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Chapter 9: Problem 29

Find the Maclaurin series for the function. \(f(x)=\cos ^{2} x\)

Short Answer

Expert verified
The Maclaurin series for the function \(f(x) = \cos^2 x\) is \(\cos^2 x = \frac{1}{2} + \frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(k)!}\).

Step by step solution

01

Maclaurin Series of \(\cos x\)

First, recall that the Maclaurin Series for \(\cos x\) is given by \(\cos x = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!}\)
02

Square of \(\cos x\)

\(\cos^2 x\) is the square of \(\cos x\), so take the square of the series found in Step 1.
03

Use of trigonometric identity

By the Pythagorean identity, we know that \(\cos^2 x = 1 - \sin^2 x\). But also, \(\sin^2 x = 1 - \cos^2 x\). So, we can simplify \(\cos^2 x\) as \(\cos^2 x = \frac{1}{2}(1 + \cos 2x)\). Thus, the series becomes \(\cos^2 x = \frac{1}{2} + \frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k (2x)^{2k}}{(2k)!}\).
04

Simplify the series

The final Maclaurin series is simplified by factoring out any constant coefficients inside the series, which gives \(\cos^2 x = \frac{1}{2} + \frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(k)!}\).

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Key Concepts

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

Series expansion
Series expansion is a fundamental concept in mathematics, involving the expression of functions as an infinite sum of terms. One particular type is the Maclaurin series, a special case of the Taylor series where the center is at zero. The key idea is that a complicated function can often be approximated by a polynomial, which is easier to work with. For example, the function \( \cos x \) can be expanded as a Maclaurin series:
  • The series is represented as \( \cos x = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k)!} \).
  • The terms of this series alternate in sign and decrease in size as \( k \) increases.
  • This representation is useful because it allows us to approximate \( \cos x \) for small values of \( x \).
Understanding series expansions is crucial for simplifying complex functions and solving calculus problems efficiently.
Trigonometric identities
Trigonometric identities simplify expressions and solve equations involving trigonometric functions. In our problem, we encountered the expression \( \cos^2 x \), which can be simplified using a well-known identity:
  • The Pythagorean identity gives us \( \cos^2 x = \frac{1}{2}(1 + \cos 2x) \).
  • This allows us to express \( \cos^2 x \) in terms of only one trigonometric function, \( \cos 2x \).
  • It simplifies the calculation of the series expansion for \( \cos^2 x \).
By understanding these identities, we can break down complex calculations and transform them into simpler expressions. This makes it easier to integrate or differentiate trigonometric functions during calculus problem solving.
Calculus problem solving
Calculus problem solving often requires the use of series, identities, and simplifications. The process involves:
  • Breaking down complex functions into simpler series expansions, such as using the Maclaurin series.
  • Applying trigonometric identities to simplify expressions.
  • Simplifying series, as seen with \( \cos^2 x = \frac{1}{2} + \frac{1}{2}\sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(k)!} \).
This approach is particularly helpful when dealing with infinite series or trigonometric functions. It's all about finding patterns and using mathematical tools to achieve more straightforward solutions. Problems like these enhance problem-solving skills and deepen understanding of fundamental calculus concepts.

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Most popular questions from this chapter

Probability A fair coin is tossed repeatedly. The probability that the first head occurs on the \(n\) th toss is given by \(P(n)=\left(\frac{1}{2}\right)^{n}\), where \(n \geq 1\) (a) Show that \(\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^{n}=1\). (b) The expected number of tosses required until the first head occurs in the experiment is given by \(\sum_{n=1}^{\infty} n\left(\frac{1}{2}\right)^{n}\) Is this series geometric? (c) Use a computer algebra system to find the sum in part (b).

Consider the sequence \(\left\\{a_{n}\right\\}=\left\\{\frac{1}{n} \sum_{k=1}^{n} \frac{1}{1+(k / n)}\right\\}\). (a) Write the first five terms of \(\left\\{a_{n}\right\\}\). (b) Show that \(\lim _{n \rightarrow \infty} a_{n}=\ln 2\) by interpreting \(a_{n}\) as a Riemann sum of a definite integral.

Use the formula for the \(n\) th partial sum of a geometric series \(\sum_{i=0}^{n-1} a r^{i}=\frac{a\left(1-r^{n}\right)}{1-r}\) Annuities When an employee receives a paycheck at the end of each month, \(P\) dollars is invested in a retirement account. These deposits are made each month for \(t\) years and the account earns interest at the annual percentage rate \(r\). If the interest is compounded monthly, the amount \(A\) in the account at the end of \(t\) years is $$ \begin{aligned} A &=P+P\left(1+\frac{r}{12}\right)+\cdots+P\left(1+\frac{r}{12}\right)^{12 t-1} \\ &=P\left(\frac{12}{r}\right)\left[\left(1+\frac{r}{12}\right)^{12 t}-1\right] \end{aligned} $$ If the interest is compounded continuously, the amount \(A\) in the account after \(t\) years is $$ \begin{aligned} A &=P+P e^{r / 12}+P e^{2 r / 12}+P e^{(12 t-1) r / 12} \\ &=\frac{P\left(e^{n}-1\right)}{e^{r / 12}-1} \end{aligned} $$ Verify the formulas for the sums given above.

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(\sum_{n=1}^{\infty} a_{n}=L\), then \(\sum_{n=0}^{\infty} a_{n}=L+a_{0-}\)

show that the function represented by the power series is a solution of the differential equation. $$ y=\sum_{n=0}^{\infty} \frac{x^{2 n}}{2^{n} n !}, \quad y^{\prime \prime}-x y^{\prime}-y=0 $$
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