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Originally posted by KazetNagorraYou won't believe this, but I already have!
Try the following Wolfram Alpha query:
Sum[1/( (X^2)(X + 1) ),{X,x,\infty}]
The only problem is, I didn't like what I got.
I got a formula with a "Digamma function".
No idea what that is, so I tried looking it up:
https://en.wikipedia.org/wiki/Digamma_function
I don't understand any of that! Its completely beyond me.
If it involves a lengthy iteration then it probably wouldn't be much use to me here anyway because I already got a lengthy iteration for it.
oh well, I can use the various workarounds to make living with there being no 'nice' formula for it acceptable.
Originally posted by humyYou could try using the method of partial fractions to split up the sum:
You won't believe this, but I already have!
The only problem is, I didn't like what I got.
I got a formula with a "Digamma function".
No idea what that is, so I tried looking it up:
https://en.wikipedia.org/wiki/Digamma_function
I don't understand any of that! Its completely beyond me.
If it involves a lengthy iteration then it probably wouldn't be ...[text shortened]... use the various workarounds to make living with there being no 'nice' formula for it acceptable.
1/N^2(N+1) = (AN + B)/N^2 + C/(N + 1) <--- Check this I'm remembering this from A'Level 25+ years ago...
multiplying out we get C = -A = B = 1
So:
1/(N^2(N+1)) = 1/N^2 - 1/N + 1/(N + 1)
Then you either have known sums you can look up or can use a generating function to get the sum.
1/N^2(N+1) = 1/N^2 - 1/N(N+1)
So
sum {r=N...infty} 1/r^2(r+1)
= lim{x->0} D^{-2} sum{r = N ... infty} (exp(rx) - (1+x)^{1 + r})
Where D^{-2} denotes antidifferentiation twice. For clarity the antiderivative of exp(rx) is exp(rx)/r, and the antiderivative of x^r is x^(r + 1)/(r + 1).
Sum = lim{x->0} D^{-2}[ sum{r = 0 ... infty} (exp(rx) - (1 + x)^(1 + r)) - sum {r = 0 ... N - 1} (exp(rx) - (1+x)^(1 + r))]
The sums are now all geometric sums and easy. The tricky bit is finding the antiderivatives.
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Originally posted by DeepThoughtThanks for that 🙂
So:
1/(N^2(N+1)) = 1/N^2 - 1/N + 1/(N + 1)
Then you either have known sums you can look up or can use a generating function to get the sum.
I confess complete ignorance on the method of partial fractions.
But I find that equation extremely useful and gives me a breakthrough, of sorts.
This is because we now have:
1/(X^2(X+1)) = 1/(X^2) + 1/(X + 1) – 1/X (using the method of partial fractions )
so:
∑ [X = x, X = +∞] 1/(X^2(X+1)) = ∑ [X = x, X = +∞] ( 1/(X^2) + 1/(X + 1) – 1/X)
Note the sum of ∑ [X = x, X = +∞] 1/X
and ∑ [X = x, X = +∞] 1/(X + 1) both unhelpfully sums up to infinity.
However the sum of ∑ [X = x, X = +∞] (1/(X + 1) – 1/X ) rather helpfully sums up to simply -1/x i.e.
∑ [X = x, X = +∞] (1/(X + 1) – 1/X ) = -1/x.
So now we have:
∑ [X = x, X = +∞] 1/(X^2(X+1))
= ∑ [X = x, X = +∞] ( 1/X^2 + 1/(X + 1) – 1/X)
= ( ∑ [X = x, X = +∞] 1/X^2 ) + ∑ [X = x, X = +∞] (1/(X + 1) – 1/X)
= ( ∑ [X = x, X = +∞] 1/X^2 ) – 1/x
so now we have a desirable simplification of the sum of the infinite series we need to evaluate.
Now explaining stuff and speaking here for the benefit of other readers that may be here who are not so knowledgeable of maths as us:
Note that, if you think about what the summation means (obviously, you would know very clearly what summation means, ∑ [X = x, X = +∞] 1/X^2 must equal what the sum would equal
∑ [X = 1, X = +∞] 1/X^2 subtract all the terms of the series of the summation from the first one in that series to the xth – 1 one in that series.
For example; if x =3, then we have:
∑ [X = 3, X = +∞] 1/X^2 = 1/(3^2) + 1/(4^2) + 1/(5^2) + 1/(6^2) ...
∑ [X = 1, X = +∞] 1/X^2 = 1/(1^2) + 1/(2^2) + 1/(3^2) + 1/(4^2) + 1/(5^2) + 1/(6^2) …
and notice most of the terms in the two above series of the summations are identical and the only difference between the two series of the summations is that the first series doesn't have the first two terms the second series does with those two terms being 1/(1^2) and 1/(2^2).
