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How We Got From There To Here: A Story of Real Analysis

Robert Rogers, Eugene C. Boman

Section 7.2 Lagrange’s Form of the Remainder

Joseph-Louis Lagrange provided an alternate form for the remainder in Taylor series in his 1797 work Théorie des functions analytiques. Lagrange’s form of the remainder is as follows.

Proof.

Note first that the result is true when \(x=a\) as both sides reduce to 0 (in that case \(c=x=a\text{.}\)) We will prove the case where \(a\lt x\text{;}\) the case \(x\lt a\) will be an exercise.
First, we already have
\begin{equation*} f(x)-\left(\sum_{j=0}^n\frac{f^{(j)}(a)}{j!}(x-a)^j\right)=\frac{1}{n!} \int_{t=a}^xf^{(n+1)}(t)(x-t)^n\dx{ t} \end{equation*}
so it suffices to show that
\begin{equation*} \int_{t=a}^xf^{(n+1)}(t)(x-t)^n\dx{ t}=\,\frac{f^{\,(n+1)}(c)}{n+1}(x-a)^{n+1} \end{equation*}
for some \(c\) with \(c\in[\,a,x]\text{.}\) To this end, let
\begin{equation*} M=\max_{a\le t\le x}\left(f^{(n+1)}(t)\right) \end{equation*}
and
\begin{equation*} m=\min_{a\le t\le x}\left(f^{(n+1)}(t)\right)\text{.} \end{equation*}
Note that for all \(t\in[\,a,x]\text{,}\) we have \(m\leq f^{(n+1)}(t)\leq M\text{.}\) Since \(x-t\geq 0\text{,}\) this gives us
\begin{equation} m\left(x-t\right)^n\leq f^{(n+1)}(t)(x-t)^n\leq M(x-t)^n\tag{1} \end{equation}
and so
\begin{equation} \int_{t=a}^xm\left(x-t\right)^n\dx{ t}\leq\int_{t=a}^xf^{(n+1)}(t)(x-t)^n\dx{ t}\leq \int_{t=a}^xM(x-t)^n\dx{ t}\text{.}\tag{2} \end{equation}
Computing the outside integrals, we have
\begin{equation*} m\int_{t=a}^x\left(x-t\right)^n\dx{ t}\leq\int_{t=a}^xf^{(n+1)}(t)(x-t)^n\dx{ t}\leq M\int_{t=a}^x(x-t)^n\dx{ t} \end{equation*}
\begin{equation} m\frac{(x-a)^{n+1}}{n+1}\leq\int_{t=a}^xf^{(n+1)}(t)(x-t)^n\dx{ t}\leq M\frac{(x-a)^{n+1}}{n+1}\tag{3} \end{equation}
\begin{equation*} m\leq\frac{\int_{t=a}^xf^{(n+1)}(t)(x-t)^n\dx{ t}}{\left(\frac{(x-a)^{n+1}}{n+1} \right)}\leq M\text{.} \end{equation*}
Since
\begin{equation*} \frac{\int_{t=a}^xf^{(n+1)}(t)(x-t)^n\dx{ t}}{\left(\frac{(x-a)^{n+1}}{n+1} \right)} \end{equation*}
is a value that lies between the maximum and minimum of \(f^{(n+1)}\) on \([\,a,x]\text{,}\) then by the Intermediate Value Theorem, there must exist a number \(c\in[\,a,x]\) with
\begin{equation*} f^{(n+1)}(c)=\frac{\int_{t=a}^xf^{(n+1)}(t)(x-t)^n\dx{ t}}{\left( \frac{(x-a)^{n+1}}{n+1}\right)}\text{.} \end{equation*}
This gives us
\begin{equation*} \int_{t=a}^xf^{(n+1)}(t)(x-t)^n\dx{ t}=\,\frac{f^{\,(n+1)}(c)}{n+1}(x-a)^{n+1}\text{.} \end{equation*}
And the result follows.

Problem 7.2.2.

