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※I'm not good at English, so please edit my question※

I read a question ,

and tried to prove $\Gamma(x) \geq x^{3}$ For all real number $x \geq 6$ using gamma function's definition and its derivative , but I couldn't prove inequality $\Gamma(x) \geq x^{3}$

This is my attempt.


For $ n \in \mathbb{N} $, $ n! = \Gamma \left ( n + 1 \right )$

Gamma Function's definiton is $\Gamma \left ( x \right ) = \int_{0}^{\infty }t^{x-1}e^{-t} \mathrm{d}t$

Let $f\left ( x , a \right ) = \int_{0}^{a}t^{x-1}e^{-t} \mathrm{d}t$

(Then $\lim_{a \to \infty} f\left ( x,a \right ) = \Gamma (x)$)

$\Rightarrow f\left ( x,a \right ) = \lim_{n \to \infty}\sum_{k=1}^{n}\left ( \frac{a}{n} \right )^{x}k^{x-1}e^{-\frac{ka}{n}}$

For partial sum$\sum_{k=1}^{n}\left ( \frac{a}{n} \right )^{x}k^{x-1}e^{-\frac{ka}{n}}$ , $ 0 \leq \frac{k}{n} \leq 1$ Because $ n \geq k \geq 1$

Let $ K = \frac{ka}{n}$ ( $ a \geq K \geq 0 $)

Then $\sum_{k=1}^{n}\left ( \frac{a}{n} \right )^{x}k^{x-1}e^{-\frac{ka}{n}}= \frac{a}{n}\sum_{k=1}^{n} K^{x-1}e^{-K}$

Let $ 0 < a < n$ , $ 1 > C = \frac{a}{n}$ ($K=kC$)

We get$\sum_{k=1}^{n}\left ( \frac{a}{n} \right )^{x}k^{x-1}e^{-\frac{ka}{n}}= C^{x} \sum_{k=1}^{n}k^{x-1}e^{-Ck}$


But, I think that isn't proper approaching

How proof inequality $\Gamma(x) \geq x^{3}$ using gamma function's definition?

I don't have any idea for proving this

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    $\begingroup$ Your inequality is clearly false as written. $\Gamma(6) = 120$ and $6^3 = 216$, so we have $\Gamma(6) < 6^3$ right away. I think you want $\Gamma(x+1)\geq x^3$ for $x\geq 6$, or $\Gamma(x)\geq (x-1)^3$ for $x\geq 7$. $\endgroup$ – Michael Lee Aug 11 '17 at 5:19
  • $\begingroup$ Given the linked question, I guess conjecture should probably be $\Gamma(x+1)\geq x^3$ for $x\geq6$. $\endgroup$ – stewbasic Aug 11 '17 at 5:21
  • $\begingroup$ Oh, It's my mistake ($\Gamma(x) \geq x^3$ for $x\geq 6$). I mean $ \Gamma(x+1) \geq x^3$ for $x\geq6$ . $\endgroup$ – G.H.lee Aug 11 '17 at 6:02
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1) Statement is false as stated: $$\Gamma\left(6\right) = 5!=120$$ $$6^{3} = 216$$

2) Proposition: $\Gamma\left(x\right)\geq x^{3}$ for all real $x\geq7$.

Proof: $$\frac{\Gamma\left(x\right)}{x^{3}}=\frac{\left(x-1\right)\Gamma\left(x-1\right)}{x^{3}}=...=\frac{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)}{x^{3}}\Gamma\left(x-4\right)$$ Then, since $\Gamma\left(x-4\right)\geq1$ for $x\geq5$: $$\frac{\Gamma\left(x\right)}{x^{3}}\geq\frac{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)}{x^{3}},\textrm{ }\forall x\geq5$$

Next: $$\frac{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)}{x^{3}}=\left(1-\frac{1}{x}\right)\left(1-\frac{2}{x}\right)\left(1-\frac{3}{x}\right)\left(x-4\right)$$ Then, $x\geq7$ implies $-\frac{1}{x}\geq-\frac{1}{7}$.

Thus:

$$\left(1-\frac{1}{x}\right)\left(1-\frac{2}{x}\right)\left(1-\frac{3}{x}\right)\left(x-4\right)\geq\left(1-\frac{1}{7}\right)\left(1-\frac{2}{7}\right)\left(1-\frac{3}{7}\right)\left(7-4\right)=\frac{360}{343}\geq1$$

Thus, $x\geq7$ implies $$\frac{\Gamma\left(x\right)}{x^{3}}\geq1$$.

Q.E.D.

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