/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Find the Taylor series generated... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=e^{x}, \quad a=2\)

Short Answer

Expert verified
The Taylor series for \( e^x \) at \( x=2 \) is \( \sum_{n=0}^{\infty} \frac{e^2}{n!} (x-2)^n. \)

Step by step solution

01

Understand the Taylor Series Formula

The Taylor series of a function \( f \) at a point \( a \) is given by the formula: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \] where \( f^{(n)}(a) \) is the \( n \)-th derivative of \( f \) evaluated at \( a. \)
02

Evaluate Derivatives of \( f(x) = e^x \)

The function \( f(x) = e^x \) has derivatives that repeat as \( f^{(n)}(x) = e^x \) for all \( n. \) Thus, \( f^{(n)}(a) = e^a \) for all \( n. \) Since we're evaluating at \( a = 2, \) then \( f^{(n)}(2) = e^2 \) for all \( n. \)
03

Substitute into the Taylor Series Formula

Substitute \( f^{(n)}(2) = e^2 \) into the Taylor series formula: \[ f(x) = \sum_{n=0}^{\infty} \frac{e^2}{n!} (x-2)^n \] This represents the Taylor series for \( e^x \) centered at \( x = 2. \)

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Key Concepts

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

Derivatives of Functions
In calculus, understanding the derivative of a function is crucial. The derivative measures how a function changes as its input changes. For the exponential function \( f(x) = e^x \), the derivative is unique in the sense that it remains the same after differentiation. That is, each derivative of \( f(x) = e^x \) is also \( e^x \).
This property makes exponential functions quite straightforward to work with when finding derivatives multiple times.
  • The first derivative \( f'(x) \) is \( e^x \).
  • The second derivative \( f''(x) \) is \( e^x \).
  • In general, all the higher order derivatives \( f^{(n)}(x) \) will be \( e^x \).
This principle simplifies the process of constructing a Taylor series, as the derivatives at the point of expansion \( a \) are simply \( e^a \). This repetition of the exponential function is an important property in calculus when dealing with series, especially Taylor series.
Exponential Function
The exponential function \( f(x) = e^x \) is one of the most important and fascinating functions in mathematics. Its base, \( e \), is an irrational constant approximately equal to 2.71828. The exponential function has unique properties that make it very useful: it grows rapidly and has a distinctive self-similarity in its derivatives.
The function \( e^x \) can be applied in various contexts such as compound interest, population growth, and natural processes that follow an exponential trend.
  • It is defined in the real domain and continues to grow exponentially as \( x \) increases.
  • As mentioned, its derivative is itself, which makes it very useful in differential equations and series expansions.
The exponential function's 'self-derivative' nature particularly impacts its Taylor series expansion. At any point \( a \), the function and each of its derivatives will evaluate to \( e^a \), significantly influencing the terms of the series.
Series Expansion
A series expansion is a way of writing a function as an infinite sum of terms. The Taylor series is a specific type of series expansion used to approximate functions locally around a point \( a \). The general formula for the Taylor series of \( f(x) \) at \( a \) is:\[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \]This means each term in the series is composed of derivatives computed at \( a \), multiplied by powers of \( x-a \), divided by factorials. When dealing with the exponential function \( e^x \), the simplicity of its derivatives allows us to write:\[ f(x) = \sum_{n=0}^{\infty} \frac{e^2}{n!} (x-2)^n \]
  • This expansion centers the series at \( x=2 \).
  • Each term \( \frac{e^2}{n!} (x-2)^n \) involves calculating powers of \( (x-2) \) and dividing by \( n! \), making it straightforward computationally once \( e^2 \) is known.
This representation is not only exact but captures the essence of how Taylor series can approximate functions smoothly across an interval. It's a powerful tool for understanding local behavior of functions.

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Most popular questions from this chapter

Sequences generated by Newton's method Newton's method, applied to a differentiable function \(f(x),\) begins with a starting value \(x_{0}\) and constructs from it a sequence of numbers \(\left\\{x_{n}\right\\}\) that under favorable circumstances converges to a zero of \(f .\) The recursion formula for the sequence is $$ x_{n+1}=x_{n}-\frac{f\left(x_{n}\right)}{f^{\prime}\left(x_{n}\right)} $$ a. Show that the recursion formula for \(f(x)=x^{2}-a, a>0\) can be written as \(x_{n+1}=\left(x_{n}+a / x_{n}\right) / 2\) b. Starting with \(x_{0}=1\) and \(a=3,\) calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain.

Is it true that a sequence \(\left\\{a_{n}\right\\}\) of positive numbers must converge if it is bounded from above? Give reasons for your answer.

Replace \(x\) by \(-x\) in the Taylor series for \(\ln (1+x)\) to obtain a series for \(\ln (1-x)\) . Then subtract this from the Taylor series for \(\ln (1+x)\) to show that for \(|x|<1\), \begin{equation} \ln \frac{1+x}{1-x}=2\left(x+\frac{x^{3}}{3}+\frac{x^{5}}{5}+\cdots\right). \end{equation}

Use the Taylor series generated by \(e^{x}\) at \(x=a\) to show that $$e^{x}=e^{a}\left[1+(x-a)+\frac{(x-a)^{2}}{2 !}+\cdots\right].$$

Which of the series in Exercises 13 46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{2}{1+e^{n}} $$

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