Problem 38 - Calculus

Question

Use Taylor series to explain the patterns in the digits in the following expansions: (a) \(\frac{1}{0.98}=1.02040816 \ldots\) (b) \(\left(\frac{1}{0.99}\right)^{2}=1.020304050607 \ldots\)

Step-by-Step Solution

Answer

Use Taylor series and geometric expansions to identify repeating decimal patterns.

Step 1: Understand the Taylor Series

A Taylor series for a function \(f(x)\) centered at a point \(a\) is given by \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \). It represents a way to approximate functions using polynomials.

Step 2: Analyze Part (a)

We have \( \frac{1}{0.98} \). We rewrite it as \( \frac{1}{1 - 0.02} \) which resembles the geometric series \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots \). By using \( x = 0.02 \), we can express \( \frac{1}{0.98} \) as \( 1 + 0.02 + 0.02^2 + 0.02^3 + \cdots \). Each term represents the digits in the expansion 1.02040816, etc.

Step 3: Explain the Pattern in Part (a)

For \( x = 0.02 \), the terms of the series give the pattern of digits. The first term is 1, the second term is 0.02, the third term is 0.0004, the fourth term is 0.000008, and so on. These contribute to the repeating pattern of digits: 1.02040816... The sequence in the exponential position indicates the repeating periodicity within this expansion.

Step 4: Analyze Part (b)

We have \( \left(\frac{1}{0.99}\right)^2 = \left(1 + 0.01\right)^2 \). Using the binomial theorem, we expand: \((1 + x)^n = 1 + nx + \frac{n(n-1)}{2}x^2 + \cdots \). For \( n = 2 \) and \( x = 0.01, \) this gives \( 1 + 2\cdot0.01 + (2\cdot1/2)\cdot(0.01)^2 + \cdots \), leading to a decimal pattern.

Step 5: Explain the Pattern in Part (b)

Calculating \( 1 + 0.02 + 0.0001 \), the powers of 0.01 result in the terms 0.000001, 0.00000001, etc., creating the repeating sequence 1.020304050607... Each successive term affects the corresponding digit, reinforcing the observed pattern in the expansion.

Key Concepts

Geometric Series

A geometric series is one of the most fundamental concepts in mathematics and is particularly useful in understanding sequences that grow based on a constant ratio. The series is represented as:

• First term: a
• Common ratio: r
• General term: ar^n

The sum of an infinite geometric series when the ratio, \( |r| < 1 \), is given by:\[S = \frac{a}{1 - r}\].
In the context of our problem, we can relate it to a Taylor series expansion for the function \( f(x) = \frac{1}{1-x} \). We used this concept to transform the expression \( \frac{1}{0.98} = \frac{1}{1 - 0.02} \) into a geometric series: \( 1 + 0.02 + 0.02^2 + 0.02^3 + \cdots \).
This series breaks down the fraction into a sum of terms that reveals a repetitive pattern in its digits, as illustrated by the repeating decimal 1.02040816... Understanding geometric series is crucial for deciphering the patterns of decimals in such expansions.

Binomial Theorem

The binomial theorem is an essential tool in mathematics that helps in expanding expressions raised to a power. This theorem is expressed as:\[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\],
where \( \binom{n}{k} \) is a binomial coefficient.
In our problem, we applied this theorem to expand \( \left(\frac{1}{0.99}\right)^2 \) by recognizing it as \( (1 + 0.01)^2 \). Upon expansion, we have:

• First term: 1
• Second term: \(2 \times 0.01 = 0.02\)
• Third term: \(0.01^2 = 0.0001\)

These terms contribute to the sequence in the decimal representation: 1.020304050607... The ordered structure of terms from binomial expansion directly affects the resulting digits, showing how such expansions can provide a clear pattern in repetitive numerical sequences.

Polynomial Approximation

Polynomial approximation involves using polynomials to estimate more complex functions. This method is particularly beneficial because polynomials are easier to manipulate and calculate than most other types of mathematical functions.
Taylor series, a form of polynomial approximation, expresses a function as an infinite sum of terms calculated from the values of its derivatives at a particular point. For example, a Taylor series expansion around \(a\) is given as:

• Zero-th degree term: \( f(a) \)
• First degree term: \( f'(a)(x-a) \)
• Second degree term: \( \frac{f''(a)}{2!}(x-a)^2 \)

And it continues for higher-order derivatives. In approximating functions like \( \frac{1}{0.98} \) or \( \left(\frac{1}{0.99}\right)^2 \), the Taylor series provides polynomial expressions that mirror these complex functions closely, revealing underlying patterns in their decimals.

Function Expansion

Function expansion is a process that allows complex functions to be expressed as infinite series, typically involving simpler functions. This concept is integral in calculus, where expansions like Taylor and Maclaurin series are used to approximate functions.In such expansions, a function \( f(x) \) is represented as a series of derivatives of \( f \) evaluated at a certain point. This series takes the form:

• Base term: Constant evaluated function \( f(a) \)
• Linear term: First derivative contributing linearly
• Quadratic term: Second derivative contributing quadratically
• Cubic term: Third derivative contributing cubically

The advantage of function expansions is their ability to simplify the analysis of functions that are otherwise difficult to handle directly. By decomposing a function into a sequence of more manageable terms, it becomes easier to study and understand their behavior, specifically in terms of growth, convergence, and patterns such as those found in digit expansions of numbers.