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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.

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It is maybe not exactly an answer to the question but may help to get an intuition:

  • First derivative is: speed = variation of distance
  • Second derivative is: acceleration = variation of speed
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  • $\begingroup$ Thank you , but I hope to get a geometric mean.a mean about the graph of function. $\endgroup$ – lanse2pty Dec 5 '14 at 14:56
  • $\begingroup$ That is why I said: "It is maybe not exactly an answer"... Probably someone more aware of geometry will provide you a better answer :). $\endgroup$ – Surb Dec 5 '14 at 15:00
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The geometric meaning of the second derivative is the rate at which the tangent of the curve is changing. A large positive second derivative intuitively means that the curve is quickly bending upwards, hence why $2x^2$ is 'more curved' around $0$ than $x^2$. At a point where the second derivative has a local maximum, the curve stops bending upwards and starts bending downwards. I would recommend reading a good chapter on convex and concave functions - I seem to remember a good appendix in Spivak's Calculus.

Edit

If you are still not convinced, consider two functions $f$ and $g$ such that $f(0) = g(0)$ and suppose that $f^{\prime\prime}(x) > g^{\prime\prime}(x)$ for all $x \in [0,a].$ Then by the fundamental theorem of calculus we have $$ f^{\prime}(a) = f^{\prime}(0) + \int_{0}^{a} f^{\prime\prime}(x) \, dx $$ $$ g^{\prime}(a) = g^{\prime}(0) + \int_{0}^{a} g^{\prime\prime}(x) \, dx $$ so that $ f^{\prime}(a) > g^{\prime}(a).$ Hence, the function $f$ with the larger second derivative has 'gotten steeper' over the interval $[0,a].$ You could interpret this as meaning that $f$ is 'more curved' than $g.$

To take an extreme example, take the function $h$ such that $h^{\prime\prime}(x) = 0$ for all $x$.You can easily show that $h$ is a straight line - it has $0$ second derivative and hence it has no curvature.

There are more nuances than I have explained here but at least this can give some intuition into what the second derivative means.

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  • $\begingroup$ I think you are wrong.. $\endgroup$ – lanse2pty Dec 6 '14 at 14:34
  • $\begingroup$ @lanse2pty I have made an edit to my answer which I hope makes you more convinced. It would be good if someone else could add to the conversation as I do not know what more to say and if I am wrong I'd like to know. $\endgroup$ – Shai Dec 7 '14 at 15:09
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One way of interpreting it is that your numerator:

$$\frac{u(x+\Delta x)+u(x-\Delta x)}{2}-u(x)$$

is the difference between $u$ at $x$ and the average of $u$ at its neighbors on each side. This means the second derivative is positive at a point if $u$ is smaller than its local averages, is zero if $u$ is equal to its local averages, and is negative if $u$ is larger than its local averages.

There are several implications of this observation. For instance, suppose $u$ is convex, that is, $u''>0$ everywhere. Then the local average of $u$ in the sense above is the same as the linear interpolant of the points $(x-\Delta x,u(x-\Delta x)),(x+\Delta x,u(x+\Delta x))$ evaluated at $x$. This means that a convex function lies below its secant lines. Similarly a concave function lies above its secant lines. Similar reasoning implies that a convex function lies above its tangent lines, while a concave function lies below its tangent lines.

As for interpreting the denominator of your expression, it just sets the "scale" at which the differences should be measured. That is, if $u$ is twice differentiable at $x$ then the numerator of your fraction is of order $\Delta x^2$ as $\Delta x \to 0$.

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