Max number of roots of radical equation

[SarahHill]

I am doing some independent study and appreciate that a polynomial (in x) of integer degree (n) can have at most n roots; many proofs to this effect exist.

My query concerns the number of roots of equations in which the powers of x are not integers (or rational numbers) but irrational numbers.

How, for instance, would one determine the total number of roots (real and complex) of this equation

x^(pi) + 3*(x^e)+x^8.99999-50*x = 100000 ? (e=2.71... pi=3.14...)

(The actual roots don't concern me, I am more interested in knowing how to determine, analytically, the total number of roots)

Many thanks for any advice.

 

[SteamKing]

I don't think there is an App for that.

What you have is a transcendental equation rather than an algebraic polynomial.

 

[SarahHill]

I was under the impression that transcendental equations contain transcendental functions (like e^x, cos(x) etc). The equations I am requesting help with contain transcendental numbers, yes, but not functions; they are of the type x raised to some numerical power (rather like a polynomial).

 

[lurflurf]

That equation has one real and infinity complex roots. It is important that the exponents are irrational, that causes there to be infinity roots. Consider x^pi=1 for a simple example. It helps to recall x^pi=e^(pi log(x)). Also consider x^(2-1/10^6)=1 and x^(2-e/10^6)=1, they have many and infinity solutions respectively despite being near x^2=1 which has 2.

 

[Curious3141]

That equation has one real and infinity complex roots. It is important that the exponents are irrational, that causes there to be infinity roots. Consider x^pi=1 for a simple example. It helps to recall x^pi=e^(pi log(x)). Also consider x^(2-1/10^6)=1 and x^(2-e/10^6)=1, they have many and infinity solutions respectively despite being near x^2=1 which has 2.

OT, but lurflurf, you have exactly $\lfloor1000\varphi\rfloor$ posts. Now don't reply to this and spoil it.

 

[SarahHill]

Thanks LurfLurf.

I agree with you that it has an infinite number of roots (real + complex). Is there any way to determine (in general) how many REAL roots such equations have, without actually trying to solve it. An upper bound on the number of REAL roots would suffice.

Is there any mileage in my 'idea' that simply rounding all powers to the next integer up and thus producing a polynomial would give an upper bound on the number of REAL roots.

So

x^(pi)+3*x^(e)=100 becomes x^4 + 3*(x^3)=100, which can have no more than 4 REAL roots.

Once again, many thanks for all of you who have taken the time to reply.

S.H

 

[lurflurf]

These type of functions are called generalized polynomials, often we are interested in them for positive values. For positive roots the values of the terms does not matter, but the number of them does, this is the Generalized polynomial theorem see http://www.bowiestate.edu/UploadedFiles/academics/departments/Math/Faculty/rs/publications/More%20on%20the%20Lost%20Cousin.PDF and here. In your example the function is increasing by the time it reaches 100000, so it cannot equal it more than once. If your function were more complicated, it could be very difficult to find the number of real roots (if we want to improve the bound).

For complex roots my point above is that
x^1.999999=1 (many roots)
x^2=1 (2 roots)
x^(2-e/10^99) (infinite roots)

The fact that exponents are integers is why we have few roots. It is like walking around a circle, we always take infinity steps, but the length of the circle is and integer multiple of the step length all the steps land in the same places.

 

[SarahHill]

Thanks for those links, I think they will lead me to a solution to my query. I have never heard of the 'lost cousin' of the fundamental theorem of algebra, it sounds funny but also kind of obvious. I think the idea of the number of terms determining the number of positive roots is just what I need.

Thank you very much!