/*! 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 13 The thinner curve is obtained us... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 13

The thinner curve is obtained using a numerical solver, while the thicker curve is the graph of the Taylor polynomial. We look for a solution of the form \\[ y(x)=y(0)+y^{\prime}(0) x+\frac{1}{2 !} y^{\prime \prime}(0) x^{2}+\frac{1}{3 !} y^{\prime \prime \prime}(0) x^{3}+\frac{1}{4 !} y^{(4)}(0) x^{4}+\frac{1}{5 !} y^{(5)}(0) x^{5} \\] From \(y^{\prime \prime}(x)=x+y^{2}\) we compute \\[ \begin{aligned} y^{\prime \prime \prime}(x) &=1+2 y y^{\prime} \\ y^{(4)}(x) &=2 y y^{\prime \prime}+2\left(y^{\prime}\right)^{2} \\ y^{(5)}(x) &=2 y y^{\prime \prime \prime}+6 y^{\prime} y^{\prime \prime} \end{aligned} \\] Using \(y(0)=1\) and \(y^{\prime}(0)=1\) we find \\[ y^{\prime \prime}(0)=1, \quad y^{\prime \prime \prime}(0)=3, \quad y^{(4)}(0)=4, \quad y^{(5)}(0)=12 \\] An approximate solution is \\[ y(x)=1+x+\frac{1}{2} x^{2}+\frac{1}{2} x^{3}+\frac{1}{6} x^{4}+\frac{1}{10} x^{5} \\]

Short Answer

Expert verified
The Taylor polynomial is: \(y(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{2}x^3 + \frac{1}{6}x^4 + \frac{1}{10}x^5\).

Step by step solution

01

Understanding the Taylor Polynomial

A Taylor polynomial approximation is used to represent a function around a point, usually zero if not specified, using its derivatives. In this exercise, we need to utilize derivatives calculated at a specific point (here, at 0) to construct the polynomial.
02

Analyzing Given Information

We are provided with the derivatives of the function at 0: \(y(0) = 1\), \(y'(0) = 1\), \(y''(0) = 1\), \(y'''(0) = 3\), \(y^{(4)}(0) = 4\), and \(y^{(5)}(0) = 12\). These values will be used in conjunction with Taylor's formula to find an approximation of \(y(x)\).
03

Formulating Taylor Polynomial

Using the Taylor polynomial formula: \[ y(x) = y(0) + y'(0)x + \frac{1}{2}y''(0)x^2 + \frac{1}{3!}y'''(0)x^3 + \frac{1}{4!}y^{(4)}(0)x^4 + \frac{1}{5!}y^{(5)}(0)x^5 \]. Substitute the given values for the derivatives to obtain: \[ y(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \frac{1}{24}x^4 + \frac{1}{120}x^5 \].
04

Simplifying and Comparing with Given Approximation

Calculate each term of the polynomial using the derivatives: \(0! = 1! = 1\), \(2! = 2\), \(3! = 6\), \(4! = 24\), and \(5! = 120\). From these computations, the Taylor polynomial simplifies to \[ y(x) = 1 + x + \frac{1}{2}x^2 + \frac{1}{2}x^3 + \frac{1}{6}x^4 + \frac{1}{10}x^5 \]. This polynomial matches the given approximate solution.

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Key Concepts

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

Numerical Solvers
Numerical solvers are essential tools in mathematics and engineering. They help solve complex equations that may not have straightforward analytical solutions. These solvers can handle equations involving derivatives, providing approximate solutions to otherwise difficult problems.

Here's why numerical solvers are important:
  • They allow for solutions to differential equations that are not easily expressed in closed forms.
  • They help in approximating functions around points, especially when functions become too complex.
  • They enable exploratory analysis, quickly testing various initial conditions or parameters.
When working with them, remember that numerical solvers often provide approximate solutions rather than exact ones. This is useful in practical applications where a close estimate is sufficient to understand the behavior of the system being studied.
Derivative Calculation
Derivative calculation plays a vital role in finding the behavior of functions. By calculating derivatives, we can understand the rates of change and slopes of functions at various points. These calculations are pivotal when constructing Taylor polynomials.

