Real Analysis – Differential Calculus III

Theorem 65 (Cauchy’s theorem) Let {[a,b]\subset\mathbb{R}} and {f}, {g} continuous such as {f;g:[a,b]\rightarrow \mathbb{R}}. If {f} and {g} are differentiable in {]a,b[} and {g'} doesn’t vanish in {]a,b[}, there exists {c \in ]a,b[} such as

\displaystyle \frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)} \ \ \ \ \ (32)

Proof: It is {g(a)\neq g(b)} since if it were {g(a)=g(b)} {g'} would vanish in {]a,b[} by Theorem 63.

Let

\displaystyle \lambda=\frac{f(b)-f(a)}{g(b)-g(a)}

and define {\varphi} as {\varphi:[a,b]\rightarrow\mathbb{R}}(differentiable in {]a,b[} and continuous in {[a,b]}) such as {\varphi=f(x)-\lambda g(x)} {\forall x \in [a,b]} it is

\displaystyle \varphi(a)=f(a)-\lambda g(a)=\ldots=\varphi(b)

Thus by applying Theorem 63 in {[a,b]} there exists {c\in [a,b]} such as {\varphi'=0}. That is

\displaystyle f'(c)-\lambda g'(c)=0 \Leftrightarrow \lambda=\frac{f'(c)}{g'(c)}

\Box

The previous theorem is perhaps more of a lemma than a theorem per se. Because it will allows us to prove more important results. Also this result can be seen as providing a method of finding (very) local approximations to functions at a given point and as such it is the same as a Taylor expansion of first order (we’ll see what this means in futures posts).

Theorem 66 (Cauchy first limit rule) Let {I \subset \mathbb{R}}, {c\in I'} and {f,g:I\setminus \{c\}\rightarrow \mathbb{R}} differentiable. Moreover {g'} doesn’t vanish in {I\setminus \{c\}} and {\displaystyle \lim _{x\rightarrow c}f(x)=\displaystyle \lim _{x\rightarrow c}g(x)=0}.

If {\displaystyle \lim _{x\rightarrow c}\frac{f'(x)}{g'(x)}} exists it is

\displaystyle \lim _{x\rightarrow c}\frac{f(x)}{g(x)}=\lim _{x\rightarrow c}\frac{f'(x)}{g'(x)} \ \ \ \ \ (33)

Proof: Let {c\in\mathbb{R}}. Since {f,g} are continuous in {I\setminus \{ c \}} and {\displaystyle \lim _{x\rightarrow c}f(x)=\displaystyle \lim _{x\rightarrow c}g(x)=0} we can set {f(c)=g(c)=0}. Let {x_n: \mathbb{N}\rightarrow I\setminus \{c\}} such as {x_n\rightarrow c^+}.

Applying Cauchy’s Theorem 65 to each interval {[c,x_n]} it is

\displaystyle \frac{f(x_n)}{g(x_n)}=\frac{f(x_n)-f(c)}{g(x_n)-g(c)}=\frac{f'(u_n)}{g'(u_n)}

with {c<u_n<x_n}.

Then {u_n\rightarrow c} by the Squeezed Sequence Theorem 17

And

\displaystyle \lim _{x\rightarrow c}\frac{f'(u_n)}{g'(u_n)}=\lim _{x\rightarrow c}\frac{f'(x)}{g'(x)}

by the definition of limit

Thus

\displaystyle \lim _{x\rightarrow c}\frac{f(x_n)}{g(x_n)}=\lim _{x\rightarrow c}\frac{f'(x)}{g'(x)}

Hence, by the definition of limit it is

\displaystyle \lim _{x\rightarrow c}\frac{f(x)}{g(x)}=\lim _{x\rightarrow c^+}\frac{f'(x)}{g'(x)} \ \ \ \ \ (34)

Analogously if {x_n} is

\displaystyle x_n\rightarrow c^-

applying Cauchy’s Theorem 65 to each interval {[x_n,c]} it is

\displaystyle \frac{f(x_n)}{g(x_n)}=\frac{f(x_n)-f(c)}{g(x_n)-g(c)}=\frac{f(c)-f(x_n)}{g(c)-g(x_n)}=\frac{f'(u_n)}{g'(u_n)}

with {x_n<u_n<c}.

