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i'm looking for an upper bound for the modified Bessel function of the first kind of a +ive real argument. It seems that it satisfies the inequality : $$I_{n}(x)\leqslant \frac{x^{n}}{2^{n}n!}e^{x}$$ But i'm not able to prove this.

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I can prove it for $0 \leq x \leq 4$ from the power series. Would that be useful to you? –  Antonio Vargas Jun 9 '13 at 23:22
    
How did you get this bound? –  Mhenni Benghorbal Jun 10 '13 at 0:20
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1 Answer

up vote 2 down vote accepted

It was proved by Yudell L. Luke in 1972 that

$$ 1 < \Gamma(\nu+1)\left(\frac{2}{x}\right)^\nu I_\nu(x) < \cosh x $$

for $x > 0$ and $\nu > -1/2$. This implies your inequality since

$$ \cosh x - e^x = -\sinh x < 0 $$

for $x > 0$ and hence

$$ \cosh x < e^x $$

for $x > 0$.

Yudell L. Luke, Inequalities for generalized hypergeometric functions, Journal of Approximation Theory, Volume 5, Issue 1, January 1972, pp. 41–65.

(Link to the article on ScienceDirect)

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