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#1 Re: This is Cool » the Mythical formula to calculate Numerical Derivatives! » 2018-02-12 22:42:05

zetafunc wrote:

These coefficients are part of a more general phenomenon called the Savitzky-Golay filter in numerical analysis. In fact the traditional five-point 'stencil' reads:

where
. You can derive this result simply by playing around with the Taylor series for
with
. It's a little tedious, but you can see the derivation here: https://en.wikipedia.org/wiki/Five-poin … he_formula

There are analogous formulae for higher order derivatives too, and several published papers about the error term.

That is very informative, zeta. Thanks for the reply.

I wonder if there is a way to calculate the midpoint differentials by 4 points around it:

dy/dx ]x=0

ddy/dx/dx ]x=0

given y[-0.5], y[0.5], y[-1.5] & y[1.5]

#2 This is Cool » the Mythical formula to calculate Numerical Derivatives! » 2018-02-11 20:48:31

George,Y
Replies: 2

I come across this method when trying to calculate the derivative of an unknown curve which I have only sample points at equal grids.

The simplest way to calculate numerical derivative is three-point formula:

dy/dx = (y[1]-y[-1]) /2/dx
ddy/dx/dx = (y[1]+y[-1]-2*y[0]) /dx/dx

But I found this one on five points :

dy/dx = ( y[-2] -8*y[-1] +8*y[1] -y[2]) / (12*dx)

ddy/dx/dx = (-y[-2] +16*y[-1] -30*y[0]+ 16*y[1] -y[2])  / (12*dx^2)

It is surprisingly accurate when I test the derivative on exp(3) using exp(2) exp(2.5) exp(3) exp(3.5) exp(4)

the result of 1st numerical derivative and 2nd are
20.04243  &   20.07127

remember the correct answer are both exp(3) = 20.08554

Can anyone prove how this works?
4th polynomial?

#4 Re: Help Me ! » Need help with building a Curriculum » 2018-01-13 23:01:02

I suggest you read books one by one.
Today's textbooks are very concise, and you can learn most topics up to undergraduate level on your own

#5 Re: Help Me ! » Arithmetic and Geometric Sequences » 2018-01-13 22:58:26

Note the common ratio is a complex number.

#6 Re: Dark Discussions at Cafe Infinity » Breathing Trouble while Meditation » 2018-01-13 22:55:58

Uh it is surprising to find so many mathematicians practice meditation!

#8 Re: Help Me ! » Matrix inverse proof » 2017-12-25 01:57:29

AX =  X diag( 入i )

V := X^(-1)

AX diag( 1/ 入i )V  =  X diag( 入i )diag( 1/ 入i ) V  = I

thus


A^(-1) =  X diag( 1/ 入i ) X^(-1)

#9 Re: Coder's Corner » Python: moving on up? » 2017-12-25 01:33:22

I believe you should use some web based GUI to ease the visit of your application.
Try JavaScript.

#10 Re: Help Me ! » Normal Probability Integration » 2014-01-01 01:07:20

bobbym wrote:

Doesn't that have two parameters not one? That is why I asked.

Normsdist in Microsoft Excel

#11 Re: Help Me ! » Normal Probability Integration » 2014-01-01 00:49:10

bobbym wrote:

Hi;

What is N?

Normal Probability Distribution Function

#13 Re: Help Me ! » Normal Probability Integration » 2013-12-31 14:23:44

bobbym wrote:

Hi;

I do not think those are doable.

Yes, they are. Recall Normal probability distribution, please.

#15 Re: Help Me ! » Normal Probability Integration » 2013-12-30 22:21:05

bobbym wrote:

Hi;

What is d?

d is the differential operator

#16 Re: Help Me ! » Normal Probability Integration » 2013-12-30 22:16:59

After this change of variable, I think now the question is easier.

#18 Re: Help Me ! » Normal Probability Integration » 2013-12-29 01:11:53

Sorry guys, I made a mistake, the question should be:

#19 Re: Help Me ! » Normal Probability Integration » 2013-12-15 15:25:05

bobbym wrote:

A different integration?

I have found a way to integrate this directly, but it is very tricky.

#20 Re: Help Me ! » Normal Probability Integration » 2013-12-14 15:59:14

There actually is closed form, but through a different integration.

#22 Re: Help Me ! » √1+(√2+(√3+(√4+... = ? Harder than it seems » 2013-12-14 15:55:18

Actually, the answer is 3
Look at this poster before 2013 Asian Football Championship, the score will be 3:0
9CRN7PE44FFD0005.jpg

#23 Re: Help Me ! » Conditional Statement » 2013-11-28 01:57:45

Let me explain conditional statement via set arguments.

If you know A is true for sure, and you can immediately infer that B is true.
A is true is sufficient condition that B is true.
We can also say that the event A is true is contained in the event B is true.
Like you are in London implies that you are in UK. But the other way around is not necessarily true.

and we can say B is necessary condition for A, for notB is in notA - you are not in UK so you are not in London.

Hope this theory make your logic class a lot easier!

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