Prove Theorem 7.2.1 for the case where \(x\lt a\text{.}\)
Hint.
Note that
\begin{equation*} \int_{t=a}^xf^{(n+1)}(t)(x-t)^n\dx{ t}=(-1)^{n+1}\int_{t=x}^af^{(n+1)}(t)(t-x)^n\dx{ t}\text{.} \end{equation*}
Use the same argument on this integral. It will work out in the end. Really! You just need to keep track of all of the negatives.
This is not Lagrange’s proof. He did not use the integral form of the remainder. However, this is similar to Lagrange’s proof in that he also used the Intermediate Value Theorem (IVT) and Extreme Value Theorem (EVT) much as we did. In Lagrange’s day, these were taken to be obviously true for a continuous function and we have followed Lagrange’s lead by assuming the IVT and the EVT. However, in mathematics we need to keep our assumptions few and simple. The IVT and the EVT do not satisfy this need in the sense that both can be proved from simpler ideas. We will return to this in Chapter 9.
Also, a word of caution about this: Lagrange’s form of the remainder is \(\frac{f^{\,(n+1)}(c)}{(n+1)!}\) \((x-a)^{n+1}\text{,}\) where \(c\) is some number between \(a\) and \(x\text{.}\) The proof does not indicate what this \(c\) might be and, in fact, this \(c\) changes as \(n\) changes. All we know is that this \(c\) lies between \(a\) and \(x\text{.}\) To illustrate this issue and its potential dangers, consider the following problem where we have a chance to compute the value of \(c\) for the function \(f(x)=\frac{1}{1+x}\text{.}\)

Problem 7.2.3.

This problem investigates the Taylor series representation
\begin{equation*} \frac{1}{1+x}=1-x+x^2-x^3+\cdots\text{.} \end{equation*}

(a)

Use the fact that \(\frac{1-(-x)^{n+1}}{1+x}=1-x+x^2-x^3+\cdots+(-x)^n\) to compute the remainder \(\frac{1}{1+x}-\left(1-x+x^2-x^3+\cdots+(-x)^n\right)\text{.}\) Specifically, compute this remainder when \(x=1\) and conclude that the Taylor series does not converge to \(\frac{1}{1+x}\) when \(x=1\text{.}\)

(b)

Compare the remainder in part a with the Lagrange form of the remainder to determine what \(c\) is when \(x=1\text{.}\)

(c)

Consider the following argument: If \(f(x)=\frac{1}{1+x}\text{,}\) then
\begin{equation*} f^{(n+1)}(c)=\frac{(-1)^{n+1}(n+1)!}{(1+c)^{n+2}} \end{equation*}
so the Lagrange form of the remainder when \(x=1\) is given by
\begin{equation*} \frac{(-1)^{n+1}(n+1)!}{(n+1)!(1+c)^{n+2}}=\frac{(-1)^{n+1}}{(1+c)^{n+2}} \end{equation*}
where \(c\in[\,0,1]\text{.}\) It can be seen in part b that \(c\neq 0\text{.}\) Thus \(1+c>1\) and so by Problem 6.1.7 of Chapter 6, the Lagrange remainder converges to \(0\) as \(n\rightarrow\infty\text{.}\) This argument would suggest that the Taylor series converges to \(\frac{1}{1+x}\) for \(x=1\text{.}\) However, we know from part (a) that this is incorrect. What is wrong with the argument?
Even though there are potential dangers in misusing the Lagrange form of the remainder, it is a useful form. For example, armed with the Lagrange form of the remainder, we can prove the following theorem.

Proof.

