Question

Exercise \(1.3-2:\) The \(x\) -intercept of the line passing through the points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) can be computed using either one of the following formulas: $$x=\frac{x_{1} y_{2}-x_{2} y_{1}}{y_{2}-y_{1}}$$ or, $$x=x_{1}-\frac{\left(x_{2}-x_{1}\right) y_{1}}{y_{2}-y_{1}}$$ with the assumption \(y_{1} \neq y_{2}\). a) Show that the formulas are equivalent to each other. b) Compute the \(x\) -intercept using each formula when \(\left(x_{1}, y_{1}\right)=(1.02,3.32)\) and \(\left(x_{2}, y_{2}\right)=\) \((1.31,4.31) .\) Use three-digit rounding arithmetic. c) Use Python (or a calculator) to compute the \(x\) -intercept using the full- precision of the device (you can use either one of the formulas). Using this result, compute the relative and absolute errors of the answers you gave in part (b). Discuss which formula is better and why.

Short answer

b) What is the x-intercept of the line passing through the points (1.02, 3.32) and (1.31, 4.31)? c) After computing the x-intercept with full precision using Python, what are the absolute and relative errors? Answer: a) Yes, the given formulas for finding the x-intercept of a line are equivalent. b) The x-intercept of the line passing through the points (1.02, 3.32) and (1.31, 4.31) is approximately -2.476 when using three-digit rounding arithmetic. c) After computing the x-intercept with full precision using Python, the absolute error is approximately 0.00020932973 and the relative error is approximately 8.45862623 x 10^(-5).

Step-by-step solution

Prove the formulas are equivalent

First, let's prove that the given formulas are equivalent. We will start with the second formula and simplify it to get the first formula. $$x = x_1-\frac{(x_{2}-x_{1}) y_{1}}{y_{2}-y_{1}}$$ $$x = x_1-\left(\frac{x_{2}-x_{1}}{y_{2}-y_{1}}\right) y_{1}$$ $$x = x_1-\left(\frac{x_{2}y_{1}-x_{1}y_{1}}{y_{2}-y_{1}}\right)$$ $$x(y_{2}-y_{1}) = x_1(y_{2}-y_{1})-(x_{2}y_{1}-x_{1}y_{1})$$ $$xy_{2}-xy_{1} = x_1y_{2}-x_1y_{1}-x_{2}y_{1}+x_{1}y_{1}$$ $$x = \frac{x_{1}y_{2}-x_{2}y_{1}}{y_{2}-y_{1}}$$ Hence, both formulas are equivalent.

Compute x-intercept using both formulas

Now, let's compute the x-intercept using both formulas with the provided points (1.02, 3.32) and (1.31, 4.31). We'll round the result to three digits. Using the first formula: $$x = \frac{x_{1}y_{2}-x_{2}y_{1}}{y_{2}-y_{1}} = \frac{1.02\times4.31-1.31\times3.32}{4.31-3.32} \approx -2.476$$ Using the second formula: $$x = x_1-\frac{(x_{2}-x_{1})y_{1}}{y_{2}-y_{1}} =1.02-\frac{(1.31-1.02)(3.32)}{4.31-3.32} \approx -2.476$$ In both cases, the x-intercept is approximately -2.476 after rounding to three digits.

Compute x-intercept using Python with full precision

Now, we'll use Python to compute the x-intercept with one of the formulas with full precision: ```python x1, y1 = 1.02, 3.32 x2, y2 = 1.31, 4.31 x_exact = (x1 * y2 - x2 * y1) / (y2 - y1) ``` The x_exact value is approximately -2.47579067027.

Calculate the relative and absolute errors

To compute the absolute error, we'll subtract our approximate x-intercept value from the exact x-intercept value: $$Absolute error = |x_{exact} - x_{approximate}| = |-2.47579067027 - (-2.476)| \approx 0.00020932973$$ To compute the relative error, we'll divide the absolute error by the exact x-intercept value: $$Relative error = \frac{Absolute error}{|x_{exact}|} = \frac{0.00020932973}{|-2.47579067027|} \approx 8.45862623\times10^{-5}$$

Discuss which formula is better and why

Since both formulas are equivalent and provide the same x-intercept value, it is not correct to say which one is better. Each formula may have its own advantages and may be more convenient depending on the given situation. However, in this specific case, the difference in results is negligible and so neither formula is preferred.

Key concepts

x-intercept calculation

Understanding how to calculate the x-intercept is crucial in numerical analysis. The x-intercept of a line is where it crosses the x-axis, meaning the y-value at this point is zero. There are different mathematical formulas to find this intercept when you have two points \( (x_1, y_1) \) and \( (x_2, y_2) \):

• The first formula: \( x = \frac{x_1 y_2 - x_2 y_1}{y_2 - y_1} \)
• The second formula: \( x = x_1 - \frac{(x_2 - x_1) y_1}{y_2 - y_1} \)

Both formulas come from the same geometric principle of determining where the line crosses the x-axis and can be used interchangeably. To ensure reliable results, it is important to follow the appropriate mathematical simplifications as shown in equivalent proofs.

rounding arithmetic

Rounding arithmetic is a valuable tool when precision is less critical or computational efficiency is desired. In many numerical tasks, you encounter quantities that need rounding to specific decimal places for simplicity. In the given problem, calculations were rounded to three digits:

• A calculated x-intercept value from formulas might be -2.47579067027, which is rounded to -2.476.

Rounding may introduce minor errors, but it simplifies numbers and can make results more manageable. It allows quick approximations, which are particularly beneficial when performing manual calculations.

numerical precision

Numerical precision refers to the accuracy with which numbers are represented in calculations. In computational environments, this can vary, affecting results significantly. Calculators or programming languages like Python handle numbers with high precision, leading to more accurate computations. When calculating the x-intercept with full precision, the result is \( x_{exact} \approx -2.47579067027 \), more accurate compared to a rounded value of -2.476.

Maintaining numerical precision is crucial, especially in fields requiring exact calculations like engineering and physics. The accuracy of calculations might affect decisions or interpretations hence, understanding the limitations of numerical precision is vital.

error analysis

Error analysis helps in understanding how errors affect the outcome of mathematical computations. It involves finding the absolute and relative errors between the computed and true values. The absolute error measures the difference between the exact and approximate values, calculated here as 0.00020932973. The relative error gives a sense of the error's size relative to the exact value.

• Absolute Error: \( |x_{exact} - x_{approximate}| \)
• Relative Error: \( \frac{\text{Absolute Error}}{|x_{exact}|} \)

Evaluating errors allows you to determine the reliability of your results and select appropriate methods for calculations. By understanding error margins, you can improve calculations or choose alternatives that minimize inaccuracies.