Calculus with polynomials

From TheBestLinks.com

Topics in Calculus

Fundamental theorem | Function | Limits of functions | Continuity | Calculus with polynomials

Differentiation

Product rule | Quotient rule | Chain rule | Implicit differentiation | Taylor's theorem

Integration

Integration by substitution | Integration by parts | Integration by trigonometric substitution | Solids of revolution | Integration by disks | Integration by cylindrical shells | Lists of integrals

Vector Calculus

Vector | Vector field | Matrix | Partial Derivative | Gradient | Flux | Divergence | Divergence Theorem | Del | Curl | Stokes' Theorem

Tensor Calculus

Tensor | Tensor field | Tensor product | Exterior power | Exterior Derivative | Covariant derivative | Manifold

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

<math>\frac{d}{dx} \sum^n_{k=0} a_k x^k = \sum^n_{k=0} ka_kx^{k-1}<math>

<math>\int \sum^n_{k=0} a_k x^k\,dx= \sum^n_{k=0} \frac{a_k x^{k+1}}{k+1} + c<math>

Hence the derivative of x¹⁰⁰ is 100x⁹⁹ and the integral of x¹⁰⁰ is x¹⁰¹/101 + c.

Table of contents

1 Proof 2 Generalisations 3 References 4 See also

Proof

Because differentiation is linear, we have:

<math>\frac{d\left( \sum_{r=0}^n a_r x^r \right)}{dx} =

\sum_{r=0}^n \frac{d\left(a_r x^r\right)}{dx} = \sum_{r=0}^n a_r \frac{d\left(x^r\right)}{dx}<math>

So it remains to find <math>\frac{d\left(x^r\right)}{dx}<math> for any natural number r. The derivative of function f(x) is given by Newton's difference quotient

<math> f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} <math>

By the binomial theorem,

<math>\left(x+h\right)^r = \sum_{k=0}^r {r\choose k} h^k x^{r-k}<math>

and therefore

<math>\frac{\left(x+h\right)^r - x^r}{h} = \sum_{k=1}^{r} {r\choose k} h^{k-1} x^{r-k}<math>

The derivative is the limit of this as <math> h \rightarrow 0 <math>

<math>\frac{d}{dx}\left(x^r\right) = \lim_{h\rightarrow 0} \left(\sum_{k=1}^{r} {r\choose k} h^{k-1} x^{r-k}\right) = {r\choose 1} x^{r-1} = rx^{r-1}<math>

which gives the claimed result.

Generalisations

<math>\frac{d}{dx} \left(ax^k\right) = akx^{k-1}<math>

is generally true for all values of k where x^k is meaningful. In particular it holds for all rational k for values of x where x^k is defined.

Similarly for integration, see table of integrals.

If one has polynomials with coefficients that are not real or complex numbers (perhaps they are integers, or numbers modulo a prime number) then one can formally define the derivative according to the rules given above. This is useful, for example, in determining whether a polynomial will have multiple roots: compute the greatest common divisor of the polynomial and its formal derivative. If this polynomial is zero, then the original polynomial cannot have any multiple roots.

References

• Calculus of a Single Variable: Early Transcendental Functions (3rd Edition) by Edwards, Hostetler, and Larson (2003) ISBN 0618226877

See also

• This result can be proved without using limits by using mathematical induction (proof).

Related links

---

Top visited 0 of 0 links

[no links posted yet]

>> place link >>

Discussion

Last posted 0 of 0 messages

[no messages posted yet]

>> post message >>

Watch

You can add this article to your own "watchlist" and receive e-mail notification about all changes in this page.