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Find the Taylor series generated by \(f\) at \(x=a\). $$f(x)=1 /(1-x)^{3}, \quad a=0$$

Short Answer

Expert verified
The Taylor series is \( \sum_{n=0}^{\infty} \binom{n+2}{2} x^n \).

Step by step solution

01

Understanding the Formula

The Taylor series for a function \(f(x)\) centered at \(x=a\) is given by:\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \] In this exercise, \(a = 0\), so we need to calculate the Taylor series at \(x = 0\).
02

Calculate the Function Value at a

Calculate \(f(0)\):Given \(f(x) = \frac{1}{(1-x)^3}\), evaluate \(f(0)\):\[ f(0) = \frac{1}{(1-0)^3} = 1 \]
03

Calculate the First Derivative and Evaluate

Find the first derivative \(f'(x)\):\[ f'(x) = \frac{d}{dx} \left( (1-x)^{-3} \right) = 3(1-x)^{-4} \] Evaluate \(f'(0)\):\[ f'(0) = 3(1-0)^{-4} = 3 \]
04

Calculate the Second Derivative and Evaluate

Find the second derivative \(f''(x)\):\[ f''(x) = \frac{d}{dx} (3(1-x)^{-4}) = 12(1-x)^{-5} \] Evaluate \(f''(0)\):\[ f''(0) = 12(1-0)^{-5} = 12 \]
05

Calculate the Third Derivative and Evaluate

Find the third derivative \(f'''(x)\):\[ f'''(x) = \frac{d}{dx} (12(1-x)^{-5}) = 60(1-x)^{-6} \] Evaluate \(f'''(0)\):\[ f'''(0) = 60(1-0)^{-6} = 60 \]
06

Assemble the Taylor Series

Using the values calculated:- \(f(0) = 1\)- \(f'(0) = 3\)- \(f''(0) = 12\)- \(f'''(0) = 60\)The Taylor series becomes:\[ f(x) = 1 + 3x + \frac{12}{2!}x^2 + \frac{60}{3!}x^3 + \ldots \]Simplify:\[ f(x) = 1 + 3x + 6x^2 + 10x^3 + \ldots \]
07

Write the General Term

Notice the pattern of the coefficients: they follow the sequence \(1, 3, 6, 10, \ldots\), which are the triangular numbers. Thus, the general term of the series is:\[ \sum_{n=0}^{\infty} \binom{n+2}{2} x^n \]

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Key Concepts

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

Function Derivatives
Derivatives are fundamental tools in calculus used to measure how a function changes as its input changes. The first derivative, denoted as \( f'(x) \), gives the rate of change or the slope of the function at any given point. Subsequent derivatives, like the second \( f''(x) \) and third \( f'''(x) \) derivatives, provide deeper insights. They help in understanding the curvature and the acceleration of the function's graph.

In our example with \( f(x) = \frac{1}{(1-x)^3} \), the derivatives were calculated at \( x = 0 \) to construct the Taylor series. The process involved using the power rule for differentiation and recognizing a pattern in the derivatives' results:
  • The first derivative is \( f'(x) = 3(1-x)^{-4} \), which simplifies to \( f'(0) = 3 \) at \( x=0 \).
  • The second derivative is \( f''(x) = 12(1-x)^{-5} \), resulting in \( f''(0) = 12 \).
  • The third derivative is \( f'''(x) = 60(1-x)^{-6} \), giving \( f'''(0) = 60 \).
Calculating these values is crucial for creating the Taylor series, as they help define each term’s coefficient.
Binomial Theorem
The binomial theorem is a powerful mathematical tool that provides a formula for expanding expressions raised to a power, such as \((a + b)^n\). This theorem suggests that each term in the expansion can be derived using binomial coefficients, given by \( \binom{n}{k} \).When creating a Taylor series, the binomial theorem often helps recognize patterns in the series' coefficients. In our Taylor series example, the coefficients directly relate to the binomial theorem through a pattern involving combinations. Specifically, each term's coefficient aligns with the binomial coefficient \( \binom{n+2}{2} \), indicating a deep connection between binomial expansion and the series.The power of the binomial theorem is in simplifying expressions and spotting underlying patterns. Recognizing these coefficients' pattern helps not only in assembling the series but also in understanding the function's behavior and relationships.
Triangular Numbers
Triangular numbers are a sequence of numbers that can form an equilateral triangle. The sequence begins with 1, 3, 6, 10, and continues onward. Each triangular number corresponds to dots arranged in a triangular grid, representing the sum of the natural numbers up to a certain point.In our Taylor series exercise, you may notice a pattern in the coefficients: \( 1, 3, 6, 10, \ldots \). This pattern directly maps to the sequence of triangular numbers. Such a pattern emerges from the structure of the function and its derivatives.Recognizing triangular numbers enables a deeper understanding of sequences and series in mathematics. For example, the nth triangular number can be calculated with the formula \( T_n = \frac{n(n+1)}{2} \). Identifying these numbers within mathematical problems like Taylor series underscores the interconnectedness of different mathematical concepts and how recognizing such patterns can simplify complex problems.

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Most popular questions from this chapter

Assume that each sequence converges and find its limit. \(2,2+\frac{1}{2}, 2+\frac{1}{2+\frac{1}{2}}, 2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}, \dots\)

A patient takes a 300 mg tablet for the control of high blood pressure every morning at the same time. The concentration of the drug in the patient's system decays exponentially at a constant hourly rate of \(k=0.12\) a. How many milligrams of the drug are in the patient's system just before the second tablet is taken? Just before the third tablet is taken? b. In the long run, after taking the medication for at least six months, what quantity of drug is in the patient's body just before taking the next regularly scheduled morning tablet?

Assume that each sequence converges and find its limit. \(a_{1}=2, \quad a_{n+1}=\frac{72}{1+a_{n}}\)

Linearizations at inflection points Show that if the graph of a twice- differentiable function \(f(x)\) has an inflection point at \(x=a,\) then the linearization of \(f\) at \(x=a\) is also the quadratic approximation of \(f\) at \(x=a\). This explains why tangent lines fit so well at inflection points.

To construct this set, we begin with the closed interval \([0,1] .\) From that interval, remove the middle open interval (1/3, 2/3). leaving the two closed intervals [ 0, 1/3 ] and [2/3, 1 ]. At the second step we remove the open middle third interval from each of those remaining. From \([0.1 / 3]\) we remove the open interval \((1 / 9,2 / 9),\) and from \([2 / 3,1]\) we remove \((7 / 9,8 / 9),\) leaving behind the four closed intervals \([0,1 / 9]\) \([2 / 9,1 / 3],[2 / 3,7 / 9],\) and \([8 / 9,1] .\) At the next step, we remove the middle open third interval from each closed interval left behind, so \((1 / 27,2 / 27)\) is removed from \([0,1 / 9],\) leaving the closed intervals \([0,1 / 27]\) and \([2 / 27,1 / 9] ;(7 / 27,8 / 27)\) is removed from \([2 / 9,1 / 3]\), leaving behind \([2 / 9,7 / 27]\) and \([8 / 27,1 / 3],\) and so forth. We continue this process repeatedly without stopping, at each step removing the open third interval from every closed interval remaining behind from the preceding step. The numbers remaining in the interval \([0,1],\) after all open middle third intervals have been removed, are the points in the Cantor set (named after Georg Cantor, \(1845-1918\) ). The set has some interesting properties. a. The Cantor set contains infinitely many numbers in [0,1] List 12 numbers that belong to the Cantor set. b. Show, by summing an appropriate geometric series, that the total length of all the open middle third intervals that have been removed from [0,1] is equal to 1

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