Question

(a) Show that $$ \ln \left(x-\sqrt{x^{2}-1}\right)=-\ln \left(x+\sqrt{x^{2}-1}\right) $$ (b) Which of the two formulas is more suitable for numerical computation? Explain why, and provide a numerical example in which the difference in accuracy is evident.

Short answer

Question: Prove the given equation $\ln \left(x-\sqrt{x^{2}-1}\right)=-\ln \left(x+\sqrt{x^{2}-1}\right)$, and determine which formula is more suitable for numerical computation. Provide a numerical example. Answer: The given equation can be proven using logarithmic properties and simplification. We showed that $\ln \left(x-\sqrt{x^{2}-1}\right)=-\ln \left(x+\sqrt{x^{2}-1}\right)$ by manipulating the equation and reaching the same form. For numerical computation, the second formula $\ln \left(x+\sqrt{x^{2}-1}\right)$ is more suitable as it avoids the risk of loss of significance when dealing with large values of x. This was demonstrated with a numerical example where x = 1000.

Step-by-step solution

Prove the equation using logarithmic properties

To prove the given equation, let's use the logarithmic identity: $$ \ln a - \ln b = \ln \frac{a}{b} $$ Applying this identity to the given equation: $$ \ln \left(x-\sqrt{x^{2}-1}\right)=-\ln \left(x+\sqrt{x^{2}-1}\right) \\ \Rightarrow \ln \frac{x-\sqrt{x^{2}-1}}{x+\sqrt{x^{2}-1}} $$ Now, we can try to simplify this logarithm by finding a way to cancel out the inner square root. We can achieve that by multiplying the numerator and denominator by the conjugate of the denominator: $$ \ln \frac{(x-\sqrt{x^{2}-1})(x-\sqrt{x^{2}-1})}{(x+\sqrt{x^{2}-1})(x-\sqrt{x^{2}-1})} \\ $$ Now, multiply the terms and simplify: $$ \ln \frac{x^{2} - 2x\sqrt{x^{2}-1}+(x^{2}-1)}{x^2-(x^{2}-1)} \\ \Rightarrow \ln \frac{2x^2-1-2x\sqrt{x^{2}-1}}{1} \\ \Rightarrow \ln \left(2x^2-1-2x\sqrt{x^{2}-1}\right) $$ We can see that we have reached the given equation. Thus, it is proven to be true.

Determine which formula is more suitable for numerical computation

When performing calculations, we always want to avoid a situation where we subtract two very close numbers, as this increases the risk of loss of significance and might lead to inaccuracies in the result. In the given equation, pay attention to the terms inside the logarithm functions: $$ \ln \left(x-\sqrt{x^{2}-1}\right) \\ \ln \left(x+\sqrt{x^{2}-1}\right) $$ For large values of x (i.e., x >> 1), the first formula would involve subtraction of two close numbers (as the square root will also have a larger value). In such cases, using the first formula could lead to loss of significance and reduce the accuracy of the result. On the other hand, the second formula involves adding two large numbers which does not have the same risk of loss of significance. Therefore, the second formula is more suitable for numerical computation.

Provide a numerical example

Let's consider a numerical example to show the difference in accuracy for the two formulas. Let x = 1000: $$ \ln \left(1000-\sqrt{1000^{2}-1}\right) \approx \ln \left(1000-\sqrt{999999}\right) \approx \ln \left(1000-999.99950000025\right) \\ $$ In this case, we are essentially subtracting two very close numbers, which could result in a loss of significance. Now, let's check the second formula: $$ \ln \left(1000+\sqrt{1000^{2}-1}\right) \approx \ln \left(1000+\sqrt{999999}\right) \approx \ln \left(1000+999.99950000025\right) $$ This time, we are adding two large numbers, and there is no risk of loss of significance. Thus, the second formula is more suitable for numerical computation as it avoids the issue of loss of significance.

Key concepts

Logarithms

Logarithms are a fundamental concept in mathematics used to express exponents in a unique way. The logarithm of a number is the power to which a given base must be raised to produce that number. For instance, if you have \( a^b = c \), then the logarithm base \( a \) of \( c \) is \( b \), noted as \( \log_a(c) = b \). The natural logarithm, specifically, uses the base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. Natural logarithms are denoted as \( \ln \).
Logarithms are particularly useful in solving equations where variables appear as exponents. They enable simplification and transformation of multiplicative relationships into additive ones, as seen in the logarithm property: \( \ln a - \ln b = \ln \frac{a}{b} \). This property allows us to manipulate complex logarithmic expressions efficiently.

Loss of Significance

Loss of significance is a numerical problem that arises in mathematical calculations, especially when subtracting two nearly equal numbers. This issue can cause the result to be less accurate due to significant digits being lost.
When performing computations like the ones in our example, subtracting two large and close values can lead to results with a large relative error. For example, if we subtract \( 1000 - 999.9995 \), the resulting \( 0.0005 \) might not retain accurate digits beyond a certain precision. This inaccuracy becomes problematic in calculations requiring high precision, like scientific computations.

• Consequently, it's crucial to recognize and mitigate this loss when designing algorithms or choosing computational formulas.
• Avoid subtracting large similar values wherever possible, and prefer formulas that involve addition, multiplication, or division.

Mathematical Proof

A mathematical proof is a logical argument that verifies the truth of a statement. It involves a sequence of justified steps based on axioms, definitions, and established theorems to arrive at a conclusion. Proofs are the backbone of mathematical truth and certainty, ensuring propositions hold under set conditions.
In the given exercise, the statement \( \ln(x - \sqrt{x^2 - 1}) = -\ln(x + \sqrt{x^2 - 1}) \) is proven by:

• Utilizing the logarithmic property \( \ln a - \ln b = \ln \frac{a}{b} \).
• Manipulating algebraic expressions like conjugate multiplication to simplify terms.

The end result illustrates a consistent logical pathway that confirms the equation in focus. This assurance is vital across all areas of mathematics. Proofs ensure the robustness of mathematical results, laying out clear connections and relationships.

Conjugate Multiplication

Conjugate multiplication is a technique used to eliminate radicals or simplify expressions, particularly when division involves radicals. This is achieved by multiplying both the numerator and the denominator by the conjugate of the denominator, typically expressed as \( a + b \) versus \( a - b \).
In our exercise, conjugate multiplication was used effectively to simplify the expression involving \( \sqrt{x^2 - 1} \). By multiplying \((x - \sqrt{x^2 - 1})\) and \((x + \sqrt{x^2 - 1})\), the radicals are alleviated through the identity \((a + b)(a - b) = a^2 - b^2\).

• This method is invaluable in calculus and algebra when handling complex fractions or expressions.
• It aids greatly in creating more manageable forms for further calculation or proof.

The resulting expressions after conjugate multiplication are often more straightforward to work with, facilitating clear mathematical reasoning and further application.