Chapter 8: Q. 91 (page 694)

Use the Maclaurin series for $e^{x}$ and $e^{- x}$ to prove that
$\sinh x = {鈭�}_{k = 0}^{鈭�} \frac{1}{(2 k + 1)!} x^{2 k + 1} .$

Short Answer

Expert verified

That, which is $\sinh x = {鈭�}_{i = 0}^{x} \frac{1}{(2 k + 1)!} x^{2 位路 1}$ is proved.

Step by step solution

Given Information

Consider the hyperbolic sine function: $\sinh x = \frac{e^{x} - e^{- x}}{2}$

Proof

Since,

$\begin{array}{rcl} e^{x} & = & {鈭�}_{k = 0}^{鈭�} \frac{1}{k!} x^{k} \\ = & 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!} + \frac{x^{3}}{5!} + 鈥� \end{array}$

And also.

$e^{- x} = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \frac{x^{4}}{4!} - \frac{x^{5}}{5!} + 鈰�$

Now, subtract the series for $e^{x}$ and $e^{- x}$

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

So,

$\begin{array}{rcl} \frac{e^{x} - e^{- x}}{2} & = & x + \frac{x^{3}}{3!} + \frac{x^{3}}{5!} + 鈥� \\ = & {鈭�}_{k = 0}^{鈭�} \frac{1}{(2 k + 1)!} x^{2 k + 1} \\ = & \sin h x \end{array}$

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91影视

Use the Maclaurin series for $e^{x}$ and $e^{- x}$ to prove that
$\sinh x = {鈭�}_{k = 0}^{鈭�} \frac{1}{(2 k + 1)!} x^{2 k + 1} .$

Short Answer

Step by step solution

Given Information

Proof

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91影视

Use the Maclaurin series for exand e-xto prove thatsinhx=鈭�k=0鈭�1(2k+1)!x2k+1.

Short Answer

Step by step solution

Given Information

Proof

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

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Probability and Statistics

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Applied Mathematics

Study anywhere. Anytime. Across all devices.

Use the Maclaurin series for $e^{x}$ and $e^{- x}$ to prove that
$\sinh x = {鈭�}_{k = 0}^{鈭�} \frac{1}{(2 k + 1)!} x^{2 k + 1} .$