Trouble with a double summation

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In summary, the inner sum is just adding 1/365 n-i number of times. so ##\frac{(n-i)}{365}##The outer sum adds over the index i, so I thought the expression is equal to ##\frac{(n-1)n-(n-1)!}{365}## but it's obviously not equal to that. where did I go wrong?
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Homework Statement
verify the last equality
Relevant Equations
verify the last equality
Screen Shot 2021-11-05 at 1.27.25 AM.png


the inner sum is just adding 1/365 n-i number of times. so ##\frac{(n-i)}{365}##

the outer sum adds over the index i, so I thought the expression is equal to ##\frac{(n-1)n-(n-1)!}{365}## but it's obviously not equal to that. where did I go wrong?
 
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  • #2
docnet said:
Homework Statement:: verify the last equality
Relevant Equations:: verify the last equality

View attachment 291798

the inner sum is just adding 1/365 n-i number of times. so ##\frac{(n-i)}{365}##

the outer sum adds over the index i, so I thought the expression is equal to ##\frac{(n-1)n-(n-1)!}{365}## but it's obviously not equal to that. where did I go wrong?
If ##i = 1## there are ##n-1## terms, if ##i =2## there are ##n -2## terms etc. Hence there are ##1 +2 \dots + n -1## terms in total.

Do an example with ##n = 4## say, and write out the sum.
 
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  • #3
docnet said:
Homework Statement:: verify the last equality
Relevant Equations:: verify the last equality

View attachment 291798

the inner sum is just adding 1/365 n-i number of times. so ##\frac{(n-i)}{365}##

the outer sum adds over the index i, so I thought the expression is equal to ##\frac{(n-1)n-(n-1)!}{365}## but it's obviously not equal to that. where did I go wrong?
PeroK said:
If ##i = 1## there are ##n-1## terms, if ##i =2## there are ##n -2## terms etc. Hence there are ##1 +2 \dots + n -1## terms in total.

Do an example with ##n = 4## say, and write out the sum.
If ##n=4##, then the sum is

##1/365\cdot \sum^3_{i=1}\sum^4_{j=i+1}1=1/365\cdot (3+2+1)=\frac{6}{365}##.

check:
##\frac{n(n-1)}{2*365}=\frac{4(3)}{2*365}=\frac{6}{365}##

so I realized that I forgot the meaning of the factorial.. oope. I somehow thought it was n! = n+(n-1)+...+2+1 . oh lord.

following the same thought process suggested by @PeroK,

##1/365\cdot sum^{n-1}_{i=1}\sum^n_{j=i+1}1=\frac{(n-1)+(n-2)+...+2+1}{365}##

which is equivalent to ##\frac{n(n-1)}{2*365}## (although I don't know how to prove it, I can see that it is true geometrically.)
 
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[tex]\sum_{i=1}^{n-1}\sum_{j=i+1}^{n}1=\sum_{i=1}^{n-1}(n-i)=n\sum_{i=1}^{n-1}1-\sum_{i=1}^{n-1}i=n(n-1)-\sum_{i=1}^{n-1}i[/tex]
Then
[tex]S:=\sum_{i=1}^{n-1}i=\sum_{i=1}^{n-1}(n-i)[/tex]
which means just changing order of addition
[tex]1+2+...+(n-1)=(n-1)+(n-2)+...+2+1[/tex]
adding RHS and LHS
[tex]2S=\sum_{i=1}^{n-1}n=n\sum_{i=1}^{n-1}1=n(n-1)[/tex]
This is equivalent to geometrical explanation.
 
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anuttarasammyak said:
[tex]S:=\sum_{i=1}^{n-1}i=\sum_{i=1}^{n-1}(n-i)[/tex]
This confused me at first, but it makes sense after thinking about it. Thank you so much.
 
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Related to Trouble with a double summation

1. What is a double summation?

A double summation is a mathematical operation that involves adding up a series of terms, where each term is itself a sum of two or more terms. It is represented by two summation symbols, with one nested inside the other.

2. How do you evaluate a double summation?

To evaluate a double summation, you first calculate the inner summation for each value of the outer summation. Then, you add up all of these inner summations to get the final result. This process can be repeated for multiple levels of nested summations.

3. What is the difference between a double summation and a single summation?

The main difference between a double summation and a single summation is that a single summation involves adding up a series of terms, while a double summation involves adding up a series of sums. In other words, a single summation has only one level of terms, while a double summation has two levels.

4. What are some common applications of double summations?

Double summations are commonly used in mathematics and statistics to represent and evaluate complex series and sequences. They are also used in physics and engineering to solve problems involving multiple variables and parameters.

5. How do you handle a double summation with different limits?

If the limits of the inner summation depend on the index variable of the outer summation, you can use the substitution method to simplify the double summation. This involves replacing the inner summation with a single term that represents the sum of all the terms in the inner summation. This allows you to evaluate the double summation as a single summation with a new set of limits.

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