4

The word "constant" is used in such expressions as "The derivative of a constant is $0$." What does it mean?

(I will post my own answer here, but I'm sure lots of others can have fun with their own answers.)

PS: I'd have thought this was obvious, but maybe not, judging by some comments: The phrase "The derivative of a constant is $0$" is Only an example, not the topic of the question.

Michael Hardy
  • 1
  • 6
    While grading an exam, I once found an assertion that $f'(5)$ is $0$ since $f'$ is a derivative and $5$ is a constant. – Michael Hardy Mar 25 '13 at 21:36
  • 1
    "The highschooler's dream" ?? – The Chaz 2.0 Mar 25 '13 at 21:40
  • @MichaelHardy I once witnessed a student, during an appeal session after an exam, arguing strongly the exact same thing! – Ittay Weiss Mar 25 '13 at 21:47
  • It is not at all unusual to see students thinking $f'(x^3)$ means the same thing as $\dfrac{d}{dx} x^3$. ${}\qquad{}$ – Michael Hardy Mar 25 '13 at 21:56
  • Ah.... students... I'm so happy to not be teaching/grading anymore. – Nik Bougalis Mar 25 '13 at 22:09
  • Apparently, you expect readers to infer key aspects of the meaning of your question from the answer you gave. This information should be provided in the question - not only in the answer. It is not clear what the intended scope of your question is (hence the algebraic perspecive I gave). – Math Gems Mar 25 '13 at 22:43
  • @MathGems : I don't expect anyone to infer the meaning of the question from anything besides the question. – Michael Hardy Mar 25 '13 at 22:52

4 Answers4

2

"Constant" always means "not depending on something" but what the "something" is depends on the context. When one says that $$ \text{If $c$ is a constant, then }\frac{d}{dx} (cf(x)) = c\frac{d}{dx} f(x), $$ then "constant" means not depending on $x$, i.e. $c$ does not change as $x$ changes.

Example:

If one wants to prove that $\dfrac{d}{dx} 2^x = (2^x\cdot\text{constant})$, one may write \begin{align} \frac{d}{dx} 2^x & = \lim_{h\to0}\frac{2^{x+h}-2^x}{h} = \lim_{h\to0}\left(2^x\cdot \frac{2^h-1}{h}\right) \\[12pt] & = 2^x\cdot\lim_{h\to0}\frac{2^h-1}{h} \text{ since $2^x$ is a constant} \\[12pt] & = (2^x\cdot\text{constant}) \text{ since the limit is a constant} \end{align}

  • In the first "since", the word "constant" means not depending on $h$. The factor $2^x$ does not change as $h$ goes to $0$.
  • In the second "since", the word "constant" means not depending on $x$. This constant remains the same as $x$ changes.

A moral: A leisurely account of what something is held not to depend on, sometimes as long as a whole sentence or maybe even two, can be worth calling the audience's attention to. It may what enables someone to understand something. One who hears only that something is a "constant" may miss the essential point.

Example: Suppose one want to prove that every function holomoprhic at a point $z_0$ in $\mathbb C$, i.e. complex-differentiable in some open neighborhood of $z_0$, is expressible as a convergent power series near $z_0$ ("Holomorphic functions are analytic."). One may start with something like $$ f(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(w)}{w-z}\, dw $$ and after some algebra and analysis one gets it to $$ \sum_{n=0}^\infty (z-z_0)^n \underbrace{\frac{1}{2\pi i}\int_\gamma \frac{f(w)}{(w-z_0)^{n+1}} \, dw} $$ The crucial thing to know about the expression over the $\underbrace{\text{underbrace}}$ is that the variable $z$ does not appear in it, i.e. it is "constant" as a function of $z$. It is not constant as a function of $n$, but that is not what matters here.

Michael Hardy
  • 1
  • The crux of it is "not changing [w.r.t. ___]" – The Chaz 2.0 Mar 25 '13 at 21:42
  • I think constants have to exist, as well! It's not a priori clear that $\lim_{h\to0}\frac{2^h-1}{h}$ is defined; I think you could argue that you've proven that "when $\dfrac{d}{dx} 2^x$ exists, it is equal to a constant times $2^x$". Or am I splitting hairs now? – yatima2975 Mar 25 '13 at 22:13
  • How one proves the existence of that limit seems somewhat off-topic here. Likewise the omitted algebraic and analytic details in the proof of analyticity. – Michael Hardy Mar 25 '13 at 22:15
  • On what context does the constancy of $\pi$ depend? I suppose, on that context which dictates that the greek letter denotes the ratio of circumference to diameter. – Kaz Mar 25 '13 at 22:34
2

A constant is one of these:

