Editing Taylor's theorem (section)

== Proofs ==

=== Proof for Taylor's theorem in one real variable ===

Let<ref>{{harvnb|Stromberg|1981}}</ref>

<math display="block"> h_k(x) = \begin{cases}
\frac{f(x) - P(x)}{(x-a)^k} & x\not=a\\
0&x=a
\end{cases}
</math>

where, as in the statement of Taylor's theorem,

<math display="block"> P(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(k)}(a)}{k!}(x-a)^k.</math>

It is sufficient to show that

<math display="block"> \lim_{x\to a} h_k(x) =0. </math>

The proof here is based on repeated application of [[L'Hôpital's rule]].  Note that, for each <math display="inline">j=0,1,...,k-1</math>, <math>f^{(j)}(a)=P^{(j)}(a)</math>.  Hence each of the first <math display="inline">k-1</math> derivatives of the numerator in <math>h_k(x)</math> vanishes at <math>x=a</math>, and the same is true of the denominator.  Also, since the condition that the function <math display="inline">f</math> be <math display="inline">k</math> times differentiable at a point requires differentiability up to order <math display="inline">k-1</math> in a neighborhood of said point (this is true, because differentiability requires a function to be defined in a whole neighborhood of a point), the numerator and its <math display="inline">k-2</math> derivatives are differentiable in a neighborhood of <math display="inline">a</math>. Clearly, the denominator also satisfies said condition, and additionally, doesn't vanish unless <math display="inline">x=a</math>, therefore all conditions necessary for L'Hôpital's rule are fulfilled, and its use is justified. So

<math display="block">\begin{align}
\lim_{x\to a} \frac{f(x) - P(x)}{(x-a)^k}
&= \lim_{x\to a} \frac{\frac{d}{dx}(f(x) - P(x))}{\frac{d}{dx}(x-a)^k} \\[1ex]
&= \cdots \\[1ex]
&= \lim_{x\to a} \frac{\frac{d^{k-1}}{dx^{k-1}}(f(x) - P(x))}{\frac{d^{k-1}}{dx^{k-1}}(x-a)^k}\\[1ex]
&= \frac{1}{k!}\lim_{x\to a} \frac{f^{(k-1)}(x) - P^{(k-1)}(x)}{x-a}\\[1ex]
&=\frac{1}{k!}(f^{(k)}(a) - P^{(k)}(a))
= 0
\end{align}</math>

where the second-to-last equality follows by the definition of the derivative at <math display="inline"> x=a</math>.

=== Alternate proof for Taylor's theorem in one real variable ===

Let <math>f(x)</math> be any real-valued continuous function to be approximated by the Taylor polynomial.

Step 1: Let <math display="inline">F</math> and <math display="inline">G</math> be functions. Set <math display="inline">F</math> and <math display="inline">G</math> to be

<math display="block">\begin{align}
F(x) = f(x) - \sum^{n-1}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k} 
\end{align}</math>

<math display="block">\begin{align}
G(x) = (x-a)^{n}
\end{align}</math>

Step 2: Properties of <math display="inline">F</math> and <math display="inline">G</math>:

<math display="block">\begin{align}
F(a) & = f(a) - f(a) - f'(a)(a - a) - ... - \frac{f^{(n-1)}(a)}{(n-1)!}(a-a)^{n-1} = 0 \\
G(a) & = (a-a)^n = 0
\end{align}</math>

Similarly, 

<math display="block">\begin{align}
F'(a) = f'(a) - f'(a) - \frac{f''(a)}{(2-1)!}(a-a)^{(2-1)} - ... - \frac{f^{(n-1)}(a)}{(n-2)!}(a-a)^{n-2} = 0
\end{align}</math>

<math display="block">\begin{align}
G'(a) &= n(a-a)^{n-1} = 0\\
&\qquad \vdots\\
G^{(n-1)}(a) &= F^{(n-1)}(a) = 0
\end{align}</math>

Step 3: Use Cauchy Mean Value Theorem

Let <math>f_{1}</math> and <math>g_{1}</math> be continuous functions on <math>[a, b]</math>. Since <math>a < x < b</math> so we can work with the interval <math>[a, x]</math>. Let <math>f_{1}</math> and <math>g_{1}</math> be differentiable on <math>(a, x)</math>. Assume <math>g_{1}'(x) \neq 0</math> for all <math>x \in (a, b)</math>.
Then there exists <math>c_{1} \in (a, x)</math> such that

<math display="block">\begin{align}
\frac{f_{1}(x) - f_{1}(a)}{g_{1}(x) - g_{1}(a)} = \frac{f_{1}'(c_{1})}{g_{1}'(c_{1})}
\end{align}</math>

Note: <math>G'(x) \neq 0</math> in <math>(a, b)</math> and <math>F(a), G(a) = 0</math> so

<math display="block">\begin{align}
\frac{F(x)}{G(x)} = \frac{F(x) - F(a)}{G(x) - G(a)} = \frac{F'(c_{1})}{G'(c_{1})}
\end{align}</math>

for some <math>c_{1} \in (a, x)</math>.