Just like in the previous steps it is

\displaystyle \lim _{x\rightarrow c}\frac{f(x)}{g(x)}=\lim _{x\rightarrow c^-}\frac{f'(x)}{g'(x)} \ \ \ \ \ (35)

From equation 34 and equation 35 it is

\displaystyle \lim _{x\rightarrow c}\frac{f(x)}{g(x)}=\lim _{x\rightarrow c}\frac{f'(x)}{g'(x)}

Finally let {c=+\infty}. Let {x=1/t} it is {x\rightarrow p\infty \Leftrightarrow t\rightarrow 0^+}. From what we proved thus far it is

{\begin{aligned} \displaystyle \lim _{x\rightarrow +\infty}\frac{f(x)}{g(x)} &= \displaystyle \lim_{t \rightarrow 0^+}\frac{f(1/t)}{g(1/t)}\\ &= \displaystyle\lim_{t \rightarrow 0^+}\frac{(f(1/t))'}{(g(1/t))'}\\ &=\displaystyle \lim_{t \rightarrow 0^+}\frac{-1/t^2f'(1/t)}{-1/t^2g'(1/t)}\\ &=\displaystyle \lim_{t \rightarrow 0^+}\frac{f'(1/t)}{g'(1/t)}\\ &=\displaystyle \lim_{t \rightarrow 0^+}\frac{f'(x)}{g'(x)}\\ \end{aligned}}

Hence, for this case it also is {\displaystyle\lim _{x\rightarrow c}\frac{f(x)}{g(x)}=\lim _{x\rightarrow c}\frac{f'(x)}{g'(x)}}.

The case {c=-\infty} can be proven in a similar way with the change of variable {x=-1/t}. \Box

Theorem 67 (Cauchy second limit rule) Let {I \subset \mathbb{R}}, {c\in I'} and {f,g:I\setminus \{c\}\rightarrow \mathbb{R}} differentiable. Moreover {g} doesn’t vanish in {I\setminus \{c\}} and {\displaystyle \lim _{x\rightarrow c}f(x)=\displaystyle \lim _{x\rightarrow c}g(x)=+\infty}. If {\displaystyle \lim _{x\rightarrow c}\frac{f'(x)}{g'(x)}} exists it is

\displaystyle \lim _{x\rightarrow c}\frac{f(x)}{g(x)}=\lim _{x\rightarrow c}\frac{f'(x)}{g'(x)} \ \ \ \ \ (36)

Proof: Left as an exercise for the reader. \Box

The two previous theorems are known by a variety of names on the mathematical literature and are one of the most used theorems in the practice of calculating limits.

A few examples will now be used to showcase their powers

Example 1 The functions {e^x} and {x} tend to infinity as {x} goes to infinity. We already saw that the exponential function goes to infinity faster than any polynomial of {x}, but Cauchy’s second theorem allows us to prove that result much faster. As always a method that proves itself to tame thorny results in a more efficient way surely is a powerful method.

\displaystyle \lim_{x\rightarrow \infty}\frac{e^x}{x} \ \ \ \ \ (37)

{\begin{aligned} \displaystyle \lim _{x\rightarrow +\infty}\frac{e^x}{x} &= \displaystyle \lim_{x \rightarrow +\infty}\frac{(e^x)'}{x'}\\ &= \displaystyle \lim _{x\rightarrow +\infty}\frac{e^x}{1}\\ &= \infty \end{aligned}}

Example 2 The functions {\cos x-1} and {x^2} tend to {0} as {x} goes to {0}. But which one of them tends to {0} more strongly?

{\begin{aligned} \displaystyle \lim_{x\rightarrow 0}\frac{\cos x-1}{x^2} &= \displaystyle \lim_{x\rightarrow 0}\frac{(\cos x-1)'}{(x^2)'}\\ &= \displaystyle \lim_{x\rightarrow 0}\frac{-\sin x}{2x}\\ &= \ldots \end{aligned}}

At the end of the last example we arrived once again at the type of limit {\displaystyle \lim_{x\rightarrow 0}\frac{f(x)}{g(x)}} where {\displaystyle \lim_{x\rightarrow 0}f(x)=\displaystyle \lim_{x\rightarrow 0}g(x)=0}.