Derivatives give us tools to:
  • Find slopes and rates of change in functions, revealing trends and movements.
  • Determine curvature and identify points of inflection.
  • Approximate complex functions using simpler polynomial forms, like Taylor and Maclaurin series.
In our exercise, derivatives at a specific point (x = 0) are calculated to find the Taylor polynomial of a function. The provided derivatives \(y'(0), y''(0),\) and so on, are calculated to provide insight into how the function behaves close to x = 0.
Polynomial Approximation
A polynomial approximation is a method of estimating a function using a polynomial equation. Specifically, the Taylor polynomial depicts a function by using a sum of its derivatives evaluated at a central point. This is useful when dealing with complex functions, allowing them to be represented more simply in the form of polynomials.

The advantages of polynomial approximations include:
  • Easing calculations for complex functions by using simpler polynomial terms.
  • Providing a clear visual representation of the function's behavior around a point.
  • Helping in numerical analysis and computer algorithms by offering easier-to-compute alternatives.
In our example, the Taylor polynomial, centered around x = 0, represents the approximation of the function. The polynomial makes use of the derivatives calculated at that point to show how the function behaves close to x = 0, resulting in a curve that approximates the function with increased precision as more terms are added.
Series Expansion
Series expansion involves expressing a function as a sum of terms, often to simplify calculations or better analyze its behavior. The Taylor series is a common example, expanding functions into infinite sums centered around a specific point.

Key aspects of series expansion include:
  • The ability to transform complex functions into a series of simpler terms for easier computation.
  • Improving accuracy in function approximation as more terms are added to the series.
  • Providing insights into the function's local behavior near the central point of the expansion.
In practical applications, like the exercise provided, series expansion helps generate numerical solutions with polynomial forms. Although only a finite number of terms from the Taylor series are used, this provides a good approximation of the function's behavior near x = 0.

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Most popular questions from this chapter

$$\text { since } x=\frac{\sqrt{85}}{4} \sin (4 t-0.219)-\frac{\sqrt{17}}{2} e^{-2 t} \sin (4 t-2.897), \text { the amplitude approaches } \sqrt{85} / 4 \text { as } t \rightarrow \infty$$.

Using the general solution \(y=c_{1} \cos \sqrt{3} x+c_{2} \sin \sqrt{3} x+2 x,\) the boundary conditions \(y(0)+y^{\prime}(0)=0, y(1)=0\) yield the system $$\begin{array}{r} c_{1}+\sqrt{3} c_{2}+2=0 \\ c_{1} \cos \sqrt{3}+c_{2} \sin \sqrt{3}+2=0 \end{array}$$ Solving gives $$c_{1}=\frac{2(-\sqrt{3}+\sin \sqrt{3})}{\sqrt{3} \cos \sqrt{3}-\sin \sqrt{3}} \quad \text { and } \quad c_{2}=\frac{2(1-\cos \sqrt{3})}{\sqrt{3} \cos \sqrt{3}-\sin \sqrt{3}}$$ Thus, $$y=\frac{2(-\sqrt{3}+\sin \sqrt{3}) \cos \sqrt{3} x}{\sqrt{3} \cos \sqrt{3}-\sin \sqrt{3}}+\frac{2(1-\cos \sqrt{3}) \sin \sqrt{3} x}{\sqrt{3} \cos \sqrt{3}-\sin \sqrt{3}}+2 x$$

For \(\lambda \leq 0\) the only solution of the boundary-value problem is \(y=0 .\) For \(\lambda=\alpha^{2}>0\) we have $$y=c_{1} \cos \alpha x+c_{2} \sin \alpha x$$ Now \(y(-\pi)=y(\pi)=0\) implies \\[ \begin{array}{l} c_{1} \cos \alpha \pi-c_{2} \sin \alpha \pi=0 \\ c_{1} \cos \alpha \pi+c_{2} \sin \alpha \pi=0 \end{array} \\] This homogeneous system will have a nontrivial solution when \\[ \left|\begin{array}{rr} \cos \alpha \pi & -\sin \alpha \pi \\ \cos \alpha \pi & \sin \alpha \pi \end{array}\right|=2 \sin \alpha \pi \cos \alpha \pi=\sin 2 \alpha \pi=0 \\] Then \\[ 2 \alpha \pi=n \pi \quad \text { or } \quad \lambda=\alpha^{2}=\frac{n^{2}}{4} ; \quad n=1,2,3, \ldots \\] When \(n=2 k-1\) is odd, the eigenvalues are \((2 k-1)^{2} / 4 .\) since \(\cos (2 k-1) \pi / 2=0\) and \(\sin (2 k-1) \pi / 2 \neq 0\) we see from either equation in (1) that \(c_{2}=0 .\) Thus, the eigenfunctions corresponding to the eigenvalues \((2 k-1)^{2} / 4\) are \(y=\cos (2 k-1) x / 2\) for \(k=1,2,3, \ldots .\) Similarly, when \(n=2 k\) is even, the eigenvalues are \(k^{2}\) with corresponding eigenfunctions \(y=\sin k x\) for \(k=1,2,3, \ldots\)

