8

One of my Calculus assignments asks how we can tell that $ y = e^{2x}$ is always positive.

The options given are:

  1. It's an exponential function with a base $> 0$.
  2. It's a square.
  3. Its derivative is always positive.

I said the only reason was 1. But I was told that number 2 is also a valid reason given that $y = e^{2x} = (e^x)^2$ is a square.

If $e$ were negative, however, wouldn't the function be negative at $x = 1/2$?

Is 2 a valid reason to conclude that $e^{2x}$ is always positive?

MITjanitor
  • 2,690
mowwwalker
  • 1,201
  • 1
    Assuming $x \in \mathbb{R}$? Because if $x = \frac { i \pi } 2$ then we know $y = -1$. – wchargin Mar 24 '13 at 21:15
  • 1
    @WChargin, edit: edit: yes. Got confused for a second there – mowwwalker Mar 24 '13 at 21:16
  • 1
    Trying to answer your last question, @Walkerneo: I don't think it is, since then we first should know that $,e^x\neq 0,$ , and for this we have to go back to (1)...again ! – DonAntonio Mar 24 '13 at 21:20
  • $$ e = 2.71828182845904523536028747135266249775724709369995... $$ http://en.wikipedia.org/wiki/E_%28mathematical_constant%29 $$$$ Why would $e$ be negative? – user13267 Mar 25 '13 at 00:25
  • @user13267, To conclude that the reason $e^{2x}$ is positive is because it's a square would mean that anything that is a square is positive, which isn't true for complex numbers. edit: Well, that's my thinking, and I don't really understand what's incorrect about it. – mowwwalker Mar 25 '13 at 00:48

8 Answers8

9

The reason is (1): any positive real number to any power is positive. Why? Definition and/or reason to define logarithms and stuff (I can't say what you know and what you don't...)

DonAntonio
  • 211,718
  • 17
  • 136
  • 287
  • That is to say that (2) isn't a valid reason? – mowwwalker Mar 24 '13 at 21:20
  • No... unless you know first that $,e^{x_0}\neq 0,$ for some value $,x_0,$ , since then you'd have $,(e^{x_0})^2=0^2=0,$ . See my comment under your question. – DonAntonio Mar 24 '13 at 21:23
  • Assuming it meant nonnegative instead of positive, would it then be valid? What I understand from the responses is that it's an incorrect reason because of the possibility of the result being $0$, but I thought it was incorrect because of the possibility of the result being negative. Is it not possible that a number $n$ to the power $2x$ is negative? – mowwwalker Mar 24 '13 at 21:31
  • again: any possitive number raised to any exponent is positive. – DonAntonio Mar 24 '13 at 21:36
  • Right, but (2) doesn't specify the sign of the number being squared. I wasn't aware prior to asking this question that, as Berci said, "sx is not defined (among reals) unless s≥0". The extent of what I've been taught about negative numbers raised to powers is that the result is positive for even powers and negative for odd ones. Is it implied that, if working with real numbers, a number that can be squared is positive? – mowwwalker Mar 24 '13 at 21:41
  • @Walkerneo, any real number can be squared and the outcome is non-negative, but not any real number can be raised to any exponent: only non-negative real numbers can be raised to any number. The number $,-1,$ cannot be raised, within the realm of the real numbers, to the exponent $,1/2,$ (why?) – DonAntonio Mar 24 '13 at 21:43
  • Because no real number squared is $-1$? I think what I'm trying to ask is: Is one of the reasons why (2) is incorrect because of the possibility of the base being negative, making the base raised to $x$ complex and its square not necessarily positive? – mowwwalker Mar 24 '13 at 21:58
6

You should be careful to note if your professor means "positive" or "nonnegative". A square (of a real number) cannot be negative, but it can be zero.

Elchanan Solomon
  • 30,807
4

The function $s^x$ is not defined (among reals) unless $s\ge 0$.

For an $s<0$, the term $(s^{1/2})^2$ is simply not defined (among reals).

