Which is greater?

Define $f(x)= \sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}$ and $g(x)= \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$ , where $x$ is a real number. Then which of $f(x)$ or $g(x)$ is greater for all $x$ ?

#Sequences #Goldbach'sConjurersGroup

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Let us start by writing the Maclaurin Series for $\displaystyle e^x$ ,
$\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$
$\displaystyle e^{-x} = \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{n!}$

Thus,
$\displaystyle f(x) = \frac{e^x-e^{-x}}{2}$ , and

$\displaystyle g(x) = \frac{e^x+e^{-x}}{2}$

It is clear that, $g(x)>f(x)$ , as the $\displaystyle e^{-x}$ term is added to the first term in $\displaystyle g(x)$ , whereas it is subtracted from the first term in $\displaystyle f(x)$ .

Anish Puthuraya - 7 years, 4 months ago

Good.

A Brilliant Member - 7 years, 4 months ago

$f(x)$ is the power series for $\sinh x$ and $g(x)$ is the power series for $\cosh x.$ Use the fact that $g(x)-f(x)=\cosh x-\sinh x=e^{-x}>0$ for all $x.$ Therefore, $g(x)>f(x)$ for all $x.$

Anish's solution is good as well; I prefer using hyperbolic trig functions for these kind of series though.

Trevor B. - 7 years, 4 months ago

Nice.

A Brilliant Member - 7 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$