Arxe Logic

Formulas for Converting Tk to Exentation Number n

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The original function defining the exponent k for an exentation n is:

\(e(n) = \begin{cases} 0 & \text{if } n=1 \(-1)^n \cdot \lfloor n/2 \rfloor & \text{if } n>1 \end{cases}\)

To find the exentation number n from an exponent k, we derive the following three conditions based on the definition of e(n):

• If the exponent k is equal to 0: This only occurs when n=1.

\(n = 1 \quad \text{if } k = 0\)

If the exponent k is positive (k>0): Positive exponents are generated when n is an even number. In this case, (−1)n=1, and the formula simplifies to k=⌊n/2⌋. Since n is even, ⌊n/2⌋=n/2. Therefore:

\(k = n/2 \rightarrow n = 2k \quad \text{if } k > 0\)

If the exponent k is negative (k<0): Negative exponents are generated when n is an odd number. In this case, (−1)n=−1, and the formula is k=−⌊n/2⌋. For an odd n, ⌊n/2⌋=(n−1)/2. Substituting this into the equation:

\(k = -(n-1)/2\)

Multiplying by −2:

\(-2k = n-1\)

Solving for n:

\(n = -2k + 1 \quad \text{if } k < 0\)

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Summary of Inverse Formulas

Combining these three conditions, the formula for finding the exentation number n given an exponent k is:

\(n = \begin{cases} 1 & \text{if } k = 0 \ 2k & \text{if } k > 0 \-2k + 1 & \text{if } k < 0 \end{cases}\)