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The natural logarithm function, \(\ln ,\) is defined so that \(e^{\ln x}=x\) for any positive number \(x\) (a) Sketch a graph of the natural logarithm function. (b) Prove the identities $$\begin{aligned} &\ln a b=\ln a+\ln b \quad \text { and }\\\&\ln a^{b}=b \ln a\end{aligned}$$ (c) Prove that \(\frac{d}{d x} \ln x=\frac{1}{x}\) (d) Derive the useful approximation $$\ln (1+x) \approx x,$$ which is valid when \(|x| \ll 1 .\) Use a calculator to check the accuracy of this approximation for \(x=0.1\) and \(x=0.01.\)

Short Answer

Expert verified
Sketch: ln(x) is increasing and passing through (1,0). Identities: ln(ab)=ln(a)+ln(b), ln(a^b)=b ln(a). Derivative: 1/x. Approximation: ln(1+x)≈x for small |x|.

Step by step solution

01

Sketch the Graph of ln(x)

The graph of the natural logarithm function, \(\ln(x)\), is a curve that passes through the point \((1,0)\) because \(\ln(1) = 0\). It is defined only for \(x > 0\). As \(x\) approaches zero from the right, \(\ln(x)\) tends to negative infinity. As \(x\) increases, \(\ln(x)\) increases but at a decreasing rate, tending towards positive infinity. The graph is always increasing, reflecting the fact that \(\ln(x)\) is a strictly increasing function.
02

Prove Identity: ln(ab) = ln(a) + ln(b)

To prove \(\ln(ab) = \ln(a) + \ln(b)\), consider the exponential form: \(e^{\ln(ab)} = ab\). Using the property \( e^{u+v} = e^u \cdot e^v\), we can rewrite this as \(e^{\ln(a) + \ln(b)} = e^{\ln(a)}e^{\ln(b)} = ab\). Since both expressions equal \(ab\), the identity is proven.
03

Prove Identity: ln(a^b) = b * ln(a)

For \(\ln(a^b) = b\ln(a)\), consider the exponentiation \(e^{\ln(a^b)} = a^b\). This can be written as \((e^{\ln(a)})^b = (e^{b\ln(a)}) = a^b\), showing that both the left side and right side simplify to \(a^b\), confirming the identity is true.
04

Prove Derivative: d/dx ln(x) = 1/x

The derivative of \(\ln(x)\) is \(\frac{d}{dx} \ln(x) = \frac{1}{x}\). This can he proven using implicit differentiation. Assume \(y = \ln(x)\) so that \(e^y = x\). Differentiating both sides, \(e^y \frac{dy}{dx} = 1\). Since \(e^y = x\), substitute back to get \( x \frac{dy}{dx} = 1\), hence \(\frac{dy}{dx} = \frac{1}{x}\).
05

Derive Approximation: ln(1+x) ≈ x

The Taylor series for \(\ln(1+x)\) around \(x = 0\) is \(x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\). For \(|x| \ll 1\), higher order terms are much smaller and can be neglected, leaving \(\ln(1+x) \approx x\). This is a common approximation when \(x\) is close to zero.
06

Check Approximation at x=0.1 and x=0.01

For \(x = 0.1\), calculate \(\ln(1.1)\) using a calculator: \(\ln(1.1) \approx 0.0953\) and compare with the approximation \(0.1\), which is quite close. For \(x = 0.01\), \(\ln(1.01)\) using a calculator gives \(\ln(1.01) \approx 0.00995\), which is also close to \(0.01\). This confirms the approximation \(\ln(1+x) \approx x\) is valid when \(|x| \ll 1\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Function
The exponential function is a vital concept in mathematics, represented as \( e^x \). Here, "\( e \)" is the base of natural logarithms, an irrational number approximately equal to 2.71828. The exponential function grows rapidly, and its rate of increase is proportional to its current value, which makes it essential in modeling growth processes. The natural logarithm, \( \ln x \), is the inverse of the exponential function. This means if \( e^y = x \), then \( \ln(x) = y \). The two functions completely mirror each other across the line \( y = x \).

