Calculus with polynomials

From Wikipedia, the free encyclopedia

In mathematics, polynomials are perhaps the simplest functions with which to do calculus. Their derivatives and indefinite integrals are given by the following rules:

$\left( \sum^n_{k=0} a_k x^k\right)' = \sum^n_{k=0} ka_kx^{k-1}$

and

$\int\!\left( \sum^n_{k=0} a_k x^k\right)\,dx= \sum^n_{k=0} \frac{a_k x^{k+1}}{k+1} + C.$

Hence, the derivative of $x 100$ is $100 x 99$ and the indefinite integral of $x 100$ is $\frac{x^{101}}{101}+C$ where C is an arbitrary constant of integration.

This article will state and prove the power rule for differentiation, and then use it to prove these two formulas.

1 The power rule
2 Proof of the power rule
3 Differentiation of arbitrary polynomials
4 Generalization
5 See also
6 References

[edit] The power rule

The power rule for differentiation states that for every natural number n, the derivative of $f(x)=x^n \!$ is $f'(x)=nx^{n-1},\!$ that is,

$\left(x^n\right)'=nx^{n-1}.$

The power rule for integration

$\int\! x^n \, dx=\frac{x^{n+1}}{n+1}+C$

for natural n is then an easy consequence. One just needs to take the derivative of this equality and use the power rule and linearity of differentiation on the right-hand side.

[edit] Proof of the power rule

To prove the power rule for differentiation, we use the definition of the derivative as a limit:

$f'(x) = \lim_{h\rarr0} \frac{f(x+h)-f(x)}{h}.$

Substituting $f (x) = x n$ gives

$f'(x) = \lim_{h\rarr0} \frac{(x+h)^n-x^n}{h}.$

One can then express $(x + h) n$ by applying the binomial theorem to obtain

$f'(x) = \lim_{h\rarr0} \frac{\sum_{i=0}^{n} {{n \choose i} x^i h^{n-i}}-x^n}{h}.$

The $i = n$ term of the sum can then be written independently of the sum to yield

$f'(x) = \lim_{h\rarr0} \frac{\sum_{i=0}^{n - 1} {{n \choose i} x^i h^{n-i}} + x^n -x^n}{h}.$

Canceling the $x n$ terms one generates

$f'(x) = \lim_{h\rarr0} \frac{\sum_{i=0}^{n - 1} {{n \choose i} x^i h^{n-i}}}{h}.$

An $h$ can be factored out from each term in the sum to give

$f'(x) = \lim_{h\rarr0} \frac{h\sum_{i=0}^{n - 1} {{n \choose i} x^i h^{n-i-1}}}{h}.$

From thence we can cancel the $h$ in the denominator to obtain

$f'(x) = \lim_{h\rarr0} \sum_{i=0}^{n - 1} {{n \choose i} x^i h^{n-i-1}}.$

To evaluate this limit we observe that $n - i - 1 > 0$ for all $i < n - 1$ and equal to zero for $i = n - 1.$ Thus only the $h 0$ term will survive with $i = n - 1$ yielding

$f'(x) = {n \choose {n-1}} x^{n-1}.$

Evaluating the binomial coefficient gives

${n \choose {n-1}} = \frac{n!}{(n-1)!\ 1!} = \frac{n\ (n-1)!}{(n-1)!} = n.$

It follows that

$f'(x) = n x^{n-1}. \!$

[edit] Differentiation of arbitrary polynomials

To differentiate arbitrary polynomials, one can use the linearity property of the differential operator to obtain:

$\left( \sum_{r=0}^n a_r x^r \right)' = \sum_{r=0}^n \left(a_r x^r\right)' = \sum_{r=0}^n a_r \left(x^r\right)' = \sum_{r=0}^n ra_rx^{r-1}.$

Using the linearity of integration and the power rule for integration, one shows in the same way that

$\int\!\left( \sum^n_{k=0} a_k x^k\right)\,dx= \sum^n_{k=0} \frac{a_k x^{k+1}}{k+1} + c.$

[edit] Generalization

One can prove that the power rule is valid for any real exponent, that is

$\left(x^a\right)' = ax^{a-1}$

for any real number a as long as x is in the domain of the functions on the left and right hand sides. Using this formula, together with

$\int \! x^{-1}\, dx= \ln x+c,$

one can differentiate and integrate linear combinations of powers of x which are not necessarily polynomials.

[edit] See also

Derivative (examples)

[edit] References

Larson, Ron; Hostetler, Robert P.; and Edwards, Bruce H. (2003). Calculus of a Single Variable: Early Transcendental Functions (3rd edition). Houghton Mifflin Company. ISBN 0-618-22307-X.

Retrieved from "http://en.wikipedia.org../../../c/a/l/Calculus_with_polynomials.html"

Category: Calculus