Berci
  • 90,745
2

Indeed $e^{2x} = (e^x)^2$ is positive because it is a square of $e^x$ which is positive.

Or $e^{2x} = (e^2)^x$ is a positive function because the base is positive.

I might go with (1) since with (2) you have to also note that $e^x$ is positive.

As a sidenote and since it is sort of related: You have to be careful with negative base numbers. With positive base numbers the exponential is always defined and you get a real number. We can of course make sense of $(-1)^7$ which is negative or even $(-1)^{\frac{1}{2}} = i$, but we don't necessarily get a real number and so saying that it is positive doesn't even make sense. Also, how would you in general define a function $s^{x}$ for any negative $s$? What is for example $(-1)^\pi$ ?

The usual rules of exponents even get messed up. See for example $$ 1 = (1^2)^{\frac{1}{2}} = ((-1)^2)^{\frac{1}{2}} = ((-1)^{\frac{1}{2}})^2 = i^2 = -1. $$ What went wrong? See this Wikipedia article for a bit more about exponentiation.

Thomas
  • 43,555
2

Hint If the option 2. is true what can we say about

$$(-1)^{1}=(-1)^{2\times\frac{1}{2}}=i^2?$$

  • 1
    I don't think we're in $\Bbb C$ here, but rather $\Bbb R$. – Pedro Mar 24 '13 at 21:30
  • @PeterTamaroff forget the last equality, for $x=\frac{1}{2}$ we have $(-1)^{1}=(-1)^{2\times x}$, does this means $-1$ positive? –  Mar 24 '13 at 21:37
  • Well, when one says a number $a$ is a square it usually means $a=r^2$ for some real $r$. Of course every nonnegative real is a square, but negative reals are discarded. I am sure you know this already, though. – Pedro Mar 24 '13 at 21:52
  • @PeterTamaroff The question doesn't tell whether x is a real or a complex number – BЈовић Mar 25 '13 at 06:50
1

Point 2 is valid by itself for the purposes of providing a way to show that $e^{2x}$ (and, indeed, $e^x$) are nonnegative.

The fact than an exponential is nonnegative everywhere is inextricably linked to the fact that when we double its argument, its value squares!

However, point 2 contains an embedded assumption: that $e^x$, for real $x$ is a real value (either positive or negative, but not complex). It should be stated.

Suppose that some real-valued function $f$ has the property that $f(2x) = {f(x)}^2$, over all real $x$. Can $f(x)$ be negative? Clearly not, and it's easy to see if we rewrite the relation as $f(x) = {f(x/2)}^2$. For all $x$, $f(x)$ is the square of some real number, and so it must be nonnnegative.

The function $e^x$ is readily identified as being such an $f$.

Furthermore, if we have a function $g$ such that $g(x) = f(2x)$, then $g$ is just $f$ scaled down by a factor of two, horizontally. If $f$ is nonnegative (or positive), then $g$ is likewise, and vice versa, since horizontal scaling has no effect on that.

Point 2, however, requires additional arguments to that $e^x$ is never zero, and thus positive.

Kaz
  • 6,859
1

Assuming a number $x$ is positive if $sgn(x) \ge 0$ where $sgn(x)$ is the signum function For Complex number's an equivalent definition would be

$$ \operatorname{csgn}(z)= \begin{cases} 1 & \text{if } \Re(z) > 0, \\ -1 & \text{if } \Re(z) < 0, \\ sgn(\Im(z)) & \text{if } \Re(z) = 0 \end{cases} $$

Consider the expression $$y = e^{2x}$$ Let $x=0+i$, so we have $$\operatorname{csgn}(y) = \operatorname{csgn}(e^{2x})=\operatorname{csgn}(e^{2i})=-1$$

Abhijit
  • 2,544
0

No, 2) is not, because being a square does not exclude the possibility it is zero somewhere. But it is enough to guarantee it nonnegative.

The_Sympathizer
  • 19,651