In practical terms:
  • For any positive number \( x \), \( e^{\ln x} = x \).
  • The exponential function is used extensively in calculus, physics, and finance to model continuous growth or decay.
Remember, understanding the relationship between the exponential function and natural logarithm is crucial for solving logarithmic and exponential equations.
Taylor Series Approximation
The Taylor series is a method of approximating functions using a sum of polynomial terms. With each terms derivative evaluated at a particular point, typically zero for the Maclaurin series substitute of Taylor series. For the function \( \ln(1+x) \), the Taylor series expands as:\[\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots\]

This series is valid for \(|x| < 1\) and provides a powerful way to approximate \( \ln(1+x) \) when \( |x| \ll 1 \). The first term, \( x \), already gives a very close estimate, which is why you might often see \( \ln(1+x) \approx x \) for small \( x \).

Using the Taylor series, we can simplify complex calculations and gain better insight into behavior near specific points by neglecting higher-degree terms:
  • The more terms we include, the more accurate the approximation.
  • This principle is the basis for many numerical methods in science and engineering.
Derivatives in Calculus
Derivatives are a primary tool in calculus that measure how a function changes with respect to its inputs. When we take the derivative of \( \ln(x) \), we find:\[ \frac{d}{dx} \ln(x) = \frac{1}{x}\]

This derivative is derived via implicit differentiation, where we express \( y = \ln(x) \) leading to \( e^y = x \). Differentiating, we get:\[ e^y \frac{dy}{dx} = 1 \ \rightarrow x \frac{dy}{dx} = 1 \ \rightarrow \frac{dy}{dx} = \frac{1}{x} \]

Understanding derivatives helps in:
  • Analyzing the behavior of functions.
  • Solving optimization problems in fields like statistics and economics.
  • Finding tangent lines and rates of change, which are crucial for interpreting real-world phenomena.
Logarithmic Identities
Logarithms come with several useful identities that simplify calculations and conceptual understanding. Two important properties are:
  • \( \ln(ab) = \ln(a) + \ln(b) \): This property helps condense products into sums, simplifying multiplication operations, especially in calculus.
  • \( \ln(a^b) = b \ln(a) \): This helps transform powers into multipliers, absolutely fundamental in solving exponential equations.
These identities stem from the properties of exponents:

Converting products and powers into sums and multiples streamlines solving exponential growth problems, understanding compound interest, and analyzing sound levels or earthquake magnitudes. By efficiently compressing complex expressions, logarithms and their identities find applications across mathematics, sciences, and finance.

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Most popular questions from this chapter

Consider a two-state paramagnet with \(10^{23}\) elementary dipoles, with the total energy fixed at zero so that exactly half the dipoles point up and half point down. (a) How many microstates are "accessible" to this system? (b) Suppose that the microstate of this system changes a billion times per second. How many microstates will it explore in ten billion years (the age of the universe)? (c) Is it correct to say that, if you wait long enough, a system will eventually be found in every "accessible" microstate? Explain your answer, and discuss the meaning of the word "accessible."

For a single large two-state paramagnet, the multiplicity function is very sharply peaked about \(N_{\uparrow}=N / 2\) (a) Use Stirling's approximation to estimate the height of the peak in the multiplicity function. (b) Use the methods of this section to derive a formula for the multiplicity function in the vicinity of the peak, in terms of \(x \equiv N_{\uparrow}-(N / 2) .\) Check that your formula agrees with your answer to part (a) when \(x=0\) (c) How wide is the peak in the multiplicity function? (d) Suppose you flip 1,000,000 coins. Would you be surprised to obtain 501,000 heads and 499,000 tails? Would you be surprised to obtain 510,000 heads and 490,000 tails? Explain.

Use a computer to produce a table and graph, like those in this section, for two interacting two-state paramagnets, each containing 100 elementary magnetic dipoles. Take a "unit" of energy to be the amount needed to flip a single dipole from the "up" state (parallel to the external field) to the "down" state (antiparallel). Suppose that the total number of units of energy, relative to the state with all dipoles pointing up, is \(80 ;\) this energy can be shared in any way between the two paramagnets. What is the most probable macrostate, and what is its probability? What is the least probable macrostate, and what is its probability?

Consider a system of two Einstein solids, \(A\) and \(B\), each containing 10 oscillators, sharing a total of 20 units of energy. Assume that the solids are weakly coupled, and that the total energy is fixed. (a) How many different macrostates are available to this system? (b) How many different microstates are available to this system? (c) Assuming that this system is in thermal equilibrium, what is the probability of finding all the energy in solid \(A\) ? (d) What is the probability of finding exactly half of the energy in solid \(A\) ? (e) Under what circumstances would this system exhibit irreversible behavior?

Calculate the multiplicity of an Einstein solid with 30 oscillators and 30 units of energy. (Do not attempt to list all the microstates.)

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