I am confused about the following:

The exponential function (by definition) is a function of the form $f(x)=a^x$ where $a>0$. However, when $a=1$, we get the constant function $f(x)=1^x=1$. Is the constant function $f(x)=1^x=1$ still considered an exponential function even though it does not have behave like an exponential function? Is the definition of the exponential function that I gave above (that I read in many textbooks) not entirely correct? Should we define the exponential function by:"a function of the form $f(x)=a^x$ where $a>0$ and $a\neq 1$"? I welcome any answer. Thanks!

  • $\begingroup$ I suppose that a lot of of the time, terminology is abused when it comes to trivial cases. $1^x$ (where $x\in\mathbb{C}$) ought to fall under the category of a polynomial, an integer, and an exponential too. But when talking about exponentials in general, we ignore trivial cases. I'm not sure though. $\endgroup$ – Myridium May 28 '14 at 16:51
  • $\begingroup$ Yes, its also sinusoidal, as it is sin(0x)+1. $\endgroup$ – Asimov May 28 '14 at 16:52
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    $\begingroup$ It depends on what properties you want an exponential function to have. If for example it is the inverse of a logarithmic function, then you certainly do not want to have to deal with $\log_1(x)$. But if all you want is continuity, $f(1)=a$ for some $a \gt 0$ and $f(x)f(y)=f(xy)$ then you may be happy to have $a=1$ as a possibility. $\endgroup$ – Henry May 28 '14 at 16:52

Well, an exponential function is one whose rate of change is proportional to its current $y$-value. i.e. one satisfying the differential equation $\frac{dy}{dx}=ky$ for some constant $k$.

$\frac{d}{dx}[a^x]=a^x\ln(a)=ka^x$ for $a \neq 0$.

Now, if $a=0$, i.e. if we want to evaluate $\frac{d}{dx}[1^x]$, we see that $\frac{d}{dx}[1^x]=1^x\ln(1)=1^x(0)=0 \neq k(1^x)$ (unless $k=0$, in which case $ \frac{dy}{dx}$ is not proportional to $y$).

So, essentially, your way is the way to go! As far as definition is concerned, an exponential function is any $f(x)=a^x$, where $a\in \Bbb{R^+} \setminus \{1\}$.

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    $\begingroup$ You give a perfectly reasonable definition, which is satisfied by the function $f(x)=1^x$, and then you spoil it by arbitrarily ruling out $k=0$. According to the definition in your first sentence, $f(x)=1^x$ is unambiguously an exponential function. $\endgroup$ – TonyK May 28 '14 at 17:09
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    $\begingroup$ @TonyK I thought that if the constant of proportionality is $0$ (i.e. if $a=b\times 0$), then $a$ and $b$ are not proportional). $\endgroup$ – beep-boop May 28 '14 at 19:01
  • $\begingroup$ That's a question of convention, again - just like the original question. I was about to post that I like thinking of $1^{x}$ as exponential BECAUSE it satisfies $\frac{d}{dx} 1^{x} = 1^{x} \ln(1) = 0$ for all $x$. $\endgroup$ – coolpapa May 28 '14 at 20:12
  • $\begingroup$ @coolpapa by that reasoning, any constant function $y=c$ should be exponential, since $\frac{d}{dx}[c]=0=0 \times c$. $\endgroup$ – beep-boop May 28 '14 at 20:14
  • $\begingroup$ I agree! $y = c$ is exponential, since it's of the form $y = c \cdot 1^{x}$. This is a semantic debate - it all depends on your definition of exponential functions. $\endgroup$ – coolpapa May 28 '14 at 20:18

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