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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 \(x\)?

Note by Paramjit Singh
3 years, 7 months ago

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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 · 3 years, 7 months ago

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@Anish Puthuraya Good. Paramjit Singh · 3 years, 7 months ago

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\(f(x)\) 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. · 3 years, 7 months ago

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@Trevor B. Nice. Paramjit Singh · 3 years, 7 months ago

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