Infinite Solutions to e^x = c?

[Frogeyedpeas]

Hello,

I have some background in complex analysis (a very minimal amount) but I did come up with a rather odd question.

Given a polynomial a + bx + cx^2 + dx^3... nx^n

There exists n or fewer solutions to the equation that each have a multiplicity of 1 to n.

Given that information suppose we take the exponential function e^x and break it down to its taylor series:

1 + x/1! + x^2/2! + x^3/3! + x^4/4!...

Doesn't that mean that there are infinite roots to the exponential function which may be equal to some type of infinity (or not)

The exponential-infinite roots of unity?

If they are positioned @ infinities then are there more "projective-like" relationships between them that allows you to differentiate between the roots of say e^x and 2^x?

 

[Sina]

polynomials are defined for for finite orders of n and so those theorems are not valid for infinite series.

 

[Frogeyedpeas]

Does that mean that theorems attached to them do not extend or generalize to infinite orders of n?

Example: The solution to the equation x³ + 1 = 0

We arrange it to:

x³ = -1

so x = [itex]\sqrt[3]{-1}[/itex] = -1 , [itex]\frac{1 - i\sqrt{3}}{2}[/itex], [itex]\frac{1 + i\sqrt{3}}{2}[/itex]

The negatives of the three roots of unity...

Generalizing this to nx³ + 1 = 0 as n [itex]\rightarrow[/itex] [itex]\infty[/itex] we know that the absolute value of the three roots of the equation approach zero but we do know that no matter the projective geometric properties of the roots (such as the angles between them) do not change...

Likewise suppose we extend the nth roots where the coefficients are becoming smaller at a rate of n!. There will be some type of circle forming where the angle between the roots becomes smaller at a very predictable (geometric) rate. So even though e^x has infinite roots, all located @ infinity. That doesn't mean that we cannot analyze and differentiate them from one another.

 

[Sina]

well I have never seen such an extension but these are not really the topics I am concentrating on so I will not make a certain statement about that :)

Also I do not see how you connected together "a polynomial with infinite order" with a polynomial with infinitely large coefficients

 

[Sina]

In any case if x is a complex number given as a+ib then you can write e^x as e^acos(b) + ie^asin(b). Now trying to solve e^x = c +id is the same thing as e^acos(b) = c and e^asin(b) = d. Then one method to solve this is look at the graph of z = e^xsin(y) and try to determine if it intersects the line z=d (and same for the other component too). In this case since your function has a sine term indeed for certain values of z it is possible to get infinite roots as it seems.

 

[Mute]

By this logic, one might expect

[tex]e^z = 0[/tex]

to have infinitely many solutions. In fact, it has none. ([itex]z = -\infty[/itex] doesn't really count, but even if it did, that would just be one solution, and [itex]1 \ll \infty[/itex]).

 

[jackmell]

Doesn't that mean that there are infinite roots to the exponential function which may be equal to some type of infinity (or not)

That's a reflection of Picard's theorems: a non-constant entire function that's not a polynomial reaches all values with at most one exception, infinitely often. In the case of e^z, that exception is zero.

 

[HallsofIvy]

In complex analysis, the exponential functions and the trig functions, sine and cosine, pretty much become the same function: [itex]e^{a+ bi}= e^a(cos(b)+ i sin(b))[/itex] and, of course, the trig functions have an infinite number of solutions (or none).