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Index shifting in summation FORMULAS

• Thread starter Thread starter Xyius
• Start date Start date Mar 13, 2010
• Tags Tags Formulas Index Summation

Click For Summary The discussion focuses on deriving summation formulas starting from different indices. The original formula for the sum of integers is correctly transformed by changing the index, but an error arises in calculating the second sum of constants. It is clarified that the sum of 1's over the new index range results in "n," not "n-1," because there are "n" terms being summed. The conversation concludes with a successful derivation of the squared sum formula, confirming that the new expression aligns with the original formula upon simplification. The participants express satisfaction with the resolution of the issues discussed. Xyius Messages 501 Reaction score 4 So I am trying to derive a formula from one of the standard summation formulas except starting at a different index. So if I have the series.. \sum i = \frac{n(n+1)}{2} Where "i" runs from 1 to n. (I don't know how to put it in the code.) If I want to make the series start from zero, I would use another index equal to, "k=i-1" And now my upper limit will be "n-1" So now I can replace "i" in the sum with "k+1", Shouldn't the new formula for the partial sums be.. \sum k+1 = \frac{(n-1)(n)}{2} + (n-1) Problem is, it isnt. A simple test will prove this. Whats wrong? Physics news on Phys.org

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jbunniii Homework Helper Insights Author Messages 3,488 Reaction score 257 Your new formula is wrong. (By the way, you can click on the equations below to see the Latex code for putting the indices on the sum.) The original sum is equivalent to \sum_{k=0}^{n-1} (k + 1) Note that the sum starts with k=0, not k=1. This can be rewritten as \sum_{k=0}^{n-1}k + \sum_{k=0}^{n-1} 1 The first sum is the same whether it starts with k=0 or k=1, but the second sum is not. Your mistake is in your calculation of the second sum. How many 1's are being added? Last edited: Mar 13, 2010 HallsofIvy Science Advisor Homework Helper Messages 42,895 Reaction score 984 You can also change the "starting point" by changing the index. Given \sum_{i= 0}^{n-1} i Let k= i+1 so that when i= 0, k= 0+1= 1 so the lower limit is i= 1. Then when i= n-1, k= (n-1)+ 1= 1 so the upper limit is k= n. Finally, of course, since k= i+ 1, i= k- 1. There's your error- your sum should be of k- 1, not k+ 1: \sum_{k=1}^n k-1 Xyius Messages 501 Reaction score 4

jbunniii said: This can be rewritten as \sum_{k=0}^{n-1}k + \sum_{k=0}^{n-1} 1 The first sum is the same whether it starts with k=0 or k=1, but the second sum is not. Your mistake is in your calculation of the second sum. How many 1's are being added?

Hmm I think I still have some issues. Since the original formula for the partial sums is.. \sum_{i=1}^{n} i = \frac{n(n+1)}{2} Since "n" represents the upper limit, the new formula for the series I am after should be just replacing all the "n" with "n-1." \sum_{k=0}^{n-1} k+1 =\frac{(n-1)(n)}{2} + \sum_{k=0}^{n-1} 1 Now you say up until this part is correct? For the second sum, shouldn't the sum of this be just the upper limit? For example.. \sum_{i=1}^{3} 1 = 3(1) = 3 Shouldn't the same thing apply for this sum? Shouldn't it therefore be.. \sum_{k=0}^{n-1} 1 = 1(n-1) = n-1 Or should it equal "n" because the index is zero? jbunniii Homework Helper Insights Author Messages 3,488 Reaction score 257

Xyius said: Since "n" represents the upper limit, the new formula for the series I am after should be just replacing all the "n" with "n-1." \sum_{k=0}^{n-1} k+1 =\frac{(n-1)(n)}{2} + \sum_{k=0}^{n-1} 1

Actually both the lower and the upper limit have changed, right? You are now summing from 0 to n-1, whereas the original formula sums from 1 to n. Now it happens that \sum_{k=0}^{n-1}k = \sum_{k=1}^{n-1}k i.e. it doesn't matter whether you start summing from k=0 or k=1 in THIS PARTICULAR SUM. That is because the summand is k, so the k=0 term is 0: it has no effect on the value of the sum. You can therefore start summing from k=1 and get the same answer.

Now you say up until this part is correct?

Yes, it's correct up to this point.

For the second sum, shouldn't the sum of this be just the upper limit? For example.. \sum_{i=1}^{3} 1 = 3(1) = 3 Shouldn't the same thing apply for this sum? Shouldn't it therefore be.. \sum_{k=0}^{n-1} 1 = 1(n-1) = n-1 Or should it equal "n" because the index is zero?

The summand is 1, and you are summing it n times. The k=0 term DOES matter in this case, because that term is not zero. Therefore \sum_{k=0}^{n-1} 1 = n Think of it as having n marbles with labels on them. It doesn't matter if the labels read 0 through n-1 or 1 through n, there are still n marbles. I could just as well write \sum_{k=1000}^{1000+n-1} 1 and the sum would still be n. Last edited: Mar 14, 2010 Xyius Messages 501 Reaction score 4 Makes complete sense! Thanks a lot! So if I wanted to derive a formula for the squared sum. The normal formula is. \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} So therefore the new sum should be.. \sum_{k=0}^{n-1} (k+1)^2 = \sum_{k=0}^{n-1} k^2-2k+1 = \frac{(n-1)(n)(2n-1)}{6} + 2\frac{(n-1)(n)}{2} + n Would this be correct? jbunniii Homework Helper Insights Author Messages 3,488 Reaction score 257

Xyius said: Makes complete sense! Thanks a lot! So if I wanted to derive a formula for the squared sum. The normal formula is. \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} So therefore the new sum should be.. \sum_{k=0}^{n-1} (k+1)^2 = \sum_{k=0}^{n-1} k^2-2k+1 = \frac{(n-1)(n)(2n-1)}{6} + 2\frac{(n-1)(n)}{2} + n Would this be correct?

Yes, it looks right to me. And I tried plugging in n = 20 and verified that the two formulas give the same answer. Of course they're really the same formula, in the sense that if you carry out the addition of the three terms in your expression by putting them over a common denominator and simplifying, you should end up with the original formula. Xyius Messages 501 Reaction score 4 Thanks a lot man! :D!

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