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A function, \(y(x)\), has \(y(1)=3, y^{\prime}(1)=6, y^{\prime \prime}(1)=1\) and \(y^{(3)}(1)=-1\) (a) Estimate \(y(1.2)\) using a third-order Taylor polynomial. (b) Estimate \(y^{\prime}(1.2)\) using an appropriate second-order Taylor polynomial. [Hint: define a new variable, \(z\), given by \(z=y^{\prime}\).]

Short Answer

Expert verified
(a) \(y(1.2) \approx 4.219\), (b) \(y'(1.2) \approx 6.18\).

Step by step solution

01

Understand the Problem

We are given the value of a function and its derivatives at \(x=1\). We need to estimate \(y(1.2)\) using a third-order Taylor polynomial and estimate \(y'(1.2)\) using a second-order Taylor polynomial with a hint to define \(z = y'\).
02

Construct the Taylor Polynomial for y(x)

The third-order Taylor polynomial for \(y(x)\) about \(x=1\) is given by:\[T_3(x) = y(1) + y'(1)(x - 1) + \frac{y''(1)}{2!}(x - 1)^2 + \frac{y^{(3)}(1)}{3!}(x - 1)^3\]Using \(y(1)=3\), \(y'(1)=6\), \(y''(1)=1\), \(y^{(3)}(1)=-1\), substitute these values into the polynomial:\[T_3(x) = 3 + 6(x - 1) + \frac{1}{2}(x - 1)^2 - \frac{1}{6}(x - 1)^3\]
03

Evaluate y(1.2) Using the Taylor Polynomial

Substitute \(x=1.2\) into \(T_3(x)\):\[T_3(1.2) = 3 + 6(1.2 - 1) + \frac{1}{2}(1.2 - 1)^2 - \frac{1}{6}(1.2 - 1)^3\]Calculate each term:\[= 3 + 6(0.2) + \frac{1}{2}(0.2)^2 - \frac{1}{6}(0.2)^3\]\[= 3 + 1.2 + 0.02 - 0.001333\]\[= 4.218667\]Thus, \(y(1.2) \approx 4.219\).
04

Define z for Constructing Taylor Polynomial for y'

Let \(z = y'\), then the Taylor series for \(z(x)\) or \(y'(x)\) around \(x=1\) is:\[T_2(x) = z(1) + z'(1)(x - 1) + \frac{z''(1)}{2!}(x - 1)^2\]From given data: \(z(1) = y'(1) = 6\), \(z'(1) = y''(1) = 1\), and \(z''(1) = y^{(3)}(1) = -1\).
05

Construct and Evaluate Taylor Polynomial for y'

The Taylor polynomial for \(z(x) = y'(x)\) is:\[T_2(x) = 6 + 1(x - 1) - \frac{1}{2}(x - 1)^2\]Substitute \(x=1.2\) into \(T_2(x)\):\[T_2(1.2) = 6 + 1(0.2) - \frac{1}{2}(0.2)^2\]\[= 6 + 0.2 - 0.02\]\[= 6.18\]Thus, \(y'(1.2) \approx 6.18\).

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Key Concepts

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

Third-Order Polynomial
A third-order polynomial is an expression involving terms with degrees zero through three. These polynomials have the form:
  • Constant term: represents the value when the variable is zero.
  • Linear term: involves the variable to the first power. This term grows proportionally with the variable.
  • Quadratic term: involves the variable squared. It grows faster than the linear term.
  • Cubic term: involves the variable to the third power. It can represent more complex growth patterns.
In the context of a Taylor series, a third-order polynomial helps approximate a function's behavior near a point. For example, if you have a function known at point \(x=1\), you can approximate its value at a nearby point like \(x=1.2\). The key idea is to use the known function values and derivatives to construct a polynomial that mimics the function's initial behavior. This approximation simplifies complex functions into polynomials, making them easier to work with.
Derivative Estimation
Estimating derivatives using Taylor polynomials is a powerful technique. When we know the derivative at a certain point and need to estimate it at another, we can use Taylor polynomials. In the given exercise, we use a second-order polynomial because it involves up to the second derivative.
  • First, the initial derivative gives us a starting point.
  • The first derivative of the polynomial adds precision as you move slightly from your known point.
  • Using the second derivative's effect on the change in slope helps refine the estimate further.
In the context above, we have \(z = y'\), and constructing a Taylor polynomial to approximate the derivative at \(x = 1.2\) leads us to use a combination of initial slope and adjustments for curving behavior (second derivative). The formula effectively captures how the derivative changes slightly as you move from \(x = 1\) to \(x = 1.2\).
Function Approximation
Function approximation with Taylor series is about expressing a complex, potentially non-linear function with easier-to-handle polynomial terms. It relies on the idea that if you know the function and its derivatives at a specific point, you can predict its values nearby.When using Taylor polynomials for approximation, there are benefits:
  • Simple to evaluate and manipulate.
  • The polynomial encapsulates key aspects of the function's behavior around the point of expansion.
  • It's a great way to handle functions when analytical solutions are difficult.
In practice, for the original exercise, the third-order polynomial constructed gives a good approximation of the function near \(x=1\). You start with the function value and add pieces that reflect the slope, curvature, and further changes captured by higher derivatives. Function approximation through Taylor series thus serves as a powerful tool in calculus for understanding and predicting function behavior.

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