MATH 409 Lecture 17

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Lecture Slides

Examples

${\mathrm{e}}^x>x+1$ for all $x\ne 0$.

Consider $f(x)={\mathrm{e}}^x-x-1$ for $x\in \mathbb{ℝ}$. Our goal is to show that $f$ is greater than $0$ for all $x\ne 0$. This function is differentiable on $\mathbb{ℝ}$ and $f^{\prime }(x)={\mathrm{e}}^x-1$ for all $x\in \mathbb{ℝ}$. We observe that $f^{\prime }$ is strictly increasing. Since $f^{\prime }(0)=0$, we have $f^{\prime }(x)<0$ for all $x<0$ and $f^{\prime }(x)>0$ for all $x>0$.

It follows that $f$ is strictly decreasing on $(-\mathrm{\infty },0]$ and strictly increasing on $[0,\mathrm{\infty })$. As a consequence, $f(x)>f(0)=0$ for all $x\ne 0$. Thus ${\mathrm{e}}^x>x+1$ for $x\ne 0$.

$\log x<x-1$ for all $x>0$, $x\ne 1$.

By above, ${\mathrm{e}}^{x-1}>(x-1)+1=x$ for all $x\ne 1$. Since the natural logarithm is strictly increasing on $(0,\mathrm{\infty })$, it follows that $\log {\mathrm{e}}^{x-1}>\log x$ for $x>0$, $x\ne 1$. Equivalently $x-1>\log x$ for $x>0$, $x\ne 1$.

Bernoulli's Inequality. ${(1-x)}^{\alpha }>1-\alpha x$ for all $x\in (0,1)$ and $\alpha >1$.

Fix an arbitrary $\alpha >1$ and consider

$f(x)=(1-x{)}^{\alpha }-1+\alpha x$

This function is differentiable on $[0,1)$ and $f^{\prime }(x)=-\alpha {(1-x)}^{\alpha -1}+\alpha$ for all $x\in [0,1)$. Since $\alpha -1>0$, we obtain that ${(1-x)}^{\alpha -1}<1$ for $x\in (0,1)$. Hence $f^{\prime }(x)>0$ for $x\in (0,1)$.It follows that the function is strictly increasing on $[0,1)$. As a consequence, $f(x)>f(0)=0$ for all $x\in (0,1)$.

${(1-x)}^{\alpha }<1-\alpha x+\frac{\alpha (\alpha -1)}{2}{x}^2$ for all $x\in (0,1)$ and $\alpha >2$.

Let us fix an arbitrary $\alpha >2$ and consider a function

$f(x)={(1-x)}^{\alpha }-1+\alpha x-\frac{1}{2}\alpha (\alpha -1)x^2$

This function is infinitely differentiable on $[0,1)$ and for all $x\in [0,1)$,

$\begin{array}{rl}{f}^{\prime }(x)& =-\alpha (1-x{)}^{\alpha -1}+\alpha -\alpha (\alpha -1)x\\ f^{″}(x)& =\alpha (\alpha -1){(1-x)}^{\alpha -2}-\alpha (\alpha -1)\end{array}$

Since $\alpha -2>0$, we obtain that $f^{″}<0$ for $x\in (0,1)$. It follows that the derivative $f^{\prime }$ is strictly decreasing on $[0,1)$. As a consequence, $f^{\prime }(x)<f^{\prime }(0)=0$ for all $x\in (0,1)$. Now it follows that the function is strictly decreasing on $[0,1)$. Consequently $f(x)<f(0)=0$ for all $x\in (0,1)$. The required inequality follows.

The function $f(x)={(1+x)}^{\frac{1}{x}}$ is strictly decreasing on $(0,\mathrm{\infty })$.

(The limit at 0 is $\mathrm{e}$)

Consider a function $g(x)=\log f(x)$ for $x>0$. For every $x>0$, we have $g(x)=\frac{\log (1+x)}{x}$. This function is differentiable on $(0,\mathrm{\infty })$, and therefore the original function $f$ is differentiable (it is the exponentiation of this function):

$g^{\prime }(x)=\frac{\frac{1}{1+x}-\log (1+x)}{x^2}$

Now we introduce another function $h(x)=\frac{x}{1+x}-\log (1+x)=1-\frac{1}{1+x}-\log (1+x)$, $x\ge 0$.

Notice that $h(x)=x^2{g}^{\prime }(x)$ for $x>0$. The function $h$ is differentiable on $[0,\mathrm{\infty })$ and $h^{\prime }(x)=\frac{1}{(1+x{)}^2}-\frac{1}{1+x}<0$ for all $x>0$. It follows that $h$ is strictly decreasing on $[0,\mathrm{\infty })$. In particular, $h(x)<h(0)=0$ for $x>0$. Then $g^{\prime }(x)<0$ for $x>0$ as well. Therefore $g$ is strictly decreasing on $(0,\mathrm{\infty })$. Since $f$ is the composition of $g$ with the strictly increasing function $y(x)={\mathrm{e}}^x$, it is also strictly decreasing on $(0,\mathrm{\infty })$.

