Editing Taylor's theorem (section)

=== Explicit formulas for the remainder ===

Under stronger regularity assumptions on ''f'' there are several precise formulas for the remainder term ''R<sub>k</sub>'' of the Taylor polynomial, the most common ones being the following.

{{math theorem
| name = Mean-value forms of the remainder
| math_statement = Let {{nowrap|''f'' : '''R''' → '''R'''}}  be ''k''&nbsp;+&nbsp;1 times [[Differentiable function|differentiable]] on the [[open interval]] between <math display=inline>a</math> and <math display=inline>x</math> with ''f''{{i sup|(''k'')}} [[continuous function|continuous]] on the [[closed interval]] between <math display=inline>a</math> and <math display=inline>x</math>.<ref>The hypothesis of ''f''{{i sup|(''k'')}} being [[continuous function|continuous]] on the [[closed interval|''closed'' interval]] between <math display=inline>a</math> and <math display=inline>x</math> is ''not'' redundant. Although ''f'' being ''k''&nbsp;+&nbsp;1 times [[Differentiable function|differentiable]] on the [[open interval]] between <math display=inline>a</math> and <math display=inline>x</math> does imply that ''f''{{i sup|(''k'')}} is [[continuous function|continuous]] on the [[open interval|''open'' interval]] between <math display=inline>a</math> and <math display=inline>x</math>, it does ''not'' imply that ''f''{{i sup|(''k'')}} is [[continuous function|continuous]] on the [[closed interval|''closed'' interval]] between <math display=inline>a</math> and <math display=inline>x</math>, i.e. it does not imply that ''f''{{i sup|(''k'')}} is [[continuous function|continuous]] at the ''endpoints'' of that interval. Consider, for example, the [[Function (mathematics)|function]] {{nowrap|''f'' : [0,1] → '''R'''}} defined to equal <math> \sin(1/x)</math> on <math>(0,1]</math> and with <math>f(0)=0</math>. This is not [[continuous function|continuous]] at ''0'', but is [[continuous function|continuous]] on <math>(0,1)</math>. Moreover, one can show that this [[Function (mathematics)|function]] has an [[antiderivative]]. Therefore that [[antiderivative]] is [[Differentiable function|differentiable]] on <math>(0,1)</math>, its [[derivative]] (the function ''f'') is [[continuous function|continuous]] on the [[open interval|''open'' interval]] <math>(0,1)</math>, but its [[derivative]] ''f'' is ''not'' [[continuous function|continuous]] on the [[closed interval|''closed'' interval]] <math>[0,1]</math>. So the theorem would not apply in this case.</ref> Then

<math display="block"> R_k(x) = \frac{f^{(k+1)}(\xi_L)}{(k+1)!} (x-a)^{k+1} </math>

for some real number <math display="inline">\xi_L</math> between <math display=inline>a</math> and <math display=inline>x</math>. This is the '''[[Joseph Louis Lagrange|Lagrange]] form'''<ref>{{harvnb|Kline|1998|loc=§20.3}}; {{harvnb|Apostol|1967|loc=§7.7}}.</ref> of the remainder.

Similarly,

<math display="block"> R_k(x) = \frac{f^{(k+1)}(\xi_C)}{k!}(x-\xi_C)^k(x-a) </math>

for some real number <math display="inline">\xi_C</math> between <math display=inline>a</math> and <math display=inline>x</math>. This is the '''[[Augustin Louis Cauchy|Cauchy]] form'''<ref>{{harvnb|Apostol|1967|loc=§7.7}}.</ref> of the remainder.

Both can be thought of as specific cases of the following result: Consider <math>p>0</math>

<math display="block"> R_k(x) = \frac{f^{(k+1)}(\xi_S)}{k!}(x-\xi_S)^{k+1-p}\frac{(x-a)^p}{p} </math>
for some real number <math display="inline">\xi_S</math> between <math display=inline>a</math> and <math display=inline>x</math>. This is the '''[[Oskar Schlömilch|Schlömilch]] form''' of the remainder (sometimes called the '''Schlömilch-[[Édouard Roche|Roche]]'''). The choice <math display="inline">p=k+1</math> is the Lagrange form, whilst the choice <math display="inline">p=1</math> is the Cauchy form.
}}

These refinements of Taylor's theorem are usually proved using the [[mean value theorem]], whence the name. Additionally, notice that this is precisely the [[mean value theorem]] when <math display="inline">k=0</math>. Also other similar expressions can be found. For example, if ''G''(''t'') is continuous on the closed interval and differentiable with a non-vanishing derivative on the open interval between <math display="inline">a</math> and <math display="inline">x</math>, then

<math display="block"> R_k(x) = \frac{f^{(k+1)}(\xi)}{k!}(x-\xi)^k \frac{G(x)-G(a)}{G'(\xi)} </math>

for some number <math display="inline">\xi</math> between <math display="inline">a</math> and <math display="inline">x</math>. This version covers the Lagrange and Cauchy forms of the remainder as special cases, and is proved below using [[mean value theorem#Cauchy's mean value theorem|Cauchy's mean value theorem]]. The Lagrange form is obtained by taking <math>G(t)=(x-t)^{k+1}</math> and the Cauchy form is obtained by taking <math>G(t)=t-a</math>.

{{anchor|Integral form of the remainder}}The statement for the integral form of the remainder is more advanced than the previous ones, and requires understanding of [[Lebesgue integral|Lebesgue integration theory]] for the full generality. However, it holds also in the sense of [[Riemann integral]] provided the (''k''&nbsp;+&nbsp;1)th derivative of ''f'' is continuous on the closed interval [''a'',''x''].

{{math theorem|name=Integral form of the remainder<ref>{{harvnb|Apostol|1967|loc=§7.5}}.</ref> |math_statement=Let <math display=inline>f^{(k)}</math> be [[absolutely continuous]] on the [[closed interval]] between <math display=inline>a</math> and <math display=inline>x</math>. Then

<math display="block"> R_k(x) = \int_a^x \frac{f^{(k+1)} (t)}{k!} (x - t)^k \, dt. </math>

}}

Due to the [[absolutely continuous|absolute continuity]] of ''f''{{i sup|(''k'')}} on the [[closed interval]] between <math display=inline>a</math> and <math display=inline>x</math>, its derivative ''f''{{i sup|(''k''+1)}} exists as an ''L''{{i sup|1}}-function, and the result can be [[#Derivation for the integral form of the remainder|proven]] by a formal calculation using the [[fundamental theorem of calculus]] and [[integration by parts]].