Can this Sum-Notation function be altered to get around the convention that $0^0 = 1$?

Question

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1. What the question is asking (plain language)

You have a compact formula

[ d_{m,n}(t)=\frac{1}{n!}\sum_{i=1}^{2^{m}} \Bigl((-1)^{S(i)}\;\max!\bigl(0,\;t-T_m(i)\bigr)^{\,n}\Bigr) \tag{1} ]

that works for all integers (n\ge 1).
When (n=0) the factor

[ \max!\bigl(0,\;t-T_m(i)\bigr)^{\,0} ]

is interpreted by most computer algebra systems as (1) even when the argument is (0) (the usual convention (0^{0}=1)). Consequently every term in the sum contributes ((-1)^{S(i)}) and the whole expression becomes a constant (the alternating sum of the first (2^{m}) Thue–Morse signs), i.e. a flat line.

What you really want for the case (n=0) is a step‑like behaviour:

on the interval ([T_m(i),\,T_m(i+1))) the term should be ((-1)^{S(i)});
on the interval ((-\infty,\,T_m(i))) the term should be zero.

In other words you would like the factor (\max(0,\;t-T_m(i))^{0}) to be (1) only when (t>T_m(i)) and (0) otherwise.
The question is: Can we rewrite (1) so that this happens without returning to an explicit piece‑wise definition?

2. Detailed solution

2.1 Replace the “max‑to‑power‑0’’ with a Heaviside (step) function

The Heaviside step function (H(x)) is defined as

[ H(x)=\begin{cases} 0, & x\le 0,\[2pt] 1, & x>0 . \end{cases} ]

(Any value at (x=0) may be chosen; we shall take (H(0)=0) to obtain the half‑open intervals required in the problem.)
Notice that for any real (x),

[ \max(0,x)=x\,H(x). ]

Hence for a positive integer (n),

[ \max(0,x)^{\,n}=x^{\,n}H(x) . ]

When (n=0) we define

[ \max(0,x)^{\,0}=H(x), \qquad\text{(definition for the present purpose)} \tag{2} ]

so that the factor is (1) only when (x>0) and (0) otherwise.

2.2 Insert the step function into the sum

Using (2) we can rewrite (1) as

[ \boxed{ d_{m,n}(t)=\frac{1}{n!}\sum_{i=1}^{2^{m}} \Bigl((-1)^{S(i)}\;(t-T_m(i))^{\,n}\,H!\bigl(t-T_m(i)\bigr)\Bigr) }\tag{3} ]

where the exponent (n) may be any non‑negative integer.

For (n\ge 1) the factor ((t-T_m(i))^{n}) already forces the term to be zero whenever (t\le T_m(i)); the extra (H) does not change the value.
For (n=0) the factor ((t-T_m(i))^{0}=1) and the whole term reduces to ((-1)^{S(i)} H(t-T_m(i))), i.e. a step that is (0) on ((-\infty,T_m(i)]) and ((-1)^{S(i)}) on ((T_m(i),\infty)).

Thus on each half‑open interval ([T_m(i),T_m(i+1))) the sum (3) contributes the single non‑zero term with index (i) and yields precisely ((-1)^{S(i)}), exactly the behaviour requested.

2.3 Verify the result for a simple example

Take (m=2) (so (2^{m}=4)) and let

[ T_{2} = (0,\,1,\,2,\,3) . ]

The Thue–Morse signs are ((-1)^{S(i)} = (1,-1,-1,1)).
Using (3) with (n=0),

[ \begin{aligned} d_{2,0}(t) &= \sum_{i=1}^{4} (-1)^{S(i)} H!\bigl(t-T_{2}(i)\bigr)
&= 1\cdot H(t-0)\;-\;1\cdot H(t-1)\;-\;1\cdot H(t-2)\;+\;1\cdot H(t-3). \end{aligned} ]

Evaluating on the intervals gives

interval	value of (d_{2,0}(t))
([0,1))	(+1)
([1,2))	(-1)
([2,3))	(-1)
([3,\infty))	(+1)

exactly the alternating pattern required, while the function is (0) for (t<0) (because all Heaviside factors are zero).

For any larger (n) the same formula (3) reproduces the original polynomial pieces because the extra (H) factor is redundant.

2.4 Final compact definition

Putting everything together, the modified definition that works for every (n\in{0,1,\dots ,m}) is

[ \boxed{ d_{m,n}(t)=\frac{1}{n!}\sum_{i=1}^{2^{m}} (-1)^{S(i)}\;(t-T_m(i))^{\,n}\;H!\bigl(t-T_m(i)\bigr) }\qquad (n\ge 0). ]

If you prefer to keep the original “max’’ notation, you can state it as

[ d_{m,n}(t)=\frac{1}{n!}\sum_{i=1}^{2^{m}} (-1)^{S(i)}\;\bigl[\max!\bigl(0,\,t-T_m(i)\bigr)\bigr]^{\,n}, \quad\text{with the convention } \bigl[\max(0,x)\bigr]^{0}=H(x). ]

Either version eliminates the unwanted constant‑line behaviour for the case (n=0) while remaining a single summation, not a piece‑wise definition.

3. Final answer

Answer:
Introduce the Heaviside step function (H) (or equivalently define (\max(0,x)^{0}=H(x))) and write

[ \boxed{% d_{m,n}(t)=\frac{1}{n!}\sum_{i=1}^{2^{m}} (-1)^{S(i)}\,(t-T_m(i))^{\,n}\,H!\bigl(t-T_m(i)\bigr) } ]

for all integers (0\le n\le m).
For (n\ge 1) this coincides with the original expression; for (n=0) it produces a step function that is zero on ((-\infty,T_m(i)]) and ((-1)^{S(i)}) on ((T_m(i),\infty)), giving the desired sequence of positive, zero, and negative segments without resorting to an explicit piece‑wise definition.

4. Common mistakes

Mistake	Why it’s wrong	How to avoid it
Treating (0^{0}=1) everywhere	This makes the (n=0) term non‑zero even when (t\le T_m(i)), destroying the intended interval structure.	Replace the power‑zero case by a step function: ( \max(0,x)^{0}=H(x)).
Using the “standard’’ Heaviside with (H(0)=\tfrac12)	The half‑open intervals ([T_m(i),T_m(i+1))) require the value at the left endpoint to be zero; a value of (1/2) would create spurious jumps.	Adopt the convention (H(0)=0) (or write the condition as (t>T_m(i)) explicitly).
Omitting the factor (H(t-T_m(i))) for (n\ge 1)	For (n\ge 1) the factor is technically unnecessary, but keeping it makes the formula uniform and prevents accidental inclusion of the (n=0) case.	Keep the step function in every term; it is harmless when (n\ge 1).
Confusing the Thue–Morse sign ((-1)^{S(i)}) with the sequence values	The sign must be applied to each term of the sum; mixing up the exponent can flip the pattern.	Remember that (S(i)) is the binary digit sum modulo 2; compute ((-1)^{S(i)}) correctly.

By watching for these pitfalls you can reliably implement the compact summation for all required values of (n).

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