Taylor's Formula

Theorem. If a function $f:I\to \mathbb{ℝ}$ is $n+1$ times differentiable on an open interval $I$, then for any two points $x,x_0\in I$, there is a point $c$ between $x$ and $x_0$ such that

$f(x)=f(x_0)+\sum _{k=1}^n\frac{f^{(k)}(x_0)}{k!}{(x-x_0)}^k+\frac{f^{(n+1)}(c)}{(n+1)!}{(x-x_0)}^{n+1}$

This function $P_n^{f,x_0}$ is called the Taylor polynomial of order $n$ generated by $f$ centered at $x_0$.

It provides information on the remainder term $r_n^{f,x_0}=f-P_n^{f,x_0}$. In many cases, this information allows us to estimate $|r_n^{f,x_0}(x)|$, the error in the estimate, or to prove an inequality of the form $f(x)<P_n^{f,x_0}(x)$ or $f(x)>P_n^{f,x_0}(x)$.

This will come in handy on the homework

l'Hôspital's Rule

(May also be spelled l'Hôpital's Rule) Helps us to compute limit of quotients in those cases where limit theorems do not apply due to indeterminacy of the form $\frac{0}{0}$ or $\frac{\mathrm{\infty }}{\mathrm{\infty }}$.

Theorem. Let $a$ be an extended real number. Let $I$ be an open interval such that $a\in I$ or $a$ is an endpoint of $I$. Suppose that $f$ and $g$ are differentiable on $I$ and that $g(x),g^{\prime }(x)\ne 0$ for $x\in I\setminus \left\{a\right\}$. Suppose further that $\underset{x\in I}{\underset{x\to a}{\lim }}f(x)=\underset{x\in I}{\underset{x\to a}{\lim }}g(x)=A$

Where $A=0\text{ or }\mathrm{\infty }$, If the limit $\underset{x\in I}{\underset{x\to a}{\lim }}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ existst (finite or infinite), then $\underset{x\in I}{\underset{x\to a}{\lim }}\frac{f(x)}{g(x)}=\underset{x\in I}{\underset{x\to a}{\lim }}\frac{f^{\prime }(x)}{g^{\prime }(x)}$.

Proof. (in case $\underset{x\to a^{+}}{\lim }\frac{0}{0}$) We extend $f$ and $g$ to $I\cup \left\{a\right\}$ by letting $f(a)=g(a)=0$. By hypothesis, $f$ and $g$ are continuous on $I\cup \left\{a\right\}$ and differentiable on $I$. By the Generalized Mean Value Theorem, for any $x\in I$, there exists $c_x\in (a,x)$ such that

$g^{\prime }(c_x)(f(x)-f(a))=f^{\prime }(c_x)(g(x)-g(a))$

That is, $g^{\prime }(c_x)f(x)=f^{\prime }(c_x)g(x)$. Since $g(c_x),g^{\prime }(c_x)\ne 0$, we obtain $\frac{f(x)}{g(x)}=\frac{f^{\prime }(c_x)}{g^{\prime }(c_x)}$. Since $c_x\in (a,x)$, we have $c_x\to a$ as $x\to a^{+}$. It follows that

$\underset{x\to a^{+}}{\lim }\frac{f(x)}{g(x)}=\underset{x\to a^{+}}{\lim }\frac{f^{\prime }(c_x)}{g^{\prime }(c_x)}=\underset{c\to a^{+}}{\lim }\frac{f^{\prime }(c)}{g^{\prime }(c)}$

The theorem includes several similar rules corresponding to various kinds of limits ($x\to a^{+}$, $x\to a^{-}$, $x\to a$ for $a\in \mathbb{ℝ}$, $x\to +\mathrm{\infty }$, $x\to -\mathrm{\infty }$), and the two types of indeterminacy ($\frac{0}{0}$ and $\frac{\mathrm{\infty }}{\mathrm{\infty }}$)

quod erat demonstrandum

Examples

$\underset{x\to 0}{\lim }\frac{1+\cos x}{x^2}$

The functions $f(x)=1-\cos x$ and $g(x)=x^2$ are infinitely differentiable on $\mathbb{ℝ}$. We have $\underset{x\to 0}{\lim }f(x)=f(0)=0$ and $\underset{x\to 0}{\lim }g(x)=g(0)=0$.

Further, $f^{\prime }(x)=\sin x$ and $g^{\prime }(x)=2x$. We obtain that the form is still indeterminate at $0$.

Even further, $f^{″}(x)=\cos x$ and $g^{″}(x)=2$. We obtain that $\underset{x\to 0}{\lim }{f}^{″}(x)=f^{″}(0)=1$ and $\underset{x\to 0}{\lim }{g}^{″}(x)=g^{″}(0)=2$. It follows that $\underset{x\to 0}{\lim }\frac{f^{″}(x)}{g^{″}(0)}=\frac{1}{2}$.

By l'Hôspital's Rule, $\underset{x\to 0}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}=\frac{1}{2}$, then by extension $\underset{x\to 0}{\lim }\frac{f(x)}{g(x)}=\frac{1}{2}$.