On Sun, May 7, 2017 at 7:51 PM, Math for Mere Mortals wrote:

>

]]>This works because your three new points *are* first derivatives. When you apply the formula to calculate the second derivative of them, you are taking a second derivative of a first derivative. That’s the same as taking the third derivative! You can apply this pattern as many times as you like, and each time you will find a higher order of derivative. When you are down to three points, apply the formula. And of course the points can be unevenly spaced in x because the handy-dandy formula accounts for that.

Here is another way to look at it. In calculus you can take derivatives in multiple steps or all at once and you will always arrive at the same result: d3y/dx3 = d/dx(d/dx(d/dx(y(x)))) What you are doing here is taking a first derivative and then finishing with a second derivative like this: d3y/dx3 = d2/dx2(d/dx(y(x)))

It is also possible to find a single formula that does all of this, but it would be too tedious and cumbersome to be of use to mere mortals. Working with the data step by step is more manageable, including in computer programs. I hope this helps answer your query.

On Thu, May 4, 2017 at 9:22 PM Math for Mere Mortals wrote:

>

]]>On Tue, May 2, 2017 at 11:25 PM Math for Mere Mortals wrote:

>

]]>On Sat, Mar 4, 2017 at 9:13 AM, Math for Mere Mortals wrote:

>

]]>Thanks for your posts on regularisation – I have learnt a lot! Please may I know how I can reference your work in my dissertation?

Regards,

Rithvik ]]>