Max number of roots of radical equation

In summary, the conversation discusses the number of roots of equations with irrational powers of x. It is mentioned that for transcendental equations, there can be an infinite number of roots, while rounding the powers to the next integer can give an upper bound for the number of real roots. The concept of generalized polynomials is also introduced, which can help in determining the number of real roots. The "lost cousin" of the fundamental theorem of algebra is also mentioned as a possible solution to the query.
  • #1
SarahHill
4
0
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.
 
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  • #2
I don't think there is an App for that.

What you have is a transcendental equation rather than an algebraic polynomial.
 
  • #3
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).
 
  • #4
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.
 
  • #5
lurflurf said:
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. :biggrin:
 

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  • #6
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
 
  • #7
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.
 
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  • #8
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!
 

FAQ: Max number of roots of radical equation

1. What is a radical equation?

A radical equation is an equation that contains a variable inside a radical symbol, such as √x or ∛y. These equations involve finding the value of the variable that makes the equation true.

2. How do you solve a radical equation?

To solve a radical equation, you need to isolate the radical expression on one side of the equation. Then, raise both sides of the equation to the appropriate power to eliminate the radical. Finally, solve for the variable using basic algebraic principles.

3. What is the maximum number of roots of a radical equation?

The maximum number of roots of a radical equation is equal to the index of the radical. For example, a square root (√) has a maximum of 2 roots, while a cube root (∛) has a maximum of 3 roots.

4. Can a radical equation have more than one solution?

Yes, a radical equation can have multiple solutions. This is because sometimes raising both sides of the equation to the appropriate power can result in extraneous solutions. It is important to check the solutions in the original equation to determine which are valid.

5. Are there any restrictions on the variable in a radical equation?

Yes, there are restrictions on the variable in a radical equation. The most common restriction is that the radicand (the expression inside the radical) cannot be negative. This is because the square root of a negative number is undefined in the real number system. Other restrictions may depend on the specific equation and must be determined by the solver.

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