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Apply a function to a vector. Given a vector \(v=(2,3,-1)\) and a function \(f(x)=x^{3}+x e^{x}+1\), apply \(f\) to each element in \(v .\) Then calculate \(f(v)\) as \(v^{3}+v * e^{v}+1\) using vector computing rules. Show that the two results are equal.

Short Answer

Expert verified
The results using both methods are equal for each component.

Step by step solution

01

Evaluate Function for Each Element

Apply the function \( f(x) = x^3 + x e^x + 1 \) to each element of the vector \( v = (2, 3, -1) \).\\[ f(2) = 2^3 + 2e^2 + 1 \] \\[ f(3) = 3^3 + 3e^3 + 1 \] \\[ f(-1) = (-1)^3 + (-1)e^{-1} + 1 \]
02

Compute Function Values

Calculate each function\'s value from Step 1. \- For \( f(2) \): \ \[ 2^3 = 8, \, 2e^2 \approx 2 \times 7.389 = 14.778, \, \] \ \[ f(2) = 8 + 14.778 + 1 = 23.778 \] \- For \( f(3) \): \ \[ 3^3 = 27, \, 3e^3 \approx 3 \times 20.085 = 60.255, \, \] \ \[ f(3) = 27 + 60.255 + 1 = 88.255 \] \- For \( f(-1) \): \ \[ (-1)^3 = -1, \, (-1)e^{-1} \approx -0.368, \, \] \ \[ f(-1) = -1 - 0.368 + 1 = -0.368 \]
03

Compute Vector Component Expressions

Use the expression \( v^3 + v * e^v + 1 \) for each vector component separately.\- \( 2^3 + 2e^2 + 1 = 23.778 \) \- \( 3^3 + 3e^3 + 1 = 88.255 \) \- \( (-1)^3 + (-1)e^{-1} + 1 = -0.368 \)
04

Compare Results

Check if the manually computed function values and the vector component-based calculations are equal.\- For \( 2 \), results: \( 23.778 \).\- For \( 3 \), results: \( 88.255 \).\- For \( -1 \), results: \( -0.368 \).\Both methods give the same results for each component, confirming they are equal.

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Key Concepts

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

Function Application
When we talk about function application, we mean using a mathematical function on each element of a given set, like a vector. This process involves substituting the elements into a function and calculating the output.
  • Suppose you have a vector \( v = (2, 3, -1) \) and a function \( f(x) = x^3 + x e^x + 1 \).
  • You "apply" the function \( f \) to each element of \( v \), meaning you replace \( x \) in the function with 2, 3, and -1 separately, then compute the resulting value.
  • This way, you evaluate the function for each individual element of the vector. This is a crucial operation in vector-based calculations where each component of the vector is processed individually.
Understanding this concept helps in managing more complex operations involving vectors in mathematics.
Exponential Functions
An exponential function is a mathematical expression where a constant base is raised to a variable exponent. In calculus and algebra, exponential functions are common because they describe growth and decay processes.
  • In our exercise, the function \( f(x) = x^3 + x e^x + 1 \) includes an exponential term: \( x e^x \).
  • The term \( e^x \) is the natural exponential function, where \( e \) (approximately 2.718) is the base. Since \( e \) is an irrational number, values of \( e^x \) grow rapidly as \( x \) increases.
  • The presence of \( e^x \) in the function changes its behavior significantly, affecting both the value and the rate of change of the function outputs for the vector elements when evaluated.
Grasping how exponential functions operate is key to understanding their influence on calculations like those demonstrated in the exercise.
Vector Components
Vectors consist of multiple components, each representing a specific dimension. In our context, we're dealing with a three-dimensional vector \( v = (2, 3, -1) \), where each number is a vector component.
  • Each component of a vector can be processed independently or in conjunction with others, depending on the operation involved.
  • In this exercise, the function \( f \) is applied to each component, producing a set of results that correspond to each vector component.
  • The resulting values are calculated by performing operations individually, which demonstrates how vector operations can be efficiently managed by treating components independently.
Understanding vector components and their processing is fundamental in performing complex vector operations across various applications in mathematics and physics.
Equality Verification
Verifying equality is about ensuring that two expressions or values give the same result. It's an essential step in mathematical problem-solving to confirm our calculations are correct and consistent.
  • In this exercise, the goal is to confirm that applying the function \( f(x) = x^3 + x e^x + 1 \) for each vector component yields the same result as computing \( v^3 + v * e^v + 1 \) directly.
  • Here, after calculating both versions, we check if the outcomes for \( 2 \), \( 3 \), and \( -1 \) are identical, which they are: \( 23.778, 88.255, \) and \(-0.368\) respectively.
  • This equality shows both methods are valid and interchangeable under the specific conditions of this problem.
Mastering equality verification ensures the reliability of mathematical computations across different approaches.

