Sigma Notation Part III

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Hckyplayer8

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I searched and couldn't find anything with a capital N in the text, so I assume it is just a typo.

The aforementioned problem is just saying sum the values of j2, starting at 212 and ending at 402.

The text wants me to use the equation [n(n+1)(2n+1) / 6] but never really explains why (If it does, its not in a way I understand.)

When I see that I see the upper bound is replaced with 40 and the lower bound is replaced with 21.

So would my equation be [40(40+21)(2*40+21) / 6]?
 
No, that's not using the formula; you can't decide on your own to replace 1 with 21, since 1 is not a variable. And it makes no difference what letter you use for a variable, so if you know that [MATH]\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}[/MATH], then you also know that [MATH]\sum_{j=1}^{N} j^2 = \frac{N(N+1)(2N+1)}{6}[/MATH]. So it could be a typo, but it is not an error.

But suppose you knew the sum of the first, say, 20 terms of a series, and the sum of the first 40 terms. What could you say about the sum of the 21st through 40th terms?
 
Dr.Peterson said:
No, that's not using the formula; you can't decide on your own to replace 1 with 21, since 1 is not a variable. And it makes no difference what letter you use for a variable, so if you know that [MATH]\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}[/MATH], then you also know that [MATH]\sum_{j=1}^{N} j^2 = \frac{N(N+1)(2N+1)}{6}[/MATH]. So it could be a typo, but it is not an error.

But suppose you knew the sum of the first, say, 20 terms of a series, and the sum of the first 40 terms. What could you say about the sum of the 21st through 40th terms?
Click to expand...

The sum of the 21st through 40th terms is the difference of the sum of the series 1-40 and the sum of the series 1-20?
 
Yes, you should know the formulas listed above. But you should also know what exactly is being added up.
Lets look at this carefully [MATH]\sum_{x=1}^{2000} x^2[/MATH]We are summing up x2. So think of f(x) = x2
Now f(1) = 12, f(2) = 22, f(3) = 32 {I will not write all the numbers up to 2000 so I will use ...} ... f(2000) = 200^2
That is [MATH]\sum_{x=1}^{2000} x^2[/MATH] = 12 + 22 + 32 + .... + 20002
The three dots, ..., are extremely important for you to understand how to deal with N
So [MATH]\sum_{j=1}^{N} j^2[/MATH] = 12 + 22 + 32 + ... + N2
N is just like 2000. Instead of stopping at 2000 we stop at N.

Is this clear?
 
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Hckyplayer8 said:
The sum of the 21st through 40th terms is the difference of the sum of the series 1-40 and the sum of the series 1-20?
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Yes!
(1+2 + 3 + ... + 20 +21 + 22 +....+40) = (1+2 + 3 + ... + 20) +(21 + 22 +....+40)

So (21 + 22 +....+40) = (1+2 + 3 + ... + 20 +21 + 22 +....+40) - (1+2 + 3 + ... + 20)
 
Jomo said:
Yes, you should know the formulas listed above. But you should also know what exactly is being added up.
Lets look at this carefully [MATH]\sum_{x=1}^{2000} x^2[/MATH]We are summing up x2. So think of f(x) = x2
Now f(1) = 12, f(2) = 22, f(3) = 32 {I will not write all the numbers up to 2000 so I will use ...} ... f(2000) = 200^2
That is [MATH]\sum_{x=1}^{2000} x^2[/MATH] = 12 + 22 + 32 + .... + 20002
The three dots, ..., are extremely important for you to understand how to deal with N
So [MATH]\sum_{j=1}^{N} j^2[/MATH] = 12 + 22 + 32 + ... + N2
N is just like 2000. Instead of stopping at 2000 we stop at N.

Is this clear?
Click to expand...

Yes. So is it incorrect to view N (or any variable as a placeholder for the upper bound) as positive infinity? Because that is what I have been doing. I figured if the upper bound was not given, then the summation just goes on and on.
 
Hckyplayer8 said:
View attachment 14448
I searched and couldn't find anything with a capital N in the text, so I assume it is just a typo.

The aforementioned problem is just saying sum the values of j2, starting at 212 and ending at 402.

The text wants me to use the equation [n(n+1)(2n+1) / 6] but never really explains why (If it does, its not in a way I understand.)

When I see that I see the upper bound is replaced with 40 and the lower bound is replaced with 21.

So would my equation be [40(40+21)(2*40+21) / 6]?
Click to expand...
Here is a good reason to immediately see that the way you used the formula can't be right. It is ALWAYS good to know that what you are doing has to be wrong.

[MATH]\sum_{j=1}^{40} j^2[/MATH] = 12 + 22 + 32 + .... 202+ 212 + ... +402 =[40(40+21)(2*40+21) / 6]

[MATH]\sum_{j=21}^{40} j^2[/MATH] = 212 + ... +402 =[40(40+21)(2*40+21) / 6]

You would have gotten the same results for both those summations but clearly the 1st one must be larger, therefore something went wrong with your logic. I try to do wrong things all the time but catch it.

As a high school student who did not want to study much I used the above method a lot. For example I caught the following mistake I was about to make all the time. I would think that 5(x+y) = 5x + y, that is just multiply the 1st term by 5. But I always realized that x+y = y+x. Then I would realize that using my method I would get 5(y+x) = 5y + x. Since got different answers I must be wrong! Then I would multiply both the x and y by 5. Mistakes happen but it is always good to catch them by any means.
 