This can also be performed for <math>(a, c_{1})</math>:

<math display="block">\begin{align}
\frac{F'(c_{1})}{G'(c_{1})} = \frac{F'(c_{1}) - F'(a)}{G'(c_{1}) - G'(a)} = \frac{F''(c_{2})}{G''(c_{2})}
\end{align}</math>

for some <math>c_{2} \in (a, c_{1})</math>.
This can be continued to <math>c_{n}</math>.

This gives a partition in <math>(a, b)</math>:

<math display="block">a < c_{n} < c_{n-1} < \dots < c_{1} < x</math>

with

<math display="block">\frac{F(x)}{G(x)} = \frac{F'(c_{1})}{G'(c_{1})} = \dots = \frac{F^{(n)}(c_{n})}{G^{(n)}(c_{n})} .</math>

Set <math>c = c_{n}</math>:

<math display="block">\frac{F(x)}{G(x)} = \frac{F^{(n)}(c)}{G^{(n)}(c)}</math>

Step 4: Substitute back

<math display="block">\begin{align}
\frac{F(x)}{G(x)} = \frac{f(x) - \sum^{n-1}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k}}{(x-a)^{n}} = \frac{F^{(n)}(c)}{G^{(n)}(c)}
\end{align}</math>

By the Power Rule, repeated derivatives of <math>(x - a)^{n}</math>, <math>G^{(n)}(c) = n(n-1)...1</math>, so:

<math display="block">\frac{F^{(n)}(c)}{G^{(n)}(c)} = \frac{f^{(n)}(c)}{n(n-1)\cdots1} = \frac{f^{(n)}(c)}{n!}.</math>

This leads to:

<math display="block">\begin{align}
f(x) - \sum^{n-1}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k} = \frac{f^{(n)}(c)}{n!}(x-a)^{n}
\end{align}.</math>

By rearranging, we get:

<math display="block">\begin{align}
f(x) = \sum^{n-1}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k} + \frac{f^{(n)}(c)}{n!}(x-a)^{n}
\end{align},</math>

or because <math>c_{n} = a</math> eventually:

<math display="block">f(x) = \sum^{n}_{k=0} \frac{f^{(k)}(a)}{k!}(x-a)^{k}.</math>

=== Derivation for the mean value forms of the remainder ===

Let ''G'' be any real-valued function, continuous on the closed interval between <math display=inline>a</math> and <math display=inline>x</math> and differentiable with a non-vanishing derivative on the open interval between <math display=inline>a</math> and <math display=inline>x</math>, and define

<math display="block"> F(t) = f(t) + f'(t)(x-t) + \frac{f''(t)}{2!}(x-t)^2 + \cdots + \frac{f^{(k)}(t)}{k!}(x-t)^k.
</math>

For <math> t \in [a,x] </math>. Then, by [[mean value theorem#Cauchy's mean value theorem|Cauchy's mean value theorem]],

{{NumBlk|:|<math> \frac{F'(\xi)}{G'(\xi)} = \frac{F(x) - F(a)}{G(x) - G(a)}</math>|{{EquationRef|★★★}}}}

for some <math display="inline">\xi</math> on the open interval between <math display=inline>a</math> and <math display=inline>x</math>. Note that here the numerator <math display="inline">F(x)-F(a)=R_k(x)</math> is exactly the remainder of the Taylor polynomial for <math display="inline">y=f(x)</math>. Compute

<math display="block">\begin{align}
F'(t) = {} & f'(t) + \big(f''(t)(x-t) - f'(t)\big) + \left(\frac{f^{(3)}(t)}{2!}(x-t)^2 - \frac{f^{(2)}(t)}{1!}(x-t)\right)  +  \cdots \\
& \cdots + \left( \frac{f^{(k+1)}(t)}{k!}(x-t)^k - \frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}\right)
= \frac{f^{(k+1)}(t)}{k!}(x-t)^k,
\end{align}</math>

plug it into ({{EquationNote|★★★}}) and rearrange terms to find that

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

This is the form of the remainder term mentioned after the actual statement of Taylor's theorem with remainder in the mean value form.
The Lagrange form of the remainder is found by choosing <math> G(t) = (x-t)^{k+1} </math> and the Cauchy form by choosing <math> G(t) = t-a</math>.

'''Remark.''' Using this method one can also recover the integral form of the remainder by choosing

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

but the requirements for ''f'' needed for the use of mean value theorem are too strong, if one aims to prove the claim in the case that ''f''{{i sup|(''k'')}} is only [[absolutely continuous]]. However, if one uses [[Riemann integral]] instead of [[Lebesgue integral]], the assumptions cannot be weakened.

