Looking for a mathematical identity for a summation 1

[Rvanpelt]

Here is the summation I am looking for an identity for:

[Σ](nidi^3/N)

I can bring out the N since that is a constant. The ^3 can go away by a variable substituion so I am essentially left with:

[Σ]XiYi

My question is can I break part that summation so I end up with Xi and Yi in seperate summations/functions?

Some background:

ni and di are two sets of i data points of liquid droplet diameters (di) and % occurance (ni). The summation of the products (nidi^3) represents a numerical average of the droplet size based on volume mean. There is a similar relationship based on the summation of (ni x di) that represents the average droplet size based on a number mean. I am looking to convert the volume mean to number mean without knowing the individual data points.

Because the values of ni would be the same in both cases, if I can isolate the summation of ni from the summation of di I can then divide by the summation of ni and be left with only the summation of di^3. At this point I know of an identity that pulls the exponent out so I can find out what the summation of di would be. The final step would be to reverese the identity I am asking about to re-construct the summation of nidi.

 

[Unotec]

Are you asking if ?XiYi=?Xi?Yi?

<<A good friend will bail you out of jail, but a true friend
will be sitting beside you saying ” Damn that was fun!” - Unknown>>

 

[rb1957]

aren't you summing over i = 1 to x?

you're right about the constant (the sum can be divided instead of dividing each term),
i think i get what you mean with the power, ie making an input of di^3, instead of cubing di (though why you want to or need to i don't know)

but isn't Sum (xiyi) for i =1 to (say)3 ...
x1y1+x2y2+x3y3 ???

 

[vc66]

Are you asking if ?XiYi=?Xi?Yi?

If that's what you're asking, I'm pretty sure that cannot be said. Try it for yourself, with some simple numbers.

V

 

[Unotec]

I agree, it cannot be done, thus my question.

I am unclear on what th OP is asking for

<<A good friend will bail you out of jail, but a true friend
will be sitting beside you saying ” Damn that was fun!” - Unknown>>

 

[zekeman]

No

 

[SomptingGuy]

I'll add "no" too. You are attempting to remove/ignore any correlation between the two variables.


- Steve

 

[prex]

Well, the answer is not so sharply: NO!
In fact, if you know the random distribution of your data (and generally a normal distribution is assumed), there is a probabilistic relationship of the kind you are looking for.
To explain that, let's take a parabolic distribution of data, as the normal distribution is quite cumbersome to handle(but it can be done). Also to make things easier we'll go in reverse from the average of diameters (sizes) to the average of volumes.
So call x_i=n_i/N the fraction of droplets with diameter d_i; the summation [&Sigma;]x_id_i is equivalent to the integral [&int;]₀¹d(x)dx and similarly for the cube.
If we now assume d(x)=A+B(x-x²) where A may be interpreted as something that's close to an average value (not exactly that, but a kind of) and B, in the same vein, as something close to a variance, we can compute the difference (integrals between 0 and 1):
[&int;]d(x)³dx-([&int;]d(x)dx)³
This is found to be equal to (if I'm not in error) 2AB²/45+19B³/7560.
Now this quantity will be relatively small if the variance is relatively small with respect to the mean value. Also this expression (or another one computed for a different distribution) can be used to correct the simple equation
d_avg=³[&radic;]([&Sigma;]x_id_i³)

prex

http://www.xcalcs.com

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[FeX32]

Why would you want to break that up?

Fe

 

[rb1957]

back at the OP ... no-one has much of a clue as to what you are trying to do (or why) ... could you restate your question more clearly pls ?

 

[IRstuff]

Your approach seems suspect. Since you already have the volume mean and you have the total volume, why not just divide one by the other?

TTFN

FAQ731-376

 

[rb1957]

how would you know the total volume of droplets without knowing the individual droplets ? unless you're collecting all the run-off ??

the original equation is clearly the mean droplet volume, but you're asking about a "number" mean (as opposed to a "volume" mean) ...

 

[IRstuff]

Ostensibly, the mean number is the total volume divided by the mean volume.

TTFN

FAQ731-376

 

[rb1957]

and the mean volume is the total volume divided by the number (of droplets) ... sounds a bit circuitious (specious?)

 

[zekeman]

Well, looking at the OP statement and noting statistical nature of the problem following Prex, I think under some basic assumptions you can get a good answer as follows:
Assume a general probability density function p(x), not necessarily Gaussian or other, and further assume it to be symmetric about the mean, xo.
then
integral x*p(x)dx = xo def of mean
(1) integral (x-xo)^2*p(x)dx= sigma^2=int(x^2-2xxo+xo^2)*p(x)dx
so that after rearranging and using definition of xo
(2) Int x^2*p(x)dx= sigma^2+xo^2
where sigma is the standard deviation
Now get
integral (x-xo)^3*p(x)dx =integral(x^3-3x^2*xo+3xo^2*x-xo^3)p(x)dx
Now, from the assumption of symmetry, the LHS =0. Rearranging the RHS, yields
integral x^3*p(x)dx= 3xo*(Sigma^2+xo^2)-3xo^3+xo^3
=3xo*(sigma^2+xo^2)
So all you must know is the average dia and its standard deviation to get average Dia^3
Please check the math.

 

[zekeman]

Correction
Found an error at the end . Should read
integral x^3*p(x)dx= 3xo*sigma^2+xo^3

 

[Rvanpelt]

Thanks for the large number of responses. I would have replied sooner but apparently e-mail notification was not on.

First, to clarify my question:

The only two pieces of information I am given is the resulting volume-mean dimater of a sample of liquid droplets, D30, and the total number of diamter ranges taken, N. I do not know the individual diameters and occurnece of each diameter range, di and ni.

I do know the general formula for calculating D30 based on a set of di and ni data points is D30 = ([&Sigma;]nidi^3/N)^1/3. Is from 1 to N.

I also know that the general formula for calculating D10, number mean, is D10=[&Sigma;]nidi/N, again i is from 1 to N.

I also can safely assume the data points follow a normal distribution about the mean. This means that values of ni for corresponding values of di will approach a maximum as di approaches D10 from either side.

The post by Prex looks very interesting and I will need to digest that more.

Essentially I am looking for a mathematical way to calculate D10 if I only know D30 and N without any knowledge of what the individual values of ni and di are.

Based on my own work, a simplified summation can be reached where D30/N=[&Sigma;]AiBi.

Where I am stuck is whether [&Sigma;]AiBi can be seperated into the respectice variables. I already know that [&Sigma;]Ai +[&Sigma;]Bi does not work and I already know that [&Sigma;]Ai X [&Sigma;]Bi does not work. I am looking to see if there are any identities or work-arounds that split the variables up that I have not found yet.

 

[zekeman]

What you are missing if this is a normal distribution is the standard deviation. Without it there is no way of relating D10 to D30 as my last eq shows.
The LHS is your D30^3 and the RHS involves D10 (my xo) and the standard deviation.
Case in point -- if sigma =0 then D10=D30 is obvious as seen in the equation.