# Infinite Solutions to e^x = c?

1. Nov 27, 2011

### 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?

2. Nov 27, 2011

### Sina

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

3. Nov 27, 2011

### 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 = $\sqrt[3]{-1}$ = -1 , $\frac{1 - i\sqrt{3}}{2}$, $\frac{1 + i\sqrt{3}}{2}$

The negatives of the three roots of unity...

Generalizing this to nx3 + 1 = 0 as n $\rightarrow$ $\infty$ 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 ex has infinite roots, all located @ infinity. That doesn't mean that we cannot analyze and differentiate them from one another.

4. Nov 27, 2011

### 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

5. Nov 27, 2011

### Sina

In any case if x is a complex number given as a+ib then you can write ex as eacos(b) + ieasin(b). Now trying to solve ex = c +id is the same thing as eacos(b) = c and easin(b) = d. Then one method to solve this is look at the graph of z = exsin(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.

6. Nov 27, 2011

### Mute

By this logic, one might expect

$$e^z = 0$$

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

7. Nov 28, 2011

### jackmell

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.

8. Nov 28, 2011

### HallsofIvy

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