3
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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 $a\in \Bbb{R^+} \setminus \{1\}$.

$\endgroup$
  • 6
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.