Don't understand the change of the index of a sequence

Hi,

In our notes, our teacher explained us how to change the index of a sequence. There are some parts that I just don't understand.

He first wrote :
$\displaystyle \sum_{i=0}^{h-1}2(h-i)2^{i}=\{{i}\leftrightarrow h-i}\}=2\sum_{i=1}^{h}i2^{h-i}$

And then he "explained" (if I can call this "explain") :

$\displaystyle j=h-i$
That, I understand.

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$\displaystyle \sum_{i=0}^{h-1}2(h-i)2^{i}$

(we simply repeat the sequence)

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$\displaystyle =2\sum_{j=h}^{1}j2^{h-j}$

I understand : we move the first "2" left of Sigma, we replace "(h - i)" by "j" and "i" by "h - j"

Q1. Where does the upper limit "1" come from ?
Q2. Where does the lower limit "j=h" come from (j = h - i, not h ?!)?


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$\displaystyle =2\sum_{j=1}^{h}j2^{h-j}$

Q3. Where does the upper limit "h" come from ?
Q4. Where does the lower limit "j=1" come form ?

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$\displaystyle =2\sum_{i=1}^{h}i2^{h-i}$

Q5. Where does the lower limit "i=1" come from ? It's upposed to be "i = j - 1", so if we previously had "j = 1", we should have "i = 1 - 1 = 0".
Q6. Why has the "j" between Sigma and "2" become "i" ?
Q7. Why the exponent "h-j" has become "h - i" ?

Am I simply dumb or our teacher forgot to write a few things ?

Thanks for your help :)

$\displaystyle =2\sum_{j=h}^{1}j2^{h-j}$

I understand : we move the first "2" left of Sigma, we replace "(h - i)" by "j" and "i" by "h - j"

Quote:

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Originally Posted by NZAU1984

Q1. Where does the upper limit "1" come from ?
Q2. Where does the lower limit "j=h" come from (j = h - i, not h ?!)?

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We have j = h - i. Now let's look at the limits on the summation.
Lower limit of i: i = 0 leads to j = h - 0 = h
Upper limit of i: i = h - 1 leads to j = h - (h - 1) = 1

Now, some of my own confusion. We currently have
$\displaystyle \sum_{i = 0}^{h - 1} 2(h - i)2^i~ = \sum_{j = h}^1 2j2^{h - j}$

The problem is that the original summation has that h is 1 or greater. The second summation has that h is 1 or less. I cannot make any headway for this part.

Quote:

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Originally Posted by NZAU1984

$\displaystyle =2\sum_{j=1}^{h}j2^{h-j}$

Q3. Where does the upper limit "h" come from ?
Q4. Where does the lower limit "j=1" come form ?

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This is merely re-writing the sum in a different order: 1 + 2 + 3 re-written is simply 3 + 2 + 1. So this is just the same series as above written backward.

Notice that we no longer have that pesky problem with the earlier summation limits. I can't explain that one either.

Quote:

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Originally Posted by NZAU1984

$\displaystyle =2\sum_{i=1}^{h}i2^{h-i}$

Q5. Where does the lower limit "i=1" come from ? It's upposed to be "i = j - 1", so if we previously had "j = 1", we should have "i = 1 - 1 = 0".
Q6. Why has the "j" between Sigma and "2" become "i" ?
Q7. Why the exponent "h-j" has become "h - i" ?

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Notice that in the original summation we can replace i with any letter we want. i is called a "dummy index" for this reason. Notice that j is also a dummy index. So we can replace it with any letter. In this case we are going to ignore the definition of i above (because it's a dummy index) and set j = i (because it too is a dummy index.) So where ever you see a j in the summation you simply replace it with an i.

-Dan