# Why does differentiating a polynomial reduce its degree by $1$?

This may seem a bit silly but I am wondering: can it intuitively be shown that the derivative of a polynomial is precisely 1 degree lower than itself? I understand the basics of calculus enough to appreciate the derivations of derivatives and I also can see why the slope of many rising functions (with many exceptions of course) kinda grow more slowly than the functions themselves (as x grows), but, frankly, I don't have the "duh" insight that you want to have.

In fact, I feel a bit unconvinced about the whole of calculus itself (though I am still at the very basics). I know that I hardly know anything but do any of you (additional question) know a nice and readable book/books which could reintroduce me to the subject on a more rigorous level?

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A prerequisite would be to have some intuition about what the degree of a polynomial is, which I certainly don't have. Do you? You did mention something about rates of divergence to infinity, so maybe one strategy would be a proof that $P'/P\to0$. – Jack M Feb 8 '14 at 20:14
The slope of a secant line through $(x,P(x))$ and $(a,P(a))$ is $\frac{P(x)-P(a)}{x-a}$. This is fairly easily seen to be a polynomial of degree $1$ less than the degree of $P(x)$, since $\frac{x^k-a^k}{x-a}=x^{k-1}+ax^{k-2}+\cdots +a^k$. But this hardly qualifies as intuitive, it is computational. – André Nicolas Feb 8 '14 at 20:20
I'd also ask about the nature of the ring from which the coefficients are taken. – hardmath Feb 8 '14 at 20:21
André Nicolas, shouldn't the last a have a power of (k-1)? – Just_a_fool Feb 8 '14 at 20:33
For your additional question I recommend Calculus by Spivak. – littleO Feb 8 '14 at 20:45

$p(x) = x^n$ is the $n$-dimensional volume of an $n$-dimensional cube with side length $x$.

If we add a thin layer of thickness $dx/2$ into each $n-1$-dimensional face of the cube, the new volume is $p(x+dx)$.

Because the $(n-1)$-dimensional volume (area if $n=3$) of each face is $x^{n-1}$, the added $n$-dimensional volume is $$p(x+dx) - p(x) \propto dx x^{n-1}$$ (RHS is just thickness $\cdot$ volume of side), so $$p'(x) \propto x^{n-1}.$$

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– lhf Feb 9 '14 at 1:14

I don't know an answer in general. But a special case seems to have some intuition behind it. A polynomial of degree n has n roots. Let our special case be when all the roots are real and unique. Then the roots of the derivative are those places where the sign of the slope changes and must be between the n roots. So it would seem that there must be n-1 roots for the derivative. Also the derivative of a polynomial is a polynomial (which I'm really only sure of algorithmically), so the derivative would be a polynomial of degree n-1. Hopefully someone can generalize this.

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Nice... this is Sturm's theorem – Mitch Jun 6 '14 at 16:58

I'm surprised no one has yet mentioned the connection to the difference operator on sequences of the form $P(1),$ $P(2),$ $P(3),$ $\ldots$ where $P$ is a polynomial with integer coefficients.

Example 1: Consider the sequence generated by $P(n) = 3n + 2,$ followed by the sequence generated by taking consecutive differences of the original sequence:

$$5, \;\; 8, \;\; 11, \;\; 14, \;\; 17, \;\; 20, \;\; 23, \;\; 26, \;\; \dots$$ $$3, \;\; 3, \;\; 3, \;\; 3, \;\; 3, \;\; 3, \;\; 3, \;\; \dots$$

Note that the original sequence is generated by a $1$st degree polynomial, while the sequence of consecutive differences is generated by a $0$th degree polynomial.

Example 2: Consider the sequence generated by $P(n) = n^2,$ followed by the sequence generated by taking consecutive differences of the original sequence, followed by the sequence generated by taking twice-consecutive differences of the original sequence:

$$1, \;\; 4, \;\; 9, \;\; 16, \;\; 25, \;\; 36, \;\; 49, \;\; 64, \;\; \dots$$ $$3, \;\; 5, \;\; 7, \;\; 9, \;\; 11, \;\; 13, \;\; 15, \;\; \dots$$ $$2, \;\; 2, \;\; 2, \;\; 2, \;\; 2, \;\; 2, \;\; \dots$$

Note that the original sequence is generated by a $2$nd degree polynomial, the sequence of consecutive differences is generated by a $1$st degree polynomial, and the sequence of twice-consecutive differences is generated by a $0$th degree polynomial.

In general, if you start with a sequence generated by a polynomial of degree $N$ with integer coefficients, then the sequence generated by consecutive differences is generated by a polynomial of degree $N-1$ with integer coefficients, the sequence generated by twice-consecutive differences is generated by a polynomial of degree $N-2$ with integer coefficients, etc.

What does this have to do with differentiation? The process above of taking consecutive differences is essentially that of taking a difference quotient where the function is the polynomial that generates the original sequence and $h=1.$ More precisely, the sequence of consecutive differences is generated by the polynomial $\frac{P(n+1) - P(n)}{1}.$ No, this is not a proof of what you're asking about, but it is certainly suggestive that differentiation might be expected to lower the degree of a polynomial by one.

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Very very roughly (for an ideal polynomial...):

• The number of roots of a polynomial is it's degree
• The derivative turns the local maximals/minimals of a polynomial into the new roots
• A maximimal/minimal point exists between two roots, so there is one fewer of them than roots
• So taking a derivative reduces the degree by 1
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This is exactly Mitchell's answer above. Why post the same thing again? – Marek Jun 6 '14 at 14:42