To solve this problem, we need to address it in a systematic manner:

Step 1: Calculate the value of $x$

Given: $x = 2^3 + 2^{-19} + 2^{-22}$ $x = 8 + \frac{1}{2^{19}} + \frac{1}{2^{22}}$ We can simplify this by calculating the fractions:

$2^{-19} \approx 1.9073486328125 \times 10^{-6}$ $2^{-22} \approx 2.384185791015625 \times 10^{-7}$

Adding these values to 8: $x \approx 8 + 1.9073486328125 \times 10^{-6} + 2.384185791015625 \times 10^{-7}$ $x \approx 8.0000019073486328125 + 0.0000002384185791015625$ $x \approx 8.0000021457672119140625$

Step 2: Determine machine numbers on the Marc-32

The Marc-32 is a hypothetical floating-point representation. If it has a precision of 32 bits, the closest representable numbers around $x$ are called "machine numbers."

Step 2.1: Determine the closest machine numbers just to the left and right of $x$.

Assume Marc-32 uses the IEEE 754 standard for single-precision (32-bit) floating-point numbers:

• The exponent has 8 bits.
• The mantissa (significand) has 23 bits.
• The precision in binary is $2^{-23}$.

Let's denote:

• $fl(x)$ as the machine number closest to $x$ (which represents the floating-point approximation).
• $x_{\text{left}}$ as the largest machine number less than $x$.
• $x_{\text{right}}$ as the smallest machine number greater than $x$.

Given $x \approx 8$, we have: $2^3 \leq x < 2^4$ In binary, this is equivalent to: $x = 8 \text{ in binary} \approx 1000.0000 \ldots$ The precision is $2^{-23}$, so we adjust the binary accordingly.

Step 2.2: Calculate $fl(x)$, absolute error, and relative error

Since $x \approx 8 + \text{small values}$, and the precision is $2^{-23}$:

• $fl(x) = 8$ (because 8 is the closest representable number).
• The absolute error is $|x - fl(x)|$. $\text{Absolute error} = |8.0000021457672119140625 - 8| = 2.1457672119140625 \times 10^{-6}$

Relative error: $\text{Relative error} = \frac{|x - fl(x)|}{|x|} = \frac{2.1457672119140625 \times 10^{-6}}{8.0000021457672119140625}$ $\text{Relative error} \approx \frac{2.1457672119140625 \times 10^{-6}}{8} \approx 2.682209014892578125 \times 10^{-7}$

Step 3: Verify that the relative error does not exceed $2^{-34}$

$2^{-34} \approx 5.820766091346741 \times 10^{-11}$.

Clearly, the relative error $\approx 2.682209014892578125 \times 10^{-7}$ does not exceed $2^{-34}$.

Thus, the machine number just to the right of $x$ would be approximately $fl(x) = 8$ (in this simplified hypothetical context), with the specified errors computed as above.

Summary

• $fl(x)$ is the closest representable machine number to $x$, which is $8$ in this case.
• Absolute error: $2.1457672119140625 \times 10^{-6}$.
• Relative error: $2.682209014892578125 \times 10^{-7}$.

The relative error does not exceed $2^{-34}$.

Would you like more details or have any further questions?

Related Questions:

1. What is the binary representation of $x = 8 + 2^{-19} + 2^{-22}$?
2. How does the Marc-32 representation compare to IEEE 754?
3. What is the significance of machine precision in numerical analysis?
4. How can underflow and overflow occur in floating-point arithmetic?
5. What are the typical rounding errors encountered in floating-point computations?

Tip:

When working with floating-point numbers, always consider the machine epsilon, which gives an upper bound on the relative error due to rounding in floating-point arithmetic.