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 \(x\)?

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TopNewestLet us start by writing the

Maclaurin Seriesfor \(\displaystyle e^x\),\(\displaystyle e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\)

\(\displaystyle e^{-x} = \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{n!}\)

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

addedto the first term in \(\displaystyle g(x)\), whereas it issubtractedfrom the first term in \(\displaystyle f(x)\). – Anish Puthuraya · 2 years, 11 months agoLog in to reply

– Paramjit Singh · 2 years, 11 months ago

Good.Log in to reply

\(f(x)\) 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. · 2 years, 11 months ago

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– Paramjit Singh · 2 years, 11 months ago

Nice.Log in to reply