But the thing is that Cauchy’s first rule (and in fact the second rule too) can be used more than one time. Hence we’ll just apply it again (we’ll start from the begining again) just so we don’t lose our train of thought

{\begin{aligned} \displaystyle \lim_{x\rightarrow 0}\frac{\cos x-1}{x^2} &= \displaystyle \lim_{x\rightarrow 0}\frac{(\cos x-1)'}{(x^2)'}\\ &= \displaystyle \lim_{x\rightarrow 0}\frac{-\sin x}{2x}\\ &= \displaystyle \lim_{x\rightarrow 0}\frac{-\cos x}{x}\\ &= -\dfrac{1}{2} \end{aligned}}

As an exercise calculate

\displaystyle \lim_{x \rightarrow 0} \frac{e^x-1}{1}

Another mathematical theorem from real analysis which is very important to Physics, in a conceptual level, is what we’ll call Lagrange’s theorem. Even though it is a theorem in Real Analysis it has a very nice interpretation in geometrical and in kinematic terms.

Theorem 68 (Lagrange’s theorem) Let {[a,b]\subset\mathbb{R}} and {f:[a,b]\rightarrow\mathbb{R}} continuous. If {f} is differentiable in {]a,b[} there exists {c\in ]a,b[} such as

\displaystyle \frac{f(b)-f(a)}{b-a}=f'(c) \ \ \ \ \ (38)

Proof: In theorem 65 let {g(x)=x} and the result follows trivially. \Box

As I was saying before the statement and proof of this theorem it can be interpreted both geometrically and kinematically. The geometric interpretation states that the secant to the function {f(x)} in the interval {[a,b]} has a given slope and that we can always find a tangent to the function {f} in the interval {[a,b]} whose slope is the same as the secant. Hence the straight lines defined by these secant and tangent are parallel.

In a kinematic sense if {x} represents time and {f(x)} represents the distance travelled this result implies that if we transverse the distance {f(b)-f(a)} in the time interval {b-a} then we have an average speed which is

\displaystyle \frac{f(b)-f(a)}{b-a}

Since in this context {f'(x)} can be interpreted as the being the instantaneous speed (or just speed for short) the previous result states that there exists a time instant {c} in which your instantaneous speed is the same as you average speed for the whole time interval.

Example 3 Show that {e^x-1>x\quad \forall x \neq 0}.

Proof: Let {f(t)=e^t}. Assume that {x>0} and apply Theorem 68 to the interval {[0,x]}.

\displaystyle \frac{e^x-e^0}{x-0}=\left( e^t \right)'_{t=c}

with {0<c<x}.

Then

\displaystyle \frac{e^x-1}{x}=e^c>1

Assume now that {x<0} and apply once again 68, but this time to the interval {[x,0]}.

\displaystyle \frac{e^0-e^x}{0-x}=\left( e^t \right)'_{t=c}

with {x<c<0}.

Then

\displaystyle \frac{1-e^x}{-x}=e^c<e^0=1\Leftrightarrow 1-e^x<-x\Leftrightarrow e^x-1>x

Notice that in the we didn’t invert the sign of the inequality while multiplying by {-x} because {x<0} and consequently {-x>0}. \Box

The last theorem has two important corollaries that we’ll state below.

Corollary 69 Let {I} be an interval in {\mathbb{R}} and {f:I\rightarrow\mathbb{R}} continuous. If {f'} exists and vanishes in the interior of {I}, then {f} is constant.

Proof: By reductio ad absurdum let us assume that {f} is not constant. Then there exists {a,b \in I} such as {a<b} and {f(a)\neq f(b)}. Since {f} is continuous in {[a,b]} and differentiable in {]a,b[}, by theorem 68 it is

\displaystyle \frac{f(b)-f(a)}{b-a}=f'(c)

with {c\in ]a,b[}.

Hence {\frac{f(b)-f(a)}{b-a}=0} which is absurd since it would imply that {f(b)=f(a)}, which is contrary to our initial hypothesis. \Box

Corollary 70 () Let {I} be an interval in {\mathbb{R}} and {f:I\rightarrow\mathbb{R}} continuous. If {f'} exists and is positive (negative) in the interior of {I}, then {f} is strictly increasing (decreasing).

Proof: Let us take the case {f'>0}. Take {a,b \in I} such as {a<b}. From theorem 68 it is

\displaystyle frac{f(b)-f(a)}{b-a}=f'(c)>0

with {c \in ]a,b[}.

Since {b-a>0} it is {f(b)>f(a)} and {f} is strictly increasing. \Box

And with these results we finish the Differential Calculus part of our course in Real Analysis. The next theoretical posts of Real Analysis will dwell on the theory of numerical series.

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