Calculus with polynomials

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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 x100 is 100x99 and the indefinite integral of x100 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.

Contents

[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) = xn 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 xn 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 h0 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

See also: power rules

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

[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.