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Chapter 1: Q20p (page 32)

$e^{x} s i n x$

Short Answer

Expert verified

The Maclaurin series, i.e., $e^{x} \sin (x) = x + x^{2} + \frac{x^{3}}{3} - \frac{x^{5}}{30} + 鈥�$

The Maclaurin series using MATLAB tool, i.e.,

Step by step solution

01

Given Information.

The given function.

$e^{x} \sin x$

02

Definition of Maclaurin series.

All of the terms in a Maclaurin series are nonnegative integer powers of the variable. It's a Taylor series expansion of a function with a value of around $0 .$

03

Find the Maclaurin series for exsinx..

Use the Maclaurin series of $\sin (x)$ and $e^{x}$

$f (x) = f (0) + \frac{f^{'} (0)}{1!} (x) + \frac{f^{''} (0)}{2!} (x^{2}) + \frac{f^{'''} (0)}{3!} (x^{3}) + 鈥�$

$e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + 鈥�$

$\sin (x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + 鈥�$

Further, multiply $\sin (x)$ and $e^{x}$

$e^{x} \sin (x) = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} + x^{2} - \frac{x^{4}}{3!} + \frac{x^{3}}{2!} - \frac{x^{5}}{2! 3!} + \frac{x^{4}}{3!} + \frac{x^{5}}{4!} + 鈥�$

$e^{x} \sin (x) = x + x^{2} + \frac{x^{3}}{3} - \frac{x^{5}}{30} + 鈥�$

04

Use MATLAB tool to find Maclaurin series.

Find first few terms of Maclaurin series.

Hence, the Maclaurin series, i.e., $e^{x} \sin (x) = x + x^{2} + \frac{x^{3}}{3} - \frac{x^{5}}{30} + 鈥�$

The Maclaurin series using MATLAB tool, i.e.,

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Most popular questions from this chapter

Prove theorem $14.3$ . Hint: Group the terms in the error as role="math" localid="1657423688910" $(a_{n + 1} + a_{n + 2}) + (a_{n + 3} + a_{n + 4}) + 鈰�$ to show that the error has the same sign as role="math" localid="1657423950271" $a_{n + 1} .$ Then group them asrole="math" localid="1657423791335" $a_{n + 1 + (a_{n + 2} + a_{n + 3}) + (a_{n + 4} + a_{n + 5}) + 鈰�}$ to show that the error has magnitude less than $| a_{n + 1} | .$

Solve for all possible values of the real numbers xand y in the following equations. $x + i y = {(1 - i)}^{2}$

$\sqrt{1 + \ln (1 + x)}$

Use the ratio test to show that a binomial series converges for $| x | < 1$

Solve for all possible values of the real numbersand in the following equations.

$\frac{x + i y + 2 + 3 i}{2 x + 2 i y - 3} = i + 2$

See all solutions

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