Problem 150 Find the Taylor series of the gi... [FREE SOLUTION]

91影视

Calculus Volume 2 (2016)

Gilbert Strang

$Math Studyset 91影视 Explanations$ Math

2016 Edition

Chapter 6: Problem 150

Find the Taylor series of the given function centered at the indicated point. $\quad F(x)=\int_{0}^{x} \cos (\sqrt{t}) d t ; f(t)=\sum_{n=0}^{\infty}(-1)^{n} \frac{t^{n}}{(2 n) !}$ at $a=0$ (Note: $f$ is the Taylor series of $\cos (\sqrt{t}) .)$

Short Answer

Expert verified

The Taylor series of $ F(x) $ is $ \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{(n+1)(2n)!} $.

Step by step solution

Understand Taylor Series Expansion

The Taylor series of a function $ f(t) $ centered at $ a=0 $ is given by $ \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n $. This formula expresses a function as an infinite sum of terms calculated from the values of its derivatives at a single point.

Note the Provided Series

We are given the Taylor series of $ f(t) = \cos(\sqrt{t}) $ as $ \sum_{n=0}^{\infty}(-1)^{n} \frac{t^{n}}{(2n)!} $. This series expansion is centered at $ t = 0 $.

Express $ F(x) $ as a Taylor Series

$ F(x) = \int_{0}^{x} \cos(\sqrt{t}) \, dt $ can be expressed as a series by integrating the given series for $ \cos(\sqrt{t}) $. Thus, $ F(x) = \int_{0}^{x} \sum_{n=0}^{\infty}(-1)^{n} \frac{t^{n}}{(2n)!} dt = \sum_{n=0}^{\infty}(-1)^{n} \int_{0}^{x} \frac{t^{n}}{(2n)!} dt $.

Integrate Term by Term

Integrate each term $ \int_{0}^{x} \frac{t^{n}}{(2n)!} dt $. The integral of $ t^n $ is $ \frac{t^{n+1}}{n+1} $ evaluated from 0 to x. Thus, the term becomes $ \frac{x^{n+1}}{(n+1)(2n)!} $.

Form the Taylor Series of $ F(x) $

The Taylor series of $ F(x) $ centered at $ a=0 $ is then \[ F(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{(n+1)(2n)!}. \] This is the Taylor series expansion of the integral function centered at 0.

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Key Concepts

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

Power Series

A power series is an infinite series of the form $ \sum_{n=0}^{\infty} a_n (x-c)^n $, where $ a_n $ are coefficients and $ c $ is the center of the series. Power series provide a way to represent functions as sums of polynomials, making complex functions easier to analyze and compute. The idea is that the series will converge to the function within a certain interval. This interval depends on the radius of convergence

A series may converge to a function, like the cosine function.
Powers like $ (x-c)^n $ expand around $ c $, simplifying functions that are analytically complex.
Power series are used in approximating functions across applied mathematics.

In our problem, we are given a power series for $ \cos(\sqrt{t}) $. From this, we aim to understand how $ F(x) = \int_{0}^{x} \cos(\sqrt{t}) dt $ can be expressed using series terms.

Cosine Function

The cosine function, $ \cos(x) $, is a trigonometric function characterized by its wave-like graph and periodicity. In mathematical terms, it can be expressed as a power series: \[ \cos(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!} \]

Cosine is even, meaning $ \cos(-x) = \cos(x) $.
It repeats every $ 2\pi $, essential for many periodic processes.
Understanding its series helps evaluate integrals and find series expansions of more complex functions.

Using its power series representation aids in tasks such as the integration of $ \cos(\sqrt{t}) $, crucial for our original exercise.

Integration of Series

The integration of series involves finding the integral of each term in a power series representation separately. This process transforms a series term by term:

For $ \cos(\sqrt{t}) $, the original series is $ \sum_{n=0}^{\infty} (-1)^n \frac{t^n}{(2n)!} $.
Integrating individually, we use $ \int_{0}^{x} t^n dt = \frac{x^{n+1}}{n+1} $.
This allows converting the series of $ \cos(\sqrt{t}) $ into that of its integral, $ F(x) $.

The result is a new power series accurate for $ F(x) $ within the original's radius of convergence.

Centered Expansion

Centered expansion refers to the point around which a function's series is developed, fundamentally affecting convergence and accuracy. For Taylor series, this point is denoted as $ a $ and in our exercise is set to zero, making it a "Maclaurin series" as a special case of Taylor's.

The choice of $ a = 0 $ simplifies polynomial expressions.
Every term $ (x-a)^n $ becomes $ x^n $, streamlining calculations.
Centering expands cosine and integral functions symmetrically for small values of $ x $.

Choosing $ a = 0 $ leverages symmetry and eases computations crucial to derive the power series representation for $ \cos(\sqrt{t}) $. This centered approach ensures maximal simplicity and efficiency when dealing with such series.

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91影视

Find the Taylor series of the given function centered at the indicated point. \(\quad F(x)=\int_{0}^{x} \cos (\sqrt{t}) d t ; f(t)=\sum_{n=0}^{\infty}(-1)^{n} \frac{t^{n}}{(2 n) !}\) at \(a=0\) (Note: \(f\) is the Taylor series of \(\cos (\sqrt{t}) .)\)

Short Answer

Step by step solution

Understand Taylor Series Expansion

Note the Provided Series

Express \( F(x) \) as a Taylor Series

Integrate Term by Term

Form the Taylor Series of \( F(x) \)

Key Concepts

Power Series

Cosine Function

Integration of Series

Centered Expansion

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