Problem 45 - Calculus: One and Several Variables

Question

Let \(P_{n}\) be the \(n\) th Taylor polynomial for the function $$f(x)=\ln (1+x)$$ Find the least integer \(n\) for which: (a) \(P_{n}(0.5)\) approximates in 1.5 within \(0.01 ;\) (b) \(P_{n}(0.3)\) approximates \(\ln 1.3\) within \(0.01 ;(c) P_{n}(1)\) approximates \(\ln 2\) within 0.001.

Step-by-Step Solution

Answer

The least integer \(n\) for each part of the problem is: (a) \(n=3\) (b) \(n=4\) (c) \(n=7\)

Step 1: Find the general expression for the nth derivative of f(x)

To find the Taylor polynomial, we first need to find the general expression for the nth derivative of the function. We know that: \(f(x) = \ln(1+x)\) Now, find the first few derivatives and look for a pattern: \(f'(x) = \frac{1}{1+x}\) \(f''(x) = -\frac{1}{(1+x)^{2}}\) \(f'''(x) = \frac{2}{(1+x)^{3}}\) Notice the pattern; the alternating signs and the factorials in the numerators. We can write the \(n\)th derivative of the function as: \(f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{(1+x)^n}\)

Step 2: Write the Taylor polynomial for f(x)

The Taylor polynomial of degree \(n\) for a function is given by: $$ P_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(0)}{k!}(x-0)^k $$ Using the expression we derived for the \(n\)th derivative of \(f(x)\), we can write the Taylor polynomial of degree \(n\) for \(f(x)\) as: $$ P_n(x) = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} x^k $$

Step 3: Find the error term for the nth Taylor polynomial for f(x)

The error term for the \(n\)th Taylor polynomial is given by: $$ E_n(x) = f(x) - P_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} $$ where \(0 < c < x\). We know the \((n+1)\)th derivative of \(f(x)\): \(f^{(n+1)}(x) = \frac{(-1)^{n}(n)!}{(1+x)^{n+1}}\) So the error term for the Taylor polynomial is: $$ E_n(x) = \frac{(-1)^n (n)! x^{n+1}}{(n+1)!(1+c)^{n+1}} $$

Step 4: Find the least integer n for the given tolerances

Now we can use the information given in each part to find the least integer \(n\). (a) We want \(P_n(0.5)\) to approximate \(\ln(1.5)\) within 0.01. This means: $$|E_n(0.5)| = \left|\frac{(-1)^n(0.5)^{n+1} (n)!}{(n+1)!(1+c)^{n+1}}\right| < 0.01$$ Since \(0 < c < 0.5\), we know that \((1+c)^{n+1} < 2^{n+1}\). We can then write: $$ \left|\frac{(-1)^n(0.5)^{n+1} (n)!}{(n+1)!}\right| < 0.01 $$ (b) Similarly, for \(P_n(0.3)\) to approximate \(\ln(1.3)\) within 0.01: $$|E_n(0.3)| = \left|\frac{(-1)^n(0.3)^{n+1} (n)!}{(n+1)!(1+c)^{n+1}}\right| < 0.01$$ Since \(0 < c < 0.3\), we know that \((1+c)^{n+1} < 1.3^{n+1}\). We can then write: $$ \left|\frac{(-1)^n(0.3)^{n+1} (n)!}{(n+1)!}\right| < 0.01 $$ (c) Lastly, for \(P_n(1)\) to approximate \(\ln(2)\) within 0.001: $$|E_n(1)| = \left|\frac{(-1)^n (n)!}{(n+1)!(1+c)^{n+1}}\right| < 0.001$$ Since \(0 < c < 1\), we know that \((1+c)^{n+1} < 2^{n+1}\). We can then write: $$ \left|\frac{(-1)^n (n)!}{(n+1)!}\right| < 0.001 $$ Solve each of these inequalities for \(n\) and find the least integer for each case: (a) \( \left|\frac{(-1)^n(0.5)^{n+1} (n)!}{(n+1)!}\right| < 0.01 ; \) Solving, we get n > 2 (b) \( \left|\frac{(-1)^n(0.3)^{n+1} (n)!}{(n+1)!}\right| < 0.01 ; \) Solving, we get n > 3 (c) \( \left|\frac{(-1)^n (n)!}{(n+1)!}\right| < 0.001 ; \) Solving, we get n > 6 Result: For each part of the problem, the least integer \(n\) is: (a) 3 (b) 4 (c) 7

Key Concepts

Taylor series expansion

Taylor series expansion is a powerful mathematical tool used to represent functions as infinite sums of terms calculated from the values of their derivatives at a single point. This concept allows us to approximate complex functions using simple polynomials, making them easier to understand and compute. The basic idea is to represent a function as a series:

• Take the function, such as \(f(x) = \ln(1+x)\).
• Find its derivatives and identify a pattern.
• Use these derivatives to write a polynomial that approximates the function.

For our function \(f(x) = \ln(1+x)\), the series expansion involves alternating signs, showing a pattern in the signs of its derivatives. The Taylor polynomial up to degree \(n\) is given by:\[P_n(x) = \sum_{k=1}^{n} \frac{(-1)^{k-1}}{k} x^k\]This polynomial provides a finite approximation of the infinite Taylor series, which helps in practical computation where infinite series cannot be genuinely handled.

Error estimation in Taylor series

Error estimation is crucial in Taylor series, as it tells us how close our Taylor polynomial is to the actual function. When we use a Taylor polynomial \(P_n(x)\) to approximate \(f(x)\), there is often some residual difference called the error, denoted by \(E_n(x)\). This difference arises because a polynomial of finite degree cannot completely capture the potentially infinite details of a function.
The error can be estimated using:

• Find the next derivative after the last one used in the polynomial, \(f^{(n+1)}(x)\).
• Use the inequality for the error term:\[E_n(x) = \frac{(-1)^n (n)! x^{n+1}}{(n+1)!(1+c)^{n+1}}\]where \(0 < c < x\).
• Assessing how small this error becomes gives us confidence in the approximation’s accuracy within specified bounds, like 0.01 or 0.001, as seen in different parts of the exercise.

This way, we ensure our polynomial provides an accurate approximation for the values within desired limits.

Natural logarithm function

The natural logarithm function, denoted as \(\ln(x)\), is a fundamental mathematical function with various applications in science and engineering. It is the inverse of the exponentiation function with base \(e\), where \(e\) is approximately 2.71828. The natural logarithm of a number is the exponent to which \(e\) must be raised to yield that number. In this context, \(\ln(1+x)\) represents a shifted version of the natural logarithm, starting at zero (as \(\ln(1) = 0\)).
This function is especially interesting due to its behavior around \(x = 0\), often used in series expansions like the Taylor series. For small values of \(x\), \(\ln(1+x)\) can be effectively approximated by its Taylor polynomial. This is because its derivatives have a predictable and decreasing pattern, leading to manageable polynomial terms.

• \(f(x) = \ln(1+x)\) has simple derivative forms, especially useful for polynomial expansion.
• Its pattern of alternating signs in derivatives leads to an equally predictable series expansion.
• Understanding this function and its approximation helps tackle complex mathematical calculations more feasibly.

Such properties make \(\ln(1+x)\) a preferred candidate in many analytical calculations, especially when nuanced precision is required.