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Find the \(n\)th Maclaurin polynomial \(P_{n}(x)\) for \(f(x)=\arctan x\).

Short Answer

Expert verified
The \(n\)th Maclaurin polynomial \( P_{n}(x) \) for \( f(x) = \arctan x \) is \( P_{n}(x) = x \).

Step by step solution

01

Calculate the zeroth derivative

The zeroth derivative of a function is the function itself. For the function \( f(x) = \arctan x \), evaluated at \( x = 0 \), \( f(0) = \arctan(0) = 0 \).
02

Calculate the first derivative

The first derivative of \( f(x) = \arctan x \) is \( f'(x) = 1/(1+x^{2}) \). When evaluated at \( x = 0 \), \( f'(0) = 1/(1+0^{2}) = 1 \).
03

Calculate the second derivative

The second derivative of \( f(x) = \arctan x \) is \( f''(x) = 0 \) for all \( x \). Therefore, \( f''(0) = 0 \).
04

Notice the pattern- calculate higher derivatives

You'll see that all higher-order derivatives will also be zero. Since the function \( \arctan x \) only has two non-zero derivatives, these are the only terms that will contribute to the Maclaurin series.
05

Write out the terms of the Maclaurin polynomial

The Maclaurin polynomial is given by \( P_{n}(x) = f(0) + f'(0)x + f''(0)x^{2}/2! + \ldots + f^{n}(0)x^{n}/n! \) Since the third derivative and all higher derivatives equal to 0 for \( f(x) = \arctan x \), the Maclaurin polynomial is \( P_{n}(x) = 0 + 1*x + 0*x^{2}/2! + 0*x^{3}/3! + \ldots \).
06

Simplify the polynomial

Since all terms from the third derivative onward are 0, they do not contribute to the sum. Therefore, \( P_{n}(x) = x \).

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Key Concepts

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

Understanding Derivatives
A derivative is a measure of how a function changes as its input changes. It gives us the rate of change or the slope of the function at a given point. In the context of Maclaurin polynomials, derivatives are crucial because they form the building blocks of the polynomial terms. We calculate derivatives at a specific point, typically zero, for these polynomials.
- **First Derivative**: This tells us the rate of change of the function itself. For example, the first derivative of the function \(f(x) = \arctan x\) is \(f'(x) = \frac{1}{1+x^2}\). For the Maclaurin polynomial, we evaluate this derivative at \(x = 0\), which simplifies to \(1\).
- **Higher Derivatives**: These derive from the first derivative, showing changes at finer levels. For \(f(x) = \arctan x\), surprisingly, the second derivative is already zero. This sets a pattern where subsequent derivatives also equal zero under these specific conditions, simplifying the polynomial dramatically.
The Arctan Function
The arctan, or inverse tangent function, is an important trigonometric function that gives the angle whose tangent is a given number. While functions like sine and cosine are often dealt with in this space, arctan is unique because of its specific rate of change and behavior at key points.
- **Behavior at Zero**: \(f(x) = \arctan x\) interestingly evaluates to zero when \(x = 0\). This simplicity at zero plays into its role in creating concise Maclaurin polynomials.
- **Derivative Simplicity**: The inverse tangent's change rate, seen in its first derivative \(f'(x) = \frac{1}{1+x^2}\), simplifies the evaluation for Maclaurin approximations. The function's derivatives beyond the first and zeroth particularly zero out, as seen in earlier examples. This property leads to elegant polynomial approximations.
Polynomial Approximation
Polynomial approximation involves representing a function with a polynomial that maintains similar properties over a small interval. Maclaurin polynomials are a specific type of such approximations, centered around zero, making them an easy-to-compute option for many functions.
- **Maclaurin Series Formula**: The general form is \(P_n(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \ldots + \frac{f^n(0)x^n}{n!}\). It's a tailored approximation using a function's derivatives evaluated at zero.
- **Simplicity for Arctan**: With \(f(x) = \arctan x\), the polynomial \(P_n(x) = x\) is surprisingly straightforward. This results because the only non-zero derivatives are the zeroth and first. Higher terms beyond the first vanish, making \(x\) a satisfactory Maclaurin approximation.

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Most popular questions from this chapter

Let $$f(x)=\frac{e^{x}-e^{-x}}{x}$$ a. Find \(\lim _{x \rightarrow 0}\left(e^{x}-e^{-x}\right) / x\) b. Use three-digit rounding arithmetic to evaluate \(f(0.1)\). c. Replace each exponential function with its third Maclaurin polynomial, and repeat part (b). d. The actual value is \(f(0.1)=2.003335000\). Find the relative error for the values obtained in parts (b) and (c).

Exercise 3 details a rather inefficient means of obtaining an approximation to \(\pi\). The method can be improved substantially by observing that \(\pi / 4=\arctan \frac{1}{2}+\arctan \frac{1}{3}\) and evaluating the series for the arctangent at \(\frac{1}{2}\) and at \(\frac{1}{3}\). Determine the number of terms that must be summed to ensure an approximation to \(\pi\) to within \(10^{-3}\).

Suppose two points \(\left(x_{0}, y_{0}\right)\) and \(\left(x_{1}, y_{1}\right)\) are on a straight line with \(y_{1} \neq y_{0}\). Two formulas are available to find the \(x\)-intercept of the line: $$ x=\frac{x_{0} y_{1}-x_{1} y_{0}}{y_{1}-y_{0}} \quad \text { and } \quad x=x_{0}-\frac{\left(x_{1}-x_{0}\right) y_{0}}{y_{1}-y_{0}} $$ a. Show that both formulas are algebraically correct. b. Use the data \(\left(x_{0}, y_{0}\right)=(1.31,3.24)\) and \(\left(x_{1}, y_{1}\right)=(1.93,4.76)\) and three-digit rounding arithmetic to compute the \(x\)-intercept both ways. Which method is better and why?

The following Maple procedure chops a floating-point number \(x\) to \(t\) digits. (Use the Shift and Enter keys at the end of each line when creating the procedure.) chop \(:=\operatorname{proc}(x, t)\) local e, \(x 2 ;\) if \(x=0\) then 0 else \(\begin{array}{l}e:=\operatorname{ceil}(\text { evalf }(\log 10(a b s(x)))) \\\ x 2:=\operatorname{evalf}\left(\text { trunc }\left(x \cdot 10^{(t-e)}\right) \cdot 10^{(e-t)}\right) \\ \text { end if } \\ \text { end }\end{array}\) Verify the procedure works for the following values. a. \(\quad x=124.031, t=5\) b. \(\quad x=124.036, t=5\) c. \(\quad x=-124.031, t=5\) d. \(\quad x=-124.036, t=5\) e. \(\quad x=0.00653, t=2\) f. \(\quad x=0.00656, t=2\) g. \(\quad x=-0.00653, t=2\) h. \(\quad x=-0.00656, t=2\)

The sequence \(\left\\{F_{n}\right\\}\) described by \(F_{0}=1, F_{1}=1\), and \(F_{n+2}=F_{n}+F_{n+1}\), if \(n \geq 0\), is called a Fibonacci sequence. Its terms occur naturally in many botanical species, particularly those with petals or scales arranged in the form of a logarithmic spiral. Consider the sequence \(\left\\{x_{n}\right\\}\), where \(x_{n}=F_{n+1} / F_{n}\). Assuming that \(\lim _{n \rightarrow \infty} x_{n}=x\) exists, show that \(x=(1+\sqrt{5}) / 2\). This number is called the golden ratio.

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