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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, 5 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 $$\displaystyle f(x)$$. · 3 years, 5 months ago

Good. · 3 years, 5 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. · 3 years, 5 months ago