Let $u(x)\in C^2(R)$ is a real function. so: $$ u'(x)=\lim_{\Delta x\rightarrow 0} \frac{u(x+\Delta x)-u(x)}{\Delta x} $$ And: $$ u''(x)=\lim_{\Delta x\rightarrow 0} \frac{u'(x+\Delta x)-u'(x)}{\Delta x} =\lim_{\Delta x\rightarrow 0}\frac{\frac{u(x+2\Delta x)-u(x+\Delta x)}{\Delta x}-\frac{u(x+\Delta x)-u(x)}{\Delta x}}{\Delta x} ~~~~~(2) $$ So: $$ u''(x)=\lim_{\Delta x\rightarrow 0} \frac{u(x+2\Delta x)-2u(x+\Delta x)+u(x)}{\Delta x^2} ~~~~~(3) $$
Let $x=x-\Delta x$, $$ u''(x-\Delta x)=\lim_{\Delta x\rightarrow 0} \frac{u(x+\Delta x)-2u(x)+u(x-\Delta x)}{\Delta x^2} ~~~~~(4) $$ So: $$ u''(x)=\lim_{\Delta x\rightarrow 0} \frac{u(x+\Delta x)-2u(x)+u(x-\Delta x)}{\Delta x^2} ~~~~~(5) $$ So: $$ u''(x)=\lim_{\Delta x\rightarrow 0} \frac{\frac{u(x+\Delta x)+u(x-\Delta x)}{2}-u(x)}{\frac{\Delta x^2}{2}} ~~~~~(6) $$
First,are the above formulas really right?
Second,what is the geometric mean of second order derivative, or how to directly and visually understand second order derivative?
Seemingly,second order derivative describe the bend of function's graph.But the denominator of formula 6 make it become complex.And I can't really understand the second order derivative.