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Most popular questions from this chapter

Plot a formula for several parameters. Make a program that reads a set of \(v_{0}\) values from the command line and plots the corresponding curves \(y(t)=v_{0} t-0.5 g t^{2}\) in the same figure (set \(g=9.81\) ). Let \(t \in\left[0,2 v_{0} / g\right]\) for each curve, which implies that you need a different vector of \(t\) coordinates for each curve. Name of program file: plot_bal12.py.

Simulate by hand a vectorized expression. Suppose \(\mathrm{x}\) and \(\mathrm{t}\) are two arrays of the same length, entering a vectorized expression, $$ y=\cos (\sin (x))+\exp (1 / t) $$ If \(\mathrm{x}\) holds two elements, 0 and 2, and \(\mathrm{t}\) holds the elements 1 and 1.5, calculate by hand (using a calculator) the y array. Thereafter, write a program that mimics the series of computations you did by hand (typically a sequence of operations of the kind we listed on page 182 - use explicit loops, but at the end you can use Numerical Python functionality to check the results). Name of program file: simulate_vector_computing.py.

Animate a planet's orbit. A planet's orbit around a star has the shape of an ellipse. The purpose of this exercise is to make an animation of the movement along the orbit. One should see a small disk, representing the planet, moving along an elliptic curve. An evolving solid line shows the development of the planet's orbit as the planet moves. The points \((x, y)\) along the ellipse are given by the expressions $$ x=a \cos (\omega t), \quad y=b \sin (\omega t) $$ where \(a\) is the semimajor axis of the ellipse, \(b\) is the semiminor axis, \(\omega\) is an angular velocity of the planet around the star, and \(t\) denotes time. One complete orbit corresponds to \(t \in[0,2 \pi / \omega] .\) Let us discretize time into time points \(t_{k}=k \Delta t\), where \(\Delta t=2 \pi /(\omega n) .\) Each frame in the movie corresponds to \((x, y)\) points along the curve with \(t\) values \(t_{0}, t_{1}, \ldots, t_{i}, i\) representing the frame number \((i=1, \ldots, n)\). Let the plot title of each frame display the planet's instantaneous velocity magnitude. This magnitude is the length of the velocity vector $$ \left(\frac{d x}{d t}, \frac{d y}{d t}\right)=(-\omega a \sin (\omega t), \omega b \cos (\omega t)) $$ which becomes \(\omega \sqrt{a^{2} \sin ^{2}(\omega t)+b^{2} \cos ^{2}(\omega t)}\) Implement the visualization of the planet's orbit using the method above. Run the special case of a circle and verify that the magnitude of the velocity remains constant as the planet moves. Name of program file: planet_orbit.py.

Plot a w-like function. Define mathematically a function that looks like the 'w' character. Plot this function. Name of program file: plot_w.py.

Experience overflow in a function. When an object (ball, car, airplane) moves through the air, there is a very, very thin layer of air close to the object's surface where the air velocity varies dramatically \(^{18}\), from the same value as the velocity of the object at the object's surface to zero a few centimeters away. The change in velocity is quite abrupt and can be modeled by the functiion $$ v(x)=\frac{1-e^{x / \mu}}{1-e^{1 / \mu}} $$ where \(x=1\) is the object's surface, and \(x=0\) is some distance away where one cannot notice any wind velocity \(v\) because of the passing object \((v=0)\). The vind velocity coincides with the velocity of the object at \(x=1\), here set to \(v=1\). The parameter \(\mu\) is very small and related to the viscosity of air. With a small value of \(\mu\), it becomes difficult to calculate \(v(x)\) on a computer. Make a function \(v(x, m u=1 E-6\), exp=math.exp) for calculating the formula for \(v(x)\) using exp as a possibly user-given exponentional function. Let the v function return the nominator and denominator in the formula as well as the fraction (result). Call the v function for various \(\mathrm{x}\) values between 0 and 1 in a for loop, let mu be \(1 \mathrm{E}-3\), and have an inner for loop over two different exp functions: math. exp and numpy. exp. The output will demonstrate how the denominator is subject to overflow and how difficult it is to calculate this function on a computer. Also plot \(v(x)\) for \(\mu=1,0.01,0.001\) on \([0,1]\) using 10,000 points to see what the function looks like. Name of program file: boundary_layer_func1.py.

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