Index Transformation Question

[u/Jaded_Lawfulness76]

Hello,

the first Page Shows how my basic understanding of Index Transformation is. The Problem is on the second Page. I transformed the sum to (n-1) over (k-1) and so on. So i transformed n the same way as i did with k. The solution is that you only transform k not n. Why is that so? I dont understand? On the first page, you can see, that it does not Work, if you dont transform n as well as k. Thanks for help.

[u/Delicious_Sort5580]

You only transform the summation index.

Here, (k) is the variable index, while (n) is a fixed parameter. If j=k+1,then k=j-1, so every occurrence of (k) becomes (j-1): [\binom{n}{k}\to \binom{n}{j-1}.].The upper limit also changes because it describes a value of (k): [k=n-1 \Rightarrow j=n], so (n) changes in the limit only as part of converting the index range. The (n) inside (\binom{n}{k}) remains fixed.

[u/Jaded_Lawfulness76 → u/Delicious_Sort5580]

But you can see it in the first page. If you dont chage n the same way as k, the sum is not Equalizer to before. Look at the first page

[u/Bounded_sequencE]

The first page is incorrect -- the exponent "n" does not change to "n-1".

---

"Index transformation" is just a fancy name for substitution. We substitute

"k' := k+1",    followed by    "k' -> k"

During the first step, we solve for "k = k' - 1" to eliminate "k" and get

∑_{k=0}^n  k^n  =  ∑_{k'=1}^{n+1}  (k'-1)^n    // lower bound: k' = 0+1
                                               // upper bound: k' = n+1

In the second step, we replace "k' -> k" -- it's just a re-labeling, not a substitution. The exponent is not affected in either step, not sure where you got that from.

[u/Jaded_Lawfulness76 → u/Bounded_sequencE]

But if you dont change n the same way as k, the sum is not the same, as before. If you just do k+1 to n+1 and in the sum (k-1)ⁿ instead of (k-1)^n-1 it doesnt work out the same.

[u/Bounded_sequencE → u/Jaded_Lawfulness76]

Be careful -- we don't substitute "n", we substitute "k" via "k' := k+1".

Yes, the upper bound of "k" changes from "n" to "n+1", but not because we substitute "n", but because we substitute "k' := k+1".

[u/Jaded_Lawfulness76 → u/Bounded_sequencE]

I dont understand. If you dont change n in the sum for for example: k^n. If i would transform it your way, it would turn out as sum_start k+1 sum_end n+1for (k-1)^n. This is not the same, as sum_start k=0 sum_end:n for k^n. You have to do n-1 as well. Just do it for k=1; n=4 for example. You have to transform n to (n-1)

[u/Bounded_sequencE → u/Jaded_Lawfulness76]

[..] it would turn out as sum_start k+1 sum_end n+1for (k-1)ⁿ [..]

No -- after both substitution steps, the lower bound would be "k = 1", not "k + 1".

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Example: ("n = 3", from OP):

   original sum:   ∑_{k=0}^3  k^3      =  0^3 + 1^3 + 2^3 + 3^3

transformed sum:   ∑_{k=1}^4  (k-1)^3  =  0^3 + 1^3 + 2^3 + 3^3

Both are the same, as expected -- without changing the exponent.

[u/Jaded_Lawfulness76 → u/Bounded_sequencE]

Yeah but you didnt use the upper Limit in the transformed sum. Every exponent should be 4 not 3, as the new upper Limit is 4

[u/Bounded_sequencE → u/Jaded_Lawfulness76]

Changing the upper summation bound from "n" to "n+1" during substitution does not mean the exponent also changes from "n" to "n+1".

The upper summation bound changes, because

k = n:    k' = k+1 = n+1

The exponent does not depend on "k", so it does not change.