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Hckyplayer8 said:
Yes. So is it incorrect to view N (or any variable as a placeholder for the upper bound) as positive infinity? Because that is what I have been doing. I figured if the upper bound was not given, then the summation just goes on and on.
Click to expand...
No No No! N is not infinity! N is some non negative integer.
Soon enough you will have a summation going up to N and then take the limit as N goes to infinity. Think about it, if N is going to infinity it is NOT infinity.

I am glad that I cleared that up for you!
 
Jomo said:
No No No! N is not infinity! N is some non negative integer.
Soon enough you will have a summation going up to N and then take the limit as N goes to infinity. Think about it, if N is going to infinity it is NOT infinity.

I am glad that I cleared that up for you!
Click to expand...

Thank you.
 
I'm sorry but in spite of all of the above, I'm still struggling to use the equation.

I know the problem is telling me to sum the squares. Why can't I do just that? What is the need for the formula? Sure summing the squares will be tedious and lead to a huge number, but I'm failing to see how this formula fits in or is a shortcut.
 
Hckyplayer8 said:
I'm sorry but in spite of all of the above, I'm still struggling to use the equation.
I know the problem is telling me to sum the squares. Why can't I do just that? What is the need for the formula? Sure summing the squares will be tedious and lead to a huge number, but I'm failing to see how this formula fits in or is a shortcut.
Click to expand...
Sorry but it is impossible for me to follow this thread. If you will simply reply with the exact sum or question, then I will answer it.
 
Hckyplayer8 said:
I'm sorry but in spite of all of the above, I'm still struggling to use the equation.

I know the problem is telling me to sum the squares. Why can't I do just that? What is the need for the formula? Sure summing the squares will be tedious and lead to a huge number, but I'm failing to see how this formula fits in or is a shortcut.
Click to expand...
Try an example. First directly add up 1^2 + 2^2 + ... + 10^2. Then apply the formula. Please show your work, so we can be sure you are doing it correctly. It seems likely that you are still misreading something.

Then you can try the same for 1^2 + 2^2 + ... + 1000^2. I presume you will not want to try adding it up directly (unless you use a spreadsheet), but will be able to quickly apply the formula (to get a very large number). Then we can discuss why you don't think it is a shortcut ...
 
Dr.Peterson said:
Then you can try the same for 1^2 + 2^2 + ... + 1000^2. I presume you will not want to try adding it up directly (unless you use a spreadsheet), but will be able to quickly apply the formula (to get a very large number). Then we can discuss why you don't think it is a shortcut ...
Click to expand...
If we want \(\displaystyle \sum\limits_{k = 1}^{1000} {{k^2}} \) there is a shortcut not a spreedsheet.
 
Hckyplayer8 said:
View attachment 14448
I searched and couldn't find anything with a capital N in the text, so I assume it is just a typo.

The aforementioned problem is just saying sum the values of j2, starting at 212 and ending at 402.

The text wants me to use the equation [n(n+1)(2n+1) / 6] but never really explains why (If it does, its not in a way I understand.)

When I see that I see the upper bound is replaced with 40 and the lower bound is replaced with 21.

So would my equation be [40(40+21)(2*40+21) / 6]?
Click to expand...
Oh, now I see that you replaced the 1 in the formula with 21. You can't do that! Now I fully understand Dr Peterson's reply about 1 not being a variable. You can only replace N with the upper limit and only then if you are starting from 1.
 
Dr.Peterson said:
Try an example. First directly add up 1^2 + 2^2 + ... + 10^2. Then apply the formula. Please show your work, so we can be sure you are doing it correctly. It seems likely that you are still misreading something.

Then you can try the same for 1^2 + 2^2 + ... + 1000^2. I presume you will not want to try adding it up directly (unless you use a spreadsheet), but will be able to quickly apply the formula (to get a very large number). Then we can discuss why you don't think it is a shortcut ...
Click to expand...

Okay. I did just that and see it works. I guess my next question is why? Nothing about the formula is intuitive to me. The upper bound times the upper bound plus one. Then multiply that by two times the upper bound plus one. Then divide that by six.

Why does this work?
 
Hmmm. I thought I was getting it/putting it all together. But apparently not. The sum of the series one through forty is 22140. The sum of the series one through twenty one is 3311. When check against wolframalpha they both check out as correct. But when I find the difference of the sum of the two series I get 18829 while wolframalpha produces 19270 when plugging in 21 and 40 as the bounds.
 
So you want to see the proof of this. This is good. Never accept anything that your math teacher say or even what someone on this forum says. Always assume we are wrong until you prove it to your satisfaction!

To see some proofs look here:
1+2 + 3 +... +n and 1^2 + 2^2 + ... + n^2 Look here
1^3 + 3^ 3 +... + n^3 Look here
 
Hckyplayer8 said:
Hmmm. I thought I was getting it/putting it all together. But apparently not. The sum of the series one through forty is 22140. The sum of the series one through twenty one is 3311. When check against wolframalpha they both check out as correct. But when I find the difference of the sum of the two series I get 18829 while wolframalpha produces 19270 when plugging in 21 and 40 as the bounds.
Click to expand...
Here is your error. Yes, 1^2 + 2^2 +...+20^2 +21^2 + 22^2 +....+ 40^2 = 22140
Also 1^2 +2^2 +...+20^2 + 21^2 = 3311.
Now if we subtract we get 22^2 + 23^2 + 24^2 +.... + 40^2 = 22140 - 3311.

Now this last line is correct. The problem is that the lhs is NOT what you wanted!

Can you fix it?
 
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