Order of approximation: Difference between revisions

Content deleted Content added
Usage in science and engineering: used exponential function as a simple example.
m cleaning up a bit
Line 14:
In the case of a [[smooth function]], the ''n''th-order approximation is a [[polynomial]] of [[degree of a polynomial|degree]] ''n'', which is obtained by truncating the Taylor series to this degree. The formal usage of ''order of approximation'' corresponds to the omission of some terms of the [[Series_(mathematics)|series]] used in the [[Series_expansion|expansion]]. This affects [[Accuracy_and_precision|accuracy]]. The error usually varies within the interval. Thus the terms (''zeroth'', ''first'', ''second,'' etc.) used above meaning do not directly give information about [[percent error]] or [[significant figures]].
 
For example, considerin the [[w:Taylor's theorem|Taylor series]] expansion of the [[Exponential_function#Formal_definition|exponential function]]:,
<math display="block">e^x=\underbrace{1}_{0^\text{th}}+\underbrace{x}_{1^\text{st}}+\underbrace{\frac{x^2}{2!}}_{2^\text{nd}}+\underbrace{\frac{x^3}{3!}}_{3^\text{rd}} + \underbrace{\frac{x^4}{4!}}_{4^\text{th}} + \mathcal O^5 </math>
 
</math>The the secondzeroth-order term is not<math>1, smaller than</math> the first -order term. is And<math>x, at</math> second-order is <math>x=^2/2,
<math display="block">e^x =
</math> and so forth. If <math>x<1,</math> each higher order term is smaller than the previous. If <math>x</math> is very small<math>(x<<1)</math>, the first order approximation,
\underbrace{1}_{0^\text{th}} +
<math display="block">e^x\approx =1+x, </math>
\underbrace{x}_{1^\text{st}} +
is often sufficient. But at <math>x=1,</math> the first-order term, <math>x,</math> is not smaller than the zeroth-order term, <math>1.</math> And at <math>x=2,</math> even the second-order term, <math>2^3/3!=4/3,\,</math> is greater than the zeroth-order term.
\underbrace{\frac{x^2}{2!}}_{2^\text{nd}} +
\underbrace{\frac{x^3}{3!}}_{3^\text{rd}} +
\underbrace{\frac{x^4}{4!}}_{4^\text{th}} +
\mathcal O^5
</math>The zeroth-order term is <math>1,
</math> the first-order term is <math>x,
</math> second-order is <math>x^2/2,
</math> and so forth. If <math>x<1,\,
</math>each higher order term is smaller than the previous. If <math>x
</math> is very small <math>(x<<1)</math>, the first order approximation,
 
<math display="block">e^x\approx 1+x,
</math>
 
is often sufficient. But at <math>x=1,
</math> the second-order term is not smaller than the first order term. And at <math>x=2,
</math> even the second-order term <math>(2^3/3!=4/3)
</math> is greater than the zeroth-order term.
 
=== Zeroth-order ===