt{23}/3))-\sqrt{2} \cos((1/3) \tan^{-1}(\sqrt{23}/3))

1+\sqrt{6} \sin((1/3) \tan^{-1}(\sqrt{23}/3))-\sqrt{2} \cos((1/3) \tan^{-1}(\sqrt{23}/3))

1+2\sqrt{2} \cos((1/3) \tan^{-1}(\sqrt{23}/3))

 

[WhiteTrout]

@uart: Quite sure it is the one in the OP.

Then please read your OP again. The one in the OP is :

28D³+36D²-41D²+4 = 28D³-5D²+4

Thank you for pointing that out. I admit that I was careless on my behalf and caused unnecessary quagmire.

@Vela: Yes, that should pretty much clear it up then. Sorry for all the trouble, but I was unsure what to make of the original question at that moment...

@LCKurtz: Ah, alright. Thanks for clearing the clouds. The "answers" look more reasonable in this light now.

@SammyS: Woah, that is very elaborate of you. Thank you for going through the hassle to formulate that, and your time too. I should consider trying to replicate that for personal interests.

Thanks to everybody for donating their time into this question.
I've got acquainted to some concepts that might be useful later; I have no regrets in making this thread even though the source of the problem itself was accidental.

 

[I like Serena]

@LCKurtz: Thanks for the confirmation of what I was already thinking.

@SammyS: Nice!

You've obviously laboured to the point where it is obvious that they are all real, and that if you add them, the sum will turn out as an integer.
I take it on faith that the other coefficients will turn out as integers as well. :)

I think it's obvious that they are all irrational.

I wonder though if they are transcendental as well...?

 

[LCKurtz]

Leave it to LCKurtz to give a nice direct method to show that indeed the roots of f(x) = 28x³+36x²-41x+4 are irrational; notwithstanding that the coefficients are integers.

I've spent some time on the following polynomial and used WolframAlpha to find its roots and then express them without using ,i, the imaginary unit .

The polynomial has the following zeros:

1-\sqrt{6} \sin((1/3) \tan^{-1}(\sqrt{23}/3))-\sqrt{2} \cos((1/3) \tan^{-1}(\sqrt{23}/3))

1+\sqrt{6} \sin((1/3) \tan^{-1}(\sqrt{23}/3))-\sqrt{2} \cos((1/3) \tan^{-1}(\sqrt{23}/3))

1+2\sqrt{2} \cos((1/3) \tan^{-1}(\sqrt{23}/3))

At this point we could leave out all the context and pose the following challenge problem (for which we, of course, have a proof ):

Given the three numbers:
p=1-\sqrt{6} \sin((1/3) \tan^{-1}(\sqrt{23}/3))-\sqrt{2} \cos((1/3) \tan^{-1}(\sqrt{23}/3))

q=1+\sqrt{6} \sin((1/3) \tan^{-1}(\sqrt{23}/3))-\sqrt{2} \cos((1/3) \tan^{-1}(\sqrt{23}/3))

r=1+2\sqrt{2} \cos((1/3) \tan^{-1}(\sqrt{23}/3))

show pqr is rational.

[Edit] I was assuming your roots were the roots of 28x³+36x²-41x+4.

but apparently they are the roots of
x^3-3x^2-3x+2

but the problem still stands.

 

[I like Serena]

Hmm, I just plugged the number into WolframAlpha who says that pqr = -2 (exact output).
(Does that count as a proof? )

However, looking at the solution, we have:
p + q + r = 3
28 = a₃ = a
36 = a₂ = -a (p + q + r) = -3a
4 = a₀ = -a pqr = 2a

Something seems to be wrong here!

 

[vela]

I tried doing what SammyS did, and I got the roots to be
\begin{align*}
x_1 &= -\frac{3}{7}+\frac{1}{7}\sqrt{\frac{395}{3}}\cos \theta \\
x_2 &= -\frac{3}{7}+\frac{1}{7}\sqrt{\frac{395}{3}}\cos (\theta+2\pi/3) \\
x_3 &= -\frac{3}{7}+\frac{1}{7}\sqrt{\frac{395}{3}}\cos (\theta-2\pi/3)
\end{align*}
where \theta = \frac{1}{3}\arctan\left(\frac{14}{3819}\sqrt{\frac{91202}{3}}\right)
These expressions numerically match what Mathematica spits out if you ask it to find the roots of the polynomial directly.