Real Analysis/Applications of Derivatives

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Real Analysis
Applications of Derivatives


Higher Order Derivatives[edit]

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)[edit]

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


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</math>. Thus, f'(c)=0

Rolle's theorem can be generalized as follows:

Theorem(Mean Value Theorem)[edit]

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



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


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)[edit]

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!}


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!}