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:
- <math>\left( \sum^n_{k=0} a_k x^k\right)' = \sum^n_{k=0} ka_kx^{k-1}</math>
and
- <math>\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.</math>
Hence, the derivative of <math>x^{100}</math> is 100x 99 and the integral of <math>x^{100}</math> is <math>x^{101}/101+c.</math>
This article will state and prove the power rule for differentiation, and then use it to prove these two formulas.
Contents |
The power rule
The power rule for differentiation states that for every natural number n, the derivative of <math>f(x)=x^n</math> is <math>f'(x)=nx^{n-1}, </math> that is,
- <math>\left(x^n\right)'=nx^{n-1}.</math>
The power rule for integration
- <math>\int\! x^n \, dx=\frac{x^{n+1}}{n+1}+C</math>
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.
Proof of the power rule
To prove the power rule for differentiation, we use the definition of the derivative as a limit:
- <math>f'(x) = \lim_{h\rarr0} \frac{f(x+h)-f(x)}{h}. </math>
Substituting <math> f(x) = x^n </math> gives
- <math>f'(x) = \lim_{h\rarr0} \frac{(x+h)^n-x^n}{h}.</math>
One can then express <math>(x+h)^n</math> by applying the binomial theorem to obtain
- <math>f'(x) = \lim_{h\rarr0} \frac{\sum_{i=0}^{n} {{n \choose i} x^i h^{n-i}}-x^n}{h}. </math>
The <math>i = n</math> term of the sum can then be written independently of the sum to yield
- <math>f'(x) = \lim_{h\rarr0} \frac{\sum_{i=0}^{n - 1} {{n \choose i} x^i h^{n-i}} + x^n -x^n}{h}. </math>
Canceling the <math>x^n</math> terms one generates
- <math>f'(x) = \lim_{h\rarr0} \frac{\sum_{i=0}^{n - 1} {{n \choose i} x^i h^{n-i}}}{h}.</math>
An <math>h</math> can be factored out from each term in the sum to give
- <math>f'(x) = \lim_{h\rarr0} \frac{h\sum_{i=0}^{n - 1} {{n \choose i} x^i h^{n-i-1}}}{h}.</math>
From thence we can cancel the <math>h</math> in the denominator to obtain
- <math>f'(x) = \lim_{h\rarr0} \sum_{i=0}^{n - 1} {{n \choose i} x^i h^{n-i-1}}.</math>
To evaluate this limit we observe that <math> n-i-1 > 0</math> for all <math>i < n - 1</math> and equal to zero for <math>i=n-1.</math> Thus only the <math>h^0</math> term will survive with <math>i = n - 1</math> yielding
- <math>f'(x) = {n \choose {n-1}} x^{n-1}. </math>
Evaluating the binomial coefficient gives
- <math>{n \choose {n-1}} = \frac{n!}{(n-1)!\ 1!} = \frac{n\ (n-1)!}{(n-1)!} = n.</math>
It follows that
- <math>f'(x) = n x^{n-1}. \! </math>
Differentiation of arbitrary polynomials
To differentiate arbitrary polynomials, one can use the linearity property of the differential operator to obtain:
- <math>\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}.</math>
Using the linearity of integration and the power rule for integration, one shows in the same way that
- <math>\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.</math>
Generalization
One can prove that the power rule is valid for any real exponent, that is
- <math>\left(x^a\right)' = ax^{a-1}</math>
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
- <math>\int \! x^{-1}\, dx= \ln x+c,</math>
one can differentiate and integrate linear combinations of powers of x which are not necessarily polynomials.
See also
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 061822307X.ko:다항식의 미적분