From this, we can reason that:
∑ [X = 3, X = +∞] 1/X^2 = ( ∑ [X = 1, X = +∞] 1/(X^2) ) – 1/(1^2) – 1/(2^2)
and the more general equation for any x is:
∑ [X = x, X = +∞] 1/X^2 = ( ∑ [X = 1, X = +∞] 1/(X^2) ) – ∑ [X = 1, X = x – 1] 1/(X^2)
But ∑ [X = 1, X = +∞] 1/X^2 of the above is equal to just a constant which is:
∑ [X = 1, X = +∞] 1/X^2 = (π^2)/6
So we now have:
∑ [X = x, X = +∞] 1/X^2 = ( ∑ [X = 1, X = +∞] 1/(X^2) ) – ∑ [X = 1, X = x – 1] 1/(X^2)
= ( (π^2)/6 ) – ∑ [X = 1, X = x – 1] 1/X^2
and thus:
∑ [X = x, X = +∞] 1/(X^2(X+1)) = ( (π^2)/6 ) – ( ∑ [X = 1, X = x – 1] 1/(X^2) ) – 1/x
so now we have a desirable change of the sum we need to evaluate from being of an infinite series to a more manageable finite series of ∑ [X = 1, X = x – 1] 1/(X^2).
And that ∑ [X = 1, X = x – 1] 1/(X^2) sum is a harmonic series (I confess I know next to nothing about the mathematics of harmonic series. So not sure where I can go from this point. But I still like the look of this! )
Originally posted by humyGood spot with the cancellation of terms, the only thing I'm wondering about is that you're effectively subtracting one divergent series from another and it's possible to get any answer you want by regrouping terms in a conditionally convergent sum. Since the sum of 1/N*(N+1) is absolutely convergent (as can be seen by bounding it with an integral) I think you're safe doing that. But I can't remember the rules for this so hopefully Soothfast or someone will confirm it's o.k.. It might be possible to get some sort of recursion relation to get the finite sum of 1/N^2.
Thanks for that 🙂
I confess complete ignorance on the method of partial fractions.
But I find that equation extremely useful and gives me a breakthrough, of sorts.
This is because we now have:
1/(X^2(X+1)) = 1/(X^2) + 1/(X + 1) – 1/X (using the method of partial fractions )
so:
∑ [X = x, X = +∞] 1/(X^2(X+1)) = ∑ [X = x, X = +∞] ( 1/(X^ ...[text shortened]... armonic series. So not sure where I can go from this point. But I still like the look of this! )
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Originally posted by humyUpdate:
Thanks for that 🙂
I confess complete ignorance on the method of partial fractions.
But I find that equation extremely useful and gives me a breakthrough, of sorts.
This is because we now have:
1/(X^2(X+1)) = 1/(X^2) + 1/(X + 1) – 1/X (using the method of partial fractions )
so:
∑ [X = x, X = +∞] 1/(X^2(X+1)) = ∑ [X = x, X = +∞] ( 1/(X^ ...[text shortened]... armonic series. So not sure where I can go from this point. But I still like the look of this! )
My equation ( i.e. ∑ [X = x, X = +∞] 1/(X^2(X+1)) ) is for a discrete probability distribution so that function of x ( x ∈ â„•) outputs probability of x i.e. P(x).
In this case, x cannot equal zero but must start at x=1.
I believe I have found a very efficient iterative formula for finding all the probabilities of x P(x) from x=1 to x=U (thus this is a function for obtaining probability mass ) where U is whatever finite Upper limit of x you think would be acceptably high enough for the particular application.
But I don’t know how to properly edit it here on this link with the standard notation for iterative formulas which involves lowered “n+1” and “n” on the right-hand side of “x”.
So, just to avoid confusion here, I show the iterative formula below by writing “n+1” and “n” in the none standard way as “[n+1]” and “[n]”:
P(x[1]) = ( (π^2)/6 ) – 1
P(x[n+1]) = P(x[n]) – ( 1 / ( ((x[n])^2) * ((x[n])+1) ) )
I don't know if it is possible to make an even more efficient iterative formula to the above (anyone? ) but, still, I am well pleased with this result so far!
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how do I edit in the equation into WalframAlpha online to ask what is the average f(x) output of some f(x) over a continuous variable input x of, say, from x=0 to x=1 ?
For example, I tried inputting " average {f(x) = 3(x^2) } 0<x<1 " but didn't get what I wanted.
What I want is it to either output a numerical value of the average or a general formula for the average -don't mind which in this case because I just want to use that to check I got my algebraic workings for deriving the various formulas for the average correct.
I am 99.9% sure that I got them all correct. But, for what I want, 99.9% sure is not good enough. I need to be 100% sure. It would be very embarrassing if I published my work that is supposed to revolutionize the world of statistics only to find it contains some silly mathematical error.
Originally posted by humyTo find the average value of a function over some interval [a,b] (b>a) simply integrate from a to b and divide the result by |b-a|.
how do I edit in the equation into WalframAlpha online to ask what is the average f(x) output of some f(x) over a continuous variable input x of, say, from x=0 to x=1 ?
For example, I tried inputting " average {f(x) = 3(x^2) } 0<x<1 " but didn't get what I wanted.
What I want is it to either output a numerical value of the average or a general formula for ...[text shortened]... to check I got my algebraic workings for deriving the various formulas for the average correct.