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Find the Taylor series generated by \(f\) at \(x=a.\) $$f(x)=3 x^{5}-x^{4}+2 x^{3}+x^{2}-2, a=-1$$

Short Answer

Expert verified
The Taylor series for \( f(x) \) at \( x = -1 \) is \(-7 + 23(x + 1) - 41(x + 1)^2 + 36(x + 1)^3 - 16(x + 1)^4 + 3(x + 1)^5 \).

Step by step solution

01

Calculate the derivatives of the function

To find the Taylor series, we need to calculate the derivatives of the function \( f(x) = 3x^5 - x^4 + 2x^3 + x^2 - 2 \). We will calculate the first few derivatives since the degree of the polynomial is 5.\[ f'(x) = 15x^4 - 4x^3 + 6x^2 + 2x \]\[ f''(x) = 60x^3 - 12x^2 + 12x + 2 \]\[ f'''(x) = 180x^2 - 24x + 12 \]\[ f''''(x) = 360x - 24 \]\[ f'''''(x) = 360 \]\[ f''''''(x) = 0 \]These calculations are sufficient, as higher derivatives will be zero for a polynomial of degree 5.
02

Evaluate derivatives at x=a

Next, evaluate the derivatives at \(x = a = -1\):\[ f(-1) = 3(-1)^5 - (-1)^4 + 2(-1)^3 + (-1)^2 - 2 = -3 - 1 - 2 + 1 - 2 = -7 \]\[ f'(-1) = 15(-1)^4 - 4(-1)^3 + 6(-1)^2 + 2(-1) = 15 + 4 + 6 - 2 = 23 \]\[ f''(-1) = 60(-1)^3 - 12(-1)^2 + 12(-1) + 2 = -60 - 12 - 12 + 2 = -82 \]\[ f'''(-1) = 180(-1)^2 - 24(-1) + 12 = 180 + 24 + 12 = 216 \]\[ f''''(-1) = 360(-1) - 24 = -360 - 24 = -384 \]\[ f'''''(-1) = 360 \]We do not need to calculate beyond this.
03

Write the Taylor series formula

The Taylor series for a function \( f \) around \( x = a \) is given by \[ T(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n \]. In our case, \( a = -1 \).
04

Substitute evaluated derivatives into the Taylor series

Using the values from Step 2, substitute into the Taylor series:\[ T(x) = f(-1) + f'(-1)(x + 1) + \frac{f''(-1)}{2!}(x + 1)^2 + \frac{f'''(-1)}{3!}(x + 1)^3 + \frac{f''''(-1)}{4!}(x + 1)^4 + \frac{f'''''(-1)}{5!}(x + 1)^5 \]Substitute the evaluated derivative values:\[ = -7 + 23(x + 1) - \frac{82}{2}(x + 1)^2 + \frac{216}{6}(x + 1)^3 - \frac{384}{24}(x + 1)^4 + \frac{360}{120}(x + 1)^5 \]
05

Simplify the expression

Simplify each term in the series:\[ -7 + 23(x + 1) - 41(x + 1)^2 + 36(x + 1)^3 - 16(x + 1)^4 + 3(x + 1)^5 \]This expression represents the Taylor series for \( f(x) \) at \( x = -1 \).

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Key Concepts

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

Polynomial Derivatives
Understanding polynomial derivatives is essential to finding a Taylor series. A polynomial is a mathematical expression formed by adding together terms consisting of a variable raised to different powers. The derivative of a polynomial function is the key to understanding how the function behaves at different points. It measures the rate at which the function's value changes as the input changes.

To compute the derivative of a polynomial, pay attention to the power of each term:
  • Reduce the power by one.
  • Multiply the coefficient by the original power.
Taking derivatives multiple times reveals how quickly a function's slope changes: - The first derivative gives the slope of the function. - The second derivative reveals concavity or the direction of the curve. - Higher-order derivatives can provide deeper insights into the function's progression. In the original exercise, the function is a fifth-degree polynomial, so you will compute up to the fifth derivative. Derivatives beyond the fifth are zero because the polynomial will have no more terms left to differentiate.
Polynomial Evaluation
Once derivatives are calculated, the next step is to evaluate these derivatives at a specific point. This point, called the center of the Taylor series, is given as \( x = a \), and here it is \( a = -1 \). Evaluating a polynomial at a point means finding the value of the polynomial when the variable is equal to this specific value.

For example, if we have the derivative function \( f'(x) = 15x^4 - 4x^3 + 6x^2 + 2x \) and need to evaluate it at \( x = -1 \), substitute \( -1 \) into every instance of \( x \) and compute:
  • \( f'(-1) = 15(-1)^4 - 4(-1)^3 + 6(-1)^2 + 2(-1) \)
  • Simplify each term step-by-step.
Evaluate each derivative that was calculated in the earlier step in a similar fashion. This gives the coefficients for the Taylor series expansion, which measure how the function changes around \( x = -1 \) at different degrees of approximation.
Factorials in Taylor Series
Factorials play a crucial role in forming the coefficients of a Taylor series expansion. A factorial is the product of all positive integers up to a given number. Represented by the symbol \( n! \), where \( n \) is the integer, the factorial can be calculated as follows:
  • \( n! = n \times (n-1) \times (n-2) \times \ldots \times 1 \)
  • Examples: \( 3! = 3 \times 2 \times 1 = 6 \); \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
In the Taylor series formula, each term is divided by the factorial of the term's degree. For instance, the term involving the second derivative is divided by \( 2! = 2 \), the term involving the third derivative by \( 3! = 6 \), and so forth. Factorials help to normalize the terms of the series, ensuring that even as the power increases, the terms decrease in size. This process reveals a more accurate approximation of the polynomial based on how quickly the derivatives change.

Ultimately, factorials ensure the convergence properties of the Taylor series, allowing it to approximate not just polynomials but also more complex functions.

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