First note that the binomial series is, in fact, the Taylor series for the function \(f(x)=\sqrt{1+x}\) expanded about \(a=0\text{.}\) If we let \(x\) be a fixed number with \(0\leq x\leq 1\text{,}\) then it suffices to show that the Lagrange form of the remainder converges to \(0\text{.}\) With this in mind, notice that
\begin{equation*} f^{(n+1)}(t)=\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)\cdots\left(\frac{1}{2}-n\right)\left(1+t\right)^{\frac{1}{2}-(n+1)} \end{equation*}
and so the Lagrange form of the remainder is
\begin{equation*} \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}= \frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)\cdots \left(\frac{1}{2}-n\right)}{(n+1)!}\frac{x^{n+1}}{(1+c)^{n+\frac{1}{2}}} \end{equation*}
where \(c\) is some number between \(0\) and \(x\text{.}\) Since \(0\leq x\leq 1\) and \(1+c\geq 1\text{,}\) then we have \(\frac{1}{1+c}\leq 1\text{,}\) and so
\begin{align*} 0\amp \leq \left|\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)\cdots\left(\frac{1}{2}-n\right)}{(n+1)!}\frac{x^{n+1}}{(1+c)^{n+\frac{1}{2}}}\right|\\ \amp =\frac{\left(\frac{1}{2}\right)\left(1-\frac{1}{2}\right)\cdots\left(n-\frac{1}{2}\right)}{(n+1)!}\frac{x^{n+1}}{(1+c)^{n+\frac{1}{2}}}\\ \amp =\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\left(\frac{5}{2}\right)\cdots\left(\frac{2n-1}{2}\right)}{(n+1)!}\left(x^{n+1}\right)\frac{1}{(1+c)^{n+\frac{1}{2}}}\\ \amp \leq\frac{1\cdot 1\cdot 3\cdot5\cdot\,\cdots\,\cdot\left(2n-1\right)}{2^{^{n+1}}(n+1)!}\\ \amp =\frac{1\cdot 3\cdot 5\cdot\,\cdots\,\cdot\left(2n-1\right)\cdot 1}{2\cdot4\cdot 6\cdot\,\cdots\,\cdot 2n\cdot\left(2n+2\right)}\\ \amp =\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\cdots\,\cdot\frac{2n-1}{2n}\cdot\frac{1}{2n+2}\\ \amp \leq\frac{1}{2n+2}\text{.} \end{align*}
Since \(\lim_{n\rightarrow\infty}\frac{1}{2n+2}=0=\,\lim_{n\rightarrow\infty}0\text{,}\) then by the Squeeze Theorem,
\begin{equation*} \lim_{n\rightarrow\infty}\abs{\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}}=0, \text{ so } \lim_{n\rightarrow\infty}\left(\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right)=0\text{.} \end{equation*}
Thus the Taylor series
\begin{equation*} 1+\frac{1}{2}x+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!}x^2+\frac{\frac{1}{2}\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)}{3!}x^3+\cdots \end{equation*}
converges to \(\sqrt{1+x}\) for \(0\leq x\leq 1\text{.}\)
Unfortunately, this proof will not work for \(-1\lt x\lt 0\text{.}\) In this case, the fact that \(x\leq c\leq 0\) makes\(\,1+c\leq 1\text{.}\) Thus \(\frac{1}{1+c}\geq 1\) and so the inequality
\begin{equation*} \frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)\left(\frac{5}{2}\right)\cdots\left(\frac{2n-1}{2}\right)}{(n+1)!}\frac{|x|^{n+1}}{(1+c)^{n+\frac{1}{2}}}\leq\frac{1\cdot 1\cdot 3\cdot 5\cdot\,\cdots\,\cdot\left(2n-1\right)}{2^{^{n+1}}(n+1)!} \end{equation*}
may not hold.

Problem 7.2.5.

Show that if \(-\frac{1}{2}\leq x\leq c\leq 0\text{,}\) then \(|\frac{x}{1+c}|\leq 1\) and modify the above proof to show that the binomial series converges to \(\sqrt{1+x}\) for \(x\in\left[-\frac{1}{2},0\right]\text{.}\)
To take care of the case where \(-1\lt x\lt -\frac{1}{2}\text{,}\) we will use yet another form of the remainder for Taylor series. However before we tackle that, we will use the Lagrange form of the remainder to address something mentioned in Chapter 4. Recall that we noticed that the series representation
\begin{equation*} \frac{1}{1+x}=1-x+x^2-x^3+\cdots \end{equation*}
did not work when \(x=1\text{,}\) however we noticed that the series obtained by integrating term by term did seem to converge to the antiderivative of \(\frac{1}{1+x}\text{.}\) Specifically, we have the Taylor series
\begin{equation*} \ln\left(1+x\right)=x-\frac{1}{2}x^2+\frac{1}{3}x^3-\cdots\text{.} \end{equation*}
Substituting \(x=1\) into this provided the convergent series \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\text{.}\) We made the claim that this, in fact, converges to \(\ln 2\text{,}\) but that this was not obvious. The Lagrange form of the remainder gives us the machinery to prove this.

Problem 7.2.6.

(a)

Compute the Lagrange form of the remainder for the Maclaurin series for \(\ln\left(1+x\right)\text{.}\)

(b)

Show that when \(x=1\text{,}\) the Lagrange form of the remainder converges to \(0\) and so the equation \(\ln 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots\) is actually correct.
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