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Let us consider a continuous function $y=f(x)$,

The first order derivative of the function $f(x)$ is defined by

$$\frac{d}{dx}f(x) = \lim\limits_{h \to 0} \frac{f(x) - f(x - h)}{h}$$

The second order derivative is

$$ \frac{d^{2}}{dx^{2}}f(x) = \lim\limits_{h \to 0} \frac{f(x) - 2f(x - h) +f(x-2h)}{h^{2}}$$ and $$ \frac{d^{3}}{dx^{3}}f(x) = \lim\limits_{h \to 0} \frac{f(x) - 3f(x - h) +3f(x-2h) - f(x-3h)}{h^{3}},$$

and so son...

Now using same reasoning, by induction how we prove the bellow expression can be used to calculate nth derivative of $f(x)$ ?

$$ \frac{d^{n}}{dx^{n}}f(x) = \lim\limits_{h \to 0} \frac {1}{h^{n}} \sum_{k=0}^{n} (-1)^{k} \binom{n}{k}f(x-kh)$$

H. Rivera
  • 61

2 Answers2

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Let's assume your formula is correct for the nth derivative. We then have \begin{align} \frac{d^{n+1}}{dx^{n+1}}f(x)=&\lim_{h\rightarrow 0} \frac{\frac{d^{n}f(x)}{dx^{n}}-\frac{d^{n}f(x-h)}{dx^{n}}}{h}=\lim_{h\rightarrow 0}\sum_{k=0}^{n}(-1)^k\binom{n}{k}\frac{f(x-kh)-f(x-(k+1)h)}{h^{n+1}}\\ &=\lim_{h\rightarrow 0}\sum_{k=0}^{n+1}(-1)^k\left(\binom{n}{k}+\binom {n}{k-1}\right)\frac{f(x-kh)}{h^{n+1}}=\lim_{h\rightarrow 0}\sum_{k=0}^{n+1}(-1)^k\binom{n+1}{k}\frac{f(x-kh)}{h^{n+1}}. \end{align} The last line follows from a recursion relation on binomial coefficients (https://en.wikipedia.org/wiki/Binomial_coefficient#Recursive_formula).

Teddy Baker
  • 886
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HINTS: First, $$\frac{d^{n+1}}{dx^{n+1}} f(x) = \lim_{h \rightarrow0}\frac{\frac{d^{n}}{dx^{n}} f(x)-\frac{d^{n}}{dx^{n}} f(x-h)}{h}.$$ Second, $$\sum_{k=0}^{n}(-1)^k\binom{n}{k}f(x-kh) = (-1)^0 \binom{n+1}{0} f(x) + \sum_{k=1}^{n}(-1)^k\binom{n}{k}f(x-kh).$$ Third, $$\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}.$$ Fourth $$\binom{n}{n}=\binom{n+1}{n+1}.$$

Changing the variable of summation and using the above relationships should easily prove the assertion.

Math Lover
  • 15,153