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Since $e^x$ can be represented by an infinitely long Taylor polynomial, and by the fundamental theory of algebra every degree-$n$ polynomial has $n$ roots, why doesn't $e^x$ have infinitely (or any) real or complex solution?

ryang
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Skully425
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    It's not a polynomial, so theorems about polynomials aren't obviously relevant. – lulu Jun 08 '23 at 20:55
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    An "infinitely long polynomial" (more commonly called a "power series") is not a polynomial, so the fundamental theorem of algebra doesn't apply – Ben Grossmann Jun 08 '23 at 20:56
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    @Skully For a function $f(x)$, solutions to the equation $f(x) = 0$ are called "roots of $f$" or "zeros of $f$", but not "solutions of $f$". – Ben Grossmann Jun 08 '23 at 20:58
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    Incidentally, it's a good exercise to pick a proof of the fundamental theorem of algebra and see exactly where it breaks down if we try to apply it to power series instead of genuine (finite) polynomials. For example, in the Liouville's-theorem-based proof that I first learned, we rely on the polynomial having a leading term in order to bound its reciprocal's magnitude outside a compact region. Power series don't have highest-degree terms! – Noah Schweber Jun 08 '23 at 21:01
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    @Skully You mean find the Picard theorems to be interesting. These tell you about the zeros of functions that can be written in the form of a power series (at least, they can be written in the form of a power series over certain regions of convergence for those series). – Ben Grossmann Jun 08 '23 at 21:01
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    This is a good question! Keep thinking about these things. – MJD Jun 09 '23 at 05:22
  • Actually polynomials have roots only over algebraically closed fields. In particular the theorem does not apply to $\mathbb{R}$. Secondly, congratulations, you just found a big difference between polynomials and "infinitely long polynomials", or what we would call: series. These are not the same, their properties do not coincide. – freakish Jun 09 '23 at 07:55
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    Also: https://math.stackexchange.com/q/2895800/42969 and https://math.stackexchange.com/q/1631886/42969. – Martin R Jun 09 '23 at 07:58
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    Indeed assuming $e^x=0$ true for some $x\in \mathbb R$ implies $e^x=0$ for any $x$. – user Jun 09 '23 at 08:21
  • $|e^z|=|e^{x+iy}|=|e^xe^{iy}|=|e^x||e^{iy}|=|e^x|>0$. – JMP Jun 09 '23 at 09:28

1 Answers1

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First, the commentaors are completely right in saying that a series wih infinitely many terms is not a polynomial, so theorems about polynomials are irrelevant. Second, your staement about "$e^x$ have no soluiton" is completely meaningless. Functions do not have solutions. Equations may or may not have solutions. You probably mean to ask about possible solutions o the equation $e^x=0$. Third suppose there is some $x$ such that $e^x=0$. Then $e^x e^{-x}=0e^{-x}=0$ but $e^xe^{-x}=e^{-x+x}=e^0=1$, a contradiction.

P. Lawrence
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    I think point 2 is making too much of a simple terminological error. OP may be trying to translate from their native language. Ben Grossman's approach in https://math.stackexchange.com/questions/4715136/why-does-ex-have-no-solution#comment9986365_4715136 seems to me to be more to the point. – MJD Jun 09 '23 at 05:21
  • So, we can make this into an elementary problem: if a function $F : \mathbb C \to \mathbb C$ satisfies $F(x+y)=F(x)F(y)$ for all $x,y$, then either $F$ is identically zero or $F$ is never zero. – GEdgar Jun 09 '23 at 07:57
  • @MJD I can not understand the answer . By definition $a^{-m}=\frac 1{a^m}$ so $e^{-x}=\frac 1{e^x}.$ Therefore, if $e^x=0$, then $e^{-x}$ is simply undefined. This turns to a circular logic : $0\cdot \frac 10=1$ A contradiction ? I would think this is not a contradiction. This operation is just undefined , I think . – lone student Jun 09 '23 at 08:00
  • @lonestudent: No, $e^{-x}=\frac 1{e^x}$ is not “by definition.” $e^x$ is defined as an infinite series (for all real or complex $x$), and so is $e^{-x}$. Then $e^x \cdot e^{-x} = 1$ can be proven using that definition, and it follows that no factor in that product can be zero. – But actually all this has been said before, the question is a duplicate IMO. – Martin R Jun 09 '23 at 08:02
  • @MartinR Thanks, but my highschool textbook says $a^{-m}=\frac 1{a^m}$ is definition of $a^{-m}$ . Now, I am confusing . – lone student Jun 09 '23 at 08:06
  • @lonestudent: Well, there are many different approaches to define $a^x$ for a general exponent $x$. One way is to define it for positive integer exponents first, then for arbitrary integer exponents, then for rational exponents, and finally for real exponents. – There are also different ways to define the Euler number, e.g. as $\sum_{n=0}^\infty 1/n!$ or as $\lim_{n\to \infty} (1+1/n)^n$. – Martin R Jun 09 '23 at 08:13
  • (Cont.) I am referring to the definition of the exponential function as $\exp(x) = \sum_{n=0}^\infty x^n/n!$, which can be shown to satisfy $\exp(x+y) = \exp(x) \cdot \exp(y)$ (and also $a^x = \exp(x \ln(x))$, which in turn can be used as the definition of $a^x$). – So it really depends on what you define in which order. – Martin R Jun 09 '23 at 08:14
  • @MartinR Thank you for your comments. I guess I understand your comments. If this were a high school-level exam question (with real exponent ), I would answer it like this : Obviously $x≠0$ . Then if $e^x=0$ then $e^{2x}=0^2=0$. This implies $e^x=e^{2x}$ . This folllows $x=2x\implies x=0 .$ A Contr. – lone student Jun 09 '23 at 08:26
  • @lonestudent I'm not sure I follow your proof. Why is it obvious that $x \neq 0$ in the beginning? – Accelerator Jun 09 '23 at 08:46
  • @lonestudent ...or that $e^x=0$ for any $x\neq 0$ – user Jun 09 '23 at 08:46
  • @Accelerator Because $a^0=1$ by definition of $a^{-m}=\frac {1}{a^m}.$ So $a^0=\frac {1}{a^0}\implies \left(a^0\right)^2=1\implies a^0=1$ . – lone student Jun 09 '23 at 08:51
  • @user yes, indeed . – lone student Jun 09 '23 at 08:53
  • @lonestudent I'm sorry, but I still don't follow. I don't know where $a$ came from, and it seems like you tried to prove that $a^0 = 1$ by supposing $a^0 = 1$ by definition? – Accelerator Jun 09 '23 at 09:10
  • @lonestudent I think it's better for you to forget about the $a$ thing because it's confusing. When you want to prove that $e^x \neq 0$ for any real (or even complex) $x$, you can't assume what you want to prove. It's like asking a friend to prove they can run faster than you because you're curious who's faster, but that friend just answers, "I can clearly run faster." I'm sure you would ask, "How? You didn't do anything." If knowing what to suppose and what to prove confuses you, you can try reading an introductory proofs textbook to understand. – Accelerator Jun 09 '23 at 09:35