=== Derivation for the integral form of the remainder ===

Due to the [[absolutely continuous|absolute continuity]] of <math>f^{(k)}</math> on the [[closed interval]] between <math display=inline>a</math> and <math display=inline>x</math>, its derivative <math>f^{(k+1)}</math> exists as an <math>L^1</math>-function, and we can use the [[fundamental theorem of calculus]] and [[integration by parts]]. This same proof applies for the [[Riemann integral]] assuming that <math>f^{(k)}</math> is [[continuous function|continuous]] on the closed interval and [[Differentiable function|differentiable]] on the [[open interval]] between <math display=inline>a</math> and <math display=inline>x</math>, and this leads to the same result as using the mean value theorem.

The [[fundamental theorem of calculus]] states that

<math display="block"> f(x)=f(a)+ \int_a^x \, f'(t) \, dt.</math>

Now we can [[Integration by parts|integrate by parts]] and use the fundamental theorem of calculus again to see that

<math display="block"> \begin{align}
f(x) &= f(a)+\Big(xf'(x)-af'(a)\Big)-\int_a^x tf''(t) \, dt \\
&= f(a) + x\left(f'(a) + \int_a^x f''(t) \,dt \right) -af'(a)-\int_a^x  tf''(t) \, dt \\
&= f(a)+(x-a)f'(a)+\int_a^x \, (x-t)f''(t) \, dt,
\end{align} </math>

which is exactly Taylor's theorem with remainder in the integral form in the case <math>k=1</math>. The general statement is proved using [[mathematical induction|induction]]. Suppose that
{{NumBlk|:|<math> f(x) = f(a) + \frac{f'(a)}{1!}(x - a) + \cdots + \frac{f^{(k)}(a)}{k!}(x - a)^k + \int_a^x \frac{f^{(k+1)} (t)}{k!} (x - t)^k \, dt. </math>|{{EquationRef|eq1}}}}

Integrating the remainder term by parts we arrive at

<math display="block">\begin{align}
\int_a^x \frac{f^{(k+1)} (t)}{k!} (x - t)^k \, dt = & - \left[ \frac{f^{(k+1)} (t)}{(k+1)k!} (x - t)^{k+1} \right]_a^x + \int_a^x \frac{f^{(k+2)} (t)}{(k+1)k!} (x - t)^{k+1} \, dt \\
= & \ \frac{f^{(k+1)} (a)}{(k+1)!} (x - a)^{k+1} + \int_a^x \frac{f^{(k+2)} (t)}{(k+1)!} (x - t)^{k+1} \, dt.
\end{align}</math>

Substituting this into the formula {{nowrap|in ({{EquationNote|eq1}})}} shows that if it holds for the value <math>k</math>, it must also hold for the value <math>k+1</math>. Therefore, since it holds for <math>k=1</math>, it must hold for every positive integer <math>k</math>.

=== Derivation for the remainder of multivariate Taylor polynomials ===

We prove the special case, where <math>f:\mathbb R^n\to\mathbb R</math> has continuous partial derivatives up to the order <math>k+1</math> in some closed ball <math>B</math> with center <math>\boldsymbol{a}</math>.  The strategy of the proof is to apply the one-variable case of Taylor's theorem to the restriction of <math>f</math> to the line segment adjoining <math>\boldsymbol{x}</math> and <math>\boldsymbol{a}</math>.<ref>{{harvnb|Hörmander|1976|pp=12–13}}</ref>  Parametrize the line segment between <math>\boldsymbol{a}</math> and <math>\boldsymbol{x}</math> by <math>\boldsymbol{u}(t)=\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a})</math> We apply the one-variable version of Taylor's theorem to the function <math>g(t) = f(\boldsymbol{u}(t))</math>:

<math display="block"> f(\boldsymbol{x})=g(1)=g(0)+\sum_{j=1}^k\frac{1}{j!}g^{(j)}(0)\ +\ \int_0^1 \frac{(1-t)^k }{k!} g^{(k+1)}(t)\, dt.</math>

Applying the [[chain rule]] for several variables gives

<math display="block">\begin{align}
g^{(j)}(t)&=\frac{d^j}{dt^j}f(\boldsymbol{u}(t))\\
&= \frac{d^j}{dt^j} f(\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a}))\\
&= \sum_{|\alpha| =j} \left(\begin{matrix} j\\ \alpha\end{matrix} \right) (D^\alpha f) (\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a})) (\boldsymbol{x}-\boldsymbol{a})^\alpha
\end{align}</math>

where <math>\tbinom j \alpha</math> is the [[multinomial coefficient]]. Since <math>\tfrac{1}{j!}\tbinom j \alpha=\tfrac{1}{\alpha!}</math>, we get:

<math display="block"> f(\boldsymbol{x})= f(\boldsymbol{a}) + \sum_{1 \leq |\alpha| \leq k}\frac{1}{\alpha!} (D^\alpha f) (\boldsymbol{a})(\boldsymbol{x}-\boldsymbol{a})^\alpha+\sum_{|\alpha|=k+1}\frac{k+1}{\alpha!}
(\boldsymbol{x}-\boldsymbol{a})^\alpha \int_0^1 (1-t)^k (D^\alpha f)(\boldsymbol{a}+t(\boldsymbol{x}-\boldsymbol{a}))\,dt.</math>