  • a literal object such as the scalar $3$ or the vector $\left[\begin{array}{ccc}0 & 0 & 0\end{array}\right]$.
  • a symbol denoting just one object, rather than a variable. For instance $\pi$.
  • a function (possibly multi-variable) which maps all domain values to the same object, essentially ignoring its arguments and producing a constant as its value
  • a formula consisting only of operations on constants
  • a multi-variable function in which some arguments are constrained to be constants, in some context, and whose value does not vary with the remaining arguments. (Sometimes constrained arguments are called parameters, and there is a semicolon notation for writing them).
  • a numeric approximation to some unchanging quantity found in nature, such as Planck's constant.
  • any one of the solutions to an equation, or the set itself. The constant $2$ can also be expressed as "the solution to $2x - 4 = 0$". (Solutions to equations are not always constants, of course).
Kaz
  • 6,859
  • I wonder if the concept of "constant" that includes such instances as $\pi$ and Planck's constant should be regarded as a separate concept from the one exemplified in my question and in my answer? – Michael Hardy Mar 25 '13 at 22:56
1

A constant is 0-ary function into some codomain.

It is often better to talk about a "constant in X" to specify which codomain X is being discussed, as that often determines it's properties with respect to the rest of an expression. A constant is often represented by the element of the codomain that the 0-ary function evaluates to.

ex0du5
  • 1,366
  • 8
  • 14
  • 1
    This seems a bit like a convention adopted for use when certain concepts are being used. – Michael Hardy Mar 25 '13 at 22:59
  • @MichaelHardy: I think this way of looking at constants prevents a lot of the confusion that has been mentioned in various other answers, and is quite generally applicable. It allows use at the level of logical expressions with quantifiers, including differentials and integrals. Function evaluations obey this definition (by currying the 0-ary functions), and generally this is a very consistent way to look at constants. – ex0du5 Mar 25 '13 at 23:28
  • I think it is more important to specify the domain than than the codomain, for reasons explained in my own answer. – Michael Hardy Mar 25 '13 at 23:52
0

Hint $\ $ TFAE for any operator $\rm\:f \to f'$ satisfying the Product Rule $\rm\:(fg)' = f'g + fg'$

$\rm(1)\quad (cf)'\! =\, cf'\ $ for all $\rm\:f$

$\rm(2)\quad c'\! = 0$

Proof $\ \ (1\Rightarrow 2)\ \ $ Putting $\rm\:f=1\:$ in $\,(1)\,$ gives $\rm\: c' = c 1' = 0\ $ by the Lemma below.
$\rm(2\Rightarrow 1)\ \ $ By the Product Rule $\rm\: (cf)'\! = c'f + cf' = cf'\ $ by $\rm\:c' = 0$.

Lemma $\rm\ \ 1' = 0.\ \ $ Proof $\ $ By the Product Rule$\rm\ \ (1 = 1\cdot 1)' \Rightarrow 1' = 1'+1'\Rightarrow\ 1' = 0.$

Math Gems
  • 19,574
  • Your two identities are true pointwise, so that if $c'(x_0)=0$ then $(cf)'(x_0)=cf'(x_0)$ even if $c$ is not constant. I'm not sure you've proved more than that, unless maybe you're doing differential algebra. At any rate, this answer is a bit (pun only slightly intended) tangential. – Michael Hardy Mar 25 '13 at 22:12
  • @Michael The point is that this equivalence holds quite generally, assuming only the Product Rule. This is indeed the meaning of "constant" in differential algebra. – Math Gems Mar 25 '13 at 22:23
  • But still it's at most tangentially related to the question. – Michael Hardy Mar 25 '13 at 22:24
  • @Michael If so, then you should be more specific in your question. You asked "what does it mean that the derivative of a constant is 0"? and you mentioned property $(1)$ as one meaning of constant in your answer. I showed how to view this very generally (purely algebraically). If you are seeking only answers in more specialized contexts then you should state that explicitly in your question. Currently it is tagged only "terminology" so the context appears quite general. – Math Gems Mar 25 '13 at 22:34
  • You're the one who's looking only at a specialized context. I certainly did not ask "what does it mean that the derivative of a constant is 0?", and I'm surprised that anyone could think I did. I intended the context to be quite general, not to be simply about derivatives of constants being $0$. ${}\qquad{}$ – Michael Hardy Mar 25 '13 at 22:37
  • I answered the only question posed in your original question. If you have other questions, or more general questions, then you should make them explicit in your question, not in your answer. – Math Gems Mar 25 '13 at 22:48
  • No, you answered something that was NOT posed. My question was explicit and I think it was clear. The phrase "such expressions as" means I'm identifying something as an example. I don't believe any reasonable person could read my question and think it's asking "what does it mean that the derivative of a constant is 0?". – Michael Hardy Mar 25 '13 at 22:51
  • The entire question is "The word "constant" is used in such expressions as 'The derivative of a constant is 0.' What does it mean?" But you give no clue whatsoever what the denotation of "such expressions" is. If a reader chooses one that does not match what you had in mind, then that is not a fault of the reader. Rather, it is due to the extreme imprecision of the question. – Math Gems Mar 25 '13 at 23:03