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Use Taylor polynomials with remainder term to evaluate the following limits: (a) \(\lim _{x \rightarrow 0} \frac{1-\cos (x)}{x^{2}}\) (b) \(\lim _{x \rightarrow 0} \frac{\log \left(1+x^{2}\right)}{2 x}\) (c) \(\lim _{x \rightarrow 0} \frac{\log (1-x)+x e^{x / 2}}{x^{3}}\) Hint: Use Taylor polynomials for the standard functions [e.g., \(\cos (t), \log (1+t)\), and \(\left.e^{t}\right]\) to obtain polynomial approximations to the numerators of these fractions; and then simplify the results

Short Answer

Expert verified
a) \frac{1}{2} \ b) 0 \ c) -\frac{1}{3}

Step by step solution

01

- Taylor Polynomial for \cos(x)

Expand\( \cos(x) \) using Taylor series around\( x = 0 \): \[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots\] Use this to approximate \cos(x) \for small \( x \).
02

- Substitute Taylor Polynomial

Now substitute the Taylor expansion \cos(x) \into the limit: \[ \lim_{x \to 0} \frac{1 - (1 - \frac{x^2}{2} + O(x^4))}{x^2} \]
03

- Simplify Expression

Simplify the expression: \[ \lim_{x \to 0} \frac{1 - 1 + \frac{x^2}{2} - O(x^4)}{x^2} = \lim_{x \to 0} \frac{\frac{x^2}{2} - O(x^4)}{x^2} = \lim_{x \to 0} \frac{1}{2} = \frac{1}{2} \]
04

- Taylor Polynomial for \log(1 + x^2)

Expand \log(1 + x^2) \ using Taylor series around\( x = 0 \): \[ \log(1 + x^2) = x^2 - \frac{x^4}{2} + O(x^6) \] Use this to approximate \log(1 + x^2) \ for small \( x \).
05

- Substitute Taylor Polynomial

Now substitute the Taylor expansion into the limit: \[ \lim_{x \to 0} \frac{x^2 - \frac{x^4}{2} + O(x^6)}{2x} \]
06

- Simplify Expression

Simplify the expression: \[ \lim_{x \to 0} \frac{x^2}{2x} = \lim_{x \to 0} \frac{x}{2} = 0 \]
07

- Taylor Polynomial for \log(1 - x) \and e^{x/2}

Expand \log(1 - x) \and \ e^{x/2} \ using Taylor series around\( x = 0 \): \[ \log(1 - x) = -x - \frac{x^2}{2} - \frac{x^3}{3} + O(x^4) \] and \[ e^{x/2} = 1 + \frac{x}{2} + \frac{(x/2)^2}{2!} + O(x^3) \]
08

- Substitute Taylor Polynomials

Substitute the Taylor expansions into the limit: \[ \lim_{x \to 0} \frac{-x - \frac{x^2}{2} - \frac{x^3}{3} + x \left(1 + \frac{x}{2} + O(x^2)\right)}{x^3} \]
09

- Simplify Expression

Simplify the expression: \[ \lim_{x \to 0} \frac{-x - \frac{x^2}{2} - \frac{x^3}{3} + x + \frac{x^2}{2} + O(x^3)}{x^3} = \lim_{x \to 0} \frac{-\frac{x^3}{3} + O(x^3)}{x^3} = -\frac{1}{3} \]

