While I was researching on case study of "Energy saving through efficient Industrial Boiler System", I found a problem as follows:
Assume there is the sequence of natural numbers below cited:
a1, a2, a3, ……….an
Where: a2 = a1+ S, a3 = a2 + L, a4 = a3 + S, a5 = a4 + L ……..an = a(n-1) + (S or L)
In fact, S and L are added to natural numbers off and on.
S and L = constant members of real numbers®
We have:
an = a1 +{(n-1)/2}(S +L) If n = 2k +1
an = a1 + 0.5 {(n-2)L + nS} If n = 2k
Do you know any easy way to calculate below series?
SUM (an) from n = 1 to n = i and “i” is a member of natural numbers
Is there any real number for: limit Sum (an) if “n” tends to infinity?
I would like to inform you that my problem has been solved by Mr. Nico Potyka on below link:
https://www.xing.com/net/mathe/general-interest-remarks-and-links-5223/energy-saving-through-efficient-industrial-boiler-system-33117616/
The answer is as follows:
Let's consider the cases s1, and si for even and uneven i.
1. s1 = a1
2. 1 < i = 2k+1. Let's decompose the sum in separate parts for a1, S and L. Then we obtain:
coefficients of a1:
i
coefficients of S:
(i-1) + (i-3) + ... + 2
= (1 + 2 + ... + (i-1)/2) * 2
= ((i-1) / 2 * (i+1) / 2) / 2 * 2
= (i^2-1)/4
coefficients of L:
(i-2) + (i-4) + ... + 1
= 1 + 3 + ... + (i-2)
= ((i-1)/2)^2
=(i-1)^2/4
So in this case we obtain si = a1*i + S * (i^2 - 1) / 4 + L * (i-1)^2 / 4
3. 1 < i = 2k.
coefficients of a1:
i
coefficients of S:
(i-1) + (i-3) + ... + 1
= 1 + 3 + ... i-1
= (i/2)^2
= i^2 / 4
coefficients of L:
(i-2) + (i-4) + ... + 2
= (2 + 4 + ... + i-2)
= 2 * (1 + 2 + ... (i-2)/2))
= 2*((i-2)/2)*i/2)/2
= (i^2 - 2i)/4
So we obtain si = a1*i + S * i^2 / 4 + L * (i^2 - 2i) / 4
Better check the result ;)
PS. A general form for i in N is:
si = a1*i + (as*S +al*L) / 4
where as := i^2 - (i mod 2)
and al := (i-1)^2 - ((i+1) mod 2)
So much to thank him for solving of my problem.
PS: I consider this quotation “Xenophanes said: The gods did not reveal all things to men at the start; but, as time goes on, by searching, they discover more and more” for below link (camel - stationary traveller):
http://www.youtube.com/watch?v=MKBwku-PsPY&feature=relat...
Best Regards
Gholamreza Soleimani
Assume there is the sequence of natural numbers below cited:
a1, a2, a3, ……….an
Where: a2 = a1+ S, a3 = a2 + L, a4 = a3 + S, a5 = a4 + L ……..an = a(n-1) + (S or L)
In fact, S and L are added to natural numbers off and on.
S and L = constant members of real numbers®
We have:
an = a1 +{(n-1)/2}(S +L) If n = 2k +1
an = a1 + 0.5 {(n-2)L + nS} If n = 2k
Do you know any easy way to calculate below series?
SUM (an) from n = 1 to n = i and “i” is a member of natural numbers
Is there any real number for: limit Sum (an) if “n” tends to infinity?
I would like to inform you that my problem has been solved by Mr. Nico Potyka on below link:
https://www.xing.com/net/mathe/general-interest-remarks-and-links-5223/energy-saving-through-efficient-industrial-boiler-system-33117616/
The answer is as follows:
Let's consider the cases s1, and si for even and uneven i.
1. s1 = a1
2. 1 < i = 2k+1. Let's decompose the sum in separate parts for a1, S and L. Then we obtain:
coefficients of a1:
i
coefficients of S:
(i-1) + (i-3) + ... + 2
= (1 + 2 + ... + (i-1)/2) * 2
= ((i-1) / 2 * (i+1) / 2) / 2 * 2
= (i^2-1)/4
coefficients of L:
(i-2) + (i-4) + ... + 1
= 1 + 3 + ... + (i-2)
= ((i-1)/2)^2
=(i-1)^2/4
So in this case we obtain si = a1*i + S * (i^2 - 1) / 4 + L * (i-1)^2 / 4
3. 1 < i = 2k.
coefficients of a1:
i
coefficients of S:
(i-1) + (i-3) + ... + 1
= 1 + 3 + ... i-1
= (i/2)^2
= i^2 / 4
coefficients of L:
(i-2) + (i-4) + ... + 2
= (2 + 4 + ... + i-2)
= 2 * (1 + 2 + ... (i-2)/2))
= 2*((i-2)/2)*i/2)/2
= (i^2 - 2i)/4
So we obtain si = a1*i + S * i^2 / 4 + L * (i^2 - 2i) / 4
Better check the result ;)
PS. A general form for i in N is:
si = a1*i + (as*S +al*L) / 4
where as := i^2 - (i mod 2)
and al := (i-1)^2 - ((i+1) mod 2)
So much to thank him for solving of my problem.
PS: I consider this quotation “Xenophanes said: The gods did not reveal all things to men at the start; but, as time goes on, by searching, they discover more and more” for below link (camel - stationary traveller):
http://www.youtube.com/watch?v=MKBwku-PsPY&feature=relat...
Best Regards
Gholamreza Soleimani
