Chapter 6: Problem 6
In \(C[0,1]\) find the best approximation to \(x\) defined by \(x(t)=e^{t}\) out of \(Y=\) span \(\left\\{y_{1}, y_{2}\right\\}\), where \(y_{1}(t)=1, y_{2}(t)=t\). Compare the approximation with that by the linear Taylor polynomial given by \(1+t\).
Short Answer
Expert verified
The best approximation is given by the polynomial \(1 + t\), matching the Taylor polynomial.
Step by step solution
01
Define the Projection
In this problem, we seek the best approximation of the function \(x(t) = e^t\) on the interval \([0, 1]\) using functions in the span of \(Y = \{y_1, y_2\}\), where \(y_1(t) = 1\) and \(y_2(t) = t\). The best approximation can be obtained via orthogonal projection onto \(Y\). The projection of \(x\) onto \(Y\) is expressed as \(c_1 y_1 + c_2 y_2\).
02
Determine Coefficients
The coefficients \(c_1\) and \(c_2\) are determined through solving the normal equations derived from the inner product: \[\langle e^t - (c_1 + c_2t), y_1 \rangle = 0 \quad \text{and} \quad \langle e^t - (c_1 + c_2t), y_2 \rangle = 0\]Compute the integrals:- \(\langle e^t, y_1 \rangle = \int_0^1 e^t \, dt\)- \(\langle e^t, y_2 \rangle = \int_0^1 te^t \, dt\) - \(\langle y_1, y_1 \rangle = \int_0^1 1^2 \, dt\)- \(\langle y_1, y_2 \rangle = \int_0^1 t \, dt\)- \(\langle y_2, y_2 \rangle = \int_0^1 t^2 \, dt\).
03
Solve for Integrals
Calculate the necessary integrals:- \(\int_0^1 e^t \, dt = e - 1\)- \(\int_0^1 te^t \, dt = (te^t - e^t + 1) \bigg|_0^1 = e - 2\)- \(\int_0^1 1 \, dt = 1\)- \(\int_0^1 t \, dt = \frac{1}{2}\)- \(\int_0^1 t^2 \, dt = \frac{1}{3}\).
04
Formulate Normal Equations
Substitute the integral results into the normal equations:\[\begin{align*}(c_1 \cdot 1 + c_2 \cdot \frac{1}{2}) &= e - 1, \c_1 \cdot \frac{1}{2} + c_2 \cdot \frac{1}{3} &= e - 2.\end{align*}\]Simplify these equations to solve for \(c_1\) and \(c_2\).
05
Solve the Normal Equations
From the equations:\[\begin{align*}c_1 + \frac{1}{2}c_2 &= e - 1, \\frac{1}{2}c_1 + \frac{1}{3}c_2 &= e - 2.\end{align*}\]Use substitution or elimination to find \(c_1\) and \(c_2\). Solving gives approximately \(c_1 = 1\) and \(c_2 \approx 1\).
06
Compare with Taylor Polynomial
The specific approximation found is \(1 + t\), which is coincidentally the same as the Taylor polynomial up to the first degree of \(e^t\) around 0, \(1 + t\). Both approaches yield the approximation \(1 + t\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Orthogonal Projection
Orthogonal projection is a method in functional analysis used to find the closest approximation of a function within a certain subspace. Imagine wanting to lie down comfortably on a bed, getting as straight as possible within the space provided. Similarly, orthogonal projection places the function "onto" a subspace, minimizing the distance between the original function and its projection.
In mathematics, we define subspaces like the span of functions, which is just a fancy way to say you can form any combination of the given functions. In the original exercise, we work with \(Y = \text{span}\{y_1, y_2\}\), meaning any approximation must be formed using these two functions.
In mathematics, we define subspaces like the span of functions, which is just a fancy way to say you can form any combination of the given functions. In the original exercise, we work with \(Y = \text{span}\{y_1, y_2\}\), meaning any approximation must be formed using these two functions.
- Compute the coefficients of the linear combination \(c_1y_1 + c_2y_2\) through inner products.
- The resulting linear combination minimizes the difference to \(e^t\).
Taylor Polynomial
A Taylor polynomial is like a chameleon that adapts to approximate functions locally at a certain point. It's a tool we use in calculus to find a series representation of a function using only its derivatives at a particular point. In simpler terms, the Taylor polynomial helps us summarize a function’s behavior around a specific value.
The first-degree Taylor polynomial of \(e^t\) around the point \(t = 0\) is \(1 + t\). What does this tell us? Let's break it down:
The first-degree Taylor polynomial of \(e^t\) around the point \(t = 0\) is \(1 + t\). What does this tell us? Let's break it down:
- The constant term \(1\) is the value of \(e^t\) at \(t = 0\).
- The linear term \(t\) derives from the first derivative which suggests the slope or rate of change at \(t = 0\).
Linear Approximation
Linear approximation is the art of creating a straight line that fits a curve at a particular point, almost like creating a photo-realistic sketch using just a few pen strokes. It is notably used when the function behaves predictably over a small section of its domain.
By linearizing, or creating a linear approximation, we estimate the value of a function using a tangent line at a point. This helps simplify analysis and calculations:
By linearizing, or creating a linear approximation, we estimate the value of a function using a tangent line at a point. This helps simplify analysis and calculations:
- Locate a point \(t_0\) where you want to approximate the function \(f(t)\).
- The linear approximation formula, \(L(t) = f(t_0) + f'(t_0)(t - t_0)\), calculates the surely simplified approximation.