Double sum of differences in x and y coordinates

[DryRun]

Homework Statement

$$\sum_{i=1}^{m} \sum_{j=1}^{n}[2(x_i-x_{i-1})-3(y_j-y_{j-1})]$$

Homework Equations

Multiple-sigma notation.

The Attempt at a Solution

I agree this seems like basic summation stuff, but i do not agree with the given answer. So, here is what I've done.

$$\sum_{i=1}^{m} \sum_{j=1}^{n}[2(x_i-x_{i-1})-3(y_j-y_{j-1})]
\\=2\sum_{i=1}^{m} (x_i-x_{i-1})-3\sum_{j=1}^{n}(y_j-y_{j-1})
\\=2(x_m-x_0)-3(y_n-y_0)$$

However, the given answer for this problem is:
$$2n(x_m-x_0)-3m(y_n-y_0)$$

 

[STEMucator]

Do you know what the partition of the interval is? Was there an interval given in the question?

 

[Dick]

Homework Statement

$$\sum_{i=1}^{m} \sum_{j=1}^{n}[2(x_i-x_{i-1})-3(y_j-y_{j-1})]$$

Homework Equations

Multiple-sigma notation.

The Attempt at a Solution

I agree this seems like basic summation stuff, but i do not agree with the given answer. So, here is what I've done.

$$\sum_{i=1}^{m} \sum_{j=1}^{n}[2(x_i-x_{i-1})-3(y_j-y_{j-1})]
\\=2\sum_{i=1}^{m} (x_i-x_{i-1})-3\sum_{j=1}^{n}(y_j-y_{j-1})
\\=2(x_m-x_0)-3(y_n-y_0)$$

However, the given answer for this problem is:
$$2n(x_m-x_0)-3m(y_n-y_0)$$

The terms that are indexed with i are independent of j. So when you apply the j sum they will get added to themselves n times. So they will pick up a factor of n. Similarly for the terms indexed with j.

 

[Mark44]

Homework Statement

$$\sum_{i=1}^{m} \sum_{j=1}^{n}[2(x_i-x_{i-1})-3(y_j-y_{j-1})]$$

Homework Equations

Multiple-sigma notation.

The Attempt at a Solution

I agree this seems like basic summation stuff, but i do not agree with the given answer. So, here is what I've done.

$$\sum_{i=1}^{m} \sum_{j=1}^{n}[2(x_i-x_{i-1})-3(y_j-y_{j-1})]
\\=2\sum_{i=1}^{m} (x_i-x_{i-1})-3\sum_{j=1}^{n}(y_j-y_{j-1})
\\=2(x_m-x_0)-3(y_n-y_0)$$

You're skipping some steps above.

After the first line you should have this:
$$ \sum_{i=1}^{m} \left(\sum_{j=1}^{n}[2(x_i-x_{i-1}] - \sum_{j=1}^{n}[3(y_j-y_{j-1})]\right)$$
The first sum isn't affected by j, so will simplify to n * the expression being summed. Something similar happens for the outer summation.

However, the given answer for this problem is:
$$2n(x_m-x_0)-3m(y_n-y_0)$$

 

[DryRun]

Do you know what the partition of the interval is? Was there an interval given in the question?

Yes, there are two intervals given, but in this case, the answer still wouldn't be correct, so i omitted that part from the stated problem in the 1st post. Anyway, here they are:

Let $P_1 = \{ x_0, x_1, x_2,..., x_m \}$ be a partition of $[a_1, a_2]$
Let $P_2 = \{ y_0, y_1, y_2,..., y_n \}$ be a partition of $[b_1, b_2]$

The terms that are indexed with i are independent of j. So when you apply the j sum they will get added to themselves n times. So they will pick up a factor of n. Similarly for the terms indexed with j.

That just cleared up all my confusion! Clear and concise explanation. Thanks, Dick!

You're skipping some steps above.

After the first line you should have this:
$$ \sum_{i=1}^{m} \left(\sum_{j=1}^{n}[2(x_i-x_{i-1}] - \sum_{j=1}^{n}[3(y_j-y_{j-1})]\right)$$
The first sum isn't affected by j, so will simplify to n * the expression being summed. Something similar happens for the outer summation.

I had intentionally skipped those steps, as all the terms in i are independent of j, and all the terms in j are independent of i. So, in the expansion, i had wrongly assumed that expanding all the i terms with respect to j and all the j terms with respect to i, would give 1. Thanks for the help.