Real Analysis/Applications of Derivatives

 Real Analysis Applications of Derivatives

Higher Order Derivatives

Let $f:\mathbb{R}\to\mathbb{R}$ be differentiable

Let $f'(x)$ be differentiable for all $x\in\mathbb{R}$. Then, the derivative of $f'(x)$ is called the second derivative of $f$ and is written as $f''(a)$

Similarly, we can define the nth-derivative of $f$, written as $f^{(n)}(x)$

Theorem(Rolle's Theorem)

Let $f:[a,b]\to\mathbb{R}$

Let $f$ be continuous on $(a,b)$ and differentiable on $[a,b]$

Let $f(a)=f(b)$

Then, there exists $c\in (a,b)$ such that $f'(c)=0$

Proof

If $f(x)$ is constant, then $f ' (x)=0$ $\forall x\in [a,b]$

Hence, let $f$ be non-constant

Let $c\in [a,b]$ such that $f(c)=\sup f([a,b])$ (this is possible due to the Minimum-maximum theorem)

Without loss of generality, we can assume that $f(c)>f(a)$

Assume if possible $f'(c)>0$. Thus, $\lim_{x\to c}\tfrac{f(x)-f(c)}{x-c}>0$ and hence, there exists $x_1\in [a,b]$ such that $f(x_1)>f(c)$ contradicting the fact that $f(c)$ is a maximum.

Similarly, we can show that the assumption $f'(c)<0$ leads to a contradiction[/itex]. Thus, $f'(c)=0$

Rolle's theorem can be generalized as follows:

Theorem(Mean Value Theorem)

If $f(x)$ and $g(x)$ are differentiable on $[a,b]$ (without both having infinite derivatives at the same point) then there exists $c$ within $[a,b]$ such that

$\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}$.

Proof

Define the function $h:[a,b]\to\mathbb{R}$ as

$h(x)=f(x)(g(b)-g(a))-g(x)(f(b)-f(a))$

Obviously, this function satisfies $h(a)=h(b)$, and by Rolle's theorem, there is a $c$ within $[a,b]$ such that $h'(c)=0$.

Theorem(Taylor's Theorem)

Let $f:[a,x]\to\mathbb{R}$

Let $f^{(n-1)}$ be differentiable on $[a,x]$

Then, there exists $c\in (a,x)$ such that

$f(x)=f(a)+\tfrac{(x-a)f'(a)}{1!}+\tfrac{(x-a)^2 f''(a)}{2!}+\ldots +\tfrac{(x-a)^{n-1}f^{(n-1)}(a)}{(n-1)!}+\tfrac{(x-a)^nf^n(c)}{n!}$

Proof

For the proof, we use the technique known as "Telescopic Sum"

Consider the function $\phi:[a,x]\to\mathbb{R}$, given by,

$\phi (y)=f(y)+\tfrac{(x-y)f'(y)}{1!}+\tfrac{(x-y)^2f''(y)}{2!}+\ldots+\tfrac{A(x-y)^n}{n!}$, where the constant $A\in\mathbb{R}$ is chosen so as to satisfy $\phi (a)=\phi (x)$

By Rolle's theorem, we have that there exists $c\in (a,x)$ such that $\phi '(c)=0$

Expanding, we have (be careful while applying the product rule!)

$0=\phi '(c)$

$=f'(c)+\left[ \tfrac{(x-c)f''(c)}{1!}-f'(c)\right] +\left[ \tfrac{(x-c)^2f'''(c)}{2!}-\tfrac{(x-c)f''(c)}{1!}\right] +\ldots$

$+\left[ \tfrac{(x-c)^{n-1}f^{(n)}(c)}{(n-1)!}-\tfrac{(x-c)^{n-2}f^{(n-1)}}{(n-2)!}\right]-\tfrac{A(x-c)^{n-1}}{(n-1)!}$

Which can be rearranged to give the telescopic sum:

$\left[ f'(c)-f'(c)\right] + \left[ \tfrac{(x-c)f''(c)}{1!}-\tfrac{(x-c)f''(c)}{1!}\right] +\ldots+\left[ \tfrac{(x-c)^{n-1}f^{(n)}(c)}{(n-1)!}-\tfrac{A(x-c)^{n-1}}{(n-1)!}\right]=0$

That is, $\frac{(x-c)^{n-1}f^{(n)}(c)}{(n-1)!}=\frac{A(x-c)^{n-1}}{(n-1)!}$, or $A=f^{(n)}(c)$

Now, we can easily see that $\phi (x)=f(x)$ and that $\phi (a)=f(a)+\tfrac{(x-a)f'(a)}{1!}+\tfrac{(x-a)^2 f''(a)}{2!}+\ldots +\tfrac{(x-a)^{n-1}f^{(n-1)}(a)}{(n-1)!}+\tfrac{(x-a)^nf^n(c)}{n!}$

but by choice, $\phi (x)=\phi (a)$ and hence we have:

$f(x)=f(a)+\tfrac{(x-a)f'(a)}{1!}+\tfrac{(x-a)^2 f''(a)}{2!}+\ldots +\tfrac{(x-a)^{n-1}f^{(n-1)}(a)}{(n-1)!}+\tfrac{(x-a)^nf^n(c)}{n!}$

QED