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It is a really great question!

The binomial tree bob uses is a common tool in stochastic calculus and the problem that you posted is similar to gambler's ruin problem.

The gambler's ruin problem is like this. A gambler has n coins and went to gamble. A win gets another coin and a lose gives away one. How long does it take for him to lose all coins?

It is also named as barrier problem.

The question you posted is more complicated - a double barrier instead of one barrier. A has to win before B wins.

I remember such problem usually involves alternative probability series to cancel out over-counting.

**George,Y**- Replies: 0

The flame is not gas, not fluid, not solid, but moving plasma.

However the speed and direction of the plasma is highly influenced by that of reactant and environment.

Examples include ejecting flame, chimney effect and dancing flame in the wind.

It is commonly neglected that the plasma itself prohibits gas diffusion.

So sometimes the combustion is incomplete - the fresh air cannot freely get into the flame and react with the fuel.

And as a flame is loaded above - a pot on flame. e.g.

this plasma barrier gets stronger, and a complete combustion could turn incomplete.

A flame on candle may not produce black smoke (a sign of incomplete combustion), but will leave black stain at the bottom of some solid above it.

I invented the idea of plasma barrier, or has it already been found?

Mathegocart wrote:

it's George Y!

Hey, long time not seeing you guys!

iamaditya wrote:

Hi George, Y.

Were you there all the time? Why did you not post anything...

Sorry guys, I went through a chaos of changing and losing jobs.

I was not in the mood for mathematics.

Otherwise the electrons will flow from areas with lower potential to that with higher potential

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_formulaThere 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]

**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?

Or Kuran online university

Today's textbooks are very concise, and you can learn most topics up to undergraduate level on your own

Note the common ratio is a complex number.

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

Happy New Year!

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)

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

Try JavaScript.

bobbym wrote:

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

Normsdist in Microsoft Excel

bobbym wrote:

Hi;

What is N?

Normal Probability Distribution Function

bobbym wrote:

Hi;

I do not think those are doable.

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

bobbym wrote:

Hi;

What is d?

d is the differential operator

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

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

bobbym wrote:

A different integration?

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

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