Using a CAS to solve the auxiliary equation \(m^{4}+2 m^{2}-m+2=0\) we find \(m_{1}=1 / 2+\sqrt{3} i / 2, m_{2}=1 / 2-\sqrt{3} i / 2\) \(m_{3}=-1 / 2+\sqrt{7} i / 2,\) and \(m_{4}=-1 / 2-\sqrt{7} i / 2 .\) The general solution is $$y=e^{x / 2}\left(c_{1} \cos \frac{\sqrt{3}}{2} x+c_{2} \sin \frac{\sqrt{3}}{2} x\right)+e^{-x / 2}\left(c_{3} \cos \frac{\sqrt{7}}{2} x+c_{4} \sin \frac{\sqrt{7}}{2} x\right)$$

(a) The boundary-value problem is \\[ \frac{d^{4} y}{d x^{4}}+\lambda \frac{d^{2} y}{d x^{2}}=0, \quad y(0)=0, y^{\prime \prime}(0)=0, y(L)=0, y^{\prime}(L)=0 \\] where \(\lambda=\alpha^{2}=P / E I .\) The solution of the differential equation is \(y=c_{1} \cos \alpha x+c_{2} \sin \alpha x+c_{3} x+c_{4}\) and the conditions \(y(0)=0, y^{\prime \prime}(0)=0\) yield \(c_{1}=0\) and \(c_{4}=0 .\) Next, by applying \(y(L)=0, y^{\prime}(L)=0\) to \(y=c_{2} \sin \alpha x+c_{3} x\) we get the system of equations $$\begin{aligned} c_{2} \sin \alpha L+c_{3} L &=0 \\ \alpha c_{2} \cos \alpha L+c_{3} &=0 \end{aligned}$$.To obtain nontrivial solutions \(c_{2}, c_{3},\) we must have the determinant of the coefficients equal to zero: \\[ \left|\begin{array}{rr} \sin \alpha L & L \\ \alpha \cos \alpha L & 1 \end{array}\right|=0 \quad \text { or } \quad \tan \beta=\beta \\] where \(\beta=\alpha L .\) If \(\beta_{n}\) denotes the positive roots of the last equation, then the eigenvalues are found from \(\beta_{n}=\alpha_{n} L=\sqrt{\lambda_{n}} L\) or \(\lambda_{n}=\left(\beta_{n} / L\right)^{2} .\) From \(\lambda=P / E I\) we see that the critical loads are \(P_{n}=\beta_{n}^{2} E I / L^{2}\) With the aid of a CAS we find that the first positive root of \(\tan \beta=\beta\) is (approximately) \(\beta_{1}=4.4934,\) and so the Euler load is (approximately) \(P_{1}=20.1907 E I / L^{2} .\) Finally, if we use \(c_{3}=-c_{2} \alpha \cos \alpha L,\) then the deflection curves are $$y_{n}(x)=c_{2} \sin \alpha_{n} x+c_{3} x=c_{2}\left[\sin \left(\frac{\beta_{n}}{L} x\right)-\left(\frac{\beta_{n}}{L} \cos \beta_{n}\right) x\right]$$ (b) With \(L=1\) and \(c_{2}\) appropriately chosen, the general shape of the first buckling mode, \\[ y_{1}(x)=c_{2}\left[\sin \left(\frac{4.4934}{L} x\right)-\left(\frac{4.4934}{L} \cos (4.4934)\right) x\right] \\] is shown below.
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