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Key Concepts

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

Taylor Series Expansion
The Taylor series expansion is a powerful tool in calculus used to approximate functions using polynomials. It involves expressing a function as an infinite sum of terms calculated from the function's derivatives at a point. For instance, the expansion for \( \cos(x) \) around zero (Maclaurin series) is: \[\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots\]This approximation becomes very useful for calculating limits, especially when the function involves trigonometric, exponential, or logarithmic components.
Another example is the expansion of \(\log(1 + x^2)\) around zero:\[\log(1 + x^2) = x^2 - \frac{x^4}{2} + O(x^6)\]The series provides polynomial approximations to functions that can simplify the process of evaluating limits when \( x \) is close to zero. Just remember, the accuracy of the approximation increases with the number of terms included from the series.
Limits in Calculus
Limits are fundamental in calculus and are used to describe the behavior of functions as inputs approach a specific value. In the context of Taylor Series, limits help simplify complex expressions by approximating them using polynomials.
The limit \( \lim_{x \to 0} \frac{1 - \cos (x)}{x^{2}} \) is evaluated using the Taylor expansion of \( \cos(x) \). Substituting the expansion into the limit expression and simplifying step by step allows us to find the result.
Similarly, evaluating \( \lim_{x \to 0} \frac{\log (1 + x^{2})}{2 x} \) involves substituting the Taylor series of the logarithmic function and simplifying. Understanding these steps is crucial because they demonstrate how applying polynomial approximations can make finding the limit more straightforward.
Numerical Analysis
Numerical analysis involves techniques for approximating mathematical procedures. When it comes to analyzing functions, Taylor polynomials provide a bridge between exact values and numerical approximations.
By using the Taylor series' several terms, we can understand the function's behavior near a specific point. This method is especially useful when simpler forms of the functions are needed for computational purposes. Computing the limit often involves dealing with cumbersome expressions, but numerical methods make it easier to handle these by breaking them down into manageable polynomial parts.
For example, approximating \( \log(1 - x) + x e^{x / 2} \) using Taylor series simplifies the limit evaluation process considerably, and this is where numerical analysis shines: reducing complex operations to simpler, polynomial-based calculations.
Polynomial Approximations
Polynomial approximations are used to estimate the behavior of functions using polynomials, which are simpler to compute and understand. Taylor polynomials offer such approximations by expressing a function as a sum of polynomials based on its derivatives.
The Taylor polynomial for a function \( f(x) \) around \( a = 0 \) is generally written as: \[T_n(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \cdots + \frac{f^n(0)x^n}{n!}\]Using these, we can approximate \( \cos(x) \) up to the second-order term to compute \(\lim_{x \to 0} \frac{1 - \cos (x)}{x^2} \), simplifying the calculation to one involving straightforward algebra.
This approach is also applied when computing \(\lim_{x \to 0} \frac{\log(1 + x^2)}{2x} \) and \(\lim_{x \to 0} \frac{\log (1-x) + x e^{x / 2}}{x^3}\). By using polynomial approximations, difficult limits become more accessible and easier to evaluate systematically.

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Most popular questions from this chapter

(a) Let the polynomial \(p(x)\) be an even function, meaning that \(p(-x)=p(x)\) for all \(x\) of interest. Show this implies that the coefficients are zero for all terms of odd degree. (b) Let the polynomial \(p(x)\) be an odd function, meaning that \(p(-x)=-p(x)\) for all \(x\) of interest. Show this implies that the coefficients are zero for all terms of even degree. (c) Let \(p(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_{5} x^{5} .\) Give conditions on the coefficients \(\left\\{a_{0}, a_{1}, a_{2}, a_{5}\right\\}\) so that \(p(x)\) is even. Repeat with \(p(x)\) being odd.

(a) Bound the error in the approximation $$ \sin (x) \approx x $$ for \(-\pi / 4 \leq x \leq \pi / 4\). (b) Since this is a good approximation for small values of \(x\), also consider the "percentage error" $$ \frac{\sin (x)-x}{\sin (x)} \approx \frac{\sin (x)-x}{x} $$ Bound the absolute value of the latter quantity for \(-\delta \leq x \leq \delta .\) Pick \(\delta\) to make the absolute value of the percentage error less than \(1 \%\).

Evaluate $$ p(x)=1-\frac{x^{3}}{3 !}+\frac{x^{6}}{6 !}-\frac{x^{9}}{9 !}+\frac{x^{12}}{12 !}-\frac{x^{15}}{15 !} $$ as efficiently as possible. How many multiplications are necessary? Assume all coefficients have been computed and stored for later use.

For \(f(x)=e^{x}\), find a Taylor approximation that is in error by at most \(10^{-7}\) on \([-1,1]\). Using this approximation, write a function program to evaluate \(e^{x} .\) Compare it to the standard value of \(e^{x}\) obtained from the MATLAB function \(\exp (x)\); calculate the difference between your approximation and \(\exp (x)\) at 21 evenly spaced points in \([-1,1]\).

Define \(h(x)=f(x) g(x)\). Let the Taylor polynomials of degree \(n\) for \(f(x)\) and \(g(x)\) be given by $$ p_{n}(x)=\sum_{t=0}^{n} a_{i} x^{i}, \quad q_{n}(x)=\sum_{j=0}^{n} b_{j} x^{j} $$ Let \(r_{n}(x)\) be obtained by first multiplying \(p_{n}(x) q_{n}(x)\) and then dropping all terms of degree greater than \(n\). (a) For \(n=2\), show that the Taylor polynomial of degree 2 for \(h(x)\) equals \(r_{2}(x) .\) (b) For general \(n \geq 1\), show that the Taylor polynomial of degree \(n\) for \(h(x)\) equals \(r_{n}(x)\). Hint: For repeated differentiation of the product \(f(x) g(x)\), use the Leibniz formula: $$ \frac{d^{k}}{d x^{k}}[f(x) g(x)]=\sum_{j=0}^{k}\left(\begin{array}{l} k \\ j \end{array}\right) f^{(j)}(x) g^{(k-j)}(x) $$

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